green functors and g-sets

Serge Bouc

Green Functors and G-sets

~ Springer

Author

Serge Bouc Equipe des groupes finis CNRS UMR 9994 UFR de Mathdmatiques Universit6 Paris 7 - Denis Diderot 2, Place Jussieu F-75251 Paris, France e-mail: [email protected], fr

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Die Deutsche Bibl iothek - CIP-Einheitsaufnahme

B o u t , Serge: Green functors and G-sets / Serge Bouc. - Berlin ; Heidelberg ; New York ; Barcelona ; Budapest ; Hong Kong ; London ; Milan ; Paris ; Santa Clara ; Singapore ; Tokyo : Springer, 1997

(Lecture notes in mathematics ; 1671) ISBN 3-540-63550-5

Mathematics Subject Classification (1991): 19A22, 20C05, 20J06, 18D35

ISSN 0075- 8434 ISBN 3-540-63550-5 Springer-Verlag Berlin Heidelberg New York

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Contents

M a c k e y f u n c t o r s 5

1.1 E q u i v a l e n t def ini t ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.1 Def in i t ion in t e r m s of subgroups . . . . . . . . . . . . . . . . . . 5

1.1.2 Def in i t ion in t e r m s of G-sets . . . . . . . . . . . . . . . . . . . . 6

1.1.3 Def in i t ion as modu le s over the Mackey a lgebra . . . . . . . . . . 7

1.2 T h e Mackey func tors M ~ M y . . . . . . . . . . . . . . . . . . . . . . 8

1.3 C o n s t r u c t i o n of H ( M , N ) and M ( ~ N . . . . . . . . . . . . . . . . . . . 9

1.4 Iden t i f i ca t ion of H ( M , N ) . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.5 Iden t i f i ca t ion of M @ N . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.6 A n o t h e r iden t i f i ca t ion of M Q N . . . . . . . . . . . . . . . . . . . . . . 16

1.7 Func to r iMi ty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.8 n-fold t ensor p r o d u c t . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.8.1 Def in i t ion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.8.2 Unive r sa l p rope r ty . . . . . . . . . . . . . . . . . . . . . . . . . 29

1.9 C o m m u t a t i v i t y and assoc ia t iv i ty . . . . . . . . . . . . . . . . . . . . . 38

1.10 A d j u n c t i o n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

G r e e n f u n c t o r s 41

2.1 Def in i t ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.2 Def in i t ion in t e r m s of G-sets . . . . . . . . . . . . . . . . . . . . . . . . 46

2.3 Equ iva lence of the two def ini t ions . . . . . . . . . . . . . . . . . . . . . 48

2.4 T h e B u r n s i d e func to r . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

2.4.1 T h e B u r n s i d e func to r as Mackey func to r . . . . . . . . . . . . . 52

2.4.2 T h e B u r n s i d e func to r as Green func to r . . . . . . . . . . . . . . 55

2.4.3 T h e B u r n s i d e f unc t o r as in i t ia l ob j ec t . . . . . . . . . . . . . . . 57

2.4.4 T h e B u r n s i d e func to r as un i t . . . . . . . . . . . . . . . . . . . 59

T h e c a t e g o r y a s s o c i a t e d t o a G r e e n f u n c t o r 61

3.1 E x a m p l e s of modu les over a Green func to r . . . . . . . . . . . . . . . . 61

3.2 T h e ca tegory CA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.3 A - m o d u l e s a nd r e p r e s e n t a t i o n s of CA . . . . . . . . . . . . . . . . . . . 71

T h e a l g e b r a a s s o c i a t e d t o a G r e e n f u n c t o r 81

4.1 T h e eva lua t ion func tors . . . . . . . . . . . . . . . . . . . . . . . . . . 81

4.2 E v a l u a t i o n and equiva lence . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.3 T h e a lgebra A(f/2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

4.4 P r e s e n t a t i o n by genera to r s and re la t ions . . . . . . . . . . . . . . . . . 85

4.5 E x a m p l e s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

VI C O N T E N T S

4.5.1 T h e Mackey a lgebra . . . . . . . . . . . . . . . . . . . . . . . . 94

4.5.2 T h e Yoshida a lgebra . . . . . . . . . . . . . . . . . . . . . . . . 95

M o r i t a e q u i v a l e n c e a n d r e l a t i v e p r o j e c t i v i t y

5.1

5.2

5.3

5.4

5.5

99

M o r i t a equ iva lence of a lgebras A ( X 2) . . . . . . . . . . . . . . . . . . . 99

Re la t ive p ro j ec t i v i t y . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

C a r t e s i a n p r o d u c t in CA . . . . . . . . . . . . . . . . . . . . . . . . . . 103

5.3.1 Def in i t ion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

5.3.2 A d j u n c t i o n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

5.3.3 C a r t e s i a n p r o d u c t in CA x CA . . . . . . . . . . . . . . . . . . . 109

M o r i t a equ iva lence and re la t ive p ro j ec t i v i t y . . . . . . . . . . . . . . . 112

P r o g e n e r a t o r s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

5.5.1 F in i t e ly gene ra t ed modu le s . . . . . . . . . . . . . . . . . . . . . 114

5.5.2 I d e m p o t e n t s and p rogene ra to r s . . . . . . . . . . . . . . . . . . 115

C o n s t r u c t i o n o f G r e e n f u n c t o r s 123

6.1 T h e func to r s H ( M , M ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

6.1.1 T h e p r o d u c t 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

6.2 T h e oppos i t e func to r of a Green func to r . . . . . . . . . . . . . . . . . 127

6.2.1 R igh t modu le s . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

6.2.2 T h e dua l of an A - m o d u l e . . . . . . . . . . . . . . . . . . . . . . 130

6.3 Tensor p r o d u c t of Green func to r s . . . . . . . . . . . . . . . . . . . . . 134

6.4 B i m o d u l e s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

6.5 C o m m u t a n t s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

6.6 T h e func to r s M | N . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

A M o r i t a t h e o r y 153

7.1 C o n s t r u c t i o n of b i m o d u l e s . . . . . . . . . . . . . . . . . . . . . . . . . 153

7.2 M o r i t a c o n t e x t s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

7.3 Converse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

7.4 A r e m a r k on b i m o d u l e s . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

Composition 8,1

8.2

8.3

8.4

8.5

8.6

8.7

167 Bise ts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

C o m p o s i t i o n and tensor p r o d u c t . . . . . . . . . . . . . . . . . . . . . . 168

C o m p o s i t i o n and Green func tors . . . . . . . . . . . . . . . . . . . . . . 170

C o m p o s i t i o n and assoc ia ted categories . . . . . . . . . . . . . . . . . . 173

C o m p o s i t i o n and modu le s . . . . . . . . . . . . . . . . . . . . . . . . . 175

F u n c t o r i a l i t y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

E x a m p l e : i n d u c t i o n and res t r i c t ion . . . . . . . . . . . . . . . . . . . . 180

A d j o i n t c o n s t r u c t i o n s

9.1

9.2

9.3

9.4

9.5

9.6

183

A left ad jo in t to t he f lmctor Z ~-+ U OH Z . . . . . . . . . . . . . . . . . 183

T h e ca tegor ies D u ( X ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

T h e func to r s Q u ( M ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

T h e func tors L u ( M ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

Left a d j u n c t i o n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

T h e func tors S u ( M ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

CONTENTS VII

9.7

9.8

9.9

T h e func to r s Ru(M) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

R i g h t a d j u n c t i o n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

E x a m p l e s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

9.9.1 I n d u c t i o n and res t r i c t ion . . . . . . . . . . . . . . . . . . . . . . 215

9.9.2 Inf la t ion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

9.9.3 Coinf la t ion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

10 A d j u n c t i o n a n d G r e e n f u n c t o r s

10.1

10.2

10.3

10.4

10.5 10.6

10.7

10.8

223 Froben ius m o r p h i s m s . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

Left ad jo in t s and t ensor p r o d u c t . . . . . . . . . . . . . . . . . . . . . . 227

T h e Green func tors L u ( A ) . . . . . . . . . . . . . . . . . . . . . . . . . 231

Lu(A)-modules and a d j u n c t i o n . . . . . . . . . . . . . . . . . . . . . . 234

R igh t ad jo in t s and tensor p r oduc t . . . . . . . . . . . . . . . . . . . . . 242

Ru(M) as L u ( A ) - m o d u l e . . . . . . . . . . . . . . . . . . . . . . . . . . 250

Lu(A)-modules and r igh t ad jo in t s . . . . . . . . . . . . . . . . . . . . . 255

E x a m p l e s and app l i ca t ions . . . . . . . . . . . . . . . . . . . . . . . . . 264

10.8.1 I n d u c t i o n and res t r i c t ion . . . . . . . . . . . . . . . . . . . . . . 264

10.8.2 T h e case U/H = �9 . . . . . . . . . . . . . . . . . . . . . . . . . 264

10.8.3 A d j u n c t i o n and M o r i t a con tex t s . . . . . . . . . . . . . . . . . . 266

11 T h e

11.1

11.2

11.3

11.4

11.5

11.6

s i m p l e m o d u l e s 275

Genera l i t i e s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275

Class i f ica t ion of the s imple modules . . . . . . . . . . . . . . . . . . . . 275

T h e s t r u c t u r e of a lgebras / i ( H ) . . . . . . . . . . . . . . . . . . . . . . 278

T h e s t r u c t u r e of' s imple modules . . . . . . . . . . . . . . . . . . . . . . 282

11.4.1 T h e i s o m o r p h i s m SH,v(X) ~-- Hom([XH] , V)~ G(H) . . . . . . . . 282

11.4.2 T h e A - m o d u l e s t r u c t u r e of SH,V . . . . . . . . . . . . . . . . . . 289

T h e s imple Green func tors . . . . . . . . . . . . . . . . . . . . . . . . . 291

S imple func tors and e n d o m o r p h i s m s . . . . . . . . . . . . . . . . . . . . 295

12 C e n t r e s

12.1 T h e cen t re of a Green func tor

12.2

12.3

305 . . . . . . . . . . . . . . . . . . . . . . . 305

T h e func to r s CA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 12.2.1 A n o t h e r ana logue of the cen t re . . . . . . . . . . . . . . . . . . 315

12.2.2 E n d o m o r p h i s m s of the r e s t r i c t ion func to r . . . . . . . . . . . . . 323

12.2.3 I n d u c t i o n and inf la t ion . . . . . . . . . . . . . . . . . . . . . . . 329

E x a m p l e s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332

12.3.] T h e func tors FPB . . . . . . . . . . . . . . . . . . . . . . . . . . 332

12.3.2 T h e blocks of Mackey a lgebra . . . . . . . . . . . . . . . . . . . . 335

Bibl iography 337

Index 339

Introduct ion

The theory of Mackey functors has been developed during the last 25 years in a series of papers by various authors (J .a . Green [8], a . Dress [5], T. Yoshida [17], J. Th~venaz and P. Webb [13],[15],[14], G. Lewis [6]). It is an at tempt to give a single framework for the different theories of representations of a finite group and its subgroups.

The notion of Mackey functor for a group G can be essentially approached from three points of view: the first one ([8]), which I call "naive", relics on the poset of subgroups of G. The second one ([5],[17]) is more "categoric", and relies on the category of G-sets. The third one ([15]) is "algebraic", and defines Mackey functors as modules over the Mackey algebra.

Each of these points of view induces its own natural definitions, and the reason why this subject is so rich is probably the possibility of translation between them. For instance, the notion of minimal subgroup for a Mackey functor comes from the first definition, the notion of induction of Mackey functors is quite natural with the second, and the notion of projective Mackey functor is closely related to the third o n e .

The various rings of representations of a group (linear, pernmtation, p-permutation...), and cohomology rings, are important examples of Mackey functors, having moreover a product (tensor product or cup product). This situation has been axiom- atized, and those functors have been generally called C-functors in the literature, or Green functors.

This definition of a Green functor t o r a group G is a complement to the "naive" definition of a Mackey fnnctor: to each subgroup of G corresponds a ring, and the various rings are connected by operations of transfer and restriction, which are compatible with the product through Frobenius relations.

The object of this work is to give a definition of Green functors in terms of G- sets, and to study various questions raised by this new definition. From that point of view, a Green fimctor is a generalized ring, in the sense that the theory of Green functors for the trivial group is the theory of ordinary rings. Now ring theory gives a series of directions for possible generalizations, and I will treat some cases here (tensor product, bimodnles, Morita theory, commutants, simple modules, centres).

The first chapter deals only with Mackey functors: my purpose was not to give a full exposition of the theory, and I just recall the possible equivalent definitions, as one can find for instance in the article of Thevenaz and Webb ([15]). I show next how to build Mackey functors "with values in the Mackey functors", leading to the functors 7-{(M, N) and M@N, which will be an essential tool: they are analogous to the homomorphisms modules and tensor products for ordinary modules. Those constructions already appear in Sasaki ([12]) and Lewis ([6]). Thc notion of r~dinear map can be generalized in the form of r~-linear morphism of Mackey functors. The

2 INTROD UCTION

reader may find that this part is a bit long: this is because I have tried here to give complete proofs, and as the subject is rather technical, this requires many details.

Chapter 2 is devoted to the definition of Green functors in terms of G-sets, and to the proof of the equivalence between this definition and the classical one. It is then possible to define a module over a Green functor in terms of (-;-sets. I treat next the fundamental case of the Bm-nside functor, which plays for Green functors the role of the ring Z of integers.

In chapter 3, I build a category CA associated to a Green functor ,4, and show that the category of A-modules is equivalent to the category of representations of CA. This category is a generalization of a construction of Lindner ([9]) for Mackey functors, and of the category of permutation modules studied by Yoshida ([17]) for cohomological Mackey functors.

Chapter 4 describes the algebra associated to a Green functor: this algebra enters the scene if one looks %r G-sets ~ suct~ that the evaluation functor at ft is an equivalence of categories between the category of representations of Cn and the category of Endc~(f~)-modules. This algebra generalizes the Mackev algebra defined by Theve- naz and Webb ([1.5]) and the Hecke algebra, of Yoshida ([17]). It is possible to give a definition of this algebra by generators and relations.

This algebra depends on the set f/, but only up to Morita equivalence. Chapter 5 is devoted to the relation between those Morita equivalences and the classical notion of relative projectivity of a Green functor with respect to a G-set (see for instance the article of Webb [16]). More generaliy, I will deduce some progenerators for the category of A-modules.

Chapter 6 introduces some tools giving new Green functors from known ones: after a neat description of the Green functors ~(/11,/1I), I define the opposite functor of a Green flmctor, which leads to the notion of right module over a Green functor. A natural example is the dual of a left module. The notion of tensor product of Green functors leads naturally to the definition of bimodule, and the notion of comnmtant to a definition of the Mackey functors 7t.4(M, N) and M(~,4N.

Those constructions are the natural framework for Morita contexts, in chapter 7. The usual Morita theory can be generalized without difficulty to the case of Green functors for a given group G.

The chapters S,9, and 10 examine the relations between Green functors and bisets: this notion provides a single framework for induction, restriction, inflation, and coinflation of Mackey functors (see [2]).

In chapter 8, I show how the composition with U, if U is a G-set-H, gives a Green functor A o U for the group H starting with a Green functor A for the group G. This construction passes down to the associated categories, so there is a corresponding functor from CAoU to Ca. This gives a functor between the categories of representations, which can also be obtained by composition with U. I study next the functoriality of these constructions with respect to U, and give the example of induction and restriction.

Chapter 9 is devoted to the construction of the associated adjoint functors: I build a left and a right adjoint to the functors of composition with a biset /14 ~ 114 o U for Mackey fnnctors, and I give the classical examples of induction, restriction and inflation, and also the less well-known example of coinfiation.

Chapter 10 is the most technical of this work: I show how the previous left adjoint

INTROD UCTION 3

functors give rise to Green functors, and I study the associated functors and their adjoints between the corresponding categories of modules. An important consequence of this is the compatibility of left adjoints of composition with tensor products, which proves that if there is a surjective Morita context for two Green functors A and B for the group G, then there is one for all the residual rings A(H) and B(H), for any subgroup H of G.

In chapter 11, I classify the simple modules over a Green functor, and describe their structure. Applying those results to the Green functor A@A ~ I obtain a new proof of the theorem of Th4venaz classifying the simple Green functors. Finally, I study how the simple modules (or similarly defined modules) behave with respect to the constructions ~ ( - , - ) and - Q - .

Chapter 12 gives two possible generalizations of the notion of centre of a ring, one in terms of commutants, the other in terms of natural transformations of functors. The first one gives a decomposition of any Green functor using the idempotents of the Burnside ring, and shows that up to (usual) Morita equivalence, it is possible to consider only the case of Green functors which are projective relative to certain sets of solvable rr-subgroups. The second one keeps track of the blocks of the associated algebras. Then I give the example of the fixed points functors, and recover the isomorphism between the center of Yoshida algebra and the center of the group algebra. Next, the example of the Burnside ring leads to the natural bijection between the p-blocks of the group algebra and the blocks of the p-part of the Mackey algebra.

Chapter 1

Mackey functors

All the groups and sets with group action considered in this book will be finite.

1.1 Equivalent definitions

Throughout this section, I denote by G a (finite) group and R a ring, that may be non-commutat ive. First I will recall briefly the three possible definitions of Mackey functors: the first one is due to Green ([8]), the second to Dress ([5]), and the third to Th6venaz and Webb ([15]).

1.1 .1 D e f i n i t i o n in t e r m s o f s u b g r o u p s

One of the possible definitions of Mackey functors is the following:

A Mackey functor for the group G, with values in the category R - M o d of R-modules, consists of a collection of R-modules M ( H ) , indexed by the subgroups H of G, to-

ll M ( H ) --+ M ( K ) whenever Ir is a gerber with maps t H : M ( K ) --+ M ( H ) and r K : subgroup of H, and maps Cc,H : M ( H ) , M ( ~ H ) for x 6 G, such that:

H K A ' H �9 If L C_ t ( C_ H, then t h - t L : t H and r L r K = r H .

�9 If x, y E G and H G G, then CyjHCx, H : Cyx, H.

XH H = 7,~H Moreover �9 If x E G and H C_ G, t h e n Cx,H tH = t . K C x , K and cx,icr K xKceG H.

e~,H = I d if x E H.

�9 (Mackey axiom) If L C H _D K, then

H H L K FL tA" E = ~LnxKCx,LXAA-FLxnA.

x E L \ H / K

H The maps tK H are called transfers or traces, and the maps r K are called restrictions.

A morphism 6 from a Mackey functor M to a Mackey functor N consists of a collection of morphisms of R-modules OH : M ( H ) --+ N ( H ) , for H C_ G, such that if

6 CHAPTER I. MACKEY FUNCTORS

K C_ H and x E G, the squares

Oi< 0~< OH M(I<) - , N (K) M(K) , N (K) M(H) , N(H)

M(H) , N(H) M(H) , N(H) M(~H) , N(~H) OH OH OzH

are commutat ive.

1 .1 .2 D e f i n i t i o n in t e r m s o f G-se t s

If K and H are subgroups of G, then the morphisms of G-sets from G / K to G/H are in one to one correspondence with the classes xH, where x E G is such that K ~ C H. This observation provides a way to extend a Mackey functor M to any G-set X, by choosing a system of representatives of orbits G \ X , and defining

M ( X ) = 0 M(G~) x6G\X

There is a way to make this equality functori~l in X, and this leads to the following

definition:

Defini t ion: Let R be a ring. If G is a (finite) group, let G-set be the category of finite sets with a left G action. A Mackey flmctor for the group G, with values in R-Mod, is a bifunctor from G-set to R-Mod, i.e. a couple of flmctors (M*, M.) , with M* contravariant and M. covariant, which coincide on objects (i.e. M*(X) = M. (X) = M ( X ) for any G-set X). This biflmctor is supposed to have the two following properties:

�9 (M1) If X and Y are G-sets, let ix and iv be the respective injections from X and Y into X [I Y, then the maps M*(ix) | M*(iv) and M.( ix) | M.(iv) are mutua l inverse R-module isomorphisms between M ( X LI Y) and M ( X ) | M (Y).

�9 (M2)~f

T ~ Y

'1 1 o Z , X

~s a cartesian (or pull-back) square of G-sets, then M*(/3).M.(a) = M.(5) .M*(7) .

A morphism 0 from the Mackey functor M to the Mackey functor N is a natural transformation of bifunctors, consisting of a morphism Ox : M ( X ) ~ N ( X ) for any G-set X, such that for any morphism of G-sets f : X --+ Y, the squares

Ox Ox M(X) ., N(X) M(X) , N(X)

M ( Y ) , N(Y) M(Y) , N (Y) Oy Oy

1.1. EQUIVALENT DEFINITIONS 7

are commutative. I will denote by MackR(G) or Mack(G) the category of Mackey functors for G over

R. Conversely, if M is a Mackey functor in the sense of this second definition, then one can build a Mackey functor M, in the first sense by setting

MI(H) = M ( G / H ) t H = M.(p~) r~ = ~vJ . . . . (pK)H' C~,H = M.(7~,H)

where pH : G / K ~ G / H is the natural projection, and 7~,H : G / H ~ G f f H is the map gH ~-* gx-I .~H = gHx -1

1.1 .3 D e f i n i t i o n as m o d u l e s over t h e M a c k e y a l g e b r a

There is a third definition of a Mackey fnnctor, using the Mackey algebra #n(G): consider first the algebra #(G) over Z: it is the algebra generated by the elements t~,

H and cx,u, where H and K are subgroups of G such that K C H, and x 6 G, with r K ~

the following relations: H K t~t L = t ~ V L C K C _ H

K H_rH V L C K C H Y L T K _ _

Cy,zHCx, H = Cyx, H W x~ y , H

t~ = rg = ch,H g h, H h E H

C~,Ht~ = t;~-c~,K g x, K, H

Cx,KVH ~H = r~KC~,H V x , K , H

E t". = E = 1 H H

rKtLH H , K C T L = ~ V K C H D L ~Kn~L x ,K~nL K~nL _ _

x G K \ H / L

any other product of r~, tK H and cg.u being zero.

A Mackey functor M for the first definition gives a module M for the "algebra" #R(G) = R | It(G) (which is not really an algebra if R is not commutative) defined by

M = | M(H) HCG

and a morphism f : M --~ N of Mackey functors gives a morphism of #n(G)-modules

It it then possible to define a Mackey functor as a ItR(G)-module, and a morphism of Mackey functors as a morphism of Itn(G)-modules: if M is a #R(G)-module, then M corresponds to a Mackey functor Mi in the first sense, defined by MI(H) = t ~M,

H and C,.H being defined as the multiplications by the corresponding the maps t~, r K elements of the Mackey algebra.

8 CHAPTER 1. M A C K E Y F U N C T O R S

1 . 2 T h e M a c k e y f u n c t o r s M ~ M y

If Y is a G-set, and M is a Mackey functor for G, then let My be the Mackey functor defined by

My(X) = M(X • Y)

and for a map of G-sets f : X --~ X', by

(My)*(f) = M ' ( f • Id)

(My) . ( f ) = M~(f • Id)

This construction is functorial in Y: if Y' is another G-set, and if g is a morphism of G-sets from Y to Y', then there is a morphism of Mackey functors Mg : My --~ My,, defined over the G-set X by

M~,x = M,(Id • ~): My(X) --~ My,(•

To see this, let f : X --* X ' be a map of G-sets. Then the square

Mg,x My(X) , My,(X)

(Mv).( f ) I I (Mz ' ) . ( f ) ) Mv(X') Mg,x, Mz,(X')

is commutative, since M , ( f • Id) o M.(]d • g) =/14. ( f • g) =/14. (Id • g) o M . ( f x ]d). The square

Mg,x My(X) , My,(X)

(MY)*(f) I l (Mv')*(f)

My (X') Mz, (X ' ) )

Mg,X'

is also commutative, because the square

X x g X • , X •

f • l f x l d

X~ • y , X' x Y ~ I d x 9

is cartesian. There is also a morphism Mg from My, to Mz defined over X by

M} = M*(Id • g) : My,(• ~ My(X)

In other words, I have defined a bifunctor from G-set to the category MackR(G) of Mackey functors for G over R, which is equivalent to #R(G)-Mod. I will check the conditions (M1) and (M2) for this bifunctor, proving that Y ~ M r is a Mackey functor with values in the category of Mackey functors.

1.3. CONSTRUCTION OF H(M, N) AND M@N 9

For the condition (M1), let Y and Y' be G-sets, and i and i' be the respective injections from Y and Y' into Y II Y'. If X is a G-set, then

" ( " ' ) ) ( . . . . '1) (M~,x|174 = M.(Idx~)| o M (Idxz)@M (Idxz = Id

and it it also clear that

(M k. @ M~) o (Mi,.,. @ Mi, x) = Id

For the condition (M2), let

T ~ Y

'1 1 ~ Z , U

9 be a cartesian square. Then for any G-set X

As the square

is cartesian, I have

M~x o M~,x = M*(Id x ~) o M.(Id x a)

Id x ~/ X x T , X x Y

[ d x 5 1 I Id•

X x Z X x U I d x ~

M*(Ia • 8) o M.(Zd • ~) - M.tId • ~) o M*(Id • -~)

and so M~x o M~,x = M~,x o M}

So I have defined a Mackey functor with values in the Mackey functors. If now A is a ring, and F is a functor from MackR(G) to A - M o d , then I will get

a Mackey functor _F with values in A - M o d , defined for G-sets X and Y, and for a map f of G-sets from X to Y, by

F(X) = F(Mx) F*(f) = F(MJx) F . ( f ) = F(Mj,x)

1 . 3 C o n s t r u c t i o n o f 7 ~ ( M , N ) a n d M @ N

From now on, the ring R will be commutative. There are two impor tant examples of the previous construction: in both cases, let ]14 and N be Mackey functors for G with values in R - M o d . For the first construction, I consider the functor

F = HomM~ck(C)(N,-)

and I can define the Mackey %nctor ~ ( N , M), with values in R - M o d , by the following

formulae 7-{(N, M)(X) = nomM~ck(a)(X, Mx)

10 CHAPTER i. MACKEY FUNCTORS

"}-{(N, M)*(f) = HomMack(G)(N , MIx) 7-{(N, M).(f) = HomMack(G)(N, MLx )

It is a kind of "internal homomorphisms" in the category of Mackey functors. For the second construction, I need to observe that the Mackey algebra admits a

natural anti-automorphism rn ~ ~ , defined by

This allows to view any left #R(G)-module as a right #R(G)-module. Now being given a Mackey functor N, I can consider the functor

F : M H M ~ , R ( a ) N

The functor _P is then denoted by M@N: it is a Mackey functor with values in R-Mod, such that for any G-set X

(M+N)(X) = M @,R(G)

I will call it the tensor product of the Mackey functors M and N.

1.4 I d e n t i f i c a t i o n of 7 / ( M , N )

If H is a subgroup of G, then there is a restriction functor from Mack(G) to Mack(H), and an induction functor from Mack(H) to Mack(G), defined over a G-set X by

(ResaM)(X) = M(Ind•X) (Ind~N)(Y) = N(ResgY)

As functors between categories of Mackey functors, they are mutual left and right adjoints (see [14]). Moreover, the isomorphism of G-sets

X x (G/H) ~_ Ind~Res~X

gives the following isomorphism of Mackey functors

Mc/H ~-- Ind~Res~M

Those remarks prove that

~ (M, N)(H) = HomM~k(a)(M, Na/H) "~ HomM,~k(H)(ResaHM, aes~4N)

I will translate the operations of transfer and restriction for 7-/(M, N) using the previous identification.

I must first state precisely the adjunction

HomM~r Na/H) ~ Homu~ck(H)(ResaM, ResiN)

i morphism r E Homu~k(H)(Res~i, ResiN) gives in particular for any subgroup K of G, a morphism r from M(K) to N(K). The adjunction now gives a morphism ~* from M to Ind~ReSaH N defined on the subgroup L of G by the morphism ~ from i ( t ) to Ind, Res /(t) n given by

* z L rr~ =

1,4. IDENTIFICATION OF H(M, N) 11

Conversely, if r E HomM~k(a)(M, Ind~Res~N), then r gives morphisms

r : M(L) ---* (~eH\a/LN(H (1 ~L)

so I get a morphism from M(L) to N(H Cl ~L). The adjunction gives a morphism ~. from ResaM to Res iN: if K C_ H, the morphism r from M(K) to N(K) is the component on the double class H.1.K = H in the above map.

Now if H D_ H' D_ K, and if m E M(K), I have

K

Restricting to NG/H,(K), I get

H * t rH, d)K(m ) E (~yeH,\a/h-g(H N Yti)

The component associated to y is obtained as follows: if x is the representative of H\G/K, then there exists h E H and/c E K such that y = hxk. The y component is

HnYK K then equal to In particular, the component y = 1 comes fi'om the component x = 1, and the morphism from Res~,M to Res~,N associated to rH,r is defined by

= =

If now H C H ' __D It', and if m E M(K), then (tH'r is an element of Na/H,(K) = | M ~K), and the component y = 1 comes from the dou- bles classes x E H\H' /K . The homomorphism from Resg, M to Resg, N associated to tH'r is then defined by

H t x tHn~KCun~i~.XrH~nK(m )

xEH\H'/K

that is H'

= tH~nKX CHn~KxrH~nK(m) xEH\H'/K

Finally, it is clear that if x E G, then the morphism xr from Resau M to ReSfHN is defined for K C_ ~H and m E M(K) by

=

The next proposition is a summary of the previous remarks:

P r o p o s i t i o n 1.4.1: Let M and N be Mackey func tors for the g roup G. T h e n for any subg roup H of G,

7-{(M, N)(H) ~_ HomM~ck(H)(Res~M, Res iN)

An e l emen t r in 7-I(M,N)(H) is defined by a sequence of m o r p h i s m s eL E Homn(M(L),N(L)), for L C H. Then:

12 CHAPTER 1. M A C K E Y FUNCTORS

�9 I f H C_ H' __D K, t h e n

H I (tH r =

�9 I f K C_ H' C_ H, t h e n

E A" -1 K

(~,r : e , , ( , J

o r

�9 I f x E G a n d KC_XH, t h e n

(xr = xeA~(x-l~)

Remark : This proposition shows that H(M,N) coincides with the construction called the exponent by Sasaki (see [12]).

1.5 Ident i f i ca t ion of M@N Let M and N be Mackey functors for G. a) The definition of M@N shows that

M@N(H) = M | NG/H

But NC/H ~-- Ind/~ResTzN, and the restriction functor from MackR(G) to Mackn(H) comes from the natural morphism i,~ from the algebra #R(H) to #R(G). In other words, if M is a Mackey functor for G, then

Res~M = i/~(l,Mg))./h r

The induction functor from MackR(H) to Mackn(G) is left adjoint to it. Then for any Maekey functor L for H, I ha.re

Ind~L = #R(G) | ~"

This proves that

M@N(H) = /~ | #R(G) | ResGH N = M | ZH(I~,MH)).N.O --

N

M@N(H) = /~i~(I~,(H)) | = Res~M | Res iN

since i~(1,,(H)) is invariant under the antiautomorphism of #n(G).

b) I also claim that

M | N _~ (~ .ccM(H) 0R N(H)) /5

where Z is the R-submodule generated by the elements of the form

t L.m | n - m | r Ln for K C i C G, m E M(K) , n 6 N(L)

z E H \ H ' / A

1.5. IDENTIFICATION OF M@N 13

' m @ n t f n - r ) ' m | for L C K CG, rnE M(K) , h E N ( L )

gT)? C?4R gn -- m @R n for Is C_ G, rn E M(K) , n E N(K) , g E G

To see this, let, P be the right hand side, and 7r be the projection from @HC_Gy(H)@R N(H) onto P. There is a map 0 fi'om ~)Uc_GM(H) | N(H) to M | N, sending

m ,~n n C M(H) | N(H) to m.~ ,R tG)~ E M Q , ~ ( c , ) N , since M(H) C_ M and

N(H) C_ N. It is easy to see that this factors through P, since for instance

L . . . . 77~'r'Is @~n(G) ~ - - rN @.R(G) FL"Tt = 0

Conversely, let 0' be the map from A4 @~/G/ N- to P which sends ra | n to the image in P of @uc_at~m @n tuur~. This map is well defined, because if a = tAzrC,. E #n(G), then

" ' " _ " H !

I have also

0 (m ~,~{o)a . , , )

and this proves that 0' is well defined. Now if ,z E M(H) and ~r ~ N ( H ) , then

K

So 0'0 is the ident i ty . And if m E M and n E ~v, then m = ~Ifc(; t~rn, and n = ~I<ca tl4r h1~" which gives

72z @~,R(G) 7), Z H = Ix = tH'tn <%R(G) tIs . . . . HC(; I:'CG

On the other hand

"]'( ? :

H C G H C G KCG

H C G

So 00' is also equal to identity, and this proves the claim.

c) Points a) and b) show that


where ,7 is the submodule generated by

tL-m | n - m | r~-n

hm | hn - m | n for

for K C L C H , m E M ( K ) , n E N ( L )

for L C K C_ H, m E M(K) , n G N ( L )

K C_H, m E M(K) , n 6 N(K) , hE H

d) In order to obtain formulae for transfer and restriction in M@N, I must first describe precisely the isomorphism 0 of a) giving M@N(H) as

| Ind~,Resi N -~ Resell M |

If K is a subgroup of H, then M ( K ) (resp. N(K) ) maps into R e s I M (resp. into

Res iN) , so there is a morphism 7r from | | N (K) to the right hand side. This morphism is surjective, because__if m 6 M ( K ) and m' 6 M(K ' ) for I f 7 ~ K',

then m | m' = 0 in Res~M | Res iN-

Now M ( K ) maps in to /~ , and N(K) maps into IndIResiN(I(), so into Ind',ResiN. So there is a morphism 7r' from |162174 to the left hand side. The above isomorphism is the given (from right to left) by

If now K is any subgroup of G, then

Na/H(K) = I n d I R e s I N ( K ) ~- @=eH\a/KN(H N =K)

If n E N ( H N =K) C NC/H(K), let n' be the element x-b~ E N ( H = N K), viewed inside NG/H(H ~ N K). In N~/H(K), I have then

K ! n = ~H~nKrt

Then if m 6 M(K) , I have the following equalities in M e,~(@ Ind~Res iN

t K , - rn @an(G) n = rn ~UR(G) H=nh "n = m.t~I=nK @un(a) n' = .. .

-= rK x r ~ m h m | n �9 H~nK m @~n(G) x - in =

and this element is in the component M ( H N =K) | N ( H N =K) of R e s I M |

Res iN. It is the image under 0 of 7r'(m | n). Those predsions give explicit formulae of transfer and restriction for M@N: let H c H' be subgroups of G, and K be a subgroup of H. If [m | n]A denotes the image in M@N(H) of m | n E M ( K ) | N(K), then 0-~([m | n]K) is the image of m | n in

2~ | I n d I R e s I N ( g ) , which is a quotient of eLCO e~H\V/L M(L) | N ( H N ~n):it is the image of the element m | n of the component corresponding to L = K and x = H.1.K. By the transfer from H to H', this element maps to the image of the element

lAX

1.5. IDENTIFICATION OF M@N 15

of the component corresponding to L = K and x = Hql.L, which in turn is the image under 0 -1 of the element [m @ n]/~- of the component K C_ H' of M@N(H'). Finally

H t

If now K is a subgroup of H', then 0-~([m | n~K) is the image of the element m @ n of the component L = K and x = H ' . t . K of M @,Mc) Na/H,. Its restriction to H is the image of

@zcH\H'/L'~ @ x1"K~nKrt

corresponding to the double classes H\H'.I.K/K. The image under 0 of this element is the element

Ix" @ xrK nKr~ @xEH\H'/KXr HxNK?7~

In other words

H t [xrH~nKrn @ xrH~nKn]Hn~K xEH\H'/K

Fina,lly, if x E G, conjugation by z from MQN(H) to MON(~H) is given by

I have proved the following:

P r o p o s i t i o n 1.5.1: Let M and N be Mackey functors for the g roup G. T h e n for any subg roup H of G

( M + N ) ( H ) ~_ (e,,-~_H M(r , ' )Q~ N(K))/J

where ,7 is the R - s u b m o d u l e gene ra t ed by

tL-m|174 f o r K C L C H , m E M ( K ) , hEN(L)

m@ntI[n--r[mORnfor L C_ K C H , m E M ( K ) , nE N(L)

hm@Rhn-m| m E M ( K ) , hEN(K), h E H

Let [m | ~]K deno te the image in MQN(H) of m @ n E M(K) | N(K) . T h e n

�9 If H _C H' _D K, then

H'(E ) r H m | n ] l , =

�9 If K C_ H _C H', t h e n

�9 I f x E G , t hen

~EH\H'Irr

16 CHAPTER i. MACKEY FUNCTORS

1.6 A n o t h e r ident i f i cat ion of M @ N

The first identification of M@N gives M'.~)N(H) as a quotient of @Kc_HM(I() @R N(K). This expression is not very useful to evalnate M@N(X) for an arbi t rary G-set X. If K is a subgroup of H, there is a corresponding "G-set over G/H", defined as the natural project ion p~ : G/K --+ G/H. More generally, a G-set Y over the G-set X is a couple (Y, 6), where Y is a G-set and ~ a map of G-sets from Y to X. A morphism from (Y, r to (Y', ~') is a morphism of G sets f from Y to Y' such that r = r The next idea is then to consider for any (;-set X

T(X) = ( , \(Y))/ j Y ~ X

where J is the R-submodule generated by tim elements

M+(f)(m ') | n - m' Q N.(f)(r+), M . ( f ) ( m ) @ n' - m @ X*(f)(n')

V f : (Y, r --+ (Y', r Vm E M(Y) , m' E M(Y'), n E N(Y) , n' E N(Y')

(Note that there is no serious set theoretical problem in this definition, since the category of finite G-sets over X admits a small equivalent subcategory).

If rn ~ n r M(Y) | N(Y) , I will denote by [rn | n]ir,+/ its image in T(X).

L e m m a 1.6.1: W i t h those definit ions

T(G/H) Mr W C_ G

Proof: There is indeed a natural map 0 from M @ N ( H ) to T(G/H), which sends [ m | n]K to [m @ n](a/,Q,~): this map is well defined, because M,(pI~ 1~.) = t,~ and

M*(p~) H whereas M , ( x ) i s conjugation by x, and M*(x)is conjugaison by x - ' = r A - ~

Conversely, let (Y,r be a G-set over G/II. As G is transit ive on G/m there is a system of representatives S of G\Y contained in r In part icular G, c_ H for all s E S. I denote by i+ the injection g.G+ ~ g.s from G/G~ to Y, and for m E M(Y) (resp. n C N(Y)) , I write m, = M*(i~)(rn) (resp. n, = N*(ij(n)) . The map m ~ %~esrn~ is an isomorphism from M(Y) to ~esM(G+). I define a map 0' from T(X) to MQN(H) by

e, = Z [ " + e n+]< sES

The map 0' does not depend on the choice of .5' inside q5-1 (H): if S ' is another system, then for any s' E S' , there exists s r .5'and h+ E H such that s ' = h~.s. In those conditions, I have ms, = hsrr+~, and fins, | n+,]a+, = [hsm~ | hsns]a+, = [m~ | n~]a, in M(~N(H). The map 0' is also well defined: if f : (Y, r --+ (Y', r is a morphism of sets over G/H, I have to check that

O'(M.(f)(m) (9 n' - rn 0 N*(f)(n')) = 0

1.6. ANOT HER fDENTIFICATION OF M<~N 17

and t h a t

O'(M*(f)(rn') @ n - m' Q N,(. f )(n)) = 0

Let S ' C_ ~ ' -~ (H ) be the sys tem of representa t ives of G\Y ' chosen for Y'. Then [o1" any s E S, there exists a unique t~ E S' and an e lement h, E H such tha t f ( s ) = b.~.G. In those condi t ions

O'([M.(f)(rn) | n'](v, r = G~,e.s,[M'(i~,)M.(f)(rn,) ~, N*(i~,)(n)]G/

Let P be the pul l -back of Y and G/G~,, such tha t the square

P ~ Y

t i c < , , Y '

is car tes ian. Then P ~_ {(g.G~,,g) E G/G~ x Y I f(Y) : g .s '} . In par t icu la r , i f s ' gt f ( Y ) , then P is e m p t y and M*(i~,)M,(f) = 0 in this case. And if f (9) = g.s', let s E ,5' C? (G..s). Then there exists gl such tha t y = g~.'s, so f ( s ) = 9~lg.s ', and this proves in par t i cu la r tha t s' = G. Conversely, if t~ = s' and y = g.s, then f ( y ) = gh~.G and (gh~G,,, y) E P. So P may be identif ied as the union of the orbi ts G.s such tha t t~ = s ~. I have the following car tes ian square

His

t s=J

G/G~, - , Y'

where % ( g . G j = gh..G.,. This shows tha t

0 ' ( [ M . ( f ) ( m ) G n'](z,,r = ~ [ M . ( % ) M * ( i j ( m ) Q N*(it.)(rt')]~:~ sES

Gt But M . ( % ) = h~ta/, and so in M~N((;~,)

* �9 * �9 / - - 1 * " / [M.(~.)M (~)(.~) 0 N (~,.)(,~)b,, = [M'(i~)(.~) o ~:'< N (~,.)(,~)b,

and moreover

a,l - | . ,

"c;~ h., N (z<.) = N*(%)N*(it ,) = N*(i<.%) = N * ( f i J = N * ( i j N * ( f )

whence final]),

O'([M.(f)(rn) • n'](r,,r = ~ [ M * ( i J ( r n ) | N*(is)N*(f)(n')]a, = . . . sE,q'

. . . . 0'([.~, c0 N*(f)(W)](v,r

18 CHAPTER 1. MACKEY FUNCTORS

A similar argument shows that

O'(M*(f)(m') (4 ,z - m' ~% N.(f)(n)) 0

and so O' is well defined. It is cleat" that 0'0 is equal to identity of M<~;N(H). Conversely,

O0'([rn | n](y,r = ~ [ M ' ( i ~ ) ( m ) G, N'(ij(n)](ClC,,e)i,) .s6S

But Y is isomorphic to the disjoint union of the GIGs, so

m = ~ M.(i.)M*(i~)(m) sES

so in T(X), I have

[ m e n](y,r = ~ [V*(i~)(m) r M*(i.)M.(it)M*(it)(n)](a/a.,r s,t6S

and as M is a Mackey functor, the product M*(i~)M.(it) is zero if s # t, and equal to the identity if s -- t. So 00' is equal to the identity, and this proves the lemma. ..

I can now give T a structure of bifunctor: let X and X ' be G-sets, and f be a morphism from X to X'. Then any set (Y, r over X defines a set (Y,.fr over X': I can define

T.(f) ([m (9 n]O.-,r : [m | nJ(z, ye~)

The map T.( f ) is well defined, because if g : (}; r + (Y', #) , if m E M(Y) and n' E N(Y') , then

T.(I) ([M.(g)(m) | n'](y,,r = [M.(g)(rn) | n'](z,,fr . . . .

. . . . [-~ o N*(g)(~')J(y,:~) : ~.(f) ([.~ o N*(g)(~')](y,r


| : r .(s)(w |

so T.(f) is well defined. It is clear moreover that this turns T into a covariant functor. Conversely, (Y', r is a set over X' , then let. (Y, r be the pull-back of X and Y', giving a cartesian square

a

Y , y '

X , X '

f The set (Y, O) is by definition the inverse image of (Y', r under f . Now let

T*(f) (Ira | n](y,,r = [M*(a)(m) | N*(a)(n)](z,r

1.6. ANOTHER IDENTIFICATION OF M@N 19

The map T* is well defined: if g is a morphism from (Y', r to (Y~', r then I have the following commutat ive diagram

Y

l \h r }1 al )

X f

, y '

~, r

,) X !

There is a map h because the squares (Y, Y', X, X') and (Y~, Y/, X, X') are cartesian by construction. Moreover

Then as (Y~, Y/, X, X ' ) is cartesian, there exists a unique morphism h from Y to I/1 such that r = r and ga = alh. The square (Y, Y', Y1, ]/1') is then cartesian, for its composition with the cartesian square (Y1, Y(, X ,X ' ) gives the cartesian square ( Y , Y ' , X , X ' ) . In those conditions, if m' E M(Y') and n'l C N(Y/), I have

T*([M~(g)(m') | nl](gl, r = [M*(al)M.(g)(m') | N*(al)(nl)](y~,r ) . . . .

. / . t . t * * t . . . . . [~/I (a) ( rn)@N (h)N (al)(nl)](y,r -= [M.(h)M (a)(m)| (al)(nl)](yl,~l) . . .

. . . . [M*(a)(m') | N*(alh)(n~l)](y,r = [M*(a)(m') @ N (ga)(nl)](y,r . . . .

. . . . [M*(a)(rn') | N*(a)N*(g)(n'l)](r,r = T*([rn' | N*(g)(n'l)](g, r

A similar argument proves that

T*([M*(g)(rrt'l) @ n'](y, r = T*([m' 1 @ N.(g)(~'~,/)](y1, r

so the map T*(f) is well defined. It is also clear that T becomes a contravariant functor.

In order to prove that the bifunctor T is a Mackey functor, I must check the conditions (M1) and (M2). Let X = Y [I Z be the disjoint union of Y and Z. I denote by iF and iz the respective injections from Y and Z to X. If (U, r is a G-set over X, let Uy = r (resp. Uz = r let eg (resp. Ca) be the restriction of r to Uy (resp. to Uz), and let uy (resp. uz) be the injection from Uy into U. Then U is the disjoint union of Uy and Uz. Let [rn | n](v,r E T(X). Since M and N are Mackey functors, I know that rn =

(M.(nv)M*(uy)+M.(uz)M*(uz))(m) and n = (N.(uy)N*(uy)+N.(uz)N*(uz))(n) . Moreover

. . . . [M*(uv)(m) | N*(uy)N,(uv)N*(uy)(n)](uv,r . . . .

20 C'HAPTER 1. M A C K E Y FUNCTORS

as the square '~ty

Uy ~ Lr

Y ~ X

is cartesian. For the same reason, l have

iy

[M.(uz)M'(uz)(m) @ A: . (~z )N ' ( ~,z)(7, )](u.,.~) = T.(iz)T*(iz)(Ira ,~:: n](u,~,))

Since finally

. . . . [ M * ( . ~ , y ) ( , , ~ . ) ~:.~ N' ( ,,z):'~'.(~Y )N' (~Y )(,O](v,..~,.i = o

I have also

and

* 1 [M.(uz)M (uz)( n) ~3 N.(uy)N*(uy)(r~)](u,,) = 0

This proves one half of (M1). ] don't need to check that T'( iz )T. ( iy) is the identity of T(Y), if I know that. (M2) is true for T: in that case indeed, this follows from the fact that since iy is injective,

then the square Id

Y ~ Y

Y ~ X

is cartesian. In the same way, the cartesian square

0 , Z

1 1 Y , X

iy

and the fact. that T(0) = 0 show thai T*(iy)T.(iz) = O. assuming (M2). To prove (M2), let

(t V ~ Y

',1 1 Z , X

d

This proves (Nil) for T,

1.6. A N O T H E R I D E N T I F I C A T I O N OF M @ N 21

be a cartesian square, let (U, 05) be a G-set over Y, and let [m@ n](u,r E T ( Y ) . Let W, 05' : W --+ V and a': W ~ U such that the square

a !

W ~ U

V , Y a

is cartesian. Then

T.@T*(a)([~ | q(<~)) = [V*(.')(~) | N*(~')(~)](~,~,)

But the square (I, I

W -- ~ U

1 Z ~ X

d is composed of the two previous ones. It is then cartesian, and

. . . . [M'(~')(.~) | N'(a')(~)](w,~,) and this proves (M2) for the functor T. So T is a Mackey functor.

To prove that T = M @ N , all I have to do is to check that when f is the natural projection p~ f rom.G/K to G / H , for K C H, then T , ( f ) is equal up to the isomorphisms of lemma 1.6.1, to the transfer t~- of M@N, and that T*( f ) is equal to the restriction H Let (Y, r be a G-set over G / K , and let S C 05-1(/s be a system of F/s representatives of the orbits of Y. Then if [m | n](y,r E T ( G / K ) , I have

e'([~ | q(~,~)) = ~ [ . ~ e ~1~ ~ M+N(K) sC: S

So

But I have also

H '(Ira| 05)) ~[ms | E M+N(H) tKO = s E S

=ha s c_ r c_ (pIr So

| q(y,+)) = | = | +,r s E S

Now if (!/, 05) is a G-set over G / H , let Z be the pull-back of Y and G / K . I have the following cartesian square

a

Z ~ Y

G / K , a / H H

PK

22 CHAPTE R 1. M A C K E Y F U N C T O R S

As pH is surjeetive, so is a. Then if $ C_ ~- I (H) is a system of representatives of the orbits of Y, and if [m | n](Y,e) E T ( G / H ) , I have

G~ 7)2 Gs ~l I

s E S s 6 S x 6 K \ H / G s

But to ~ E S and x E K\H/G~ , I can associate the couple (K, xs) E Z, and as a is surjective, it is easy to see that I obtain that way a system of representatives S' C (r of orbits of Z. Then

T*([m | n](y,r = [M*(a)(m) | N*(a)(l~)](Z,b )

whence O'T*([m | n](y,r = E [M*(a)(m)~, | N*(a)(n),,]G~,

s'6S'

Here M*(a)(m)~, = M*(i~,)M*(a)(m) = M*(ai~,)(m), denoting by i'~ the injection from G/G~, into Z associated to s'. If s' = (K, xs), then G~, = K N ~G~ and

ai~,(gG~, ) = a(gs') = a((gK, gxs)) = gxs = i~(gxG~ ) = i~xp~c~G~ (gG~, )

�9 " ca~ and this gives so aZs, = ZsXp (xnG~,

r G ' * " rrl G , * " O'T*([m | n](y,r = E Ix A-=na=M ( z = ) ( ) | xrK=naN (z=)(n)]Kn=a, (h,xs)es'

and finally, I have

I still have to check the compatibility of the isomorphisms 0 and 0 ~ with G-conjugations. Let H C_ G and x E G. If ( K r is a G-set over G /H, let X c_ r be a system of representatives of G \ Y . Let m~ be the conjugation gH ~ (gx-~)~H from G / H to G f f H . Then if s E S, I have

m~r = m~(xH) = ~H

so xS = {xs I s E S} is such that m~r C_ ~H, and then

s' 6xS

But if s' = xs, then G~, = ~Gs, and m~, = M*(i~,)(m), where z~, is the injection gG~, ~ g J from G/G~, to Y. But

i~,m~:(gG~) = i~,(gx-lG~, ) = gx- l s ' = g s = i~(gG~)

.t So i',m~ = i,. But then % = i~(m,) -1 and

m~, = M*((m~)- l )M*( i s ) (m) = M. (mx) (m, )

It follows that

O'T.(m~)([m @ n](v,r = ~-~[M.(m~)(m~) | N.(m~)(n~)l~a, s E S

1.6.

On the other hand

ANOTHER IDENTIFICATION OF M@N 23

a

y ' ~ Y

X' , X g

is cartesian. Then

( M@N)*(g)([m | n](y,r = [M*(a)(m) @ N*(a)(n)](y,,r

Remark : This proposition shows that the tensor product of Mackey functors is the same as the box product defined by Lewis (see [6]).

sES

SO

(M@N).(mx) (0'([m @ n](r,r = ~[M.(m=)(m=)| N.(m.)(n=)]=a. sES

and this shows that

| n > +,) = (M N).(mxI (O'([m | @ +,) )

I finally proved the following:

Proposi t ion 1.6.2: Let M and N be Mackey functors for the group G. If X is a G-set, then

(M~N)(X) ~- ~e= ( * .(Y)eRu(Y))/g

where J is the R-submodule generated by

M.(f)(m) @. n' - m | N*(f)(n') for f : (Y, r --+ (Y', r m E M(Y), n' E N(Y')

M*(f)(m') @R n'-- m @n N.(f)(n) for f : (Y,r -+ (Y', r m' E M(Y'), n E N(Y)

If [m | n](y,r denotes the image in M~N(X) of m @ n E M(Y) @R N(Y), for

Y - ~ X , t h e n

�9 I f g : X --+ X ' , t hen

(M6N).(g) (Ira | hi(r,+)) = [ m | n](y,=r

�9 I f g : X' --+ X, let Y', r and a such tha t the square

24 CHAPTER 1. MACKEY FUNCTORS

1.7 Functoriality

The previous constructions have obvious functorial properties: the correspondence

( M , N ) ~ 7~(M,N)

is a functor in two variables, contravariant in M and covariant in N, and the correspondence

(M, N) ~ M~N is a covariant functor in M and N: if f (resp. g) is a morphism of Mackey functors from M ' to M (resp. from N to N'), and if Y is a G-set, I can define a morphism 7-~(f,g)y from ~(M,N)(Y) to 7-((M',N')(Y) by

7-((f,g)z(h ) = g(y) o h o f for h C ~ ( M , N ) ( Y ) = HOmMack(a)(M, Ny)

where g(y) is the morphism from Ny to N~ defined by

( ~ ( r ) ) ~ = g ~ r : N~(X) = N ( X Y ) ~ x ~ ( x ) = X ' ( X Y )

It is easy to check that if u : Y ~ Y' is a map of G-sets, then the square

?-t(M, N).(u) 7-t(M,N)(Y) , N(M,N)(Y')

~( f ,g)g[ [ 7-t(f, g)z,

) ~(M',N')(Y) ~(M',N').(u) ~(M',N')(Y')

is commutat ive, as well as the square

N(M, N)*(u) 7t(M, N)(Y') ~ 7-t(M, N)(Y)

7-t(f,g)Y, I l T-t(f,g)Y , 7{(M', N')(Y) n(M',N')(g') n(M',X')*(u)

So I have a morphism of Mackey functors 7-((f, g) from 7-t(M, N) to ~ ( M ' , N') .

Now if f (resp. g) is a morphism from M to M' (resp. from N to N') , then I can define a morphism (f@9)y from M@N(Y) to M'QN'(Y): if (Z, r is a G-set over Y, let

( f ~ g ) z ([m | n](z,r = [fz(m) | gz(n)](z,r If f and g are morphisms of Mackey functors, then these maps are well defined, and if u : Y -~ Y' is a morphism of G-sets, then the squares

(M+N).(u) M + N ( Z ) , M+N(Y ' )

( f ~ g ) z I I(fQg)Y' + M'QN'(Y) (M'~N').(u) M'~N'(Y')

1.8. N-FOLD TENSOR PRODUCT 25

and (M+N)*(u

M+N(Y') M+N(Y)

(f@g)r' 1 l (f@g)Y M'@N'(Y') M'@N'(Y)

(M'+N')*(u) are commutative, turning f@g into a morphism of Mackey functors from M@N to M@N'.

1.8 n-fold tensor product

1.8.1 Definition

The second identification of the tensor products can be extended to the case of the n-fold tensor product, defined as follows: let n be an integer, greater or equal to 2, and let Mi for i E {1, . . . ,n} be Mackey functors for the group G. If X is a G-set, let

(MI+...+Mn)(X)= | (Ml(Y)|174 <,.Till YZX

where fli is the R-submodule generated by the elements of the form

m' 1 @...m',_, @ M~,.(f)(m~) | re'i+ , @... @ m L - . . . * t * t * t * I

. . . M; ( f ) (m 1 ) | @ M;_ 1 (f)(m~_l ) | m~ @ M;+, (f)(m,+ a ) | | M: (f)(m=)

for f : (Yr --* (Y',r m~ C Mj(Y') i f j # i and mi G M(Y). If (Y,r is a G-set over X, and if m~ E Mi(Y), for i E {1 , . . . , n} , I denote by [m~ | . . . |162 the image of m~ @. . . @ mn in (MI@...@M,O(X). Then:

P r o p o s i t i o n 1.8.1: Le t Mi, i C { 1 , . . . , n } be M a c k e y f u n c t o r s for G. Le t L1 = M~, and for i C { 2 , . . . , n } , let Li = Li-I@Mi. T h e n for any G-set X, t h e r e is an i s o m o r p h i s m

L~(X) ~_ (MI@...@M,O(X)

which t u r n s MI@. . . @M~ in to a M a c k e y f u n c t o r , in t he fo l lowing way:

�9 I f f : X ~ X ' is a m o r p h i s m of G-sets , if (]I,r is a G-set over X, and i f m i C Mi(Y), for i E { 1 , . . . , n } , t h e n

(MI@... @M~).(f)([m, | | m~](y.r = [ml | | m,~](z,lr

�9 I f g : X' ~ X, let Y', a and r such t h a t t he squa re a

y ' , Y

X ' , X g

is c a r t e s i an . T h e n

(MI@. . . @M,O*(g) ([~1 | | ?TtnI(Y,$)) = [M;(a)(m,)| . . | M:(a)(m=)](y,,r


P r o o f : By induction on n, the case n = 2 being a consequence of the identification of M@N. By definition of L~ = Ln-I@M~, I have

Ln(X) = ( (~ L n - l ( r ) @ i n ( Y ) ) / J YZX

where J is the submodule generated by

in-l . .( f)([) @ rn' - l | M~(f)(Tn')

i*~_~(f)(l') @ m - l' | M,~,,(f)(m)

for f : (Y, 05) --+ (Y', 05'), m E M(Y), m ' r M(Y"), 1E Ln_~(Y), 1' E L~-I(Y') . By induction hypothesis, I know that L~-I(Y) is generated by the elements

[ml @ . . . | 7"rtn-1](Z4b)

where (Z, ~b) is a G-set over Y. I define a map 0 : L,~(X) --+ (M~@... @MrO(X) in the following way: if (Y, 05) is a G-set over X, if m E M(Y), if (Z, ~b) is a G-set over Y, and if I -- [ml @. . . | m~-l](z,v,), let

0([I @ ~Ttl(y,~)) = [/rt 1 @ . . . @ 7T/n_ 1 @ Mn(//))(fr/)](ZKbg, )

I must check that this map is well defined: let f : (Y, 05) -~ (Y', qS') be a morphism of G-sets over X. Then, by induction hypothesis,

Ln_l,.(f)(1) = Ira, | | m,~-l](z,i~)

So for m' E M,~(Y')

O(L,~_l,.(f)(l) @ m') = [ma @.. . @ rn,~_l @ M*(f~)(m')](z,r

whereas

. . . . M* M* m' e(l @ i~ ( f ) (m ' ) ) [/Ytl @ @ m~-i @ ~(~b) ( f ) ( ) ] ( z .+~ )

It follows that O((in_l).(f)(l) @ m t) : O(l @ M*(f)(gt'lt))

since 05'f = 05 and M*(f~b) = M~(~)M~(f). Similarly, if g : (Y', 05') --+ (Y, aS), let Z', a and ~b' such that the square

a Z' ) Z

y ' ~ Y g

is cartesian. Then by induction hypothesis, I have

L;_l(g)(1 ) = [M;(a)(rnl) | | M~_l(a)(m~_l)](z,,p, )

1.8. N-FOLD TENSOR PRODUCT 27

and so

O(L*_i(g)([ ) @ m') = [Ml(a)(mi) @.. . @ Mn_l(a)(mn_i) @ M,~(r162162

But in (Mlc~... @Mn)(X), the right hand side is equal to

[rnl | @ m=-i | M~,.(a)M:( ~')(rn')](z,e~)

On the other hand

O(1 @ M,~,.(g)(m')) = [m~ @.. . @ rnn_~ @ M:( @ )(M,O.(g)(m')](z,r

and I have

o(~:_i(g)(,) | = 0(~ | M . ( g ) ( < ) since M~,,(a)M~(r = M:(r in the above cartesian square. Thus 0 is well defined.

I can now define a map O' in the other direction: if

[m~ | | "~=](r,r e M , @ . . . @M=(X)

let l = ml | . . . | r an - l , so I can consider the e lement [l](y,r E L,~-I(Y), and let

o'(i~ | 1 7 4 ~,>.,) = [c,l~..~)| ~ L~(.O

This map is well defined: if i is an integer lower or equal to n - 1, and if u E J i , then u is a l inear combina t ion of elements of the form v - w, with

! I ! ! = ?Tt I @ . . . ?Yti_ 1 @ M i , . ( f ) ( m i ) | miq_ 1 @ . . . @ ~ n

* l * l ,~ f w = M;(f)(m'l) @. . . | M~_l(f)(mi_,) @ mi | Mi+l(f)(mi+l) | | M*(f)(m,~)

for f : (Y, r --+ (Y', r and e lements m', C Mj(Y') if j r i, and rni E Mi(Y). I l i I But then if l ---- ?'D. 1 @ . . . mi_ 1 @ Mi,,(f)(m.i) | mi+~ @. .. | m,~_~, and if

t" = M ; ( f ) ( m l ) | | M~Cl(/)(m'~_,) | "Z~ | M~*+l(f)(m{+~) | | Mg(/)(m'~_~)

I have

0'(i<(~,.,) = [I,">,., | M:(~)(<)](~,. ,

and in L,~(X), this e lement is equal to [Ln_l,.(f)([Uq(Y, Id)) @ T/Zn](y,,&,). But

L,~_,,.(f) ([l"](Y,,d)) = [/"](r,/)

and in L,~_I(Y'), this e lement is equal to [l'](y,dd ). Final ly

o,(E<~Y.,) -- [II'l(Y,,d, | ~'](y, . , , = O'(E<(Y,.,,)

Now if / = n, and if

! I V = N2,1 ( ~ . . . @ 7Ttn_ 1 (~ M n , * ( f ) ( r n n )

28

let

I have

But as the square


and so

CHAPTER 1. MACKEY FUNCTORS

* # * # w = M~ (f)(rn~) | | M~_,(f)(rn,~_~) @ rn~

# l ' = m ' l | . . . | m ~ _ 1

l " = M ~ ( f ) ( m ; ) | O M~_l(.f)(m'~_~)

0,(E~]<y,~)) ; [~,"l,Y,d)o~d<~ 0~

f Y , I/"

1 Y ~ 1/"

f

" u ) O ' ) " L~_~ /](Y,,sd) = [l ](r, Sd)

o'(Ewl<,,,,~,} = [~:_,(s)(Cn<Y,,,~))|

In Ln(X), this e lement is equal to

[[l'](Y',ld) @ Mn,.(f)(TT~n)l(y,,dy , = 0/([V](y,,qS,,)

So the map 0' is well defined. Now let (Y, 05) he a a-set over x , and (Z, q$) be a a - se t over Y. If mi E Mi(Z) for i _< n - 1 and mn E M,~(Y), let l = rnl @ ...-@ rn,~_l. I have

0,0([i,1(,,+, | ~o](y,+,) = 0,(E, | M:(+)(mo)l(,,++,) = [I'l<z,,d, | M:(+)(~n)] <z,~

and in L,~(X), this is equal to

[in-I,*(~))([l](z,Id)) @ ~n] (y,r

Finally, as L~_l,.(~b)([l](z, la)) = [/](z,~), I have 0'0 = Id. Conversely, if mi C Mi(Y) for i _< n, and if l = rnl | . . . | rnn_l , then

and this proves tha t 0 and 0' are mu tua l inverse isomorphisms, which completes the proof of the proposit ion. �9

1.8. N -FOLD T E N S O R P R O D U C T 29

1.8.2 Universal property

The analogy between the tensor product of Mackey functors and the usual tensor product will become clear with the notion of n-linear morphism, defined as follows:

Definit ion: Let M1, . . . , M~ and P be Mackey functors for the group G. An n-linear morphism L from M 1 , . . . , M,~ to P is by definition a correspondence which to any n-tuple of G sets X 1 , . . . , X~ associates an n-linear map

L~ ...... xo : M , ( X ~ ) • • M~(X~) ~ P ( X , • x X~)

in a functorial way: if fi : Xi ~ X~ for 1 < i < n are morphisms of G-sets, then the squares

Ml, . ( f , ) x . .

and

M1 (fl) •

MI(X1) X. . . • Mn(X~)

x M~,.(A) [ M , ( X ; ) x . . . x M~(X'~)

MI(X1) • • Mn(Xn)

�9 • M:(f ,~) ] I

M,(X~) x . . . x M,~(X')

are commutative.

LX1 ,...,Xn )

)

Lx;,...,x~

P(X1 • . . . • Xn)

P . ( f l x f.~) N . . .

P(X; • • X'~)

Lx1 ,...,Xn

)

Lxi,...,x,~

P ( X l x . . . • Xn)

P*(fl x . • $ m

P(X~ • • x ' )

I denote by s M~; P) the set of n-linear morphisms from M1,. . . , M~ to P, with its natural structure of R-module. This construction is clearly functorial, contravariant in/141,... M~, and covariant in P. If X1,. . . ,X~ are G-sets, I define a map

7rx ...... x~ : MI(X1) • .. x M~(X~) --~ (M~@. . .4M~) (X~ x , . . x X~)

by the formulae

~ 'X . . . . . . Xn(r~l,..., rn~) = [ M ; ( p l ) ( m l ) | | M*(pn)(mn)] (Xl•

where for all i, the map pi is the i-th projection from X1 x .. . x X~ onto 3//. Then

Propos i t ion 1.8.2" The previous equalities define an e lement

7r E s

which has moreover the following universal p roper ty : if P is a Mackey func tor for the group G, and if f 6 s then there exists a unique m o r p h i s m of Mackey functors f : M1Q...@Mn ~ P such tha t for any G-sets X 1 , . . . , X~

f x l , . . . , X n : - f lY1X. . .XXn' /~Xl , . . . ,X~

30 C H A P T E R i. M A C K E Y F U N C T O R S

Converse ly , if f is a m o r p h i s m of M a c k e y functors f rom M I Q . . . @M~ to P, then this formula defines an n-l inear m o r p h i s m from M1, . . . , M~ to P, and this c o r r e s p o n d e n c e induces an i s o m o r p h i s m o f / ? - m o d u l e s

HomM~ckR(a)(Ml@... @M,~, P) ~_ s ]~I~; P)

which is natural wi th respect to M1 . . . . . k'L~, P.

Proof." 1) First I have to check that 7r is an n-linear morphism from M1, . . . , M.~ to Ml@. . . QM.,~. I suppose given for an integer i a morphism fi : X~ --+ X~. Let then f = (Idx~• • fii • ([dx,+,x .... xn), and A = ( M ~ ) . . . @ M n ) . ( f ) . I have

aTcx~,...,x~(rnl,..., rn,~)-- A ([M~(pi)(rn,)U)...@M:(p~)(,n~)](x~•215 . . . .

. . . . [M;(p,)(~,~) o , . . e M.a(p~)(.,~)](~,•215

On the other hand, if p'~ is the projection from X1 • . . . • X[ x . . . • X'~ onto its i-th component, I have

7t"2 . . . . . . X[,...,Xn( ITL1, "'@'~]{,*(Si)(gni), ''',Trzn ) . . . .

. . . . [M;(pi)(ma)|174 (p~)(m.~)] (x ..... •

But the square

f X I • 2 1 5 , X I • 2 1 5 2 1 5 2 1 5

Xi > X[ fi

is cartesian, so M[(p~)Mi, . ( f i ) = Mi, , ( . f )M[(pi) , and then

. . . . . . , . . . .

. . . . [Ml ( f )M~(pl ) (m~) | . .|174 . .|215 )

in (MI@.. .@M~)(X1 • • X[ • • Xn). Finally as for all j r i, I havep}f = pj, it follows that

[Ml(Pl)(rnl ) @. . . @ M](pi)(rrti) @. . . @ M*(pn)(fnn)] •215 which proves that 7r is covariant with respect to the i-th factor.

Conversely, if ra'~ E Mi(X[) , I have

. . . . . . . . . .


. . . . [M*(pl) (ml)@.. .@~/ i~(pi)M*(f i ) (m~i)@.. .@Mn(pn)(mn)](Xlx. . .xX~,Id)

On the other hand, let A' = ( M , ~ . . . @Mn)*(f). I have

AITTX ....... ~:,...,Xn (m'l '""" ' 17Z1,,..., ?TZ n) . . . .

As the square

X] x . . . X X n

Xl x . . . x Xn

f X1 x . . . x X[ x . . . x X~

X 1 X. x X ~ x . , . x X n

is cartesian, this is also equal to

* �9 �9 * i Mt ( f )M1 (p~)(ml) | | MI ( f )Mi (pi)(mi) | | M;(f)M~(p,,)(m,~)] (xlx...•

Finally as p j f = pj for j r i, and a.s pi f = f i P i , it follows that ~r is contravariant with respect to the i-th factor, and hence it is an n-linear morphism.

2) I must now prove the universal property: let P be aMackey functor, and f E s M,,; P). If X is a G-set, I define a morphism f x from (MI@. . . @M~)(X) to P ( X ) in the following way: if (Y, r is a G-set over X, and if mi E Mi (Y ) for all i, let

fx ([rnt @. . . @ rn~](y,4,)) : P,(r ,ran)

where 5,,~,y denotes the diagonal injection y ~ ( y , . . . , y) from Y into Y~. First I have to check that f is well defined. I suppose that g : (Y, r --+ (Y', r is a morphism of G-sets over X, that rn} E Mj(Y') for j r i, and that mi �9 M~(Y). Then

YX ( [/7Z11 @ . . . @ ]} /~i , . (g) (Tr l i )@.- .@ 'gztn] (y,,r . . . .

. . . . P,(@')P*(~n,Y,) fY, , . . . ,Y , (/'gill,-.., M i , * ( g ) ( I Y t i ) , ' ' ' , ?TZtn)

Since f is an n-linear morphism, setting h = Id i-1 x g x Id n- i - l , I have

( ' ') ~,,...,y, r n t ~ . . . , M i , . 7hi , . . . , r n n = . . .

P ~ ( ( , , , ,) . . . . h)fy,, . . ,r, ,r,r, , . ,r, m l , . . . , m i _ l , m i , m i + l , . . . , m , ,

Let k be the map from Y to y,~- tyy , ,~- i defined by

k(y) = (g(Y), . . . , g(Y), Y, g(Y),- . . , g(Y))

The square h y , i - l y y . ~ - i ~ y.~

Y , y ' g

is cartesian, hence P*(~5,~,y,)P.(h) = P.(g)P*(k). Setting 1 = f - ~ x Id x g,~-i-1 I have k = 15n,v, so P*(k) = P*(s Using again the n-linearity of f , I have

. , , , . ~ , ) _- P (l)fr,,...,y,,y,v,,...,y,(m> �9 . . ~mi_l ,mi~mi+l~ . . . . . .

( . . . . . , . ,)) Mi+ 1 (g)('r/ti+ 1 ), . . , M,~ (g)(m n M; (g)(mi),m~, �9 = fy,...y M1 (g)(rr/1), . . . . ,

and denoting by F this expression, I have finally

27x ([m', | | M<.(g)(m 0 | | m:](v, r ) = P.(r163 . . . .

. . . . 2 7 X ( [ M 1 ( g ) ( m ' 1 ) ~ ~ ~M.~1(g)(~tz' i~1)|174 ~ ~|176162

which proves that f is well defined.

3) I must now check that j 7 is u morphism of Mackey functors. Let g be a morphism of G-sets from X to X' . If (Y, r is a G-set over X, and if mi E M~(Y) for 1 < i < n, then

( M I + . . . + Mn ).(g ) ( [ml @. . . @ rnn](y,r ) = [77ll (~ . . . | 7D, n](Y,9r

SO

)Tx, ( ( M ~ . . . ~M~).(g)([rn~ | | rn,~](v,r = P.(g)P.(r ,m,,)

that is

f x t ( ( / 1 @ . . . @J~rn)*(g) ([77Z1 | 1 7 4 77~n](Y,r ~-" P * ( g ) f x ([TIZi | | mn](Y,r

which proves that f is covariant. If now 9 is a morphism from X' to X, let Y', r and a such as the square

a y ' , Y

X ' , X g

is cartesian. Then

| =

The image under 27x, of this element is then

P.(r163 ...,r, (M; (a ) (ml ) , . . . , M~(a)(m,O)

and as f is n-linear, I have

fY,,...,z, (M; (a ) (ml ) , . . . , M,~(a)(m,O) = P*(a'~)fv,...,v(ml,..., m, 0

Moreover a~6n,z, = 6,~,va, and P.(C')P*(a) = P*(g)P.(r in the previous cartesian square. Thus

27X' ( ( M I @ . . . @an)*(g)([~ ' t l @ . . . @mn](y,r ) . . . .

1.8. N-FOLD TENSOR PRODUCT a3

. . . . P*(g)P.(r ) fy , . . . y (m, , . . . , m,~) = P*(g)fx ([m, | |162

showing that / i s contravariant, hence a morphism of Mackey functors.

4) Finally, let X1, . . . ,X,~ be G sets, and m~ E M~(X~) for 1 < i < n. Then, setting X = X 1 x . . . x X ~ , I h a v e

. . . . P*(5,~,x)fx,...,x ( M ; ( p , ) ( m , ) , . . . , Mg(pn)(m~,))

The n-linearity of f gives then

fx,...,x ( M ; ( p l ) ( m l ) , . . . , Mg(p,~)(m,,)) = P*(p, • • p=)fx~,...,x,~(ml,..., m,~)

and as (Pl x . . . x P,~)5,~,x = Idx, I have

fx1x . . .xXn 71"x ...... Xn = fXl,...,X,

5) Conversely, if f is an n-linear morphism from M ~ ) . . . QM~ to P, and if I define fx~,...,x~ by the above formulae, I get an element f of s M~; P): to see this, let i E {1 , . . . ,n}, let gi : Xi ~ X~ be a morphism of a-sets, and let m 5 r Mj(Xj) for 1 _< j _< ~. Setting then X~ = X 9 and gj = Id for j r i, and denoting by X (resp. X ' ) the product X~ x . . . x X~ (resp. X[ • x X;) , and pj (resp. p~) the projection from X (resp. X ' ) onto Xi (resp. onto X[), I have

fX~ ...... g- ( M l , . ( g l ) ( m l ) , . . . , an,.(gn)(m,z)) . . . .

. . . . ?x , ([Ml(Pl l )ml , . (gl ) (ml) | a:(p:)an,.(gn)(mn)](x,,zd) )

Since p} = pj and gj = Id for j r i, this is also equal to

k , ( [ M r ( p , ) ( m , ) | | M:(p:)Mi,.(gi)(rni) | . . . | "*(pn)(ra~)](x,jd) )

Denoting by g the product gl x . . . x g,~, the square

g X , X '

X i ' X~ gi

is cartesian (because gj = Id for j # i), so

Mi (p~)M{,.(gd : M,~..(g)M: (p{)

On the other hand, setting rn~ = M](pj)(mj), for 1 < j < n, I have the following equality in (MI@. . . @M.,~)(X')

[, ,] = ~ , | o M~,.(g)(m',) e . . . e - < (x, ,~) '

34 C H A P T E R 1. M A C I ( E Y F U N C T O R S

. . . . * ' ""Oa/[i-l(g)(?rzi-')@')'i@Mi+l(g)(;'ni+l)O" "'@M'~((J)(mn) (x,a)

Moreover

Ma(g)(m;) = M;(g)Ma(pj ) (mj) = M;(p jg) (rn j )= M;(p j ) (mj )

if j ~ i, because then pig = p~. Finally

JX~ , . . . ,Xr I, ( M l , * ( g l ) ( T g t l ) , "" Mn,,,(.(Jn)(T~ : .?:.y, ( [ M l ( P 1 ) ( ? 7 ) . l ) ( ~ . . . @ J ~ / I : ( p n ) ( g Y t n ) ] ( X , 9 ) )

But [M;(p,)(rrh) 0 - �9 @ M~(P,,)(,,z~)](.v,o ) . . . .

^ / . * . . . . ( , , ~ . ~M,,)(~)([M~ (,,~)(,,,~)0...0 M:(~)(,<],,~,) and .f is a morphism of Mackey functors. Thus

fx i ,..,,v'~ (M, , . (g , ) (mr ) , . . . , M,,,.(g,~)(m,,)) . . . .

7~ . / * . . . . P . (~ )~x ( [M1 ( p l ) ( T F / 1 ) ( ' ~ . - .~)]~/In(pn)(TYI.n)J(X.[d) ) -- P . ( g ) f x ..... . X n ( / T Z l , ' " " ' T/Zn)

which proves that f is covariant with respect to the i-th factor. Now if gi is a morphism from X[ to X~, and keeping the other notations above, I have

~*(~)~x:, ,~.(,,,,...,,,~): e*(~)f,. ([M;(~)(,,,)o...o M:(~,,)(,~)] (x,,~))

and as j7 is a morphism of Mackey functors, this is equal to

The square g

X' , X

Id I I ld

X ' ~ X ,q

being cartesian, I have

. . . [ M ; ( g ) ~ q ( p l ) ( , , ~ ) 0 . . . o M : ( g l M : ( p ~ ) ( ~ n ) ] ( x , H )

But for 1 _< j _< n, I have pig = gJP.i. Hence

P*(g) fx ...... x~(rrq O .. . O mn) . . . .

�9 . .~, ' ,([Ml(Pl)A./[l(gl)(ml)O...r = . .

. . . . k;, . . . ,~ (Ml(g , ) ( , , , ) , . . . , M;(~)(-~))

1.8. N - F O L D T E N S O R P R O D U C T 35

which proves that f is contravariant with respect to the i-th factor, and that it is an n-linear morphism.

6) To complete the proof, it remains to see that the above correspondences between s M~; P) and HomMa~k(a)( M l ~ . . . @Mn, P) , which are clearly natural with respect to M 1 , . . . , M~, P, are bijective. Let f E s It corresponds to f E HomM~k(c)(M~@.. .@M~,P) , which in turn correspond to f ' E s . . . . , M~; P) , and I must prove that f = f ' . Let X 1 , . . . , X ~ be G-sets, and mi E M~i(Xi), for 1 < i < n. Then, setting X = Xl • 2 1 5

fi" ...... X,, ( ~ 1 ' " " " 'Tr~n) = ~X)TX ...... Xn( /T~I , " �9 " , ?T~r~) . . . .

. . . . f x ( [ / l ( P l ) ( ' D ~ t l ) ~ ) . . . ~ 2 ) ]~/In(Pn)(mn)](X,id,) . . . .

. . . . P*((~n,x)fx,...,x ( / l ( P l ) ( T i ' z 1 ) , . - . , Mn(Pn)(~n)) Since f is n-linear by hypothesis, this is also

P*(5,~,x)P*(pl • 2 1 5 P~)Ix ...... Xn(TI'll, "" " , ' I n )

and a.s (Pl x . . . x P~)5~,x is the identity of X, I have

f~x~ ,...,x, ( m l , . . . , rn,~) = f x . . . . . . .J~n ( ~'nl, �9 �9 �9 , TnT~)

and f = f ' . Conversely, if f is an element in HonaM~k(G)(ma(~... @M~, P), it corresponds to f E

s M~; P) , which in turn corresponds to .f' E HomM~k(a ) (Ml+ . . . (~M~, P) , and I must show that f = f ' . Let X be a G-set, let (K r be a G-set over X, and for any i E { 1 , . . . , n } , let m~ ~ M~(Y) . Then

f~( ([rrq 0 . . . @ m~](y,r = P.(r ran) . . . .

. . . . P . ( r ,m,~) . . . .

. . . . P . ( r M~(pn)(mn)](y,~,td) )

Since f is by hypothesis a morphism of Mackey functors, this is equal to

g.(@)fy ( ( M l + . . . @ M n ) * ( ( ~ n , y ) ( [ i l ( P l ) ( n ~ l ) @ . . . @ ]Vf:(Pn)(mn)](yn,ld,))

But the square (Sn,Y

Y ~ y ~

Y ~ y ~ ~n,Y

is cartesian. So

(MI . . . .

36 CHAPTER1. MACKEY FUNCTORS

. . . . [M;(6~,z)M;(p,)(rn~) | | M:(~5,,,y)M,:(p,~)(m,~)](v,~d)

and for all i r {1,. . . ,n}, I have piSn, Y =- Id. Finally

f~( ([ml | | m~](r,,)) = P.(~b)fy ([ml @.. . | mn](y, ig))

Using again the fact that f is a morphism of Mackey functors, I get

k(In , | = k ( ( - , + . . | mn>,,d,)) . . . .

which proves that f ' = f , and completes the proof of the proposition. ,.

Let f E s ,M.,~; P) be an n-linear morphism, associated by the previous proposition to a unique morphism of Mackey functors ffrom MI~... ~M~ to P. The formula

.fx ([grZl @.. . @ WtnJ(Y,~)) z P.(~)P*(5~,y)fy,...,y(ml,..., rn,~)

shows that f is entirely determined by the knowledge, for any G-set Y, of the s-linear map fz : M~(Y) x . . . x M,~(Y) ~ P(Y) defined by

/ r ( - ~ , . . . , - ~ , ) = P ' (< ,Y)s , -~,)

Indeed, knowing f , I can recover a 7 by

This formula will give the conditions on the maps fy so that the map f is well defined and is a morphism of Mackey functors. In order f to be well defined, if g: (Y', qS') --+ (Y, qS) is a morphism of G-sets over X, if ro, j E Mj(Y) for j # i and m{ C Mi(Y'), the elements

['tTZ 1 @. . . 0.~ ~'~i-1 @ ]~i,*(g)(grtli) @ mi+l @.. . @ mn](Y,~) . . . .

. . . . [M~(g)(rnl)@...@M$_l(g)(rrZi_l)@rrz'i@M*+l(g)(rr~i+l)@...@M~(g)(rnr~)](y,,~, )

must have the same image under f , and since P.(qS') = P*(~)P*(g), this gives

P*( O ) fY (~'Ttl,.-., 17Zi-1, Mi,*(g)(wt:), r i l l+l , . . . , ~ln) . . . .

= P.(~)P.(g)fy, (M;(g)(rnl) , . . . , M;l(g)(mi_,) , ""i, M;+l(g)(rrzi+l),'", Mn(g)(r~

This equality will result from the case X = Y and r = Id, which says that for any g: y'--~ y

(A,) ?~ (,~1,..., ~,_,, V,,.(g)(<),,,~+~,..., ~.~) . . . .

. . . . P.(g)fz, (Ml(g)(rr/1), . . . , J]///*_l (g)(mi-1 ), re'i, V ; + l ( g ) ( m i + l ) , . . . , M~ (g) (ran))

If equali ty (Ai) is true for all i E { 1 , . . . , n}, then 2 7 is well defined. In those conditions, it is clear moreover that f is covariant: if g : X --+ X ' is a morphism of G-sets, then for any a - se t (Y, r over x and for any rnj C M j ( Y ) , I have

(]~/i1@''" @Mn).(g)(['rr/1 @ ' ' ' @ rr/n](Y,&)) = [7"721 @ ' ' ' @ ?Ttn](Y,gr

so that

27X' ( (MI@.. .@Mn).(g)([Ti21 @ . . . @ 7r/n](Y,r ---- P*(g)e*(r . . . .

. . . . P . ( S x ( [ ~ l | | ~n](~,~)) I still have to express the fact that 27 must be contravariant. Let g : X ' --+ X be a morphism of G-sets, and suppose the square

a y ' ~ Y

X ' ~ X g

is cartesian. If rni E Mi(Y) for 1 < i < n, then

Taking the image under fx , gives

P.(~gt ) fy , (Ml (a)(?T~l),. . . , i ,~(a)(~l ,n))

that must be equal ' to P*(g)P.(r ~n)

Since moreover P*(g)P.(r = P.(r this equality will result from the case X = Y , X ' = Y ~ , r 1 6 2 This gives

(B) ] y , (M; (g ) (m l ) , . . . ,Mn(g)(mn)) = P*(g)fy(rn, l, . . . ,mn)

for any g : Y' ~ Y. The next proposition is a summary of conditions (A~) and (B) in the case n = 2

Proposition 1.8.3: L e t M, N a n d P b e M a c k e y f u n c t o r s for t h e g r o u p G. T h e fo l l owing d a t a a r e e q u i v a l e n t :

�9 A b i l i n e a r m o r p h i s m f f r o m M, N to P .

�9 F o r a n y G-se t Y, a b i l i n e a r m o r p h i s m f y : M ( Y ) x N ( Y ) ---* F ( Y ) , such t h a t for a l l g: Y ' ~ Y , m �9 M ( Y ) , n �9 N ( Y ' ) , m' E M ( Y ' ) , a n d n' E N ( Y ' )

ii)

iii) fy, (M*(g)(m), N*(g)(n)) = P*(g)jy(~, n)

38 CtfAPTER 1. MACKEY FUNCTORS

1.9 Commutativity and associativity

It should be clear from the previous sections that the tensor product is commutative and associative:

Propos i t ion 1.9.1: Let M, N, and P be Mackey functors for the group G. Then there are i somorphisms of Mackey functors

M~N ~_ N(~M

(M+N)_~P ~ M(~(N+P)

which are natural in M, N and P.

Proof: Let X be a G-set, and (Y, r be a G-set over X. It is clear that the map

[m | n](z,r e M+N(X) ~ In | m](v,,) E N+M(X)

is well defined, and induces the desired isomorphism between M~N and N~M. The second assertion can be proved by observing that, for any Mackey fnnctor Q, morphisms from each of its sides to Q are in one to one correspondence with trilinear morphisms from M, N, P to Q. The desired isomorphism is then a consequence of this universal correspondence. Another (less canonical) way to see this isomorphism, is to use commutativity:

M+(N+P) ~_ (N4P)~M = NQP+M

and to observe that the n-fold tensor product is clearly independent, up to isomorphism, of the order of its factors. "

1.10 Adjunction

I have moreover the following adjunction properties:

Proposi t ion 1.10.1: Let M, N, and P be Mackey functors for the group G. Then there exists an i somorphism

~(M+N,P) ~_ I t(N,~(M,P))

natural in M, N, and P.

Proofi Let X be a G-set. Then

?-t(N,?'t(M,P))(X) = HomM~ck(G)(N,~(M,P)x)

On the other hand

whereas

/'f(M, P)x(Y) = 7-t(M, P)(YX) = HomM~k(a)(M, Pzx)

~(M, Px)(Y) = HomM.~k(G)(M, (Px)z)

1.i0. ADJUNCTION 39

But (Px)y(Z) = Px(ZY) = P(ZYX) , and those isomorphisms are easily seen to induce a natural isomorphism of Mackey functors (Px)~" -~ PYx. Those isomorphisms of Mackey functors induce in turn an isomorphism

It follows that

But since

7-{(M,P)x ~- 7-t(M, Px)

7-t(N,~(M,P))(X) ~_ HomM~k(G)(N,~(M, Px))

~-~(M@N, P)(X) = Hon'lMack(O )(M~N, PX )

it is enough to have a natural isomorphism

HomMar P) ~- HomM~k(a)(N, n(M, P))

This is equivalent to say that the functor N H M ~ N is left adjoint to the functor P ~-+ 7if(M, P). But a morphism from M ~ N to P corresponds naturally to a bilinear morphism from M, N to P. To prove the proposition, it is enough to build a natural bijection between HomMack(a)(N, 7-t(M, P)) and s N; P)I

But a morphism a from N to 7-t(M, P) is determined by a collection ay , indexed by G-sets, of morphisms from N(Y) to

n(M, P)(Y) = Hom~a~k(c.)(M, Py)

The morphism ay is determined by morphisms (ay(n))x from M(X) to Py(X) = P(XY) , for any G-sets X, and any n C N(Y). I have then a morphism 6~x,y from M(X) OR N(Y) to P(XY), defined by

ax,Y(-~ | ~) = (~(~))~(m)

and it is easy to see that a is a morphism of Mackey functors from N to ~ (M, P) if and only if the element 5 defined that way is a bilinear morphism from M, N to P.

Conversely, if I have u bilinear morphism 5 from M, N to P, reading from right to left the above formula, I can define a morphism

which induces a morphism of Mackey flmctors from N to ~ (M, P).

Those constructions are clearly inverse to each other, and natural in M, N, and P. This completes the proof of the proposition. "

Chapter 2

Green functors

2.1 D e f i n i t i o n s

The classical definition of a Green functor is "a Mackey functor with a compatible ring structure": a Green functor A for the group G over the ground ring R is a Mackey functor, such that for any subgroup H of G, the R-module A(H) has a structure of R-algebra (associative, with unit), which is compatible with the Mackey structure, in the following sense:

�9 If z E G, and K is a subgroup of G, then the conjugation by x is a morphism of rings (with unit) from A(/s to A(~K).

�9 If H C_ K are subgroups of G, then r~ is a morphism of rings (with unit) from A ( K ) to AtH) .

�9 In the same conditions, if a E A ( K ) and b E A(H) , then

a . t ~ ( b ) = tI" ~ K a . ( ~ H ( )" )

t~(b).a = t~(b.rK(a))

There is an evident notion of morphism of Green functors: a morphism ~b from the Green functor A to the Green functor B is a morphism of Mackey functors such that, for any subgroup H of G, the morphism CH is a morphism of rings. The morphism r is said to be unitary if the morphism •H preserves unit for ~11 H. It is actually enough that the morphism r preserves unit, since

CH(1A(H) ) G G . G rHOG(1A(G)) = = = Cs(rs lA(o)) = rH1B(C ) 1B(H)

A module over the Green functor A, or A-module, is defined similarly as a Mackey functor M for the group G, such that for any subgroup H of G, the module M ( H ) has a structure of A(H)-module (with unit). This structure must be compatible with the Mackey structure, in the following sense:

�9 If x E G and K C G, let m ~ ~'m be the conjugation by x from M ( K ) to M(~K) . If a �9 A ( K ) and m �9 M ( K ) , then ~(a.m) = ~(a).~(m)

�9 If H C If are subgroups of G, if a �9 A ( K ) and m �9 M ( K ) , then r~(a.rn) = " f , ( a ) .

42 CHAPTER 2. GREEN FUNCTORS

A morphism r from the A-module M to the A-module N is a morphism of Mackey functors from M to N such that fox" any subgroup H of G, the morphism Cs is a morphism of A(H)-modules.

An important example of Green functor is obtained for any Mackey functor from the isomorphism

"H( M, M)( H) ~_ HOmM~k(H)( Res~ M, ReSaH M)

In particular, the composition of morphisms turns TI(M, M)(H) into a ring with unit, and "H(M, M) is actually a Green functor.

P r o p o s i t i o n 2.1.1: I f M is a Mackey func to r for the group G, then/-t(M, M) has a na tu r a l s t r u c t u r e of Green functor .

Proof : Let r be an endomorphism of the Mackey functor Res~M. Then r is determined by morphisms eL : M(L) -~ M(L), for s C_ H. The product of r and ~b is defined by

(r : CL~L If K is a subgroup of H, and if L is a subgroup of K, then

It is then clear that r H is a morphism of rings (with unit) from EndM~ck(u)(Res~M) G to EndM~ck(h-)(ResKM ).

Moreover, if x 6 G, and if L C_ XH, then denoting by c~ the conjugation by x, I have

(=r : ~xr162

and it is also clear that the conjugation by x is also a morphism of rings with unit from EndMa~k(H)(aes~M) to EndMack(.H)(ReSa~HM). If now H c_ It', if a e EndM~ck(~.-)(Res~-M) and if b ~ EndM.~k(u)(Res~M), then for any subgroup L of G

a.t~(b ~ a~o t ~ , ~ : a ~ o E ~ -~ ~ x6L\K/H

As a is an endomorphism of Res~M, as L C_ K, and as x 6 K, I have

L L aL o tLn .Hr x = ~Ln~HCxaL oH

( a'tK(b )) L : E tLvl=HCxaLxclHbLxclHcxL -1 FLvIxHL . . . . xeL \K/H

�9 In the same conditions, if a E A(K) and ro E M(H), then

a.tl~(rt)O 1s IV : t.(,'H(a)..~)

and if a ~ A(H) and ~ ~ M(I ( ) , then

SO

2.1. DEFINITIONS 43

hence a.t~(b) = t~(r~i(a).b ). The equality t~(b).a = t~(b.r~r(a)) follows similarly from the equality

--I L 0 -I L C x rLN~ H a L = a L x n H C ~ ?'LASH

and this proves the proposition. ,,

When G is a group and k a ring, a kG-module is a k-module with a morphism of rings kG ~ Endk(M). Reformulating this assertion in the case of Green functors, replacing "ring" by "Green fnnctor" and Endk(M) by ~ ( M , M), I get the following proposition

P r o p o s i t i o n 2.1.2: Le t M be a M a c k e y f u n c t o r , and A a G r e e n f u n c t o r for t he g r o u p G. T h e n it is equ iva len t to give M a s t r u c t u r e o f A - m o d u l e , or to give a u n i t a r y m o r p h i s m of G r e e n f u n c t o r s f r o m A to ~ ( M , M).

P r o o f : l ) Let M be an A-module. If L C_ H are subgroups of G, and if a E A(H), I define an endomorphism CH(a)L of M(L) by setting

=

and I claim that I obtain that way an endomorphism CH(a) of Res~M. Indeed, if L C K _C H, and if m E M(K), I have

and this shows that CH(a) commutes with restrictions. Similarly, if x E H, if L C H, and if m E M(L), then

H a x

since if x E H and a E A(H), then ~a = a. Hence CH(a) also commutes with conjugations by elements of H. Finally, if L C_ K C H, and if m E M(L), then

and this shows that CH(a) is an endomorphism of Res~M. It is clear that CH(1A(H)) is the identity endomorphism of ResaM, since rLH(1A(H)) = 1A(L), and that CH is a morphism of rings with unit from A(H) to EndM~k(H)(Res~M), since moreover for a, a' E A(H)

(r162 : Cg(a)L(r~(a').m) . . . .

. . . . rH(a)rH(a').rn = rH(aa').m = CH(aa')L(m)

I also claim that the maps CH define a morphism of Mackey functors from A to ~ ( M , M): indeed, let L C H ' C H be subgroups of G, let a E A(H), and m E M(L). Then

H r = r~L rH,(a).m = rLH(a).m = CH(a)L(rn)


and this shows that r = r for any subgroup L of H' , or that eH,(r~,a) =

r/~,r Now if x E G, if L C_ H _C G, if a E A(H) and m E M(L), then

: a ) . " = =

and ~ ( r r Finally if L C_ H ' __D H, if a E A(H), and m �9 M(L), then

H' H' H'

x6L\H'/H

But

where I denote by v ~ v ~ the conjugation by x -1. This equality can also be written a s

L x H

and finally H r

= t L n x H C x r x Ln~H x E L \ H ' / H

which proves that H ~

:

so 4) is a morphism of Mackey functors. So to any A-module M, I know how to associate a morphism of Green functors from A to Tt(M, M).

2) Conversely, let M be a Mackey functor, and %b be a morphism of Green functors from A to ~ ( M , M). Then for any subgroup H of G, I have a morphism of rings with unit ~)H from A(H) to EndM~k(H)(Res~M). Then, for any a E A(H) and any subgroup K of H, I have an endomorphism OH(a)r," of the R-module M(K). Then for m E M(H) and a E A(H), I set

I obtain a structure of A(H)-module (with unit) on M(H): indeed

1A(H).m = ~H(1A(H))H(rn) = m

because ~p preserves the unit. Moreover, the product (a, rn) ~ a.m is distributive with respect to addition in M(H) , because ~H(a) is an endomorphism of M(H). The product is also distributive with respect to addition in A(H), because eH is a morphism of R-modules from A(H) to EndM~ck(H)(ResaHM). Finally, as ~pH is multiplicative, it follows that for a ,a ' E A(H) and m E M(H), I have (aa').m = a.(a'm). Those structures of A(H)-module on M(H) are compatible with the Mackey functor structures of A and M: indeed, if x E G, if K C_ G, if a E A(K) and if rn E M(K), then

'2.1. D E F I N I T I O N S 45

But ~,~K(~'a) = a:(r because ~b commutes with conjugations. But for any subgroup L of "K, I have

and for L = "K this gives

Similarly, if H C_ I f , if a E A ( K ) and m E M ( H ) , then

~ . 4 ( ~ ) ' i,- = t?s<(a)KttHm ) = tHkgJh'ta,hrtrn,)

because r is an endomorphism of the Mackey functor a Res1< M. But

A" = (,/.,<%,))).

because f, is a morphism of Mackey functors from A to ~ ( M , M). Finally,

a.tI~(m) = tI~ ( (g,H(r~(a) ))H(m)) = t ~ ( r ~ ( a ) . m )

Now if a C A ( H ) and m r M(K) , then

tg~ ( a ) .m = ,/,,, (~,~ ( < ) , , / , , , t = , , , , , , , h " : ( v , , ( a ~ , . ( ,~

and moreover

Finally

E Ix -1 1,2 K K ti,'na.HC:r~ H(a)K~nHC~: rKnZH = t H ~ H ( a ) H F H

K K

and this proves that M is an A-module.

3) The last observation is that the above correspondences are mutually inverse: if M is an A-module, I associate to M the morphism 6 defined by

and starting kom 6, I define an A-module structure by

and I recover the initial A-module structure. Conversely. if g, is a. morphism of Green fimctors from A to ~ ( M , M), I associate to 4' the A-module structure defined by

and this in tm'n corresponds to the morphism b of Green functors defined by

and this completes the proof. �9


2.2 Def in i t i on in t e r m s of G-sets

The previous section leads to a definition of Green functors and their modules in terms of G-sets. Indeed, if A is a Green functor and M an A-module, then by the previous proposition there is a morphism of Green flmctors

Using adjunction, and commutativity of the tensor product, I get a morphism

A+M -~ M

which in turn corresponds to a bilinear morphism AM E s M; M), determined by maps

Ax,v : A(X) x M(Y) ~ M ( X • Y)

for any G-sets X and Y. I will use intensively the following multiplicative notation, for a E A(X) and m E M(Y)

a • m = L ~ y ( a , m )

Those remarks and notations lead to the following definition, akin to the classical definition of an algebra:

Definition: Let R be a commutative ring. A Green functor A (over R) for the group G is a Mackey functor (over R) endowed for any G-sets X and Y with bilinear maps

A(X) x A(Y) ~ A ( X x Y )

denoted by (a,b) ~-+ a x b which are bifunctorial, associative, and unitary, in tile following sense:

�9 (Bifunctoriality)Iff : X --+ X' and g: Y --+ Y~ are morphisms of G-sets, then the squares

X A(X) • A(Y) , A(X x Y)

A , ( f ) x A,(g) I I A , ( f x g)

A(X') x A(Y') ---* A(X ' x Y') X

X A(X) x A ( Y ) , A ( X x Y )

A'(f) • A*(g) l l A*(f x g)

A(X') x A(Y') .-~ A(X' x Y') •

are commutative.

�9 (Associativity) f i X , Y and Z are G-sets, then the square

A(X) x A(Y) x A(Z) I

( x ) x IdA(z) I 4-

A ( X x Y) x A(Z)

IdA(x) • ( x ) )

)

•

A(X) x A (Y x Z)

A ( X • Y • Z)

is commutative, up to identifications (X x Y) x Z ~_ X x Y x Z ~_ X x (Y x Z).

2.2. DEFINITION IN TERMS OF G-SETS 47

�9 (Unitmqty) f f . denotes the G-set with one element, there exists an element e r A( . ) such that for any G-set X and for any a 6 A(X)

A.(px)(a • ~) : a = A . ( q x ) ( e • a)

denoting by Px (resp. qx) the (bijective) projection h'om X x * (resp. f rom

. x X ) t o X .

If A and B are Green functors for the group G, a morphism f (of Green functors) from A to B is a morphism of Mackey functors such that for any G-sets X and Y, the square

is commutative.

X A(X) x A(Y) , A(X x Y)

f x x.f~, I [ f xxY

B ( x ) • B(~") , B ( x • Y) X

If moreover f . : A(*) ~ B(o) maps the unit of A to the unit of B, then [ will say

that f is unitary. I will denote by GreenR(G) or Green(G) the category of Green [unctors for G over R, and unitary morphisms between them.

R e m a r k s : 1) T h e expression "up to identif ications (X • Y) • Z _~ X • Y • Z ~_

X x (Y x Z)" means more precisely that the diagrams

- A(X)xA(Y)• ,A(X)•215

t,d•215 A( X x Y) x A(Z) A( X) x A(Y x Z)

A*(flx,<z) A(cYxY)xz) A(x• ,

are c o m m u t a t i v e . Here c~xy, z is the canonical bi ject ion ((z,y), z) ~ (x ,y ,z) from (X x Y) x Z to

X x Y x Z, and flxy, z is the bi ject ion ( x , ( y , z ) ) ~ ( z , y , z ) f rom X x (Y x Z) to X x Y x Z. From now on, I will always forget those isomorphisms, and I will wri te

d i rec t ly the c o m m u t a t i v e square of the definition. 2) Similarly, I will always wri te a x e = a = e x a, identifying (X x . ) and ( . x X )

with X .

There is an analogous definit ion for a module over the Green functor A:

D e f i n i t i o n : Let A be a Green functor over l~ for the group G. A module M over A is a Mackey functor M (over It), endowed for any G-sets X and Y with R-bilinear maps

A(X) x M ( Y ) --, M ( X x Y)

denoted by a x m ~ a x m, which are bifimctorial, associative and unitary in the fol]owing sense:


�9 (Biflmctoriality) If f : X ~ X ' and g : Y --* Y' are morphisms of G-sets, then the squares

•

A(X) • M(Y) M(X • Y)

A,(f) x M.(g) l ~ M,( f x g)

A(X') x M ( Y ' ) M ( X ' x Y') X

•

A(X) • M(Y) M(X • Y)

A*(f) x M * ( g ) I T A*(f xg)

A(X') • M ( Y ' ) M(X' x Y') )4

are commutative.

�9 (Associativity) f iX , Y and Z are G-sets, then the square

A(X) x A(Y) • M(Z) !

(• • [dM(z) [

A(X • Y) x M(Z)

IdA(x) x ( x ) +

•

A(X) x M ( ' / • Z)

M(X x Y x Z)

is commutative, up to identifications (X x Y) • Z _~ X • Y x Z _~ X • (Y x Z).

�9 (Unitarity) For any G-set X and any rn C M(X)

M.(qx)(~ • .~) = ,~

If M and N are modules over A, a morphism (of A-modules) f from M to N is a morphism of Mackey functors such that for any G sets X and Y, the square

•

A(X) • M(Y) , M ( X • V)

I d x fz [ l Jx•

A(X) • N(Y) , N ( X • Z) X

is commutative. I will denote by A-Mod the category of A-modules.

R e m a r k : Jus t as before, I will identify ( . • X) and X, and write c • m = m.

2 . 3 E q u i v a l e n c e o f t h e t w o d e f i n i t i o n s

Let A be a Green functor in the classical sense. It is clear tha t A is also an A-modu le in the classical sense, since the module A(H) has a natura l s t ruc ture of A ( H ) - m o d u l e .

T h e n proposi t ion 2.1.2 proves tha t there exists a morph i sm from A to "If(A, A), hence

2.3. EQUIVALENCE OF THE TWO DEFINITIONS 49

by adjunction a morphism from A ~ A to A, and also then a bilinear morphism from A, A to A: to be concrete, if H and K are subgroups of G, the maps

A(G/H) • A ( G / K ) ~ A(G/H • G/K) ~_ | A ( G / ( H N ~I()) z6H\G/K \

are defined for a 6 A(G/H) = A(H) and b < A(G/K) = A(K) by

H ~K z/) a X D = e rHn+K(a).rHC~h-( )

a'EH\GtK

This follows easily from the proofs of propositions 1.10.1 and 2.1.2: proposit ion 2.1.2 shows that A is a "H(A, A) module. If H is a subgroup of G, and if a E A(H), then a defines an endomorphism of the Mackey functor Res~A by

VL c_ H, b E A(L) ~ rH(a).b E A(L)

By adjuncdon, I can associate to a ~ morphism from A to Ind~Res~A : AO/H defined for K C_ G by

r H ~h" x G G * b E A(K) ~ | Hn.i,.(a).rgn,~,.(b) E IndHReSHA(E) = ~ , A(H N *K) :~EH\O/K ~HXa/*"

and the formula for a x b follows.

Similarly, if M is a module over A in the classical sense, and if m E M(K) , I have

H , , ~I~ ' ~ m ' (2.1) a • m = @ rHn.TA-(a).rHnXK[ ) xEH\G/B"

By proposit ion 1.8.3, I know that such a bilinear morphism is determined by R-bil inear maps A(Y) • M ( Y ) -* M(Y) , that I will denote by (a ,m) ~ a.m, defined by

a.7~ = M * ( ~ 2 , y ) ( a • m )

This notat ion is coherent, for in the case Y = G/H, the map

A*(5~,v/,) : A ( G / H • G/H) ~- | A(H N ~H) + A(H) z6HkG/H

is precisely the projection on the component associated to the double class H.1.H = H. Formula (2.1) now gives

M*(f2,y)(a x m) = H H

The ring structures of the A(H), and the A(H) rnodule structures of the M(H), admit a unique extension to maps A(Y) x M ( Y ) --* M(Y) defined for any G-sets Y, a.nd having propert ies i), ii), and iii) of proposition 1.8.3: indeed, if Y is a G-set, let [G\Y] be a system of representatives of orbits of G on Y. For x E [G\Y], let i~ be

the injection from G/G~ into Y defined by

i~(.qG~) = .qx

Then as Y is the disjoint union of its orbits, for any a E A(Y) and any m E M(Y) , I h a v e

a = ~ A.(ix)A*(i~)(a), m = ~ M.(i~)M*(i~)(m) ~E[a\Z] ~E[a\~']

50 C H A P T E R 2. GREEN F U N C T O R S

Now condition i) of proposition 1.8.3 shows that for x ,y E [G\Y], a' E A(Gx) and m' C M(Gv)

. . . . . . )( t) A ( ~ ) ( a ) . M ( z , ) (m) = M.(i~. a ' .M.(ix)M*(iy)(m'

But the product M.(i~)M*(iy) is zero if x • y, and equal to the identity of M(G~) otherwise. It follows that

a . m = ~ M,(i~.)(A*(i,:)(a).M*(i~,)(m)) (2.2) ze[G\Y]

and the product (a, m) --* a.m is entirely determined (in particular, the above formula does not depend on the choice of the system of representatives of the orbits of Y).

(~) N o t a t i o n : / / " f i s a m a p f i ' o m a s e t X to a set Y, I will denote it by f = 7(~) . For

the ~rst projection ~,om ~ ~ , onto ~,~ < l ~e ~enoted ~, ( 'V' I o~ ( :: ), instance,

a~d the ~econd ,,'ojec,io. by ( 7 )

With this notation, propositions 1.8.2 and 1.8.3 show that the product (a, m) ~ a x 777 can be recovered from the product (a, m) ~ a.m using the following formula

a • m = A" (a).M" (m) (2.3)

Associativity of the product • now follows from associativity of the product ".", and from condition iii) of proposition 1.8.:3, since for a r A(X) , b r A(Y), and m E M ( Z ) , I have

\ y z /

But condition iii) shows that

\ yz / \ yz / y

hence finally

For the same reason, I have

2.3. EQUIVALENCE OF THE TWO DEFINITIONS 51

The existence of unit for the product • also follows from formula (2.3), since denoting by c the unit of A(G) = A(G/G) = A(.) , I have

But formula (2.2)shows that er = A* (Y) is a unit for the product ".", since

e y . m = Z M,(i~,, (A*(i,:,A* (:).M*(ix,(rn)) xe[a\Yl

But now ( : ) = A" an,,

s~176176 t~ G/G, andthenA*((Y,)i~) =re G~ Since F G G~(e) is the unit of A(Gx), I have

A similar argument shows that cy is also a right unit for the product "." of A(Y). Conversely~ I have seen that

a.m=M*(52,v)(axm)=M*(Y)(aXm)yy (2.4)

and it is easy to deduce from this equation that bifunctoriality and associativity of the product x imply associativity of the product ".", since for instance

whereas by a similar computation

And now ife is the unit for the product x, then gr is a unit for the product ".", since


A similar argument proves that sy is also a right unit for the product "." of A(Y). �9

So the definitions of a Green functor and of a module over a Green functor in terms of G-sets are equivalent to the classical ones. It also follows from equalities (2.3) and (2.4) that the definitions of morphisms of Green fnnctors and of morphisms of modules over a Green functor coincide: for example, if f is a morphism in the classical sense from the Green functor A to the Green functor B, then fH is a ring rnorphism from A(H) to B(H), for any subgroup H of G. It follows easily that for any G-set X, the morphism fx : A(X) - , B(X) is compatible with the product ".". Then formula (2.3) gives for M = A, a E A(X) and a' E A(Y)

Taking image under fXxY of this equality gives

f x x y ( a • (A* x~Y)(a).A*(;Y)(a')) . . . .

But as f is a morphism of Mackey functors, I have

and

Thus

]~- • y A* " = B* .fy

and f is a morphism in the sense of the new definition. Equality (2.4) proves similarly that a morphism in the sense of the new definition is also a morphism in the classical sense.

2.4 T h e B u r n s i d e functor

In this section, I suppose for simplicity that R = Z. All the results can be extended to the case of an arbi t rary commutat ive ring by "tensoring everything with R ~', and replacing tensor products over Z by tensor products over R.

2.4.1 T h e B u r n s i d e func tor as M a c k e y func tor

The classical definition of the Burnside functor b (over Z) is the following: for a subgroup H of G, the module b(H) is the Grothendieck group of the category of H-sets, for relations given by decomposition into disjoint union. Restriction and induction of H-sets give a Mackey functor structure on b. The direct product of sets

2.4. THE BURNSIDE FUNCTOR 53

turns b(H) into a (commutative) ring, and it is easy to check that b is actually a Green functor.

But this definition does not give the structure of b(X), when X is an arbitrary G- set. The next proposition answers this question. If X is a G-set, I denote by G-setJ.x the category with G-sets over X as objects, and maps of G-sets over X as morphisms.

L e m m a 2.4.1: Le t H be a s u b g r o u p of G. T h e f u n c t o r m a p p i n g a G-set (I/,r over G/H t o t he H-se t r is an equ iva l ence of c a t ego r i e s f r o m G-set~a/H to H - s e t . T h e inverse equ iva lence is t he i n d u c t i o n f u n c t o r f r o m H-se t to G-set~G/H.

P r o o f : Let (Y, qS) be a G-set over G/H. Then it is clear that the inverse image r of the element H < G/H is an H-set. If f : (Y,r --+ (Y' , r is a morphism of G-sets over G/H, then r = r so

f ( r C_ r

and so I do have a functor F from Gset$c/H to H-set .

Conversely, if X is an H-set, then there exists a unique morphism of H-sets (~) : X --+ .. The image of this morphism under the induction functor from H to G gives a G-set

r r ' ( x ) = Ind~,X -+ IndT. ~_ G / H

over G/H, and it is clear that F' is a functor from H-set to G-set~a/H. I will now check that F and F ' are mutual inverse equivalences of categories. Let me first recall that Ind~/X is the quotient set of G • X by the equivalence relation identifying (gh, x) and (g, hx) for any h E H. The group G acts on the left by g'.(g, x) = (9'9, x), where I denote by (g, x) the equivalence class of (g, x). The above map r is then given by r = gH. If f is a morphism of H-sets from X to X' ,

then F'(f) is given by

: ( g , s ( x ) )

But then r = H if and only if g E H. It it then clear that the map from

r r ' ( x ) = {(h,x) I h e H, x �9 x } to x which sends (h,x) to hx is an isomorphism from the fnnctor FF' onto the identity functor of H-set . Conversely, if (I/, r is a G-set over G/H, then the map from I n d ~ r to Y sending (g, y) to gy is surjective, because any orbit of G on Y meets r It is also injective, for if gY = g'Y' with r = r = H, then gH = g'H, and there exists h �9 H such that g' = gh, and gy : ghy', i.e. y = hy'. Then (g',y') = (g, hy') = (g,Y). It is easy to see that this is an isomorphism of functors from F'F to the identity functor of G-set~a/H, and this completes the proof of the lemma. "

As a consequence, if H is a subgroup of G, then b(H) is isomorphic to the Grothendieck group of G-setIG/H, for relations given by decomposition into disjoint union of G-sets over G/H. This is actually true for any G-set:

P r o p o s i t i o n 2.4.2: Le t X be a G-set . T h e n b(X) is i s o m o r p h i c to t he G r o - t h e n d i e e k g r o u p of G-set.Lx, for r e l a t ions g iven by d e c o m p o s i t i o n in to dis- jo in t un ion . M o r e o v e r , if (Y, r is a G-set over X,


�9 If f : X ~ X' is a m o r p h i s m of G-se t s , t h e n

~,((Y, ~)): (v,j~/

�9 If f : X ' ---* X is a morphism of G-sets, then b, ((Y, r is the pull-back (Y', ~') of (Y, r along f , obtained by filling the cartesian square

a

X' , X f

Remark: In this proposition, I identify (Y, r with its image in the Grothendieck group.

P r o o f i I can consider for a while this proposition as a definition of b. It is clear that the definitions of b. and b* can be extended by linearity to group homomorphisms between b(X) and b(X'), turning b into a bifunctor defined on G-se t , with values in abelian groups (i.e. Z-modules). If I can prove that b is a Mackey functor, and that it coincides as a Mackey functor with the Burnside functor defined on transit ive G-sets, the proposit ion will follow.

First it is clear that the Grothendieck group of G-se tSx is addit ive with respect to X: if (U,r is a G-set over X L I Y , then U is the disjoint union of r and r which are G-sets respectively over X and }'. Conversely, the disjoint union of a G-set over X and a G-set over Y is a G-set over X [I Y. Addi t iv i ty of b foIlows.

Now let 7

T , X

'1 1 ~ Y , Z

fl be a cartesian square, and (U, r be a G-set over X. Let V, a and b such that the square

V , U

T , X "7

is cartesian. Then b*(7)((U, r = (V, b), so

b.(~)b'(~)((U, ~)) = (V,~b)

On the other hand, I have ~.( .)((U, r = (U, ~ ) , and as the square a

V , U

~b I a r

Y , Z


is cartesian, because it is composed of the two above squares, I have

((U, = (V,

which proves that b*(t3)b.(a') = b.(~)b*(@, and so b is a Mackey functor. Now let H C__ K be subgroups of G, and L be a subgroup of H. I denote by FH

and F~, the equivalences between the categories associated to H in lemma 2.4.1. Then the image of the H-set H/L under the functor F~ is the G-set

G/L , G /H

over G/H, denoting by ~rL H the natural projection from G/L onto G/H. The image of that set under b.(Tr~) is the set

7rKTr H H L

GIL , GIK

that is the set (G/L, 7r)) over G/K. The image of that set under the functor F~,- is the set K/L, which is the induced set of H/L from H to K. If now L is a subgroup of K, then the image of K/L under F[,- is the G-set (G/L, Trf), which is mapped under b*(~r~) to the set (P, r filling the cartesian square

P > G/L

+l 1+ G/H > G/If

K 7r H

Then P II C/(H nXL)

xEH\I '[ /L

and the image of this under FH is

II H/(H A ~L) ~ Res~If/L x e H \ K / L

Finally, if x E G and if L is a subgroup of K, then the image of K/L under F[~- is the set (G/L, rrI[). Let cx be the conjugation by x, from G/K to G/'~K. The image of the previous set under b.(cx) is (G/L, cxTr~L ~) = (G/L, ~r~Kcx), which is mapped by F~K to XK/~:L, and this completes the proof of the proposition. �9

2 . 4 . 2 T h e B u r n s i d e f u n c t o r a s G r e e n f u n c t o r

The Green functor structure of the Burnside functor is also very natural:

P r o p o s i t i o n 2.4.3: Le t X and Y be G-sets . I f E -- (U, r ( resp . F = (V,~b)) i s a G-set over X (resp. over Y), I d e n o t e by E x F t he G-set (U x V,r x r over X x Y. T h e n t he p r o d u c t x can be e x t e n d e d by l inea r i ty t o a p r o d u c t f r o m b(X) | b(Y) to b(X x Y), and th is t u r n s b in to a G r e e n f u n c t o r , w i th t h e se t ( . , I d ) over �9 as un i t c E b(*) .

56 C H A P T E R 2. G R E E N F U N C T O R S

Proof: It is clear that the product ((U, 6), (V~ ~)) is left distr ibutive with respect to disjoint union of G-sets over X, and right distr ibutive with respect to disjoint union of G-sets over Y. It follows that the product x can be extended by bil ineari ty to a product b(X) | b(Y) --+ b(X x Y). This extended product is cleariy associative, and admits as a unit the trivial G-set over itself. The product x is also clearly bifunctorial: if f : X --+ X ' and g : Y --+ Y' are morphisms of G-sets, then

b.(:) ( (v, , ) ) • b.(g)((v, v:)) = (v, : ~ ) ~< (v, g,:) : (u • v, :,~ • gv:) . . . .

. . . . (u • v , ( : • ~)(+ • ~:,)) : ~. ( f • . ) ( ( v , r 2 1 5 (v, v:))

Now if f : X' ~ X and g : Y' ~ Y are morphisms of G-sets, and if the squares

f / gl U ~ ) U V' ) V

X ~ ) X Y' ~ Y f g'

are cartesian, then the product square

f ' • g' U' x V' ~ U x V

X ' x Y ' X x Y f •

is also cartesian, and then

b*(f • g)((V, ~) • (<~))

This shows that the product x is bifunctorial, and proves the proposition.

This product • gives in turn a product "." from b(X) @ b(X) to b(X), defined for two G-sets (U, r and (V, g)) over X by

~)) \ / \ ZZ

Let P be the pull-back of (U,~b) and (V , f ) , such that the square

P ~ V

U , X r


is cartesian. Then the square

P ~ U •

X ~ X •

x x

is also cartesian, and it follows that

This equality proves that the product "." coincides for X = G/H with the product of H-sets, up to identifications of ]emma 2.4.1: indeed, if K and L are subgroups of H, then Fb( H/K ) = (G/L, 7r H) and F~,(H/L) = (G/L, ~rH), and the pull-back of those two sets is

I I V / ( l i A ':L) xEI'( \H/L

which is mapped under FH to the set

I I H / (K N XL) "-~ (H/K) • (H/L) ~6K\H/L

2 . 4 . 3 T h e B u r n s i d e f u n c t o r a s i n i t i a l o b j e c t

The Burnside functor plays the same role for Green functors as the ring Z does for rings with unit: it is an initial object of the category of Green functors:

Propos i t ion 2.4.4: Let A be a Green functor for the group G. Then there exists a unique (unitary) morphism of Green functors from b to A.

Proof: Let f be a unitary morphism from b t o A, and X b e a G-set. If (U,r is a G-set over X, then denoting by eb the unit of b, I have

Indeed, the square

(:) U m ) �9

1 U ) �9

(:)


is cartesian, and its right column is equal to e. Hence b* ( : ) ( e ) = (U, Id), which proves the above formula. Now if f is a unitary morphism of Green functors, I must have

and this proves that f is unique. Conversely, this formula defines a unitary morphism of Green functors: first it is a morphism of Mackey functors, for if g : X ~ X ~ is a morphism of G-sets, then

Thus

b , ( ( u , , ) ) = (u,g~)

Similarly, if g : X ' ~ X is a morphism of G-sets, then b* ((U, r is obtain by filling the cartesian square

Then

a

y ' , Y

X ' , X g

(:) (:) f x , b ' ( ( u , ~ ) ) = f x , ( ( u ' , ~ , ) ) = A. (~ ' )A" (~A) = A . ( ~ ' ) A ~ ( a ) A * ( ~ ) . . . .

It is also clear that f . ( ( . , Id ) ) = ~ . Finally, if Y is a G-set, and (V, r a O-set over Y, then

But up to identification of �9 x �9 w i t h . , I have

and as the product x is bifunctorial

For the same reason, I have then

. . . . sx ((u, •

2.4. THE B U R N S I D E F U N C T O R 59

which proves that f is a morphism of Green functors, and completes the proof of the proposition. �9

In particular, any Mackey functor M is a b-module, since there is a unitary morphism from b to ~ ( M , M). So Mackey functors can be identified to modules over the Burnside functor. This structure of b-module for a Mackey functor M is quite easy to describe: if X and Y are G-sets, and if (U, r is a G-sets over X, I have seen that

So if m E M ( Y ) , as the product • must be bifunctorial, I must have

( ) ( ) ( (U,r • r .y

2 .4 .4 T h e B u r n s i d e f u n c t o r as uni t

The next proposition confirms the analogy between the Burnside functor and the ring Z:

P r o p o s i t i o n 2.4.5: Le t M be a M a c k e y f u n c t o r for t he g r o u p G. T h e n t h e r e ex is t s i s o m o r p h i s m s o f M a c k e y f u n c t o r s

bQM ~- M ~ b ~- M

7{(b, M) _~ M

which a re n a t u r a l in M.

P r o o f : The first isomorphism is a consequence of the second one and of proposition 1.10.1: indeed, if 7-{(b, M) _~ M, then for any Mackey functor N, I have

"H(N,M) ~_ "H(N, 'H(b,M)) ~_ n ( b ~ N , M )

and those isomorphisms are natural in M and N. Evaluating this isomorphism at the trivial G-set gives the functorial isomorphism

HomMa&(~)( N, - ) ~- HomMa&(c)( b~ N, - )

and Yoneda's lemma now proves the first isomorphism. To prove the second one, I observe first that

HomMa~k(G)(b, M) ~_ M( . )

Indeed, I saw in the previous proof that if (U, r is a G-set over X, then in b(X), I have

So if 8 is a morphism of Mackey functors from b to M, I have

(:)0.(<


and 0 is then determined by the element 0.(gb) of M( . ) . Conversely, if m E M( . ) , then it is easy to cheek that the equality

defines a morphism 0 from b to M. This remark shows that for any G-set X

?t(b, M) (X) = HOmMack(a)(b, Mx) ~- M x ( ' ) ~- M ( X )

and it is easy to see that this isomorphism is actually bifunctorial in X, and so it is an isomorphism of Mackey functors from 7-g(b, M) to M. �9

R e m a r k : The previous argument shows also that the isomorphism 0 : b@M ~ M is obtained in the following way: if X is a G-set, if (1/r is a G-set over X, if (Z, ~b) is a G-set over Y, viewed as an element of b(Y) , and if m C M(Y) , then

Ox ( [ (Z, r | m](y,r = M.(r

C h a p t e r 3

T h e c a t e g o r y as soc ia ted to a G r e e n func tor

3.1 E x a m p l e s o f m o d u l e s over a G r e e n f u n c t o r

Let A be a Green functor for the group G, and X be a G-set. If M is an A-module, then M is in particular a Mackey functor. I know then how to build the Mackey functor Mx, and I can try to turn it into a module over A: if Y and Z are G-sets, if a E A(Y) and rr, E M x ( Z ) = M ( Z X ) , then the product a • m is in M ( Y Z X ) ~_ M x ( Y Z ) . I can view that product as a definition of an A-module structure for Mx:

L e m m a 3.1 .1: Let A be a G r e e n func tor , le t M be an A - m o d u l e , and X be a G-se t . I f Y and Z are G-se t s , t h e p r o d u c t

A(Y) • Mx(Z ) --+ M x ( Y • Z ) : (a,rn) ~-+ a • m

t u r n s Mx into an A - m o d u l e . T h e c o n s t r u c t i o n M ~ M x is an e n d o f u n c t o r f r o m t h e c a t e g o r y A - M o d of A - m o d u l e s , w h i c h is i ts own left and r ight adjo int .

Proof : The first assertion comes fi'om the f~ct that M is a module over the Green functor A. Moreover, the construction M ~+ Mx is clearly functorial in M. For the adjunction property, let N be an A-module, and 0 be a morphism of A-modules from Mx to N. Then for any G-set Y, I have a morphism 0r from M x ( Y ) = M ( Y X ) to N(Y). So I have a morphism 0yx from M ( Y X 2) to N ( Y X ) . I deduce a morphism 0~ from M(Y) to N ( Y X ) by setting for rn E M(Y)

O'v(m) = Oyx M. \ y z z

Those definitions turn 0 ~ into a morphism of Mackey functors fi'om M to Nx: if f : Y --+ Z is a morphism of G-sets, then

Nx,.(f)O~(m) = N. \ f ( y ) x / Oyx M. \ yx:c

As 0 is a morphism of Mackey Nnctors, I have

N. f(Y),~" Oyx = OzxM. \ f ( y )x~x2]

62 C H A P T E R 3. THE C A T E G O R Y ASSOCIATED TO A GREEN F U N C T O R

and SO

Nx, . ( f )O~(m) = OzxM. f ( y ) x x (m)

On the other hand

But the square

Y X , Z X

Y ~ Z

f

~s cartesian, ~o M" ( 7 ) M . ( S ) : M. (~,~;)M" (~;), a,d

O'zM.( f ) (m ) = OzzM. <zxxJ M. f(y),'r M* (m) . . . .

. . . . OzxM. f ( y ) x x M" (m) = N x , . ( f ) @ ( m )

hence O~ is covariant in Y. Similarly

N]( f )O' z : N" yx OzxM, M" - y)x z z x

As 0 is a morphism of Mackey functors

N* f ( y ) x Ozx = O y x M " \ f ( y ) x l x z J

But the square

Y X ~ Y X 2

\ f ( y ) x l x 2 ] ZX ~ ZX 2

zxm

is cartesian, so M* ( yxlx2 ] M, = M, ~ \f(y)xlx~] zx~: ](y)= '

f ( y ) x O z x = O r x M , yx M* M* = . . . yxx f ( y ) x

vxx f(y

3.1. E X A M P L E S OF MODULES OVER A G R E E N F U N C T O R 63

On the other hand

and this proves that 0' is a morphism of Mackey functors. Moreover, if U is a G-set, and if a ~ A(U), then

O[rxy(a • m)= Ogyx-']. ( uy," I ~I* (uyx I (a• m) \uy , rx / \ uy i

As M is an A-module

and

\ uy I

\ u y x x / y x x

Hence

Finally as 0 is a morphism of A-modules, I ]lave

and 0' is a morphism of A-modules. 1 (..onversel:r, if 0' is a morphism of A-modules from M to 1~:, I define for any G-set

Y a map 0"y from M ( Y X ) to N(Y) , by setting for m E M ( Y X )

A similar argument shows that 0" is a morphism of A-modules from M x to N. More- over, the constructions 0 H O' and 0' H 0 are mutual inverse isomorphisms between Hom4_Mo,~(Mx, N) and HomA-Moe(M, Nx): indeed, if 0 is a morphism of A-modules from M'x to N, associated to a morphism 0' from M to Nx, associated in turn to 0", then for any C-set Y

As 0 is a morphism of Mackey functors

Orx 2 = OyxM" y,~:x \ y x l x l X 2 /

a n d for t h e s a m e r e a s o n

64 CHAPTER 3. THE CATEGORY ASSOCIATED TO A GREEN FUNCTOR

and finally

O"y =OyM. (yXlX21 ]l/I" ( ~l'rl:C2 ) j'lJ. ( Y:r!x2 I M* (YxIX2) \ yX2 / \y,~hxl:r2 \y:ClX2X2/ \ yXl

But the square

YX > } " X 2

(::01 l y x 2 y X :~

1.J ;Z? 11" 2 ~2" 12?2 :t: 2

is cartesian, so

\y2glZlX2/ \y.7:l:r2:v2] \y:cx yxx/ and it follows that

O"Y:OY~/I*(YXl;2"2)]'~/[~(J/3?IJ':I~( ' I ] a ? I j ' I = ( ~ I $ I ' I ' 2 1 = O Y \ ~/X 2 \Yj:I:ff:] ]j273"] \ ?JaL" 1 j

since (Y~=~ (v~;) and (Y;::~) (y~:) are both equal to identity. \ yx2 /

A similar argument shows that if I start with a morphism 0' from 31 to R'x, associated to the morphism 0" from Mx to N, then 0' is the morphism associated to 0" by the first construction. Now the lemma follows, because the previous constructions are clearly functorial in M and N. �9

L e m m a 3.1.2: Let A be a G r e e n f u n e t o r , and M be an A - m o d u l e . T h e n

HomA_moa(A, M) ~- M(~

Proof : Let f be a morphism of A-modules from A to M. Then, for any G-set X, the square

X A(X) • A(o) , A(X)

Id• fo I [ fx

A(X) x M(~ -, M(X) X

has to be commutative. In particular, for all o E A(X), I must have

fx(a) = fx (a • c) = a • f .(c)

so f is determined by the element ] , (s) ~ M(~ Conversely, if m C M(~ it is clear that the equation

.L,, '((O = . , • - ,

defines a morphism f a morphism f r o m A to A4. �9

R e m a r k : I have already proved this lemma in the case A = b, by a slightly different method.

3.2. THE CATEGORY CA 65

P r o p o s i t i o n 3.1.3: Let A b e a Gr een func tor .

�9 If M is an A - m o d u l e , and X is a ( ; -set , t h e n

HomA-Mod(Ax~ M) "" M ( X )

�9 In par t i cu lar , if X and Y are G-sets , t h e n

HOmA-Mod(Ax, Ay) ~_ A ( X x Y)

Proof: The second assertion is a consequence of the first one, and of the fact that A y ( X ) = A(X • Y). And the first assertion follows from the previous temmas, since

HOmA-Mod(Ax, M) ~_ HOmA-Mod(A, Mx) ~- M x ( � 9 ~- M ( X )

3.2 The category CA Let f ~ c~(f) be the isomorphism of the previous proposition. Translating through f the composit ion product

HOmA-Mod(Ay, Az) x HomA-Mod(Ax, Ay) ~ HOmA-Mod(Ax, Az)

leads to the product oy

oy : A ( X x Y) • A (Y • Z) ~ A ( X x Z)

defined by setting, for a E A(X x Y) and a' E A(Y x Z)

a oy a '= el(a-~(a')o a- l (a))

With this definition, I have

L e m m a 3.2.1: Let a E A(X • Y) and a' C A(Y x Z). T h e n

\ xz / \ x y y z /

Proof: Let a E A ( X x Y). Then a determines a morphism h of A-modules from A to Axy , such that if U is a G-set and b r A(U), then

hu(b) = b x a E A ( U X Y ) = Axy(U)

I need to state precisely the homomorphism f = o:- l (h) from Ax to Ay. It is determined by its evaluations on G-sets, and the previous lemmas show that for a G-set U and b C Ax(U) = A(U x X), I have

fu(b) = Ay,. A;, uxx I t

66 C H A P T E R 3. THE C A T E G O R Y ASSOCIATED TO A GREEN F U N C T O R

Now it follows that

\ uy ) \ u x x y )

Similarly, any a' E A(Y x Z) is associated to a morphism .f' from A r to Az defined for b' E A(U • Y) = Az(U) by

\ uz I \ u y y z /

The composite f' o f is such that

\ ~z I \uyy.=/ uy / \uxa'y/

which may also he written, since A is a Green functor

( f lo f ) l - f ( " ) :A* ( t l~ lZ lA* ( IA~ lZ) 'U,Z / U,y'~IZ \ ?.,Yly2Z ] \'lzXxylY22J

But the square

is cartesian, and

( '{~xyz i u x y y z / U X Y Z , U X y 2 Z

( ::: )1 1 (<:':::) \ u y y z /

u y y z / \ uyly2z } k uyz / k u x y y z /

Finally

~l'os/~,~: ~. C"~q ~. (~"~:~-~1 ~" ( '"~ / A" ( ~'~'y~- ~ a , ~ . . . . \ uz I \ uyz J \ u x y y z J uxxyl~lTz

\ u x x Y Y Z l

The morphism f" = f ' o f from Ax to Az is then associated to the morphism h" = c~(f' o f ) from A to A x z defined for c E A(U) by

h';u(c) = f " u x A . \( "'r / ( u x )

and h" is associated to the element a" = h"o(e:) of A(XZ). But

h " ' ( c ) = f " x A * ( X ) A ' ( : )

3.2.

Final ly

THE CATEGORY CA

a"--A,(XlX2YZ~A*( Xlx2yz I (A.(xXx)A'(:)(~)XaXa ') \ XlZ / \XlX2x2YyZ/

As A is a Green functor, I have

But the square

: d . ( XlX2YlY2Z IA*(XlX2YlY2Z)(aXa ') \xlxIx2yly2z] \ x2yly2z

X Y Z ) X2Y2Z

1 X 2 Y Z ( xlx2yz 5 X3Y2Z

\xlx2x2yyz

is commutat ive. So

A* ( XlX2yz I kX,X2x:yyz/

It follows that

xlz / \ x x y z /

xlxlx2yly2z k x x y z / \ x x y y z /

\ x x y y z / \ x2yly2z /

,, xz / \ x y y z /

which proves the Iemma.

The previous lemma leads to the following definition:

67

�9 If X and Y are G-sets, then

HomeA(X,Y) = A(Y x X)

�9 f i X , Y, and Z are G-sets, i f a r A (Y x X) = HomcA(X,Y) and ira' r A(Z x Y) = HomcA(Y, Z), then the composite morphism a' o a E HomcA(X, Z) is defined by

a 'oa = a'oy a

This definition is made on purpose so that the following proposition holds:

�9 The objects of CA are the finite G-sets.

Defini t ion: Let A be a Green functor for the group G. I denote by CA the category defined as follows:

68 CHAPTER 3. THE CATEGORY ASS'OCIATED TO A GREEN FUNCTOR

P r o p o s i t i o n 3.2.2: T h e c a t e g o r y CA is an R-add i t ive ca t egory , and t h e cor- r e s p o n d e n c e w h i c h m a p s a G-set X to t h e A - m o d u l e Ax is a fu l l y fa i th fu l c o n t r a v a r i a n t f u n c t o r f r o m CA to A - M o d .

P r o o f : First the category CA is a pre-additive category (or "Ab-category" in the sense of Mac Lane [10]), in that for any objects X and Y of CA, the set HomcA(X,Y) = A(Y • X) has a natural structure of abelian group, for which the composition product is biadditive, and even R-bilinear. Moreover, the category CA has a zero object, namely the empty set: for any G-set X, the product X • ~ is empty, and as A(0) = {0}, I have

HomcA(r X) = HomcA(X, ~) = {0}

The existence of biproducts in CA (in the sense of Mac Lane [10] VIII.2) will then follow from the axioms of Mackey functors, and from the following lemma:

L e m m a 3.2.3: Le t f : X ~ Y be a m o r p h i s m of G-sets . I set

f . = A. f (x )x (c) E A(Y • X)

( x )A I:I f * = A . xf (x) ( r •

I d e n o t e by L (resp. [*) the c o r r e s p o n d e n c e which m a p s a G-set X to t h e o b j e c t X of CA, and t he m o r p h i s m f : X ~ Y in G-set to f . E HomcA(X, Y) ( resp. to f* E HomcA(Y,X)). T h e n I. (resp. I*) is a covar i an t (resp. con- t r a v a r i a n t ) f u n c t o r f r o m G-set to CA.

Admitting this lemma for a while, I see indeed that if ix and i r are the respective injections from X and Y to Z = X L [ Y , then in CA, I have

Moreover

i * x O N x , . : N * x o z N x , . : A . ( 2 : I Z X 2 ) A * ( 2 2 1 Z Z 2 ~ ( N * x X N x , . ) \ 321Z2 \Z IZZX2/

z x • x ix(x (:/ (x)A.(:) A* (e) x A. ix(x)x (e) . . . .

. . . . A. \ x l i x (xl)ix (x2):r2

X ) 32 X

X ~ X ~

) x z . (xl x ; xz2x

\~T1ZZX2/

But as ix is injective, the square

3.2. THE CATEGORY CA 69

is cartesian, and then

A*( xlZx2 ~A.( XlX2 ~ ( x )A , (X ) ",x,zzx2/ \x , ix (x , ) ix (z~)x2] = A. xix(x)x xx

so that

i~.oix,=, A.(XlZX2~A.(\ XlX2 / XiX(x)xX )A,(Xxx )A* (xlox2)(~): A*( xzx )A*(:)(~)

and this is the unit 1A(X 2) = Id. = Id* of EndcA(X). Similarly

\ Z1Z2 / \ZlXXZ2/

Moreover

ix,. x z x = A. ix(x)x xix(x

XlX2

As the square

(x:) X , X 2

( ix . (x) ; ix(x)) I I [ x ,x2 \ iX(X,)XlX2iX(x2) ]

\ZlXXZ2/

is cartesian, I have also

A.(z,xz2 ~A.~ x,x2 ~ ( x )A.I~ ) \ZlXXZ2/' kix(Xl)XIX2ix(X2)] = A, i x ( x ) x i x ( x ) =x

and it follows that

z , z2 J i x ( ~ : ) x i x ( x ) x x (~) . . . .

( �9 . . . . A. ix(x) ix(x) zz ix(x)

The same computation shows that

iy,. o ~y = A. A. (c) zz iy(y)

But as A is a Mackey functor, I have


Since A*(ix)A* ( : ) = A* ( : ) and A*(iy)A* ( : ) : A* (•), and since ix : (ix~x)), I finally have

.. .. ( ) ( : ) ix,. o z x q- iy,. o z r = A. z A* (c) = 1A(Z 2) Z Z

which proves that the disjoint union of G-sets is a biproduet (or direct sum) in CA. �9

P r o o f of l e m m a 3.2.3: Let X, Y and Z be G-sets, let f be a rnorphism of G-sets from X to Y, and let 9 be a morphism from Y to Z. By definition of f . , g. and their product in CA, I have

zx / \ z y y x / g(y)y (c) x A. f ( x ) x (e)

As A is a Green functor, the expression inside hooks can also be written

Moreover, since the square

f(x ) x)x X I Y X

Z Y X Z Y 2 X ) (z xl zyyx ]

is cartesian, it follows that

\ zx / g f ( x ) f ( x ) x f ( x ) x (c) . . . .

= . . . . m. (e) (go f ) . g/(x)x To prove that I . is a covariant functor, is suffices then to observe that the identity morphism of Ax is mapped to the morphism h = c~(Id) from A to A x 2 defined by

hu = IduxA* ( UX ) A* (UXu ) = A* ( UX ) A* uxz

And the rnorphism h corresponds in turn to the element h.(e) of A(X2). But

proving that 1A(X2 ) is the identity morphism of X in CA. Similarly

\ xz / \ x y y z /

3.3. A-MODULES AND REPRESENTATIONS OF CA 71

As the square

\ xz / \ x y y z / x f(x)yg(y) (~)

X

x f (x ) X X Y

X Y Z X Y 2 Z

xyyz /


\ xz / x f ( x )g f ( x ) A* x f (x )

. . . . A. xgf(x)

and this proves that I* is a contravariant functor, and completes the proof of the lemma. �9

3.3 A-modules and representations of CA Defini t ion: If R is a commutative ring, and C is an R-additive category, a representation of C over R is an R-additive functor from C to R-Mod. If F and F' are two representations of C, a morphism from F to F' is a morphism of functors from F to F'. I denote by Functn(C) the category of representations of C over R.

Let M be an A-module. By composition of the contravariant functor Homn-Mod(--, M) with the additive contravariant functor X ~-* Ax from CA to A-Mod, I obtain a representation FM of C,; over R, defined for a G set X by

FM(X) = HOmcA(AX, M)

But I have seen (see prop 3.1.3) that if M is an A-module, then HomA(Ax, M) identifies with M(X) : a morphism f from Ax to M is determined by the element

of M(X) : if U is a G-set, and i f a E Ax(U) = A(UX), then

f g ( a ) = M . ( 7 ) M * ( u z ) m) (3.1)

Then if X and Y are G-sets, and if r C A(Y x X) is a morphism in CA from X to Y, it is associated by proposition 3.t.3 to a morphism ~b from Ar to Ax defined for a r A(U x Y) = Ay(U) by

eu(a ) = Ax,. Ax uyy \ ux / k u y y x /

Let m 6 M(X) . It is associated to a morphism from Ax to M defined by equa tion (3.1). By composition with r I get a morphism f r from Ar to M, which is determined by the element

and m' is also by definition the image of m under FM(~)). As ~bzA. ( ; ) A* (.Y)(e)is

the element determining the morphism r equation (3.1) gives

This is the expression of the morphism FM(r : M ( X ) ~ M(Y) , image of r under

FM. The analogy of those formulae with the definition of the composition product ox

in CA explains the following notation:

N o t a t i o n : Let A be a Green functor, and M be an A-module. l f X and Y are G-sets, i f a G A ( Y • X ) and m E M(X) , I set

With this notation, I see that

Fu(r : o ox

and now say that FM is an R-additive functor is equivalent to say that this product ox is bilinear, associative and unitary (in an obvious sense). R e m a r k : The previous notation is coherent with the initial one: if M is the functor Az, for a G-set Z, then for a E A ( Y X ) and b E M ( X ) = A ( X Z ) I find

a o x b = A z , . Az yxx <yz ) \ y x x z /

and this coincides with the initial definition of the product a ex b Now to any A-module M, I have associated a representation FM of CA. This clearly

defines a functor from A - M o d to Functn(CA): if f : M --~ N is a morphism of A- modules, composition with f induces a natural transformation from Homm-Mod(--, M) to HomA-Mog(--,N), and so a natural transformation from FM to FN. Conversely, let F be a representation of CA over R. If X is a G-set, I set

ME(X) = F ( X )

It is an R-module. If f : X ~ Y is a morphism of G-sets, I set

MF,.(f) = F( f . ) M[:(f) = F( f . )

Those definitions turn MF into a bifunctor over G-set, with values in R-Mod . More- over as F is additive, it follows that MF has property (Mt) of Mackey functors, i.e. that MF transforms disjoint unions into direct sums. The following lemma shows that My has also property (M2), hence that it is a Mackey functor:

3.3. A-MODULES AND REPRESENTATIONS OF Ca 73

L e m m a 3.3.1: Let 7

T , Y

Z , X

be a c a r t e s i a n square of G-sets . T h e n in CA, I have

fl* o 0~. =- ~5. o ",/*

Proof: By definition

f l*~ =fl*~ = A * ( z x y l A * ( zxy l ( f l* ] \ z x x y /

and moreover

( ) ( : ) (~)~,/:),~, . . . . z A* (e) x A, a(y)y f i * x a , = A , zfl(z)

. . . . A. zf l(z)a(y)y

By hypothesis, the square

is cartesian, and so is the square

which gives

7 T , Y

~1 1 ~ Z " ~ X

fl

( ~1) T , Z Y

\ zxxy /

k z x x y / zf l(z)a(y)y = A, 5( 0 f16(t) 7(O 5(t)~(t)

It follows that

\ zy / 5(0 ZS(t) 7(t) 5(t)7(t)

. . . . A. 6(t)~/(t) (c)

But the square

Moreover

& x 3 , ' = A . 6(t)t

is cartesian. Hence

\z t ty] A. \6(ti)t,t~7(tJ] 6(t)tT(t ) tt

and it follows that

)A.(')A-(":') ( ' )A .<: )< . , \ zy / 6(t)t~/(t) tt (~) = A. 6(t)~/(t )

which proves the lemma. ,,

<,.) (,)A.<,.),,, .... A* (e) • A. tT(t)

t~t2 ~ A*

(;) T ~ T 2

\ 6(t,)tlt25(tj / )

\ ztty ]

Finally, I can turn MF into an A-module: is X and Y are G-sets, if a 6 A(X) and m E MF(Y) = F(Y), then the element A . ( ~ ) A * ( ~ ) ( a ) is in A(XY 2) = HorncA(Y, XY), and I can consider its image under F , which is a morphism from F(Y) to F(XY). Then I set

( ( x Y ) A . ( x : ) ( a ) ) ( m ) E F ( X Y ) = M F ( X Y ) a x m = F A. xyy

L e m m a 3.3.2: T h o s e d e f i n i t i o n s t u r n M r in to an A - m o d u l e .

P r o o f : I must check that the product is bifunctorial, associative, and unitary. To simplify the proofs, I will first check that in the case F = FN for an A-module N, then MF is equal to the A-module N.

It is clear indeed that for any G-set X, I have

M~(X) =FN(X) = N ( X )

On the other hand

\ zy / \zt ty]

3.3. A-MODULES AND REPRESENTATIONS OF CA

On the other hand, if f : X --* Y is a morphism of G-sets, and if n E N ( X ) , then

But

( x ) A * ( : ) ( XlX2 ~ N" (XlX2)(n) f . • n = A, f ( x ) x (~) • n = N . \ f ( x l ) X l X 2 ] \ x2 /

(x) x x

X , X 2

\ f ( X l ) X l X 2 ] Y X , Y X 2

y x x

As the square


y x x \ f (Xl)XlX2 ] = N. f ( x ) x xx

and so

MFN,.(f)(n) = N. N. f ( x ) x xx '. x2 /

. . . . N. f (x

If now n C N ( Y ) , I have

Moreover

( x ) A * ( : ) ( x y ) N . ( 7 ) f* x n = A. x f ( x ) (e) x n = N. x f ( x ) y (n)

But the square (x)) xf(x

X ~ X Y

X Y ~ X Y = (-) xyy

75


is cartesian. It follows that

xyy xf(x)y xf(x xf(x

and then

(7) (x)~.(x )~.(7),:~, .... M~N(f)(n ) = N. N. xf(x) xf(x)

Now for the product, if a C A(X) and n C MFN(Y) = N(Y), then

( ( x Y ) A * ( ? ) ( a ) ) ( n ) : A . (xY)A*(Xgz) (a )ovn . . . . a x n = F N A . xyy xyy

. . . . N , (xy lY21N*( xylY2 I [ A , ( X Y ) A * ( ? ) ( a ) • ] \ xyl / \xyly2y2/ xyy

The expression inside hooks is also equal to

N.('TylY21N*(xylY21(aXrt ) \ xylYlY2 / \ xy2 /

As the square


(xy) xyy

X Y ~ X Y 2

(:~)1 1 (:;:=) XY2 ( xyly; i XY3

\ xylYlY2

\ xylyxy2 / k xyly2y2 / xyy xyy

Hence

a x n = ~ ' ( x ~ l Y 2 1 ~ ' ( x ~ ) N * ( x ~ ) N ' ( x ~ l Y 2 1 ' a x e ' , xyl J x~ x~ , xy2 j . . . .

. . . . N. xy xy

So I recover the initial product for the A-module N. Let me record this result in the following lemma:

3.3. A-MOD ULES AND REPRESENTATIONS OF CA 77

L e m m a 3.3.3: L e t A b e a G r e e n f u n c t o r , a n d N b e an A - m o d u l e . I f X a n d Y a r e G- se t s , if a C A(X) a n d n C N(Y), t h e n

a x n = A. xyy

I must now prove that the product I have defined on ME is bifunctoriah Let f : X ---* X' and g : Y ~ Y ' b e m o r p h i s m s o f G - s e t s . I r a C A(X) a n d r e E M(Y), then by definition of ME

A.(f)(a)xM~,.(g)(m) = F m. A" m.( / ) ( a ) (F(g.)(m)) . . . .

x'y' ~ A* (x'y'~ . . . . F(A*(x 'y 'y ' ] \ x' ]A.(f)(a)oy, g.)(m)

On the other hand

MF, . ( f •215 A. xyy

( ( X Y ) A ' ( X : ) ( a ) ) (m) . . . . F (f x g). oxy A. xyy

So I have to show that if I set

r = A. \x'y'y'] \ z' ] n.(f)(a) OF, g.

= (f x g). oxy A. xyy

then F (C) (m) = F(g?)(m) for any R-addit ive functor f on CA and any m E F(Y). This is equivalent to say that for any F, I have F ( r = F ( r and the following lemma proves that this is equivalent to say that r = r

L e m m a 3.3.4: L e t X a n d Y b e G-se t s . Le t r r ~ b e d i s t i n c t e l e m e n t s of A(YX) = Hornca(X,Y). L e t F b e t h e f u n c t o r FAx. T h e n F ( r r F ( r

P r o o f : Let F be the functor FAx. Let IA(X 2) C F(X). Then

F(r = r 1Ar = r r r = F(r

and so F ( r r F(~b). ..

Now to prove that ME is covariant, I must prove that

( x'Y' ) A ' ( x ' Y ' ~ ( x Y ) A * ( X : ) ( a ) r \x'y'y'] \ x' ]A.(f)(a)oy, g . = r xyy

but this equation is the case F = FAy,. And in that case, I have seen that ME = Ay,, and the product x on this functor is eovariant. It follows that the product x on ME is eovariant for any F .


Similarly, to prove that the product on MF is contravariant, I consider in the same conditions elements a' E A(Y') and m' E MF(Y') = F(Y'). I have to show that

A*(f)(a') • MT~(g)(m') = M~(f • g)(a' • m') (3.2)

The left hand side is equal to

whereas the right hand side is

( ) . . . . F ( f x g)* ox,y, d . \x 'y 'y '] \ x' ] (a') (m')

The previous lemma now proves that equality (3.2) is a consequence of the following o n e

A. zyy f(x) (a') oz, g* = ( f • g)* ox,y, A. \x 'y 'y '] \ x' ] (a')

which is the case F = FAy. As MFAy = Az, and as the product on Ay is bifunctorial, this equali ty holds, and the product on My is contravariant.

Similarly, to check the associativity of the product on My, I consider three G-sets N, Y, Z, and elements a E A(X), b e A(Y), and m �9 ME(Z) = F(Z). Then

. . . . F(A.(\xyzyz/XYZ I (xYZ)(a))F(A*(yYzZz)A*(Y;)(b)) (m) . . . .

On the other hand

( axb ) x m = F A . \ x y z z / x y /

Now lemma 3.3.4 shows that associativity of the product on My for any F follows from equali ty

A.(kxyzyz/XYZ I ( X Y x Z ) ( a ) ~ (b):A*(xyzkxyzz]lA*(xyzl(axb)\ xy /

3.3. A-MODULES AND REPRESENTATIONS OF CA 79

which expresses associativity of the product on MFaz = Az. So the product on ME is associative.

Now concerning unit, if X is a G-set, and if m E MF(X) = F(X), then

and this completes the proof of lemma 3.3.2.

The previous lemmas lead to the following theorem:

T h e o r e m 3.3.5: T h e cons truc t ions which to an A - m o d u l e M assoc ia te the representa t ion FM of CA, and to the representat ion F of CA assoc ia te the A - m o d u l e ME a r e m u t u a l inverse equivalences of categor ies from A - M o d to F~nctR(CA ).

Proof: I have already proved that for any A-module N, the A-module MFN identifies with N. I need to prove that the construction E ~ ME is functorial in E, and that for any representat ion E of CA, the representation FME is natural ly isomorphic to E.

If 0 : E --+ E ' is a morphism of representations of CA, i.e. a natural t ransformation of functors, then 0 is determined by giving, for any G-set X, a morphism of R-modules Ox from E(X) to E'(X). If Y is a G-set, and if a E A(YX) = HomcA(X,Y), then

the square E(~)

E(X) , E(Y)

o,:[ l oy

E'(X) E'(a) E'(Y)

must be commutat ive. By definition, I have Mu(X) = E(X), and to prove that E ~ ME is functorial in E, I must check tha t the morphisms Ox define a morphisra of A-modules from ME to ME,. If f : X --+ Y is a morphism of G-sets, then the square

Mg,.(f) = E(f.) ME(X) , ME(Y)

i )

ME,(X) Mz,,.(f) : E'(f.) ME,(Y)

is the above square for a = f . . Similarly, if f ' : Y --+ X is a morphism of G-sets, then

M~(f') = E(f'*) ME(X) , ME(Z) 0x I I ~

, M ~ , ( Z ) ME,(X) M*c,(f') = E'(f'*)

the square


is the same square for a = f'*. Hence the maps Ox induce a morphism of Mackey functors from ME to ME,. Moreover, for any G-set Y, for all a ~ A(Y) and all m E ME(X), I have

so 0 is a morphism of A-modules. Finally, by definition, for any E and any X I have

F.~(X) : M~(X): E(x)

Moreover, if a r A(YX) = HomcA(X, Y), and if rn r E(X), then

FMs(a)(rn) :ME'* ( Y ; ) M~ ( yx l (a . . . .

yx yx * [A . ( yxlx: I A , ( y x l x 2 1 ( a ) ] ( m ) \yxlx2x2/ \ yxl ]

. . . . E ( ( Y ~ ~:) ogx (YX l * o g x 2 A . ( y x l x 2 1 A * ( y x l x 2 1 ( a ) ) ( m ) �9 \ y x x / \yx~x2x2/ \ yxl ]

I set

u = = A. (e) E y . yyx

( ) yx yx A* v = = A. (r r A ( Y X Y X 2) yxx yxyxx

and xlx2 IA ( xix2)(al A(yX3

\yxlx2x2/ \ yx~ The elements u, v, and w do not depend on F , and moreover

FM~(a)(m) = E(~ o~ v o~ ~)(m) (3.3)

If I take for E any fnnctor FN associated to an A-module N, then I know that ME ~-- N, and the previous equality is

FN(a)(m) = FN (u oyx v oyx~ w)(m)

Then for any N, I have

and temma 3.3.4 proves that

U Oyx V Oyx2 W ~ a

Changing equali ty (3.3) according to this proves that for any functor E

FME(a)(rn) = E(a)(m)

and the theorem follows. ..

Chap ter 4

The algebra functor

assoc iated to a Green

4.1 T h e eva lua t ion functors

Theorem 3.3.5 states an equivalence between A-modules and representations of CA, so it is possible to use the general results on representations of categories (see [3]). In particular

P r o p o s i t i o n 4.1.1: Le t A be a G r e e n func to r . I f X is a G-set , t h e f u n c t o r Ex of e v a l u a t i o n at X (def ined by Ex(F) = F(X) ) f r o m FunctR(CA) to EndcA(X)- M o d has a left ad jo in t Lx,- : V H Lx,v def ined by

Lx,v(Y) = HomcA(X, Y) | = A ( Y X ) | V

I f f E HomcA(Y, Z) = A(ZY) , t h e n t he image u n d e r Lx,y( f ) of t h e e l e m e n t a | v of Lx,v(Y) is equa l to

Lx,y( f)(a + v) : ( f oy a) | v C Lx,v(Z)

I t follows t h a t ExoLx,_ is t he i den t i t y f u n c t o r on A(X2) -Mod, wh ich t h e r e b y m a y be ident i f ied to a full s u b c a t e g o r y of A - M o d .

R e m a r k : There are two different products on A(X2): one is the product ".", the other is the product ox. The algebra structure on A(X 2) in the proposition corresponds of course to the second one.

This adjunction provides in particular a co-unit r/x, which for any functor F on CA induces a morphism ~X,F : Lx,p(x) ~ F defined for a G-set Y by

a | v E LX,F(x)(Y) ~ 7]X,Fy(a @ v) = F(a)(v) E F(Y)

This defiIfition makes sense because if a C A ( Y X ) = HorncA(X,Y), then F(a) is a morphism of R-modules from F(X) to F(Y) , and so F(a)(v) E F(Y) .

82 C H A P T E R 4. THE A L G E B R A ASSOCIATED TO A GREEN F U N C T O R

4.2 Evaluation and equivalence

Now a natural question is to know if there are G-sets f~ such that Eo is an equivalence of categories between Functn(CA) and A(f~2)-Mod. To answer that question, I need the following definitions:

Definitions: 1) Let f~ be a G-set. If n is a non-negative integer, I denote by n~2 the G-set obtained by taking the disjoint union of n copies of f~, and I say that nf~ is a multiple of fL 2) If X and Y are objects of CA, I say that X is a direct summand of Y in CA (or that X divides Y in CA) if there exists a split monomorphism from X to Y in CA, that is if there exists an element a E A ( Y X ) and an element fl E A ( X Y ) such that

fl Oy Oe = 1A(X2 ).

In particular, if i is an injective morphism from the G-set X into the G-set Y, then lemmas 3.2.3 and 3.3.1 show that i* or i. = IA(X2). Indeed the square

Id X , X

X , Y i

is cartesian, and it follows that

i* oy i. = Id. ox Id* = 1A(XZ ) o X 1A(X2 ) = 1A(X2 )

Thus injections in G-set are mapped by the functor I . to split monomorphisms.

Proposition 4.2.1: Le t A be a Green functor over R for t h e g r o u p G, and f~ be a G-set . T h e following conditions are equivalent:

1. T h e evaluation functor Ea f r o m FunctR(CA) to A(g~e)-Mod is an equivalence of categories.

2. For any G-set X, t h e r e exis ts an in t ege r n such t h a t X is a direct s u m m a n d of nf~ in CA.

Proof." If Ea is an equivalence of categories, then the inverse equivalence is equal to its adjoint La,-. I already know that E~ o L a - is the identity functor. Now say that Ea is an equivalence of categories is equivalent to say that the co-unit qa is an isomorphism. This means that for any functor F, the morphism rTa,F is an isomorphism. In particular, if X is a G-set, and if F is the functor Fax associated by theorem 3.3.5 to the A-module Ax , then the morphism

is an isomorphism. But A x ( X ) = A ( X 2) and

4.2. EVALUATION AND EQUIVALENCE 83

so the morphism r/a,n x,x is deduced from the composition

A(Xa) | A(f~X) ~ A(X 2)

In particular, this morphism must be surjective, so there exists an integer n, and elements c~i r A(f~X) and/3i r A(Xf~), for 1 < i < n, such that

}~'/3/oa c~ = 1A(X~) (4.1) i = 1

Now the sum of c~i's (resp. the sum of/3i's) is an element a of A((na)X) (resp. a n

element/? of A (X(na))), and equality (4.1) may be written

/5 ona cz = 1A(X2 )

Thus X is a direct summand of nPt in CA. Conversely, if hypothesis 2 of the proposition holds, and if X is a G-set, then

there exists an integer n and elements c, r A((nf~)X), and /5 E A(X(nft)) such

that ~ ona a = 1A(X~). So there exists elements a/ E A(•X) and /~i E A(XF~), for 1 < i < n, such that equality (4.1) holds. Then if F is a functor on CA, let �9 = r/n,F,X. I denote by O the morphism from F(X) to A(Xf~) | F(X) defined by

n

O(u) = ~ / 3 / | F(cc;)(u)

With those notations, I have

Oe)(a | v) = /3/| F(ai)F(a)(v) = ~/3 / | F(c~i o x a)(v) i = 1 i = 1

B u t v e f ( a ) , and the action of element A �9 A(ft 2) on F(ft) is defined by ),.v = F()~)(v). As a / o x a is in A(a2), I have also

r~

Oe#(a @ v) = E fl' @ (c~i ox a).v . . . . i = 1

. . . . /3ioa(oqoxa) | /3ioact/)oxa | i = 1

and this proves that @~ is the identity. Conversely

) r = F ( g d F ( ~ d ( ~ ) = r ( 9 / o ~ ~/)(~) = r ~, o~ ~/ (~) = / = 1 i = 1

Thus O and ~ are mutual inverse isomorphisms. As the G-set X was arbitrary, this proves that r/a is an isomorphism, and Ea is an equivalence of categories. This completes the proof of the proposition.

84 C H A P T E R 4. THE A L G E B R A ASSOCIATED TO A G R E E N F U N C T O R

4.3 The algebra A(f 2) Condition 2 of proposit ion 4.2.1 holds in part icular if I take for ft the disjoint union of G / H , for all subgroups H of G: indeed, if X is a G-set, let E a system of represen tat ives of orbits G \ X , viewed as a G-set with trivial C-action. Let i be the map from X to f~ x E defined as follows: if x E X, then there exists a unique xo in G.x N E, and an element g E G such that z = 9.zo. Then denoting by G~0 the stabilizer of x0, the class gG~.o E f t does not depend on the choice of g, and I set i (z) = (gG~:o, xo). The map i is injective: if i(x) = i(x'), then x0 = x; and gGxo = g'G,:' o, so {x} = gGxo.Xo = g'G~,o.x' o = {x'}. Moreover, it is a morphism of G-sets: it is clear that i(h.x) = (hgO~o, zo) = h.i(x). As G acts t r ivial ly on E, I see that ft • E _~ IElfl, and then any G-set is a subset of a disjoint union of copies of fL Hence, in CA, any G-set X is a direct summand of a mult iple of ft. Those remarks show that the evaluation functor Ea is an equivalence of categories from Fune*R(CA) to A(f l2) -Mod. If M is an A-module, theorem 3.3.5 gives M(f t ) an A(fl2)-module structure, defined for a E A(f~ 2) and m r M(f l ) by

\ Ca) 1 J \O.) la ,12~ 2 J

Finally, I have proved the first assertion of the following theorem:

T h e o r e m 4.3.1: L e t A b e a G r e e n f u n c t o r for t h e g r o u p G. L e t

ft = II G/H H C G

T h e n t h e f u n c t o r w h i c h m a p s t h e A - m o d u l e M to t h e A ( f t 2 ) - m o d u l e M(f t ) is an e q u i v a l e n c e of c a t e g o r i e s f r o m A - M o d to A( f t2 ) -Mod . T h e i n v e r s e e q u i v a l e n c e m a p s t h e A ( f l 2 ) - m o d u l e V to t h e A - m o d u l e Lv d e f i n e d b y

L v ( X ) = A ( X f t ) | V

I f f : X ---* Y is a m o r p h i s m of G-se t s , if a C A ( X f t ) , b ff A(Yf~), t h e n

Lv,.(f)(a@v)=(f. oxa)@v=A, f(x)w (a)|

| (S" ox b / | v : A"

F i n a l l y , i f Z is a G - s e t , a n d if c r A(Z) , t h e n

P r o o f : To complete the proof, is suffices to keep track of identifications of theorem 3.3.5, which prove that

Lv, . ( f ) (a • v) = ( f , ox a) | v L~( f ) (b • v) = (f* ox b) | v

4.4. P R E S E N T A T I O N B Y G E N E R A T O R S A ND R E L A T I O N S 85

But I have seen that for any A-module M and any m E M ( X ) I have

f . ox m = M.( f ) (m)

It suffices now to take M = Aa to conclude that

f . ox a = Aa, . ( f ) (a) = A. f ( z ) w

and a similar argument proves the corresponding equality for f*. Finally, theorem 3.3.5 shows that

c x ( a | = A. (c) o x a | z x x

and by lemma 3.3.3 A * ( Z X ) A * ( ~ )

4.4 Presentation by generators and relations

Theorem 4.3.1 shows that the category A - M o d of modules over the Green functor A is equivalent to the category of modules over the algebra A(f~2). The following

proposition, where et,- = A* (g[~) (e) denotes the unit of A ( K ) = A ( G / K ) , gives the structure of this algebra:

P r o p o s i t i o n 4.4.1: Le t A be a G r e e n f u n c t o r for t h e g r o u p G, and let #(A) t h e a l g e b r a over R def ined by t he fol lowing g e n e r a t o r s and re la t ions :

�9 T h e g e n e r a t o r s of #(A) are:

T h e e l e m e n t s t H a nd H for K C H C G . - r K ~ _ _

- The elements Cx,H for x E G and H C_ G.

- The elements AK,~ for K C_ G and a E A(K).

�9 The relations of #(A) are:

- The relations of the Mackey algebra for r H, t H, and Cx,H, i.e.

H K = t H K H = r H V L C K C H (1) tKtL L, rL rK -- --

Cy,~ H Cx,H = Cyx,H V x , y , H (2)

= = c ,H v h c (3 )

Cx,HtH = "~H txKCx,K, Cx,K FH : r~xHcx,H VX, ts H (4)

E tg = E rH = 1 (5) H H

H H : .K r L V K C H ~ L (6) rK tL E gKA~L cx,K~nL K xnL - - - -

x E K \ H / L

t h e o t h e r p r o d u c t s o f rKH, tK and Cg,H b e i n g zero .

86 C H A P T E R 4 THE A L G E B R A ASSOCIATED TO A G R E E N F U N C T O R

- T h e addi t iona l following relat ions:

AI~,~ + AK,~, = Aa',~+~,, AK,~AK,~' = AX,',~, Va, a' E A (K) , VK C_ G (7)

rHAH,. = AZ,-,~f,.(~)r H Va E A(H), V i i C_ H C_ G (9)

/~H,a]H = H t.-t~ i,,.,{( "~) VaE A(H) , VI i C_H C_G (10)

H H tK/~K,ar K : AH,t~(a ) Va E A (K) , V K C_ H C_ G (11)

A~H,~,I~(~)C~,H = Cx,HAH,~ Vz E G, Va E A(H) , VH C_ G (t2)

T h e n the a lgebra A(f~ 2) is i somorphic to fz(A).

Proof." First it is clear that any non-zero product of generators of #(A) involving U W only the generators r H, t H, and C~,H can be put in the form tv%,vrvg , for suitable

subgroups U, V and W of G and an element g E G: relations (4) and (6) allow to H put the generators t H on the left and the generators r~,- on the right, and relation (1)

allows to reduce the products of tH's and the products of rH's. On the other hand, H , H' the product r K / ~ H ,a is zero if # H: indeed r H = rKrHH H by relation (1) and

g t AH,,a = lH, ,et f jAU,,a = t u , AH,,a

by relations (7) and (8), and H§ H'. rH~ H, = 0 if H ~ Then

H H H~Hr~ rKAH, ,a : r K r H ~ H , A H , a = 0

Now relation (9) allows to put the H, r K s on the right of the AH,='s. Similarly, the product C,,HAU,,~ is zero if H ' # H, since Cx,H = C~,HCl,H = C~,Ht H by relations (2) and (3). Now relation (12) allows to put the Cx,H'S on the right of the AHffS. Hence any non-zero product of generators of #(A) obtained by multiplying a product of generators r H, t~, and C~,H on the right by a generator AH,= can be put in the following form

for suitable subgroups U, V, W and an element g of G, and an element a of A(V) . Now I claim that those elements generate #(A) as an R-module: to see this,

it suffices to check that the R-submodule M of #(A) they generate contains the generators of #(A), and is closed 2or the product. But

t H H K

L

7,H K H L H K H t L) K - - = tKAR' ,~KCI,KrK

L

~XH C r H ~H H

L N

H H H H /~U,a --~ /~H,vH'~H,aAH,eH = tHIgH,at u : tHAH, aCl,HrH

4.4. PRESENTATION B Y GENERATORS AND RELATIONS 87

So it remains to check that M closed for the product. First I observe that setting

V

relations (2) and (3) show that g ~ % is a morphism from G to the group of invertible elements of #(A), i.e. that #(A) is an interior a-algebra (in Puig's sense [11]). With this notation, I have the following relations:

cg~H gH H gH = tgh 'Cg c g r K = r g K c g CgAH,a = "kgH,gaCg c h t H = tH = r H H = r H c h if h E H

and I can rewrite fi in the form

/3 U W ~U ~ U N g W x U g n W W = ~ v ~ V , aCgFv9 = I.UNgWb V AV, a E g r v 9 F U g A W = . . .

~U . u n g w ~ U n g W W U W . . . : ~UNgWI~V AV, aT" V C97"UgNW ~ t U n g W / ~ U n g W , tUvngW(a)CgrugnW

so any element of the form fl is one of the following

: t U n g W / ~ U n g W , b C g r u g n W ~3U, W,g,b U W

where U and W are subgroups of G and g is an element of G, and b an element of A(U N gW). Now let P be the product

P : f iS ,T ,x ,a~U,V,y ,b

for subgroups S, T, U, V and elements x and y of G, and elements a E A(S N ~T) and b E A(U N VT).

This product can also be written as

It is zero if U r T, and if U = T, then by relation (6)

r T t U . S ~ N T p T o Y V S ~ n T U n v V E : g S . n T n z y v e z S x * n T N ~ V

z E S * A T \ T / T n Y V

Relation (4) allows to exchange c~ with ,s*nr ~s*nTn~V so P can be written

P = ~ tSSn~TISn*T,at~n:TTn**~gCzzrTn[VTn~vATn~V, bcyrV~nv z E S ~ n T \ T / T N ~ V

Now relation (10) allows again to exchange ASn~T,~ and ~Sn~Tn*,~V'*Sn~T Relations (9) and (4) allow to exchange rTn~Vs~*nTn~V and Arn~V,b%. The use of relations (1) gives

P = ~ t S x xzy ,~ x xzy SnXT . C x z / ~ x z n T N Y V e T n y V hCvr.qxzYnTYnvV S n T n V S n TN V, rsnXTnxzy V . . . . . , -$ZznTnY V . . . . . . . . . z E S x N T \ T / T c W V

Now relation (12) allows to exchange c~ and A ~ n ~ v r~v -, and it follows that _ .,~11 ,rSXZNTAYVO

p ~ s y = t S n x T n ~ Y V ) ~ S N X T ~ • a / ~ S n X T n X Z Y V r T n y V .. (Xzb~CxzCyrsx z~NTYNV SnXToXzyv ~ SnXTnxzyv i

z E S * N T \ T / T N Y V


Finally, relations (7) and (2) allow to group together the terms in A and those in c, and then

p ~ s v = tSn~Tn~z~V]Sn~Tn~YV' rS2:Tn~Yva ' rT2~V . . . . . x ' b C x z y r s ~ n T Y n V

z E S X n T \ T / T n y V

So P is a sum of terms of the form/3: more precisely, relation (11) gives finally

0 i f U T ~ T / 3 S ' T ' x ' a / 3 U ' V ' y ' b = ~ / 3 o V " s n x z Y v , SClXT TOY . . . . . otherwise

z E S X A T \ T / T n y V o, , x z y , ~ s n X T n x z y v t r s n x r n x z y v a . r s n x T n X Z y V o)

( M ) It follows that the elements/3s,T ..... for S and T subgroups of G, for x E G and a E A ( S N ~T), generate #(A) as R-module. This set of generators is actually redundant: if u E U and w E W, then ~b E A(U n ~gW) = A(U n ~g~W), and

Since

/3U, W, ugw,~b U W = t U n ~ g W ~ U n ~ g W , ~ b C ~ g w r u ~ n W

W W C u g w r u g w n W = C u g C w F U g W N W = . . ,

W W W �9 . . . . : . C u 9 r U g N W C w = C u g F U g A W = C u C g r U g A W

I have also

/3U,W, ugw,Ub U W = tUn~gw/~Un~gW,~bCuCgrugnW = . . .

. . . tUn~,wCu/~UngW, bCgT.Wgnw U W = : C u t u n g w A U n g W , b c g r u g n W

and finally /3v, w , ~ , o b = /3v, w,~,b ( X )

This equality shows that elements /3U, W,g,b, as U and W run through the set of subgroups of G, as g runs through a set of representatives of double classes U \ G / W , and b runs through A(U N gW), generate #(A). Actually

P r o p o s i t i o n 4.4.2: T h e a l ge b ra #(A) is i s o m o r p h i c to t he a lgeb ra u(A) de- f ined by t he fo l lowing g e n e r a t o r s and re la t ions :

�9 T h e g e n e r a t o r s of u(A) are t he e l e m e n t s /3S,T .... w h e r e S and T are s u b g r o u p s of G, w h e r e x E G and a E A(S n XT).

�9 T h e r e l a t ions of u(A) are the fol lowing:

/3u, w,~g~,~b =/3U, W,g,b for u E U, w E W (N)

flS,r,x,~/3V, V,y,b = ~Vr Z /3~V . . . . . ~ . . . . . ,.so~T o .~ . . . . . ~, , ~, , w ~ ~ , . S n X T n X z y v w S n X T n X z y v - . - S n X T n X Z y V v~ zESXnT\T/TAYV

( M ) w h e r e @,T = O if U # T and @,T = l if U = T.

Zfls, T .... + Z'/3S, T,x,a, = /3S, T ..... +z'~' for z, z' E R (L)

4.4. PRESENTATION B Y GENERATORS AND RELATIONS 89

P r o o f i To prove this proposition, it suffices to show that with these relations, and setting

~ = ~/~,~,~.

Cx, H : /~zH,H,z,ex H

/~H,a = /~H,H,I,a

relations (1) to (12) hold in ~(A). But

tH+K K~L : /~H,K,I,etr : E /~H,L,l,rf(sK)-rL<eL) = tilL

Ix" H = T*L Fit" = /~L,K,I,sL/~K,H,I,eK E /~L,H,I,rL(eL).rf(eK) = TH

z = I C K \ K / K

So relation (1) holds in z,(A). Similarly

z=16xH\xH/XH

whence rel~tio~ (2). Relation (a) follows f~om

and from Ch,H = /~H,H,h,eH = /~H,H,I,(eH) h = t H

Relation (4) follows from

c=,Ht~ : 3~I+,I~,=,~<=.>3H,+<,1,~,~ :

~ H t : K C x , K = /~zH,zK,I ,e:K/~:K,K . . . . . K :

whereas

Cx,K r H = fl .K,K,x,e(=Io/~K,H,I,eK ~-

Similarly

T:Hcx ,H : /%K,~H,I,exK~:~H,H . . . . . H =

whereas

E ~zH,K,x,Xe K z = I E H \ H / K

E /~xH,K,x,xeK z=IEXK\XK/xK

E /~:~K,H,x,Xel,: z = I E K \ K / K

E flXK,H,x,ex K z=IEXK\~H/~H

To check relation (5), I set e = EL tL. Then e = EL r) and moreover

~s,T,=,o~ = E 5S,~,=,ogL,~,I,~L . . . . L

... = = :~T x

z=Ie(Sn=T)\T/T

Similarly

G/~S, T4',a -~" E ~L,L,I,eL/~S,T . . . . "= ~S,S,I,es/~S,T . . . . = L z=Z ES\S/(Sa=T)

9O

so e is the unit of u(A). Finally for the Mackey axiom

H H rKtL = PK,H,I,eKPH,L,I,~L =

Moreover

CHAPTER 4. THE ALGEBRA ASSOCIATED TO A GREEN FUNCTOR

E xEK\H/L

/• - t< x L x

E /~B,',H,I,rH(a) = /~K,H,I,rH(a)

E flK,H,I,rH(a) = flK,H,I,rH(a) z=lEK\K/K

E /~H,K,I,rH(a) = ~H,K,I,rH(a) z=IEH\H/K

E ]~H,K,l,r~:(a) = /~H,K,I,rH(a) z=IEK\K/K

E /~H,K,I ,a = ~ H , K , I , a z=IEK\K/K

E ~H,H,I,tH(a.eK) = ~H,tH(a) z=IEK\K/K

/ ~K , rH(a ) r H = ~K,K,I,rH(a)hK,H,I,el~. =

proving (9). Similarly

~l=I,=t~ = 3H,H,I,al~H,K,I,eI,: =

whereas

tH g , r H ( a ) = /~H,K,I,eK~K,K,I,eK =

which proves (10). For relation (11), I write

tH.~K,a = ]~H,K,I,eK/~K,K,I,a =

so that

H H tK~K,arK = ~H,K,I,ahK,H,I,eK =

and relations (L) prove (8). Moreover

FH ~H,a = ~Ix',H,l,et,:/~H,H,l,a = z=IEK\H/H

whereas

L tKn~LCx,K~nLrK~nL = ~K,KAzL,I,eKnxL~KAXL,L,x,XvKznL = . . .

. . . . E ~h',L,x,~KnXL z=IE(h" n~L)\(Kn~L)/(Kn~L)

The other products of t H, r H, and Cx,g vanish, because

flS, T,x,~flU, V,y,b = 0 if U # T

Relation ~/~-,. + ~K,a' = .~K,~+r is a trivial consequence of relations (L). Moreover

"~K,a/~K,a' = ZK,K,I,a[~h',K,I,a' = E ~K,K,1 .... ' = .~K,a.a' z=IEK\K/K

whence relation (7). Now equality

4,4,

Finally, for relation (12)

~zH,xaCx,H : ~:~H,:~H,I,Za/%H,H . . . . . H -'~

whereas

C:r,H IH,a = /~H,H . . . . . tr =-

T h i s proves (12), and proposition 4.4.2.

P R E S E N T A T I O N B Y G E N E R A T O R S AND RELATIONS

z=l EXH\~H/~H

E /~:~H,H,x,3:a = 3~H, H,a,za z=IEH\H/H

91

(9 A(U ~ ~w) --~ ,(A) U, WC_G

geu\a/w

Moreover, if ft = I_[H_Ca G/H, then

Thus

a II G/(Hn K) H,KC_G

A(fl 2) ~- @ A(H N ~K) (I) H,KCG

xEH\G/K

and the previous morphism induces a surjective R-module homomorphism from A(f~ 2) to #(A).

In order to build a morphism from #(A) to A(Ft~), and to prove that it is a morphism of algebras, I must state precisely the above isomorphism (I). If H and K are subgroups of G, and x E H \ G / K , I denote by iH,x,K the injection from G / ( H N ~ K ) into f12 defined by

and I denote by CH,~,K = A*(iH,,:,K) the associated morphism from A ( H M ~K) to A(f~:). If S and T are subgroups of G, if x E G and if a C A (S M ~T), I set

~/S,T .... = Cs,x,T(a) = A.(is,~,T)(a)

It is an element of A(Ct2). Let Q be the product

Q = "[S,T,x,a O9 "/U,V,y,b

By definition of the product oa, I have

k. a)iCd 3 / \O)iCd2a.)2~,d3/

Moreover "/S,T .. . . X "/U.V,y,b = A. (is,~:,T X @,y,V) (a • b)

To complete the proof of proposition 4.4.1, I observe that relations (N) and relation (7) and (8) show that the map b E A(U N gW) H /3u, w,g,b e #(A) is R-linear, and induces a surjective morphism of R-modules

92 CHAPTER 4. THE ALGEBRA ASSOCIATED TO A GREEN FUNCTOR

~1 ~a2ca3 Let Z be the pullback of 93 and G/(S N~T) x G/(U x YV) over the maps ( . . . . . . . . ) and is,z,T • iU,y,V: the set Z is the set of couples

(g(S n ~T),h(V n ~V)) ~ V / ( S n ~T) • G/(U • ~V)

such that gxT = hU. Denoting by i the injection from Z into G / ( S N ~ T ) x G / ( U x Y V ) , and j the map from Z to f~3 defined by

I have a cartesian square

Z

~3

i ) r n • C/(U •

1 ~S,x,T • ZU,y,V

fP

so that A(Z) ~_ ( ~ A(S N XT n =~YV)

zE(S#nT)\T/(TnyV)

Now the product G/(S G ~T) x G/(U x YV) identifies with

]_I G/(S n ~T n ~YV) zE(S~AT)\G/(TnYV)

and then

A ( G / ( S N ~T) x G/(U x YV)) = @zc(S~nT)\a/(Tn,v)A(S fl ~T N ~*~V)

}

k Oi l 022 a,)2 CO 3 /

and the product Q can also be written

(2 = A. (wlw2w3~ A.(j)A*(i)(a • b) \ 021023 /

But equality gzT = hU implies T = U. Hence if T # U, the set Z is empty, and Q is zero. And if T = U, then the map

(g(S n ~T),h(U n W ) ) 6 Z H z - ' g-~h C (S ~ n T ) \ T / ( T n ~V)

is well defined, and induces a bijection

G\Z ~- (S ~ n T ) \ T / ( T N ~V)

The orbit corresponding to z E (S x n T ) \ T / ( T N YV) is the orbit of the couple

((s n xz(r n

the stabilizer of which in G is the group S n ~T N xzYV. So the set Z identifies with

Z ~_ I I G/ (S n ~T n ~V) zE(S~AT)\T/(TnYV)

4.4. P R E S E N T A T I O N B Y G E N E R A T O R S A N D R E L A T I O N S 93

By construction of the product x, I know that

a X b @ .SnXT i ~ ~Toxz~V / x z l \ = 7 Sn~Tn~zyvka).7 Sn~Tnx~uV[ O)

zE(S~nT)\a/(TnYV)

Up to those identifications, the map A.( i ) is the projection on the components corre- r W2 W3 sponding to the elements z of (S x A T ) \ T / ( T fl YV). Now the map ( . . . . ) 3 sends

the element g (S VI ZT fl xzYv) to

(('UI('U2CU3~ j ( g ( S A Z T A x z Y V ) ) : (Cdl~ ( g S , g x r , g x z y V ) : ( g ~ , g x z y V ) 021023 / \ CUIOJ 3 /

This shows that it factors as the projection

followed by is,x~y,v. Finally, I get

zE(SXnT)\T/(TnYV) and

"/S,T,,T,a Ofl ~/U,V,y,b : ~U T E "/~ V ~sn~Zyv CrSnXT ( ~ rXTn~YV (x,b~ ~ ~, ,XZY,~SClXTnXZyV~ SnXTClxzyv~al " SnXTnxzyv ~ l] zE(SxnT)\T/(Tn~V)

(M') The analogy between those formulae and the formulae (M) giving the product in #(A) shows that the map

"/S,T . . . . E A(f~ 2) ~ /~S,T . . . . E p(A)

induces ~ surjective algebra homomorphism from A(f~ 2) to #(A). Moreover, it is easy to see that the identification relations

"/S,T,sxt,,b = "[S,T,x,b pour s E S, t E T (N ~)

analogous to relations (N) hold: indeed, if a~ denotes the map

~, : g(S N XT) E G/(S n ~T) -~ gs- '(S N "=T) E G/(S A "=T) = G/(S A "=iT)

then

iS,sxt,TOZs(~l(,%'~XT)) = is,sxt,T(gs-l(Sr '}SXT)) = (g~q- l~ ,g$- lNxtT) . . . .

. . . . (gS, g T) = n T)) So is,~=t,TC~ = iS,~,T, and taking image under A. gives relation (N').

Now I can define a morphism from u(A) to A(f~ 2) by mapping/3S,T .... to "/S,T ..... and this is equivalent to define a morphism from #(A) to A(fl ~) by

t~ ~ t'i, H- = 7H,K,I,~,,-

H TIH FK ~ K ~ "[K,H,I,eK

Cx, H e-~ Clx,H ~ "[xH,H,x,e(~H )

/~H,a ~ /V H,a ~ ~/H,H,I,a

Proposition 4.4.2 shows indeed that since relations (M' ) and (N ' ) hold, and since the ! ;H map a ~ AH,a = A.(iH,~,H)(a) is R-linear, then relations (1) to (12) hold for t/,-, r '~ ,

! c ~ , . , ),~,,~.

Those morphisms between #(A) and A(fF) are clearly mutual inverse, and this completes the proof of proposition 4.4.1. �9


4 . 5 E x a m p l e s

4 . 5 . 1 T h e M a c k e y a l g e b r a

In the case when A is the Burnside functor b. then the algebra b(f~ 2) is isomorphic to the Mackey algebra #(G): the relations of Mackey algebra hold in b(f~2), so there is a natural morphism ~5 from #(G) to b(f~2), mapping the generators of the Mackey algebra to the generators of b(f~ 2) having the same name. Conversely, there is a morphism O from b(f~ e) to #(G) defined by

G o(t~) = t?, o(~f) = ~,, Otc~,) = c~ ,

The image of -~u,~ for a E b(H) is defined by linearity from

O(Ai~,u/r~) . H = t K r K

This definition is made in order to satisfy relation (12): indeed, as H / K = t H ( K / K ) in b(H), I have the following equality in b(~ 2)

H H . H - K H H H ~ H , H / K = ~H,tH(el() = t h ' A K , s ~ c r K = ~h '~KFK = t K r K

It is not difficult then to check relations (7) to (12). So it is clear that ~ and O are mutual inverse algebra isomorphisms.

P r o p o s i t i o n 4.5.1: T h e algebra b(fi 2) is i s o m o r p h i c to the Maekey algebra.

Remarks: 1) The identification of b(X) given in propositions 2.4.2 and 2.4.3 is a way to recover the construction of Lindner ([9]) (see also the article of Th~venaz and Webb [15]): if X and Y are G-sets, then an element of b ( X Y ) is the difference of two G-sets over X Y . Now a G-set V over X Y is determined by a morphism f from V t o X and a m o r p h i s m g from V t o Y . I f W i s the G-set over YZassoc ia ted to the morphism h : W + Y and to the morphism k : W + Z, then by definition of composition in Cb, I have

,, xz / \ x y y z /

where V • W is the product V and W, viewed as a G-set over X Y 2 Z by the morphism l (v ,w) = ( f ( v ) , g ( v ) , h ( w ) , k ( w ) ) . Proposition 2.4.2 says how to compute

~Y~ W): I must fill the cartesian square b* (~yyz) (V •

F U , V x W

)

I \ xyyz /

Now U clearly identifies with the pull-back

u = {(v,w) c v • w t 9 ( , ) = h(w)}

4.5. EXAMPLES 95

with moreover r(v, w) = (f(v), g(v), h(w)) and s(v, w) = (v, w) E V x W. With those identifications

/ \

b" ( ) ( v x w)--(u,r) \ x y y z /

and proposition 2.4.2 now gives

\ xz / \ x y y z / \ yz /

In other words, the product V or W is obtained from the diagram

X Y Z

where the top square is cartesian.

2) An easy counting argument on the ranks over R in the isomorphism

#(G) ~- b(fl 2) ~- @ b(H N XK) H,KCG

~:eH\5/K

shows that the elements tilL CxrL~K, for H and K subgroups of G, for x E H \ G / K , and L subgroup of H fl XK up to conjugation by H N XK (i.e. for a well defined element (H N xK) /L of b(H N ~L)), are a basis over R of the Mackey algebra: it is proposition (3.2) of Th~venaz and Webb ([15]).

4.5.2 The Yoshida algebra

The second example of an algebra A(Ft 2) is the case when A is the fixed points functor FPn (see [14]), defined as follows: for any subgroup H of G, the module FPn(H)

H is equal to R. The restriction maps r K are all identity, and the transfers t~ are multiplication by [H : K]. It follows that denoting by [X] the permutation R-module associated to the G-set X, I have for any G-set X

FPR(X) = ItomG([X], R)

If f : X ~ Y is a morphism of G-sets, then If] is a morphism fi'om IX] in [Y], giving by composition the map

FP~(f) : FPn(Y) ~ FPn(X)

The map FPR,.(f) is defined by identifying X to a basis of IX] and Y to a basis of [Y], and setting for r E FPn(X) = Homa([X], R) and y E Y

FPn, , ( f ) (r ~_, r ~e./-~(~)

96 CHAPTER 4. THE ALGEBRA ASSOCIATED TO A GREEN FUNCTOR

The Green functor structure of FPR is obvious: if ~ E FPR(X) -- Home(IX], R) and if ~ E F P a ( Y ) = Homa([Y] ,R) , then r x ,~ is the element of Homa([X • Y],R) defined by

If X and Y are G-sets, the module FPp(YX) identifies with the G-homomorphisms from IX] to [Y] by the map sending r E Homa([YX], R) to the map

yE}"

In other words, the element r is determined by its matrix @(y,.r))~ez.~ex, which

must be G-invariant (i.e. such that r = r for all 9 E G, y E Y and xcX).

If Z is another G-set, and if 15 E FPR(ZY), then the composite 15 OF r is defined by

(?5 t \ z yyx /

Here r x r is the morphism from [ZY2X] to R defined by

so A* ( ~Y~ "} (15 x 6) is the morphism from [ZYX] to R defined bv \ z y y x ]

A" zyyx (15 •162 ~(z,y)r

Finally, the element 15 OF r of FPR(ZX) is defined by

(z,~,,~-)e( 72)- ' (~,~) ~,eY

so the product OF corresponds to the matrix product. It. follows that for any X, the algebra A(X 2) is the algebra of endomorphisms of the RG-module [X]. In particular

P r o p o s i t i o n 4.5.2: If A = FPm then the algebra A(~ 2) is the Heeke algebra

This algebra was defined by Yoshida in [17]. The modules over FPR are the cohomological Mackey functors (see [15] prop. 16.3), i.e. the Mackey functors such that

H H t ( ]m tA-rr~.(m) = [ H :

for any K C_ H and any m E M(H). It is also clear that the functor X ~ IX] is an equivalence of categories from CFpR

on the full subcategory of the category of RG-modules formed by finitely generated permutat ion modules: this functor is indeed fully faithful by definition of CFpp, and

4.5. EXAMPLES 97

essentially surjective, since any finitely generated permutat ion module is the module associated to a (finite) G-set. Theorem 3.3.5 now says that cohomologicM Mackey functors are exactly the additive functors on that category: this is a theorem of Yoshida ([17]).

The functor FPR is also the functor H~ - , R) = Ex t r a (R , R). It is also ~ subfunc tot of the Green functor H e ( - , R) = | - , R) (the standard notation is g * ( - , R), but I prefer not to use it here). A similar argument proves the

Proposition 4.5.3: I f A is t h e f u n e t o r H e ( - , R ) = ~ i H ~ ( - , R ) , t h e n

Chapter 5

Morita equivalence projectivity

and relative

By its very definition, the algebra A(fF) is only defined up to Morita equivalence: it has indeed been chosen so that the category A(f~2)-Mod is equivalent to A-Mod . A natural question is then to know if it is possible to replace f~ by another G-set.

5.1 Morita equivalence of algebras A(X 2) The next proposition gives a way to know if the algebras A(X 2) and A(Y 2) are Morita-equivalent:

P r o p o s i t i o n 5.1.1: Le t A be a G r e e n f u n c t o r for G, and let X and Y be G-sets . T h e n A(YX)= HomcA(X,Y) is an A(Y2)-module-A(X2), and A(XY) is an A(X2)-module-A(Y2). T h e fol lowing cond i t i ons a re equ iva len t :

1. T h e b i m o d u l e s A(YX) and A(XY) i nduce m u t u a l inverse M o r i t a equ iva l ences b e t w e e n A(X 2) and A(Y2).

2. T h e r e exis ts in t ege r s n and m such t h a t X d iv ides nY and Y d iv ides mX in CA.

P r o o f : First it is clear that composition induces bimodule morphisms

~: A(XY) | A(YX) ~ A(X2) : a | ~ a oy/~

and @ : A(YX) | A(XY) ~ A(Y2): f l | a ~ / 3 o x a

Those morpbisms are balanced in the sense of Curtis-Reiner ([4]): as composition is associative, I have

and similarly 0 ( 9 | ~) oy 9' = 9 ox ~ ( a | 9') = ~ ox ~ or ~'

100 CHAPTER 5. MORITA EQUIVALENCE AND RELATIVE PROJECTIVITY

They form a Morita context (see [4] (3..54)). As the image of ~ (resp. O) is a sub- bimodule of the corresponding algebra, i.e. a two sided ideal, the map ~ (resp. O) is surjective if and only if its image contains the unit. Thus �9 and O are surjective if and only if there exists elements c~i E A(XY) and /3i E A(YX) , for 1 < i < n such that

~ c ~ oz/3~ = 1A(X 2) i = 1

and elements/~} C A ( Y X ) and c~} E A(XY) for 1 _< j _< m such that

,rt~ !

j = l

The first equality means precisely that X divides nY in CA, and the second that Y divides rnX, and the proposition follows. �9

Def in i t ion : Let X and Y be G-sets. I will say that X and Y have the same stabilizers if

VHC_G, 3 x E X , G ~ = H ~ B y E K G y = H

This is equivalent to say that if

X ~_ I I nH(G/H) V ~_ I I mH(G/H) H C G H C G

t h e n n H ~ Oifand only i fmH ~ O. Now i f m i s an integer greater or equal to all quotients mH/nu (for nH ~ 0), I have ranH >_ rail, and Y is isomorphic to a sub-G- set of reX. Now by lemma 3.2.3, the set Y divides m X in CA. Similarly, if n is an integer greater or equal to all nH/mH (for mH :fi 0), then X divides nY in CA. Thus

P r o p o s i t i o n 5.1.2: I f X and Y are G-sets h a v i n g t he s a m e s tab i l izers , t h e n A(X 2) and A(Y 2) are M o r i t a - e q u i v a l e n t .

5.2 Relative projectivity

Let X be a candidate G-set to replace ~. Then in particular, there must exist an integer n such that �9 divides nX in CA. The next proposition describes this situation:

P r o p o s i t i o n 5.2.1: Le t A be a G r e e n f u n c t o r for G, and let X be a G-set . T h e fo l lowing cond i t i ons are equ iva l en t :

1. T h e r e ex is t s an i n t ege r n such t h a t �9 d iv ides nX in CA.

2. T h e set �9 d iv ides X in CA.

3. T h e m a p A, ( : ) is sur jec t ive .

4. T h e i m a g e of t he m a p A, (~) con ta ins C.

Poof : The proof of this proposition uses the next lemma:

5.2. RELATIVE PROJECTIVITY 101

L e m m a 5.2.2: Let A be a Green func to r for G, and let f : X -~ Y be a m o r p h i s m of G-sets. Then A*(f) is a m o r p h i s m of rings (with uni t ) f rom

(A(Y), .) to (A(X), .), and the image of A(X) under A.(f) is a two-s ided ideal ] %

of (A(Y), .).

Proof : The product "." can be recovered from the product x by the formula

a'b= A* ( X ) (a •

for a, b E A(X). Now for a, b E A(Y), I have

xx f(x) (a)• f(x) (b) . . . .

. . . . A*( x ) A * ( xlx2 ~ ( a x b ) = A ' ( x ) (a• . . . . xx \ f(zl) f(x2)] f(x)f(x)

and the first assertion follows, since moreover

R e m a r k : It is actually a reformulation of assertion iii) of proposition 1.8.3.

For the second assertion, if a C A(X) and b E A(Y), then

(x) f(x)z

X ~ YX

y ~ y2

yy

As the square


x ) (b• b.A.(f)(a) = A.(f)d* f(x)x


A*(f)(a)'b= A*(f)A* ( xxf(x)) (a x b)

102 CHAPTER 5. MORITA EQUIVALENCE AND RELATIVE P R O J E C T I V I T Y

so the image of A . ( f ) is a two-sided ideal of A(Y).

P r o o f o f p r o p o s i t i o n 5.2.1: It is clear that 2) implies 1). Now if 1) holds, then there exists elements c~i E A ( X . ) = A(X) and fli E A ( . X ) = A(X) , for 1 < i < n, such that

n

Y~ Bi ox c~i = 1~(.2) i= l

But A( . 2) = A(.) and

(.)A.(:) 1A(o~) = A. (e) = c e Q

Moreover (:)(x) (x) p~ ox c~i = A. (B~ • c~) = A. (Bi • c~) = A. (b~.a,)

\ O X X O / X X

This proves that

e = A. Bi.c~i

so 1) implies 4). By the previous lemma, the image of A. (~) is a two-sided ideal. So 4) implies 3).

~inal ly if 3) holds, let ~ e A ( X ) such that A. ( : ) ( . ) = ~ I can view ~ as an e lement

of A ( X . ) and ex as an element of A( .X ) . Then by the previous computation

and this proves that �9 divides X in CA. �9

Def in i t ion : Let A be a Green functor for the group G. If X is a G-set having the equivMent properties of proposition 5.21, the functor A is said to be projective relative to X .

This definition coincides with the usual one (see for example Webb [16]). In the case X = G/H, the map A. (~) is the transfer t ~ : A(H) --* A(G) = A(.) , and this is the classical notion of relative H-projectivity.

P r o p o s i t i o n 5.2.3: Le t A be a G r e e n f u n c t o r for G, and let X and Y be G-sets .

1. I f A is p r o j e c t i v e re la t ive to X, a nd if HomG-set(X, Y) 7 L O, t h e n A is p r o j e c t i v e r e l a t ive t o Y.

2. T h e f u n c t o r A is p r o j e c t i v e re la t ive to X x Y if and o n l y if it is p r o j e c t i v e r e l a t ive to X and re la t ive to Y.

3. I f A( . ) has a u n i q u e m a x i m a l two-s ided ideal (for i n s t a n c e if it is a local r ing) , t h e n A is p r o j e c t i v e re la t ive to X LI Y if and on ly if it is p r o j e c t i v e r e l a t ive to X or re la t ive to Y. In th i s case, if H is a m i n i m a l s u b g r o u p of G such t h a t A is p r o j e c t i v e r e l a t ive to G/H, t h e n A is p r o j e c t i v e re la t ive to X if and on ly if X H r O. I n p a r t i c u l a r , t h e s u b g r o u p H is u n i q u e up to G - c o n j u g a t i o n .

5.3. CARTESIAN PRODUCT IN CA 103

Proof." 1) Let .f : X --+ Y be a morphism of G-sets. Assertion 1) follows from the equality

2) If A is projective relative to X x Y, as there exists morphisms of G-sets fi'om X • Y to X and Y, assertion 1) implies that A is projective relative to X and relative to Y. Conversely, if A is projective relative to X and relative to Y, let c~ C A(X) and /3 E A(Y) such that

Then

A.

and this proves that A is projective relative to X x Y. 3) Let Z = X I_[ Y, and ix and iy the injections from X and Y into Z. It follows fi'om assertion 1) that if A is projective relative to X or relative to Y, then A is projective

Conversely, as A(Z) = A . ( i x ) (A(X) ) �9 A . ( iy ) (A(Y) ) , and as relative to Z.

I see that if A is projective relative to Z, then

Now A(Z) is the sum of two of its two-sided ideals. If A(e) has a unique maximal two-sided ideal, this forces one of them to be the whole of A(Z), thus A is projective relative to X or relative to Y. Let H be a minimal subgroup such that A is projective relative to G/H: such a subgroup certainly exists, since A is trivially projective relative to G/G = o. Since X g identifies with Homa_s~(G/H, X), it follows from assertion 1) that if X u ~ O, then A is projective relative to X. Conversely, if A is projective relative to X, it follows from assertion 2) that A is projective relative to (G/H) x X. Now the first part of assertion 3) shows that there exists x E X such that A is projective relative to (G/H) • (G/Gx). This product is isomorphic to the disjoint union of sets G/(H n eG,:) = G/(H N Gg,~), for suitable elements g E G. Then there exists g such that A is projective relative to G/(H N Gg,:), and now the minimality of H implies that H C_ Gg~, i.e. gx E X 14. �9

5.3 Cartesian product in CA

5.3.1 Def in i t ion

Let A be a Green functor, and U be a G-set. On the category A-Mod, I have the functor M H Mu (see lemma 3.1.1). As

(Ax)u(Z) = Ax(ZU) = A(ZUX) = Aux(Z)

104 C H A P T E R 5. MORITA EQUIVALENCE AND R E L A T I V E P R O J E C T I V I T Y

it follows that ( A x ) u ~- Aux , and that the above functor preserves the subcategory of A - M o d formed by the modules Ax. This subcategory is isomorphic to CA, so it is possible to define the functor X H U X on CA:

L e m m a 5 . 3 . 1 : T h e c o r r e s p o n d e n c e

X H U X

E HomcA(X, Y) ~ 1A((Uy)2) Ov ct = A. \ u y u x ] \ yx

is a f u n c t o r U x - f r o m CA t o CA.

P r o o f : It suffices to make the isomorphisms of proposition 3.1.3 explicit: an element c~ of A ( Y X ) defines a morphism from Ay to Ax , the evaluation of which at a G-set Z is given by

/3 E Ay(Z) = A(ZY) ~ /3 oy c~ �9 A(ZX) = Ax(Z)

This morphism from Ay to A x defines in turn a morphism from Auy = (Ay)u to A u x = (Ax )u , evaluating at Z as

/3 C A u y ( Z ) = A ( Z U Y ) ~ /3 oy c~ E A ( Z U X ) = A u x ( Z )

Now this morphism is determined by the image of the element 1A((Uy)2) E A u y ( U Y ) , i.e. by the dement

1A((UyF ) Oy Ct of A ( U Y U X ) . Now it suffices to write

1A((Uy)2)OyO~:A,(ttlYl"tt2Y2XlA.( ?.l, lYltl2y2x / [A, ( uy ) A.(7)(g) xoL 1 \ UlYlU2X ] \u]ylu2y2y2x] uyuy

The product inside hooks is equal to

As the square

d . ( uYly2x Id*(tlYlY2Xl(o~) \ uy luy ly2x] \ y2x /

uyx I u y y x /

U Y X - - + UY2 X ( ? I ' yX l l I ( ~ Y I Y 2 X l

\ uyl uylY2X ] \ u y u y x ] ) (Uy)2x ( UlYltL2Y2X ) UYUy2X

\'~1 ylU2y2u2y2X / is cartesian, I have

\ulylu2y2y2x \uyluyay2z \ u y u y x ) \ u y y x ]

5.3. C A R T E S I A N P R O D U C T IN CA 105

and then

\ u ly lu2x / \ u y u y x / \ u y y x / \ y2x /

. . . . A . \ uyux / \ yx /

proving the lemma. �9

In other words, this lemma means that if X, Y and Z are G-sets, if a E A ( Y X ) and fl E A( Z Y ) , then

This equality can he "proved" by the following argument: let

As the product o is associative, I also have

Observing now that 1A((Uy)2 ) is a unit for the product ovg, it follows that

P = [1,4((UZ)2 ) o z ,8] Oy o~

and using again associativity of the product o, l get

P = 1A((UZ) 2) ~ ( 8 0 z c~)

The only trouble with that argument is that I never proved that the product o is associative in those conditions. It is the reason for the following lemma

L e m m a 5.3.2: Let X, Y, Z, T and U be G-sets. Le t a E A ( X Y ) , b r A ( Y U Z ) , and c E A ( Z T ) . T h e n

(a oy b) oz c = a oy (boz c) �9 A(XUT)

P r o o f : By definition

So

\ xuz / \ x y y u z /

k xu t ] \ x u z z t / \ xuz / \ x y y u z /


\ X'aZlZ2t ,] \ x y y u z x z e t ]


As the square xyuzt

xyuzzt] X Y U Z T , XYUZ2T

1 xuzt / \ xuzlz2t / ) XUZT ( zuzt ~ XUZ2T

\xuzzt] is cartesian, it follows that

\ xuzz t ] \ xuzlz~t ] \ xuzt ] \ xyuzz t ]

and then

(oo~b)o~c: A. (x~z~ ~. ( x ~ A" ( ~ ' ~ ~ A* ( ~ ~ . . . .

\ xut / \ xuzt / \xyuzz t] \xyyuzlz2t]

. . . . ~. ( x s ~ A" ( x ~ ~ (a ~ ~ • c) \ xut / \xyyuzzt]

On the other hand

\ xut / \xyyut] \ yut ] \yuzz t]


\ xyly2ut ] \xyly2uzzt]

As the square xyuzt

xyyuzt] X Y U Z T ~ XY2UZT

(~z~ (~lS~ )

\ zyyut]


\ xyyut ] \ xyly2ut ] \ xyut ] \ xyyuzt ]

and then

aoy(bozc ) = A. (xyut~ A. (xyuzt~ A* ( xyuzt ~ A* ( xyly2uzt ~ (axbxc) . . . .

. . . . A. (x~z t~ ~. ( xy~z~ ~ (o x b • c) \ xut ] \xyyuzzt]

proving the lemma.


5.3.2 Adjunction

The previous lemma leads to a product on the right - • U:

P r o p o s i t i o n 5.3.3: Let A be a G r e e n func to r for G, and U be a G-set . T h e c o r r e s p o n d e n c e

X ~ X U

a r A ( Y X ) H ce Ox 1A((XV)~) r A ( Y U X U )

is a f u n c t o r - x U f r o m CA to CA, w h i c h is r ight adjo int to t h e f u n c t o r U • - .

Proof i Say that - x U is a functor is equivalent to say that if X, Y, and Z are G-sets, if a E A(YX) and/3 E A(ZY) , then

This follows from lemma 5.a.2 and from the fact that 1A((y~):) is a unit for the product oyu. It is clear moreover that

1A(x=) • U = 1a(x:) Ox 1A((X~7)~) = 1A((XU)2)

SO -- • U is a functor. To prove that this functor is right adjoint to the functor U x - , I observe that

Homc, (X, YU) = A ( Y U X ) = Homc,(UX, Y)

and it suffices to check the functoriali ty of this equality with respect to X and Y. So let Z and T be G-sets, and f C A ( X Z ) and g E A(TY) . I must check that the squares

Hornca(X, YU) = A ( Y U X )

1 HomcA(Z, TU) = A(TUZ)

Id

Id

A(YUX) = nom .(UX, Y)

A(TUZ) = t tornc . (UZ, T)

are commutat ive. That is for h E A(YUX) , check that

(g x U) oyu h ox f = g oy h oux (U x f )

But

(g x U)oyu hox f = (goy 1A((YU)2))Oyu ho X f = gog (1A((YU)2)Oyu hox f ) = goy ( hox f )

whereas

goyhoux (U x f ) = goy houx (1A((UX)=)Ox f ) = (gOy houx la((ux):)) ox f ) = (goy h )ox f

Now equali ty follows from lemma 5.3.2.

R e m a r k : By a similar computat ion as in lemma 5.a.1, it is possible to show that if a C A ( Y X ) , then

a • U = A. \ y u x u l \ yx /

108 CHAPTER 5. MORITA EQUIVALENCE AND RELATIVE P R O J E C T I V I T Y

N o t a t i o n : / f X is a G-set, I denote by atr,x the natural bijection from U X to X U switching the components, and by Ou, x = (cru,x ). the associated morphism from UX to X U in CA. It follows from lemma 3.2.3 that

Ou,x oux Ox,u = (au,x ). oux (~x,u ). = (au, x o ~rx,u). = Id. = 1A((UXF)

so Ou,x is an isomorphism.

P r o p o s i t i o n 5.3.4: I f X and Y are G-sets , and i f a C A ( Y X ) , t h e n t h e s q u a r e

OU,X UX ~ X U

Uxc~} l a x U

UY ~ YU OU,y

is c o m m u t a t i v e in CA, and t h e m o r p h i s m s Ou, x d e f i n e an i s o m o r p h i s m f r o m t h e f u n c t o r U x - t o t h e f u n c t o r - x U.

P r o o f : By lemma 5.3.2, I have

OU, Y Ouy (U x oL) = Ou, Y ouy (IA((Uy)2) Oy 0~) = OU, y Oy a

Whereas (~ x U) oxu Ou,x = (c~ ox 1A((xu)2)) oxu Ou,x -- c~ ox Ou,x

This product is equal to

. . . . A.(yXlUlU2X21A* ( yXlttltt2X2 ~ [a• A*(UX)(c) ] \ yulu2x2 / \yXlXlUlU2X2]

The product inside hooks is

A , , ( yxittx2 ~ d * ( y x l u x 2 1 ( o ~ ) , ,uxlx2uux2/ \ yxl ]

As the square

yxu I yxux]

Y X U , Y X U X

l 1 I

\ y x l x lu lu2x2 /


A* ( yxlulu2x2 ~ A, ( yxlux2 ~ = A. ( yxu I A, ( yxu l \ yx lx lu lu2x2] ,,UXlX2UUX~/ \ y x u u x / \ y x u x ]

5.3. C A R T E S I A N P R O D U C T IN CA 109

and then

CtOxOu.x:A.(yXltll?A2X21A. ( yX,t IA.(YXt~lA.(yxl?AX21(ol) \ yUlU2X2 / \ y x u u x / \ y x u x ] \ yxl /

Finally

a ox @,x = A. \ y u u x / \ yx /

On the other hand

. . . . A,(YltAltA2y2zlA* ( ~]lttltt2~22? IA. ( t/Yly2x )A.(ttylY2Xl(ol ) \ y lulu2x / \y lu lu~y~y2x] \ y l u u y l y 2 x \ y~x /

As the square yxtt ]

uyyx / > Y X U U Y 2 X

yuuyx ] \ yl uuyl y2x ] >

YU2YX ( ~1~1 tt2~]2,T / Y U 2 y 2 X


A.( y, ,u2 2x/A.( iA (yx I \ y lu lu2y2y2x] \ y l u u y l y 2 x ] \ y u u y x ] \ u y y x ]

sO

' \ ylulU2x / \ y u u y x / \ u y y x / \ y2x ]

o r

Ou, r o v a = A. \ yuux / \ yx /

and this proves the proposition.

5 . 3 . 3 C a r t e s i a n p r o d u c t i n CA • CA

In view of the previous paragraphs, one may try to define a cartesian product functor on CA • CA, mapping two G-sets X and Y to X • Y, and morphisms f C A ( X ' X ) and g C A(Y'Y) to the morphism (denoted by ,f • g to avoid confusion) defined by

f | = ( f • Y') oxv , ( X • g) E A ( X ' Y ' X Y )

However, this definition is generally not functorial in X and Y:

P r o p o s i t i o n 5.3.5: Let A be a Green functor for the group G. T h e fo l lowing c o n d i t i o n s are equivalent:


1. T h e a b o v e de f in i t i ons turn t h e c a r t e s i a n p r o d u c t into a f u n c t o r f r o m CA X C A to CA.

2. For any G-se t s X and Y, and any c~ E A(X) and /3 E A(Y),

/Txc~=A'( '~ :Y) ( ~

R e m a r k : I will see later that this is equivalent to say that A is commutative (i.e. that for any G-set X, the ring (A(X), .) is commutative).

Proof : Say that the cartesian product is functorial is equivalent to say that if f E A(X 'X) , if g E A(Y 'Y) , if f ' E A (X"X ' ) and g' E A(Y"Y ' ) are morphisms in CA, then

( f ' | g') ox,y, ( f | g) = ( f ' ox, f ) Q (g' oy, g)

The case f ' = 1A(X a ) and g = 1A(y2 ) now gives

(X' • g') ox,y, ( f x Y) = f @ g' = ( f • Y ' ) ~ (X • g')

In other words, for any X, Y, X', Y', and any f E A(X 'X) , g C A(Y 'Y) , I must have

(X' • g) ~ ( f • Y) = ( f • Y') ~ (X • g) (C)

The left hand side P is equal to

t ~ly'zy~ ) ' . . . . . \ X l g X 2 g l X 2 Y l X y 2


A. ( ) A" • : )

As the square

X ' Y ' X Y ( x'y'xy

x'y'x'yxy] 1 X ' Y ' X ' Y X Y


t, xiVx'~y,x'~y, xy2 ) SO

A,

x'y'xy x'y'yx'xy]

)

)

I X # /X # X \ lY 2yl y~ I t,x;Vz'~y,x'~y,zy2 )

X ' Y ' Y X ' X Y

t, x;y'~lylx'~y~xy2 ) X ' Y ' ( X ' Y ) 2 X y

\ x;Vxy~ ) \z'y'z'y~s) / X t ] Xt X X

t,x'y'~x'~y] \ y'ylZ~X )(g•


\x,y,xy) \~ 'Sx) \y'yx,x) The right hand side of (C) is

\ X;~lX2Y2 ] \ xtylxly'xlyIx2y2) [ ( f • Y') • (X • g)]


' ' " (x'x~y~x~y~y ~ ( f • g) A. r x XlYlX2Y2Y ~ A* IX# IX #X IX I # ! \ Yl ly l 2Y2 2YJ \ x xly2y ]

As the square

X ' X Y ' Y

x'y'xy'xy ] [ X ' Y X y t X y


x'xy'y x'xy'xy'yJ

) ( X'ylXlytx2y2 \ X'ylxly'xIy'x2y2 ]

X ' X Y ' X Y ' Y

t x'yix~ylx~Ix:y ) X ' Y ( X Y ' ) : X Y

A, ( x,,~,x,~,x2y2 ~ A, ( x,x,~x2~:~ ~ = A, ( x,xy,y ~ A, ( x,xy,y )

and then

IXIX # x # \ Q : A, (x#ylXly"x2Y2~ A, ( xtxy#y ~ A* xtxyly ~ A* ( lYl 2Y2Y~ ( f x g ) . . . . k :~'y~:~y~ ) \,~'y'~y'xy) x'x~'~:y'y) k :~'~,y'~y )

. . . . ~. (~,~,~) ~. ( ~ , . , ~ (~ ~ ~) : ~. (x,~,.~ \x'y'xy \x'xy'yJ \x'y'xyJ (f • g)

Finally, equation (C) can be written

\y 'yx'x] \x 'y 'xy] ( f • g)

rc#!4txy Taking its image under A. (y,y~,z), which is the inverse isomorphism of A* t~,yx,~](~'~'~'~, I get

~ i : ~.- -tx,.~,~ \y 'yx'x] ( f x g)

and the case X ~ = Y' = �9 is assertion 2). Conversely, assertion 2) implies the previous equality, so it implies equality (C).

Then

(f ' | | = (f' • Y")ox,r, ,(X' x g')ox,y,(f x Y ' )oxr , (X x g) . . . .

. . . . (f ' • Y")ox,r, ,( f x Y " ) o x y . ( X x g')oxy,(X x g) = (f ' ox, f)|

since X x - and - • Y" are functors. Thus 2) implies 1), and this proves the proposition. �9

112 CHAPTEt~ 5. MORITA EQUIVALENCE AND RELATIVE P R O J E C T I V I T Y

5.4 Morita equivalence and relative projectivity

The existence of a cartesian product functor in CA has the following consequence:

Propos i t ion 5.4.1: Let A be a Green functor for the g r o u p G, a n d Y b e G-se t . Let

: II G /H H C G

The following condit ions are equivalent:

1. The evaluation functor M ~ 34(f~y) is an equivalence of categories from A-Mod to A(ft~ , ) -Mod.

2. The functor A is projective relative to }'.

Proof: If 2) holds, I know that �9 divides Y in CA. Taking images by the functor f~ • - , I see that f~ divides f~ • Y in CA. I also know that any G-set divides a mult iple of f~ in CA. So any G-set divides a mult iple of fl • Y in CA..

But if (w, ~1) C ft • Y, then its stabilizer G'(~,y) in G is contained in Gy, so yGt~,y) r 0, and then G(~,y) is the stabilizer in G of the element G(~,y I of f~r-. Conversely, if H is a subgroup of G such that yH r ~, and if ~2 E yH, then H is the stabilizer of the element (H, y) of f~ • Y. It follows that [~ • Y and f i r have the same stabilizers, so f~ • Y is a subset of a multiple of f~r, hence it divides a mult iple of f~v in CA.

Now any G-set divides a mult iple of [Iv in CA, and proposition 4.2.1 shows that the evaluation functor M ~ M(f~r) is an equivalence of categories from A - M o d to A(f~}) -Mod.

Conversely, if this functor is an equivalence of categories, then any G-set divides a mult iple of f~y in Ca. As Ftv and f~ • Y have the same stabilizers, any G-set divides a mult iple of f~ • Y. In part icular �9 divides such a multiple, and A is projective relative to Ft • Y.

But as �9 ~_ G/G is a snbse~ of ft, it divides ft in CA, so A is projective relative to fL Now proposit ion 5.2.3.2) shows that A is projective relative to ft • Y if and only if it is projective relative to Y, and this completes the proof of the proposition. ..

Remark: There is a natural inclusion i from f~v into ft, so an inclusion f = A.( i x i) from A ( ~ ) to A(fF) . If x E G, then x defines an endomorphism as of the G-set ft by

a~(g.H) = gHx --1 = gx-l .XH

and this endomorphism stabilizes f t r (as a set). Now the map x ~ (ax). turn the algebras A(f~ 2) and A(f t}) into interior G-algebras, and f is compatible with those structures. The following lemma shows that f is actually an embedding of interior algebras (see Puig [11]):

L e m m a 5,4.2: Let i be an injective morphism of G-sets from X to Y. Then A.(i • i) is an embedding from A(X 2) into A(Y2).

5.4. M O R I T A EQUIVALENCE AND R E L A T I V E P R O J E C T I V I T Y 113

Proof: If i is an inject ive morph i sm from X to Y, then i . is a split m o n o m o r p h i s m from X to Y in CA, since i s o i . = 1A(X2). Then it is clear tha t the map

f E A ( X 2) = Endc~(X) H i . o f o i* E A(Y 2) = EndcA(Y)

is an embedd ing of algebras. So to prove the lemma, it suffices to prove that for any f E A ( X 2)

i . ox f o x i* = A.( i x i ) ( f )

(~) A" (:) But i . = A. i(~)x (c), so

\ Y 3:2 J \yZlXlX2J /(X)X (C) x f . . . .

As the square

\ yx7 ] \yXiXlXT/ \ i (x , )x ixTx3) , :rTz3 i


XlX2 XlXl21:2]

X'2 > X 3

( XlX2 ~ ( X l X 2 X 3 ~

)

\YXlXlX2J

A.( ,x,x2 x,x x3 ( x,x A.(x,x2 ) , ,yx,~,x~l k i ( x , ) z , x ~ x : U = A" ki(x,)~,~U , x , x , z ~

whence

\ yx~ / t i(x,)x,x2) ,,.~-,:,,,~:~,, ,, x~x~

. . . . A. \ i ( x , ) x 2 ] A* ( f ) = A. ( f )

Then

i,<oxfoxi. Ax (Y lXY21d , ( y l xy2 ) [d . ( XlX2 x x (C)]

The product, inside hooks is

A. \i(ml)z2z3i(x3) \ m1:c2 ,


But the square I XlX2

X 1X2-T2 / X ~ ~ X 3

§

\ylxxy~ /

is cartesian. So

A.(YlXY21A,: XlX2X3 ylxxy2/ \i(xi)x2x3i(x3)/

and finally

i , o x f o x i * = A , ( y i x y 2 ) A, \ Y~Y2 /

= A ,

which proves the lemma.

xix: )A*(X'x:~ i(,h)x2i(x2) kXlX2X2/

XlZ2 ~A*( Xl'Z2 ~A*(2glX2X3~(,f) . . . . i(xi)x2i(x2)] \ x , x :x2 / \ XlX2 /

( /13~2 ~ A* (XlX2~ ( f ) : A,( i• i)(f) = A. \i(x~)i(x2)] \XzX2/

R e m a r k : If A is projective relative to Y, proposition 5.4.1 means moreover that

A(~2)f(1A(a~))A(fl 2) = A(~2 2)

so the algebra d(ft~.) is a kind of "source algebra" of d ( fF ) (cf [11]). The next section is a refinement of this analogy.

5.5 P r o g e n e r a t o r s

5.5.1 Finitely generated modules Proposition 4.2.1 provides a way to define the notion of a finitely generated module over a Green functor A: an A-module M is t~nitely generated (oi" of / in i te type) if M(Y~) is a finitely generated A(fi2)-module. This notion depends only on the category A-Mod , by the following argument (Benson): a module M is finitely generated if and only for any set I , the map

(~i)iEI e O HomA(M, M) ~ | �9 Homa(M, ( ~ M) I I

is an isomorphism. The following lemma shows that this definition actually does not depend on f~:

L e m m a 5.5.1: Let A b e a G r e e n f u n c t o r for t h e g r o u p G, and M b e an A - m o d u l e . T h e fo l lowing c o n d i t i o n s a re equ iva l en t :

1. T h e m o d u l e M(f~) is a f in i te ly g e n e r a t e d A ( f F ) - m o d u l e .

5.5. PROGENERATORS 115

2. T h e r e exis t a (f inite) G-set X such t h a t M is a q u o t i e n t of Ax.

R e m a r k : The word "finite" is unnecessary, since the module Ax is not defined for an infinite X.

P r o o f : Suppose 1) holds. It means that M(f~) is a quotient of a direct sum of a finite number of copies of A(fl2). Taking inverse images under the equivalence of proposition 4.2.1 shows that M is a quotient of a direct sum of a finite number, say n, of copies of Aa. As nAa ~- Ann, condition 2) holds for X = nfk

Conversely, if 2) holds, then as X divides a multiple nf~ of fl in CA, the module Ax is a direct summand of A~a -~ nAa. Now M(f~) is a quotient of nAa(f~) = nA(f~2), so 1) holds. �9

5 . 5 . 2 I d e m p o t e n t s a n d p r o g e n e r a t o r s

Let A be a Green functor for the group G. By definition, an A-module P will be called a progenerator (of A-Mod) if it is projective, of finite type, and if any A-module M is a quotient of a direct sum of copies of P. Proposition 4.3.1 shows that this is equivalent to say that P(f~) is a progenerator of A(Ft2)-Mod.

Since Aa(fl) = A(ft 2) is a free module over A(fF), it follows that Aa is a progenerator. More generally, proposition 3.1.3 shows that if Y is a G-set, then Av is a projective A-module, and proposition 4.2.1 shows that Av is a progenerator if and only if any G-set X divides a multiple of Y in CA. A natural question is now to know if any direct summands of Av are also progenerators.

As the algebra of endomorphisms of Av is A(Y2), it is equivalent to consider idempotents of this algebra. I will use the following notation:

N o t a t i o n : If j is an idempotent of A(Y2), I will denote by Av.j the associated direct summand of Av.

By definition, if X is a G-set, then

(Ay j ) ( x ) = A(XY) o~ j

This is the set of morphisms (in CA) from Y to X which factor through j . The module Av.j is projective, and for any A-module M, I have

HomcA(dv.j, M) ~- jov M(Y)

The question is now to find couples (Y,j) such that Av.j is a progenerator.

Proposition 5.5.2: Let A be a G r e e n f u n c t o r for G. Le t Y be a G-set , and j be an i d e m p o t e n t of A(Y~). T h e fol lowing cond i t i ons are equ iva len t :

1. T h e m o d u l e Ay.j is a p r o g e n e r a t o r .

2. A n y G-set X d iv ides a mu l t i p l e of Y in CA, and t he two-s ided ideal o f A(Y 2) g e n e r a t e d by j is equal to A(Y~).

116 CHAPTER 5. MORITA EQ UIVALENCE AND RELATIVE PROJECTIVITY

P r o o f : If Ay.j is a progenerator, as it is a direct summand of Ay, the module A},- is also a progenerator. Now by proposition 4.2.1, any X divides a multiple of Y. Moreover, the module Ay is a quotient of a direct sum of copies of Ay.j, so it is a direct summand, which means that the composition morphism

is surjective. As the image of this morphism is A(Y:) oy j OF A(Y:), it follows that the two-sided ideal of A(Y:) generated by j is the whole of A(Y;).

Conversely, if 2) holds, then Ay is a progenerator (proposition 4.2.1), and equality

A(Y 2) oz j oz A(Y 2) = A(Y ~)

proves that Ay is a direct summand of a sum of copies of Ay.j . So Az. j is a progenerator, and this completes the proof of the proposition. �9

L e m m a 5 . 5 . 3 : L e t A b e a G r e e n f u n e t o r for G, and X b e a G - s e t . T h e n ~. ( . : ) i s a mor,hism of rin~ (with u~) from (~(X/, ) to (~(~),o~).

P r o o f : Let a and b be elements of A(X). Then

A. (a) ox A. (b) . . . . X X XX

\ X l X 3 / \ X l X 2 X 2 2 f 3 /


A . ( X l X 2 ") (a X b) X X l X l X 2 X 2 /

and the square

(x) X X

X , X 2

1 Ix,:'&) )

x~ ( x~x~,~ ~ ~ \ X l X 2 X 2 X 3 ]

is cartesian. So

\ X l X 2 X 2 Z 3 ] \ X l X l X 2 X 2 /

and

;I~XX 3fX

(x) (x) (xlx x3 A. (a)oxA. (b) = A. A.

X X X X \ X l X 3 / XX22 X X

This proves the lemma, since moreover A. (~:)(sx) : 1A(X2).

N o t a t i o n : If X is a G-set, and ira E A(X), I will set

a = A* ( X )

If Y is a G-set, and if j is an idempotent of (A(Y),.), then j is an idempotent of

A(Y2). So f~ x j is an idempotent of A(f~ • Y). Then:

P r o p o s i t i o n 5.5.4: W i t h t hose n o t a t i o n s , i r e E A. (Y. ) (A(Y) . j .A(Y)) , t he

m o d u l e Aaxy.(f~ • ~) is a p r o g e n e r a t o r .

P r o o f : The hypothesis implies that A is projective relative to Y, and that there exists elements ai and /3i in A(Y), for 1 < i < n, such that

i=1 This expression can be transformed thanks to the following lemma:

L e m m a 5.5.5: I f a E A(Y), let c x a (resp. a • c) be t h e e l e m e n t a o f A(Y), v i ewed as HomcA(Y,*) (resp. as HomcA(*, Y)). T h e n for a and fl in A(Y)

(e x a) oy (/3 • e) = A . ( i ) ( a . ~ )

• o y = •

P r o o f : By definition

and the first equality follows. For the second

~oy(o:Mc)~-A.(YlY21A* ( Yly21[A.(Y ) (/~) x oL] \ yl / \YlY2Y2/ YY

The product inside hooks is

As the square

A, ( YlY2 I (~ • a ) k ylyly2 /

Y

y2

YY ) y2

ya

118 CHAPTER 5.

is cartesian,

MORITA EQUIVALENCE AND RELATIVE PROJECTIVITY

I have

\YlY2Y2] \YlYlY2/ YY YY

and so

which is the element (/5.c,) • e. A similar computation proves the last assertion. �9

P r o o f of proposi t ion: It follows from lemma that

and equality (5.1) can also be written

n

i = 1

Taking the image of this equality under the functor f~ x - gives

i=1

This proves that the unit of da factors through Aa• x j), so the latter is a progenerator. �9

P ropos i t ion 5.5.6: If H is a subgroup of G, let

KCH

I denote by 7r the map from f~H to G/H defined by ~r(xK) = a~H. If j is an

i d e m p o t e n t of A(H), I denote by J = A*(~r)(j) the associated i dempo ten t of A(ft}). Then :

1. If e C t~(A(H).j.A(H)), then Aau.J is a progenera tor .

2. I f A(.) has a unique maximal two-sided ideal, if H is min imal such tha t A is projec t ive relative to G/H, and if Aau.J is a p rogenera to r ,

then e 6 tau(A(H).j.A(H)).

Proof: The inclusion from f~H into fl induces an embedding from A(f~) into A(ft2), which can easily be described: indeed, the isomorphism of proposition 4.4.1

#(A)"oA(a ~)= (~ A(KNXL) K,LCG

xeK\G/L

provides an identification of the element a E A(K M ~L) with

K L tKn~L~KnxL,aCxrK~nL

Now the image of A ( ~ ) identifies with

A(~2H) ~-- @ A(KA~L) K , L C _ H

x E K \ G / L

and the embedding is just an inclusion of components. Moreover, by definition of ~r

A*(~)(j) @ H �9 = rL (3) E A(aH) = (~ A(L) L C H L C H

It follows that the element J = A*Or)(j) is equal to the element

J E n n = t n ~ n , r H ( j ) C l r L

LCH

corresponding to the double class L.1.L. If

A(~2)JA(i~ 2) = d(a 2)

then the identity morphism of ~ (in CA) factors through Aa, .J , so the latter is a progenerator.

But the unit 1A(n:) is equal to

tK ~ r K 1A(~2) = E K A K , ~ K c l K

KCG

I fz E t~(A(H). j .A(H)) , then there exists elements ~, and iS, of A(H), for 1 < i < n, such that

n =

i=1

Now for any subgroup K of G

eI< rG(g) = ~ tK [xrH i X xrH , .XxrH ,oxh = h'n:H~ KXnH(Oq)" K~nH())" KXnH(Pi)) i = l

x E K \ G / H

Now setting K~ = K~AH, replacing this expression of sK in 1A(a2), and using relations of proposition 4.4.1, I get

1A(fl2 ) ~- ~ x~ - A x K x r H , x C x T ' , - . g , - A , . - H , . , r , , "~K A n . r H , ~ , c - l r K z , . - �9 ~x x , 1(x(O~i) lXx IXx l X x l r t ( x (~ ) lXx x ~xx~ K x ( P t ) 32 1% x i=1

xEK\G/H

Setting finally

x x, Kx ~ !

= tK ~ -1 K V~,K,x ~'~AK~,T~(~dCx r~K=

I have the following equality

1A(~2) : ~ u i , K , ~ J v i , K , a - i=1

K C G zEK\G/H

proving that 1A(ft2 ) is in the two-sided ideal generated by J, so that Ao~.J is a progenerator.

Conversely, if AaH.J is a progenerator, then the identity of A(fl 2) factors through Aa~.d, and then

A(a2)JAtfl 2) = A(fl 2)

Now let ai and/3i, fox" l < i < n, such that

Then

1A(a~) = ~ chJfli l<i<n

G G G G G G t G r a t G A G , s r a E -: = t GO~ i J /3 ir G

i<_{<_~

The elements t~ai are linear combinations of elements of the form

K A K . a C f l ' K ~

for suitable subgroups K, and elements a 6 A(K) and x E G. Moreover

~./~i~2 aexrI~:Xx G K x ~,G . K x , : CxtK~/~A-~,a~7"K~ : b K x A I l ' x a X r K x

and the elements t~a{ are linear combinations of elements of the form

G K t K A K , ~ r K

Similarly, the elements ~irg are linear combinations of elements of the form

K G t K / ~ K , a r a

Moreover, as

It follows that

j ~ L L : tL /~L , rHz ( j ) r L

L C H

tG ~ K �9 K ' ~ G K A K , a ~~ J F K ' AK',a' FK,

is zero is K # K', or if K = K' ~ H. Finally, there exists element aa and ba of A(U), for U C_ H, such that

G G G U U U U G t G A G , e F G E : " " " = tuAu, aurutuAu,rH(j)rutuAu, buru

U C H

.,G~ rG

U C H

G G = tVAG,~uc" t~(au.r~(j).b.)rG


Now proposition 4.4.1 shows that

UCH

But the images t~(A(U).r~(j).A(U)) are two-sided ideals of A(G) = A(.) . If A( . ) has a unique maximal two-sided ideal, then there exists a subgroup H of G such that

e E t~(A(U).r~(j).A(U))

Now the minimali ty of H implies U = H, and

eEtg(A(H).j .A(H))

which proves the proposition. ,,

R e m a r k : If A( . ) is a local ring, then there exists elements a and b of A(H) such that

c = t~4(a.j.b )

If Rosenberg's l emma holds in A(H) (for instance if R is a complete local noctherian ring, and if A(H) is of finite type over R), and i f j is primitive, then this condition entirely determines j up to conjugation by an invertible element of A(H) (the argument of Puig in [11] can be applied here without change).

Chapter 6

Construction of Green functors

6 . 1 T h e f u n c t o r s : H ( M , M )

If M is a Mackey functor, I know that ?f(M,M) is a Green functor. To describe its product x, I observe that the identity morphism from H(M,M) to itself must turn M into an ~(M, M)-module. Let X and Y be a-sets. If f E 7-i(M,M)(X) = HomM~k(a)(M, Mx), and if m E M(Y), I must find an element f • m of M(X x Y). But f is determined by morphisms

f z : M(Z) ~ Mx(Z) = M(ZX)

In particular, I have the element f z (m) in M(Y • X), so the clemcnt M. ( ~ ) (fz(rn)) in M(XY) . This is the elcment I am looking for:

P ropos i t ion 6.1.1: Let M be a Mackey functor , and let X and Y be G- sets. The Green functor s t ruc tu re of ~ ( M , M ) and the ? / (M,M)-module s t ruc tu re of M are given by the following products :

�9 I f f E H(M,M)(X) and g E 7-/(M,M)(Y), then f • g is the m o r p h i s m f rom M to Mzz defined for a G-set Z and rn E M(Z) by

( f • g)~( .~) = M . ~ x y

�9 If f E %I(M,M)(X) = HomMa~k(G)(M, Mx) and m E M(Y), then f • m is the e lement of M ( X Y ) defined by

= o fy(~) f • M. xy

In par t icular , if moreover M is an A-module for the Green func tor A, then the image of the e lement a E A(X) under the morph i sm from A to 7/(M, M) is the m o r p h i s m from M to Mx defined by

yx) (a • m) m E M(Y) H M. xy

124 CHAPTER 6. CONSTRUCTION OF GREEN FUNCTORS

Proof : Let A be a Green functor, and let M be an A-module. In the case when X = G/H and Y = G/K for subgroups H and K of G, I know that the product of a E A(X) and m E M(Y) is the element of M(G/H • G/K) defined by

/.H ?~K x a x ~ = @ . o . , . ( a ) . . n ~ , - ( ' ~ )

xEH\G/K

up to identification of G/H x G/K with I-[zEH\G/K G/(H N ::K) by the inverse of the map g.(H M ::K) H (gH, gxK).

On the other hand, the element a defines an endomorphism r of ReSaH M by

mE M(L) Hr for L_CH

By adjunction, this element defines a morphism from M to MG/H = Ind~Res~M given by

m CM(I<)e-+ @ ea,Hn'K (/'HK~/x" ( xrO )) = @ rHn'I'2(a)'1"HK~K("vm) xEHiG/K xEH\G/K

up to identification of G/K x G/H with LIxEH\~/K G/(H N ~K) by the inverse of the map g.(H M = K) ~ (gxK, gH).

It follows that a does map to the morphism from M to Mx defined by

m E M ( Y ) H M * ( x Y ) ( a x m ) y x

In the case d = ~(M, M), this formula gives for f E ~(M, M)(X) and m E M(Y)

fY(m)= M" (xY) (f •

Taking the image under M. ( : i ) gives

(~x) sY(,,,) f x m = M , xy

This formula gives a way to compute f xg , for f E "H(M, M)(X) and g E 7-/(M, M)(Y): indeed, if m E M(Z), I have

(z 1 = o f r z o M. o gz(m)

yz \ x y z / yz

But f is a morphism of Mackey functors. So

and then

s x / ~ / = - . ~ x _ . . / ~ _ o s ~ o ~ / ~ / = - . os~o~(~/ . . . .

\ x y z / \ y z x / xyz

6.1. THE FUNCTORS H(M,M) 125

kxyz j \ z x y /

This expression must be equal to

( f • g) • m - - M. (zxY I o ( f x g)z(m) \ x y z /

and it follows that

( f • g)z M. ( z Y x l = o f z z o gz \ z x y /

which proves the proposition. ..

6.1.1 The product 5

The previous proposit ion leads to the following definition:

Definition: Let M, N and P be Mackey fimctors. If X and Y are G-sets, if a C 7-I( M, N ) ( X ) and b r 7-L( N, P ) ( Y ), I denote by b 5 a the element of ~ ( M, P ) ( Y x X) defined on a G-set U by

(b ha)u : P. (uxy I obuxoau : M(U) ~ M ( U Y X ) : Mzx(U) \ u y x /

Actually, I will only know that b 5 a is in ~(M, P) (YX) after proving the next proposition:

Proposition 6.1.2: The product 5 defines a bilinear morphism from ~(N, P), ?/(M, N) to 7{(M, P), which is moreover associative.

Proof: Firs t it is clear that b 6 a is a morphism of Mackey functors from M to Pzx: indeed, if f : U --+ V is a morphism of G-sets, then the following diagram is commutat ive:

au bv x M(U) , N(UX) , P (UXY)

M(V) , N(VX) , P ( V X Y ) a v b v x

\ u y x / , P ( U Y X )

P . ( f l z I x ) 1 )

\ v y x /

The two squares on the left are commutat ive because a and b are morphisms of Mackey functors. The right square is commutat ive because its the image by P. of a commutat ive square. It follows that the products along the edges are equal, so

Pzx,.( f) o (b 5 a)u = (b 5 a)v o M.( f )


The diagram obtained with M*, N* and P*, and by reversing the vertical arrows, is �9 u x y * u y x also commutatave, for the same reasons (and also because _P ( ] = P ( "~). Thus

�9 \ u y x ] \ u x y /

b 5 a is a morphism of Mackey functors. The product 5 is moreover bifunctoria] in X and Y: if a : X ~ X t and

/3 : Y ---+ Y' are morphisms of G-sets, then the following diagram, where I set

is commutative:

a u

M(U) , N(UX)

[d I (1) N,(Iua) l

U.(~)(a)v M(U) , N(UX')

Id I (4) Id I

= ~(M,N) K = ~(N,P)

~.(~)(a)~ M(U) , N(UX')

bux \uyx/ , P(UXY) , P(UYX)

(2) P.(Zualy) l (3) P.(IuIya) l

p, (%x'y'] bux, \ uyx' J

, P(UX'Y) , P(UYX p)

(5) P,(IuIx,/3) 1 (6)P.(Iu/3Ix,) l

Y~.(/3)(b)ux, P" \uy'x'] , P(UX'Y') , P(UY'X')

The squares (1) and (5) are commutat ive by definition of ?-t,(a) and of K~,(/3). The square (2) is commutat ive because b is a morphism of Mackey functors from N to PY. The squares (3) and (6) are commutat ive because they are the image under P. of a commutat ive square. Finally the square (4) is trivially commutative.

The two possible products from M(U) to P(UY'X') along the edges of that diagram give the equality

~ ( N , P) , (9) (b) 5 ~ ( M , N) . (~) (a ) = ~ ( M , P),(/3 • ~)(b 5 a)

which proves that the product 5 is covariant. If now a : X ~ -~ X and /3 : YP ~ Y are morphisms of G-sets, then the similar

commutat ive diagram

a u

M(U) ~ N(UX)

Id I N*(Iua) [

W(~)(a)~ M(U) ~ N(UX')

'd I 'd I W(~)(a)~

M(U) ) N(UX')

I bu x \ uyx /

, P(UXY) , P(UYX)

P'(I~zY) 1 P*(,dY~) 1 p. (%x'y)

bux, \ wz ' ] P(UX'Y) , P(UYX')

P'(IuIx,/3) I P*(Iu,Ix,) 1

(~x'y') ic*(/3)(~)~, P* \~'~'/

, P(UX'Y') , P(UY'X')

6.2. THE OPPOSITE FUNCTOR OF A GREEN FUNCTOR 127

shows that the product 5 is contravariant, hence that if defines a bilinear morphism from ~ ( N , P), 7-{(M, N) to 7-t(M, P).

I must still show that the product 5 is associative: so let Q be a Mackey functor, let Z be a G-set, and let c be an element of 7-(.(P, Q)(Z). Then

[ uy x z \ ( uXY ~ obu xoa[ t

Since c is a morphism of Mackey functors from P to Qz, I have

and so

0 = 0 CUXYZ \uycc/ \ u y x z /

I o \ u z y x / \ u y x z ]

u x Y z I . . . . Q* u z y x ] o c u x Y z o bux e au

On the other hand

\ u z y x /

uxzy} ~176176 = Q. ocuxgzobuxoau \ u z y x /

This shows that the product 5 is associative, and proves the proposition. , '

6.2 T h e o p p o s i t e functor of a Green functor

D e f i n i t i o n : Let A be a Green functor for the group G. I calI opposite [unctor of A, and I denote by A ~ the Green functor equal to A as Mackey functor, and endowed with the product x~ defined for G-sets X and Y and elements a E A~ = A(X) and b E A~ by

a •176 b = A" xy

and the unit c E A~ = A( , )

The reason for this definition is the following proposition:

P r o p o s i t i o n 6 . 2 . 1 : L e 2 A be a G r e e n func to r for t h e grou p G. T h e n A ~ is a G r e e n f u n c t o r . M o r e o v e r for a ny ( ; -set X

(AoP(X), .up) = (A(X), .)~

P r o o f : I must show that the product •176 is bifunctorial, associative, and unitary. So let f : X --+ X ' and 9 : Y --+ Y' be morphisms of G-sets. Then for a E A(X) and b G A(Y) , I have

(~'x '~ (y ' z '~ A~215176176 = A. (A.(g)(b)• = A. A . ( g x f ) ( b x a ) \ x,~,,) ix,y,)

But

SO

o ( f • g) = (g x f ) o \ v ' x ' ) v,~

A.(g • f ) = A. \ \ y ' x ' ) o ( f x g) o = A. A . ( f x g)A. \ < , / / / \ y':c' ] xy

and

Aop(f)(a) x~176 = m. \ x ' y ' ) A= \ y ' x ' ] A=(f x g ) d . xy

. . . . A . ( f x g)(a x ~ b)

proving that x ~ is covariant. Similarly, if a' E A(X ' ) and b' C A(Y') , then

A~ ') x ~ A~ ') = A. xy

( Y Z ) A * ( g x f ) ( b ' x a ' ) . . . = A . xy

Since

A*(g x f ) = A* yz A*(f x g)A* xy \ y ' x ' )

Xly I ylX# a,nd since A* (.,~,) = A. (x,,,), I have

() () u yz A* yx A * ( f x g ) A . (b 'xa ' ) . . . . A~ ') x~176 ') = A. xy xy \ * ' y ' )

. . . . A*(f x 9)(a' x ~ b')

which proves that x~ is contravariant. The product x~P is also associative: if X, Y, Z are G-sets, if a E A(X) , b ~ A(Y) ,

and c C A(Z) , then

= = c x A . ( b x a ) . . . . \ x y z / x y z / xy

. . . . A . ( Z X Y t A . ( z Y X l ( c • ) \ xyz / \ zxy / \ xyz )

Similarly

a x ~ 1 7 6 ( ( b x ~ P c ) • ( A * ( z Y ) ( c x b ) \ x y z ) yz . . . .


. . . . A , ( Y Z X l A . ( Z Y X l ( c • ) k x y z / \ y z x / kxYZ/

which proves that x ~ is associative. Finally if a C A(X) , then

and it is clear similarly that a x ~ e = a. So e is a unit for A ~ To complete the proof, I compute the product 't~ if a, b E A(X) , then

a.~ A* ( x ) (a •176 : A* ( x ) m. CXlX2~ (b • xx xx \ x 2 ; c 1 /

As

XX \ X 2 Z 1 / XX \ X l X 2

and a s

X2Xl 0 ~-.

kXlX2/ XX ZX

I have a'~ A* ( X ) (b x a) =

and this proves the proposition.

R e m a r k : It follows that condition 2) of proposition 5.3.5 is equivalent to the commuta t iv i ty of all the rings (A(X), .), or to the commuta t iv i ty of all the rings A(H), for H a_ G.

6.2.1 Right modules

D e f i n i t i o n : Let A be a Green functor for G. A right module M for A, or module-A, is a Mackey functor for G, together with bilinear maps

M ( X ) x A(Y) -4 M ( X x Y)

denoted by (m, a) ,-~ ra x a for any G-sets X and Y, which are bifunctorial, associative, and unitary (in the obvious sense). A morphism 0 of modnles-A from M to N is a morphism of Mackey functors from M to N, such that for any G-sets X and Y, and any m E M ( X ) and a E A(X)

O~x~( ,~ • a) = 0~( ,~ ) x a

I can now speak of the category of modules-A, that I will denote by M o d - A . As in the case of ordinary rings:

Proposition 6.2.2: T h e c a t e g o r y M o d - A is e q u i v a l e n t to t h e c a t e g o r y A ~ M o d .

130 C H A P T E R 6. C O N S T R U C T I O N OF GREEN F U N C T O R S

Proof: If M is a module-A, I turn it into an A~ by setting, for a C A ( X ) and m E M ( Y )

a X ~ 797 ~ M . x~j

Now the proof of proposition 6.2.1 can be carried word for word here, to prove that this product is bifunctorial, associative and mfitary. Conversely, if M is art A~ I can turn it into a module-A by setting in the same conditions

m x a = M * ( x Y ) ( a x ~

It is clear that those correspondences are functorial, and inverse to each other. The

proposition follows.

6 . 2 . 2 T h e d u a l o f a n A - m o d u l e

D e f i n i t i o n : If M is an A-module, the dual of M is the Mackey functor M ~ defined on the G-set X by

M~ = Hom,~ (M(X), R)

If f : X ~ Y is a morphism of G-sets, then

( M ~ = tM*( f ) (M~ = t M . ( f )

where tl is the transpose map of I. I define a right product on M ~ by setting, for G-sets X and }", for r C M ~

for a e A(Y), and rn E M ( X x Y)

(r x a)(,~) = r o~~ , ~ )

where by definition

\ y x y /

Proposi t ion 6.2.3: These definitions turn M ~ into a module-A.

Proof: First the product is bifunctoriah if f : X ~ X ~ and g : Y ~ Y" are morphisms of G-sets, if r E M ~ if a E A(Y) , and ~7~' E M ( X ' x Y') , then

((Mo).(s)(+) x (M~ . . . .

. . . . cM*(f) (A.(g)(.) o7;-,') Moreover

/ ~ m ' = M. " M* (d . (g)(a) • m') . . . . d.(g)(a) Oy, \ x' ] \y'x'y']

(~,'y') ( Sy') . . : M. (xm"/M* ~,u ( . • . ,) �9 \ x ' ) \ y ' x ' ~ , ) \g(y)x'y'


But the square x~y

yx'g(y)) X'Y > YX'Y'

( x'y ~ ( yx'y' ~ x'g(y)] [ I \g(y)x'y']

)

kfx'y') is cartesian. So

op (x'y'~ ( x ' y A.(g)(a) o r, ~ ' = M. t x' ) M. \ x ,g (y ) ) M*

It follows that

( ( M ~ 1 6 2 x A.(g)(a))(m') . . . .

. . . . CM*(f)M. \ x' ] M. \x,g(y)j M* \yx'g(y)J (a x mr) . . . .

= r (x'Y~ M. ( x'y ~ (a \ ~ ' ) Sg(y))

The square

X Y

X

is cartesian. So

((M~162 x A.(g)(a))(m')

x~y yx,g, f JI (a • 7321 )

(,/%) I X'Y

i x , ) ) X ~

f

x m')

f(x)y M* (a xm ' ) . . . . \yx'g(y))

. . . . + ~ . / ~ ) ~" (x"//o~'~,x~'~/=.x~ +/oo~*~,~'~/ . . . .

. . . . (r x a)M*(f x g)(m') = (M~ x g)(r x a)(m')

and this proves that the product is covariant. If now qY E M~ if a' E A(Y'), and if m E M(X x Y), then

((M~162 ') x A*(g)(a'))(m) = (M~162 m) . . . .

132

Moreover

A*(g)(a') oF m = M. yxy

CHAPTER 6. CONSTRUCTION OF GREEN FUNCTORS

= r p m)

g(y)xy (a' •

So

. . . . r f (x) g(y)xy (a' • m)

On the other hand

(M~ • g)(r • a')(m) = (r • a')M.(f • g)(m) = r oy, ~ M . ( f • g)(m))

But

a oy, M . ( f • = M. \ x' ] M* \y'x 'y ']

(x'y'~ M* ( x'y' ~ ( y'xy . . . . M. \ x' ] \y 'x 'y '] M. \y ' f (x)g(y)] ( a ' • m)

As the square

g(s)xy

is cartesian, this is also

X Y Y I X Y

)

\y'x'y']

a oy, M . ( f • \ x' / M. ( f • g(y)xy (a ' xm) . . . .

. . . . M. f (x) g(y)xy (a' x m)

So I have

(M~ • g)(r • a')(m) = r f(x) g(y)xy (a' • m) . . . .


. . . . ((M~162 ') x A*(g)(a'))(m)

which proves that the product is also contravariant.

I must still prove that the product is associative and unitary. So let X, Y, and Z be G-sets. If r E M~ if a e A(Y), and b E A(Z), then for m e M ( X • Y x Z), I have

( ) ( o~ m)) ~ ~ ) = r ~ 07 (b oz (r • a) x b (m) = (r x a)(b oz

But

a~ (b~176 = M* (XY)M* ( xY ) (a x (b~176 Moreover

a x (bo~ m) =a x M. (xYZl M* ( xYZ t ( b x m ) . . . . \ xy ] \ z x y z /

. . . . M , ( y ' x y 2 z l M * ( y ' x y 2 Z l ( a x b x m ) \ ylxy2 ] \ y l zxy2z /

As the square

yxyz / ) X Y Z Y X Y Z

( x:: ) 1 1 ( ::: : ) X Y ~ Y X Y

yxy is cartesian, I have

yxy \ ylxy2 ] \ xy / k y x y z / and then

\ xy ] \ y x y z / \y lzxy2z]

. . . . M . ( x : Z ) M . ( xyz ~ ( a • 2 1 5 \ y z x y z /

On the other hand (~ • (a • b ) ) ( ~ ) : ~((a x b/o~z~ ~)

Since ( a • b )op ( x ; z ) ( xyz ~ Ovz m = M. 114. (a x b x m)

\ y z x y z / the product is associative.

Moreover, if r C M~ and if m E M(X), then

(r • ~ ) ( ~ ) = r o:~ ~)

As

(?)M" .x. the product is unitary, which completes the proof of the proposition.

6.3 Tensor p r o d u c t of Green functors

If A and B are Green functors for the group G, then their tensor product A@B can be given a structure of Green functor:

Defini t ion: If X and Y are G-sets, if ( U,r ) is a G-set over X and (V,r ) is a G-set over Y, ira E A(U), b E B(U), c E A(V) and d E B(V), I define the product of the dement [a @ b](u,r of A@B(X) by the dement [c | d](v,r of A@B(Y) by setting

[a | b](u.r x [c | d](v.r = [(a x c) | (b x d)](vxv,r215162 E (A+B)(X x Y)

If CA and r are the units A and B, I set

P r o p o s i t i o n 6.3.1: Let A a nd B be G r e e n f u n c t o r s for the g r o u p G. T h e n the product x t u r n s A@B into a Green functor, wi th unit e.

Proof: First I must check that this product is well defined. So let f : (U, r ~ (U', r be a morphism of G-sets over X, let a E A(U) and b' E B(U'). Then

. . . . [(A.U x '<O/o x el)| I r x . . . .

. . . . [(a x ~) | B ' ( f x id)(b' • ~) ]~ •162 . . . .

. . . . [ ( ax c) | (B ' ( f ) (b ' ) x d)](Uxy,r162 ,

which does coincide with the definition of

[a | B* ( f ) (b ' ) ] (u ,<9 x [c | d](v,r

Similar arguments show that the product x on A@B is well defined. This product is moreover bifunctorial: if f : X --~ X' and g : Y --+ Y' are

morphisms of G-sets, if u = [a | b](u,r is an element of A@B(X) and v = [c| d](v,r an element of A@B(Y), then

( A@B ).( f)(u) x ( A@B).(g)(v) = [a|162 [c| = [(a x c)| d)]wxV, lr )

On the other hand

(A@B).(f x g)(u x v) = [(a x c) @ (b X d)](uxv,(fxg)(r162 )

and equality f r x g r = ( f x g)(r x ~) proves that the product is covariant. If now f is a morphism from X ' to X and g is a morphism from Y' to Y, then to

compute (A@B)*(f)(u) and (A@B)*(g)(v), I fill the cartesian squares

c~ 2 P > U Q > V

X' > X y ' > y f g

6.3. TENSOR PRODUCT OF GREEN FUNCTORS 135

so that

(A@B)*(f)(u) = [A*(c~)(a)@B*(c~)(b)](p,;) (A~B)*(g)(v) = [A*('7)(c)@B*(7)(d)](Q,~)

The product (A@B)*(I)(u) x (A@B)*(g)(v)is then equal to

which can also be written as

[(A*(c~ x @ ( a x c ) ) | (B*(a • 7)(bx d))J(pxQ,Zx5 )

As the square ~ x ~

P x Q , U •

,8x5 1 [ r

X ' x X ~ X x Y f x g

is also cartesian, I have

• g)( x )(ax c))| )(bx )

which proves that the product x on A@B is contravariant, hence bifunctoriah It is clear from the definitions that the product x is associative. Finally, the

element c is a unit, since

[a @ b](g,r • [CA | eB](o,Id) = [(a • CA) @ (b x eB)l(uxo,r = [a | b](g,r

[CA @ eB](o,za) x [a | b](v,r = [(EA • a) | (eB • b)](.•162 = [a | b](u,r

and this completes the proof of the proposition. "

The tensor product of Green functors is functoriah

L e m m a 6 . 3 . 2 : : I f f : A ~ A' and g : B ~ B' are m o r p h i s m s o f G r e e n f u n e t o r s for t h e g r o u p G, t h e n f @ g : A @ B ~ A @ B ~ is a m o r p h i s m of G r e e n f u n c t o r s . I f f and g are un i tary , so is f @ g .

Proof : The morphism f (resp. the morphism g) is determined by morphisms fu (resp. gu) from A(U) to At(U) (resp. from B(U) to B'(U)), for any G-set U. Moreover, the image of the element [a | b](a,r of (A4B) (X) under (f@9)x is given by

(f@g)x ([a | bl(u,r = [f~(a) | gu(b)l(u,r

The lemma follows easily, since if f and g are morphisms of Green functors, then

fu• • c) = fu(a) x fv(c) gu• • d) = gv(b) • gv(d)


A special case of this lemma is the case when f is the identity morphism of A, and g is the unique (unitary) morphism of Green functors from b to B. since A@b ~- A, what I get is a unitary morphism ~A of Green functors from A to A@B, and it is easy to see that this morphism is given by

a �9 A(X) ~ "TA,X(a) = In @ r a @ B* (r (X,Id)

Similarly, there is a morphism ?B from B to A@B defined by

bE A@B(X) (Xdd)

The notion of opposite functor of a Green functor leads to the following definition:

Definition: Let A be a Green functor for the group G. If X and Y are G-sets, I will say that the element a �9 A(X) commutes with/3 �9 A(Y) i ra x fl = a x ~ /3, i.e.

a x / 3 = A , xy

Similarly, I will say that a subset P C A(X) commutes with a subset Q c_ A(Y) if any element of Y commutes with any element of Q. If M is a sub-Mackey functor of A, and N a sub-Mackey functor of B, I will say that M commutes with N if M ( X ) commutes with N (Y ) for any G-sets X and Y.

It is clear that if a commutes with fl, then fl commutes with a, since taking the image of the above equality under A. ( ~ ) exchanges the roles of X and Y and of a and/3. I will also say that in a more symmetric way that a and/3 commute.

L e m m a 6.3.3: The image of 7A in A~B c o m m u t e s with the image of "~B.

Proo f i Let X and Y be G-sets. If a �9 A(X) and b �9 B(Y) , then

7A,X(a) • 7B,g( b) = [a @ r x [C A,y @ 5](Y, Id) = [ (a X ~ A,Y ) @ (CB,X X b)](XxY, id)

Moreover a X C A , y = a x A * ( Y ) ( r

I have eB,X x b = B* (~Y)(b). Then Similarly,

~/A,X(a) X'TB,Y(b)= [A*(X;J ) (a ) |215 )

On the other hand

"/By(b) • ~/A,X(a) ---- [gAy @ b](Y, Id) • [a @ gB,X](X,ld) = [(CA,y • a) | (b | s

or

7S,y(b)• [A*(YxX) (a ) |215 )


The image under ( A~B) . (~;) of this element is

Since moreover

it is also

[A* ( ? ) (a)@'* ( : : ) ~" (Y:) (b)](XxY, id ) I have also

so finally

B. xy yx y

• ~A,x(a) • ~/.,y(b) = ( A 6 B ) . xy 7A,X(a)) which proves the lemma. �9

This temma leads naturMIy to the universal property of the tensor product of Green functors:

P r o p o s i t i o n 6.3.4: Let A, B and C be Green func tors for the g roup G. I f f ( resp. g) is a m o r p h i s m of Green func tors f rom A to C (resp. f rom B to C), and if the image of f c o m m u t e s wi th the image of g, t hen the re exis ts a un ique m o r p h i s m of Green functors h f rom AQB to C such t h a t f = ho"/A a n d g = h o T B .

Converse ly , if h is a m o r p h i s m of Green func tors f r o m A@B to C, t h e n f = hOTA (resp. g = ho~B) is a m o r p h i s m of G r e e n func to r f rom A to C (resp. f r o m B to C), and the images of f and g c o m m u t e .

M o r e o v e r h is u n i t a r y if and only if f and g are.

Proof : If (U,r is a G-set over X, if a E A(U) and b E B(U), then

Moreover

k Ul / \ 71"2 / (UxU, Id) The image of this element under (A@B)* (~\) , which is the product 7A(a)."/B(b)in

A@B(U), is obtained using the cartesian square

U ~ U x U

U , U x U


So it is equal to

vA(a)."/B(b) = [A*( u ) A * (ulu2~ (a)~) B* ( u ) B* (u lu~ (b)] = [a| zd) UU \ ~t 1 / UU \ U2 / (U, Id)

Now if h is a morphism of A~B to C, I have

hx([a * b](u,r : C.(r * b](u, id)) = C.(r (uL ) hu• • VB(b))

Whence

hx([a | b](u,r = C.(r (uL) "hu'TA(a) • huvB(b))

and h is determined by h o 7A and h o "~B. Then if f is a morphism from A to C and g is a morphism from B to C, there is

at most one morphism h from A~B to C such that f = h o ~'A and g = h o 7B: the morphism h is given by

hx([a|162 = C.(r (uL) (fv(a) • gu(b))

Now proposition 1.8.3 shows that a morphism h from A~B to C is determined by the bilinear morphisms ]~u from A(U) • B(U) to C(U), defined for any G set U by

(a,b) E A(U) x B(U) H C* ( uuu) hU([a|

and h is a morphism of Mackey functors if and only if the morphisms ]~u satisfy conditions i), ii) and iii) of proposition 1.8.3. But here

hu(a, b) = fu(a).gu(b)

So let k : U' ---+ U be a morphism of G-sets, and let a E A(Y), b E B(Y), a' C A(U') and b' C B(U'). Condition i) can be written as

]~u(A.(k)(a'),b) = C.(k)fu,(a',B*(k)(b))

o r

fvA.(k)(a').gu(b) = C.(k)(fu,(a').gu, B*(k)(b)) Since f is a morphism of Mackey functors, the left hand side is also equal to

C.(k)fu,(a').gu(b)

But the product "." on C is the map associated by the proposition 1.8.3 to the bilinear morphism from C, C to C defined by the product of C. So relation i) holds for this product, which gives

C.(k)fv,(a').gu(b) = C.(k)(fv,(a').B*(k)gv(b))

and this proves that i) holds for h. A similar argument proves relation ii). Now relation iii) can be written as

]~u, (A*(k)(a), B*(k)(b)) = C*(k)]tu(a, b)


which gives here

fu, A*(k)(a).gu,B*(k)(b) = C*(k)(fu(a).gu(b))

The left hand side is also C*(k)fu(a).C*(k)gu(b)

and since relation iii) holds for the product ". ' , I have

C*(k)fv(a).V*(k)gu(b) = C*(k)(fu(a).gu(b))

(in other words, the maps C*(k) are ring homomorphisms for the product ".", which is also a consequence of lemma 5.2.2). Thus the maps ] satisfy i), ii), and iii), and it follows that h is a morphism of Mackey functors.

To prove that h is a morphism of Green functors, I must check that

The left hand side is

hx• • c)| (bx d)](uxv,r162 = C . ( r r x c).gu• x d)) (6.1)

and the right hand side is

C.(r x C.(r = C.(r x r x fy(c).gy(d)) (6.2)

Equality of (6.1) and (6.2) for all r and r is equivalent to equality

f v • • e)mv• • d) = fv(a)mv(b) • f v (c )mv(d)

The left hand side is


~ ' ( ~ ~,~ ) ( ~ o ) ~ g~/~/) ~ ~" ( ~ ) ( s ~ ) ~ ~ ~ / ~ ) . . . .

. . . . C*(uuvvUV ) (fv(a)• gu(b)• f v ( c ) • gv(d)) (6.4)

If the images of g and f commute, I have

Then

fu(a) x gu(b) • fv(c) • gv(d) -- C. (u,v~u2v2~ (fu(a) x fy(c) • gu(b) x gv(d)) k Ul ~,t2'U1 ~2 /

But since C, (UlVlU2V2~ = C. (UlU2VlV2~

\ U l ~ d 2 v l v 2 / \?.llVll~2v2/


the right hand side of (6.4) is equal to

C*( uv )~y,(?~lU,2ylv2~ (fu(a) • fv(c) xgu(D)).(gv(d) ) . . . . ~ ?//O V \ ?~1 Vlt/21)2 /

which is the right hand side of (6.3). Thus h is a morphism of Green functors. Conversely, if h is a morphism of Green functors from A@B to C, the images of

f = h o 7A and 9 = h o 7B must commute: indeed, the images of 7A and 7B commute, and moreover, I have the following lemma

L e m m a 6.3.5: Le t X and Y be G-sets , and 0 be a m o r p h i s m of G r e e n func- to r s f r o m D to C. I f a E D(X) c o m m u t e s wi th ~ C D(Y), t h e n Ox(a) c o m m u t e s w i t h 0y(fl).

P r o o f : It suffices to write

yx) (~ • 9) . . . . 0r(Z) x 0x(~) = 0yxx(Z x ~) = 0y• xy

. . . . C, xy xy

The lemma follows. �9

To complete the proof of the proposition, I must still observe that if f and g are unitary, then so is h: indeed, the unit of A~B is leA | eB](oJd), and its image under h is

( ' ) . .

which is equal to the unit of C if f and g are unitary. Conversely, if h is unitary, then f = h o "/A and 9 = h o 7B are composition products of unitary morphisms, so they are unitary. �9

P r o p o s i t i o n 6.3.6: Le t A and B be G r e e n f u n c t o r s for t h e g r o u p G. Le t M be an A - m o d u l e and N be a B - m o d u l e . T h e n M ~ N has a n a t u r a l s t r u c t u r e of A Q B - m o d u l e , def ined as follows: if X a nd Y are G-sets , if (U, r is a G-set over X, and (V,r is a G-set over Y, if a E A(U), if b ~ B(U), if rn C M(V) and n C N(V) , t h e n

[a | b](~,~) • [~ e ~](.,r = [(a x ~ ) | (b • ")](~xv,+•

P r o o f : To prove that this product is well defined, associative, and unitary, one just has to mimic the proof of proposition 6.3.1, replacing A by M and B by N in suitable places. �9

6.4. BIMODULES 141

6.4 B i m o d u l e s

The notion of right-moduie over a Green functor leads naturally to the notion of bimodule:

Definition: Let A and B be Green functors for the group G. If M is an A-module, which is also a module-B, [ will say that M is an A-module-B if for any G-sets X , Y, and Z, and any elements a �9 A(X), m �9 M(Y) and b r B(Z), I have

( a • x b = a x ( m • in M ( X x Y x Z )

A morphism of A-modules-B from M to N is a morphism of Mackey functors from M to N, which is also a morphism of A-modules and a morphism of modules-B.

With those definitions, I can speak of the category of A-modules-B, that I will denote by A-Mod-B .

Proposition 6.4.1: The category A-Mod-B is equivalent to the category A@B~

Proof: To give M a structure of AQB~ is equivalent (see proposition 2.1.2) to give ~ unitary morphism of Green functors fl'om AQB ~ to ~ ( M , M). By proposition 6.3.4, this is equivalent to give unitary morphisms from A and B ~ to 7~(M,M), the images of which commute. In particular, the module M is an A-module, and a B~ i.e. a module-B.

If X is a G-set, and if a E A(X), then a defines an element

by

Aa ~ HomM~k(G)(M, Mx)

/ \ M. [xy } (a • m ) E M ( Y • X ) = M x ( Y ) m E M( Y ) ;%(m)

\ ] y x

If Z is a (;-set, and if b r B(Z), then b determines an element Pb of ~(M, M)(Z) = HomM~ck(a)(M, Mz), by

m c M ( Y ) ~ - ~ P b ( m ) = M * ( z Y ) ( b • 1 7 6 . . . .

Now Aa and Pb commute if and only if

Pb X Aa = ~ (M 'M)* ( x z ) (A~ •

where the products x are in C = ~ ( M , M). But if a is an element of C(X) , determined by morphisms

av : M(Y) ~ M ( Y X ) = Mx(Y)

and if/3 is an element of C(Z), determined by morphisms

5y : M(Y) ~ M ( Y Z ) = Mz(Y)


then the product c~ x fl is the element of C(X x Y) determined by the morphisms

(o~ x ,~)z = M. (YZXl o azz o ~z \YXZl

The product A~ x Pb is then determined by the morphisms (A~ • Pb)Z = M, (yzx~ o \ yxz / /~a,Y Z 0 Pb,Y :

\ y x z / \ y zx \ y x z /

On the other hand, the product pb x A~ is determined by

: 0 Pb,YX o Aa,y:?:Q ~ M . (.]~/I. ((t, X 7"ft) X b/) \ y z x / \ y z x / yx

Now 1a and Pb commute if and only if

(Pb • :~o)Y('~)= M. ( y x z ] (Ao • pb)~,(-~) \ y z x )

or equivalently

M*CYxzl ( M * ( x Y ) ( a • 2 1 5 = M * ( yxz l ( M * ( x y z l yx yzx) \ y x z )

This is also

\ y z x ) \ y x z ) \ y z x ) \ y x z /

thus, since M. (,:yz] is bijective \ yzx ]

a x (m• b) = (a x m ) •

and this proves the proposition. ,'

Propos i t ion 6.4.2: Let A and B be Green functors for the group G. If M is an A-module , and N a B-modu le , then the product 5 and the morphi sms from A to H(M, M) and from B to H(N,N) induce a natural structure of B - m o d u l e - A on ~(M, N).

Proof." By proposition 2.1.2, to say that M is an A-module (resp. that N is a B-module) is equivalent to give a unitary morphism of Green functors from A to 7-/(M, M) (resp. from B to ~ (M, M)). So it suffices to give ~(M, N) a structure of 7-/(N, N)-module-7"/(M, M).

So let X, Y, and Z be G-sets. If a C "H(M,M)(X), if f E 7t(M,N)(Y) , and b E 7~(N,N)(Z), I have seen that if U is a G-set, then b 5 f 5 a is the element of ~(M, P ) ( Z Y X ) defined on the G-set U by

I (b 5 f 5 a)u = IV. \uzyx ) o buxz o fux o au

6.5. C O M M U T A N T S 143

The product 5 turns "H(M,N) into a bimodule, because it is bifunctorial and associative (proposition 6.1.2), and also unitary: if b is the unit of "H(N, N), i.e. the identi ty morphism from N to N. = N, then it is clear that

= o IdN(ux) o f u = f u k'~ �9 X /

Similarly, if a is the identi ty of M, then m 5 a = m. This completes the proof of the proposition. ..

6.5 C o m m u t a n t s

D e f i n i t i o n : Let A be a Green flmctor for the group G, and M an A-module-A. I f U is a G-set, and o~ 6 A(U), [ set for any G-set X

\ / ~/x

Similarly, i f P is a subset of A(U), I set

CM(P)(X) = {m e M(X) I ~ • m = M* (XU) (m • Va6P}

More generally, i f L is a sub-Mackey functor of A, I set

CmL)(X)= {m�9 I~• (x~,](m• VU, WeL(U)} KUX/

L e m m a 6.5.1: Let be the s u b - M a c k e y functor of A generated by P (i.e. the in tersec t ion of the s u b - M a c k e y functors L of A such that L(U) D_ P). T h e n for a n y X, I have C M ( P ) ( X ) = C M ( ) ( X ) .

Proof: It is clear that CM( ) ( X ) _C C M ( P ) ( X ) . Conversely, it is easy to see that for any G-set X, I have

 ( Y ) = ~ < A . ( g ) A * ( f ) ( p ) > p e p C A ( Y ) ] : U ~ Z 9:Z--Y

so that any element c~ of (Y) can be writ ten as

c~ = E A.(gi)A*(fi)(p{) i

for suitable elements g{, fi and Pi E P. But if m 6 C M ( P ) ( X ) , then

o~ • m = ~ d . (g i )g*( f i ) (p i ) • m = ~ M.(gi • Id )M*( f i • Id)(pi • m) . . . . i i

• p{) i t l x

And for f : U ---+ Z and g : Z --+ Y, I have

KJUX K X ~ L /


. . . . M*(gxld)M* (ZX) M*(Idx f) M*(gxId)M* (XZ) M*(Id• . . . .

. . . . . ' l ' (XY) M * ( I d • f)

Thus

i yx

y x


• gi)M*(Id x f/)(m • Pi) . . . .

M (g)M*(fi)(p )) X * i i = M , zy) (m x c~) yx

P r o p o s i t i o n 6.5.2: The p r e v i o u s de f in i t i ons t u r n C M ( L ) i n to a s u b - M a c k e y f u n c t o r of M, ca l led t he c o m m u t a n t of L in M.

Proof: If f : X ~ Y is morphism of G-sets, and if m r CM(X), then for a r L(U), I have

~ • V . ( f ) ( - O = M.( ld • f ) ( a • M.(IU • f )M. ( x ~ ( ~ • \ J ~ x

. . . . M. uy uy

and this shows that M~(f)(CM(L)(X)) C C. (L) (Y) . Similarly, if m E M(Y)

a x M*(f)(m) = M*(Id x f)(a • m) = M*(Id x f)M. uy

. . . . M. M*._. ~ • X?2

which shows that M*(f)(CM(L)(Y)) c_ CM(L)(X), hence that CM(L) is a sub- k /

Mackey functor of M. ,,

In the special case when M is the functor A, viewed as an A-module-A, there is a little more:

P r o p o s i t i o n 6.5.3: Let L be a s u b - M a e k e y f u n c t o r of the Green functor A. T h e n CA(L) is a s u b - G r e e n f u n c t o r of A.

Proof: I must check that if X, Y and (J" are G-sets, if a E A(X) and /? E A(Y) commute with l E L(U), then ~ • also commutes with I. But

\uxy / \uxy /

6.5. COMMUTANTS 145

) Ix< = ~ • A . (1 • ~) = A . A . ( ~ • 2 1 5 . . . . \ u x y / yu \ u x y / \ x y u /

\ u x y / \ x y u ]

\ u x y / \ x y u / \ x u y / \ u x y /

which proves that CM(L) is closed under x. Since is it clear that ~ C CM(L)(*), the proposition follows. �9

N o t a t i o n : If M and N are modules for the Green functor A, I denote by

~A(M, N)

the commutant of A in the bimodule H( M, N).

By definition, if I set 7-I = H(M, N), then for any G-set X

H A ( M , N ) ( X ) = { f e H(X) I c~ x f = H. (xu~ ( m x ~ ) VU, Va C A(U)} \ / t tX

Let r~ (resp. l~) be the element of H(M, M)(U) (resp. of H(N, N)(U)) associated to a. Then c~ • f is the morphism from M to Nux defined for m E M(Z) by

(~ x f ) z (m) (I~ 5 f ) z (m) = N. ( zxu~ = o l~,zx o f z ( ~ ) . . . . \ Z U X /

Similarly, the morphism f x a from M to Nxu is given by

o fzg o r~,z(m) = N. , z u x = o fzu o M. (a x m) \ Z X U / \ Z X ? s Z'U

Since f is a morphism of Mackey fnnctors, I have

and then

o iV. o fuz(o~ • m) = N. o fuz(oe • m) \ Z X ' a / \ Z ' t l X / \ Z X t t /

Thus f is in t t A ( M , N ) ( X ) if and only if for any U and any a E A(U), I have

\ z u x / \ z u x / \ z x u / \ z u x /

which reduces to

• f ~ ( m ) = f ~ ( ~ • .~) Thus HA(M, N) (X) is just the set of A-module homomorphisms from M to Nx.

P r o p o s i t i o n 6.5.4: Let A be a G r e e n func to r for t h e grou p G. If M and N are A - m o d u l e s , and X is a G-set , t h e n

HA(M, N) (X) = Homa-Mod(M, Nx)

I f m o r e o v e r M = N , t h e n t h e p r o d u c t 5 t u r n s ~A(M, M) in to a G r e e n f u n c t o r .


6.6 The functors M |

Let A and B be Green functors for the group G. If M is an A-module-B, and N an A- module, I have built 7~A(M, N), which is a priori a Mackey functor, Actually, in that case, it is a B-module: i f X and Y are G-sets, ifb E B(X) and f E HomA(M, Nz), I denote by fib the element of HomMack(a)(M, Mz) deduced from the action of B on the right on M. Then f 5 Pb C ~(M, N ) ( Y X ) . It is the morphism of Mackey functors from M to Nzx defined on the G-set U by

( f S Pb)u(m) = N. k u y x / o fux o Pb,u(rn) = N. kUyX/ o fux (m x b)

It follows easily that if I set

\ u x y /

I obtain a morphism from M to Nxy. Moreover, it is clear that this turns St(M, N) into a B-module. Finally, if f is a morphism of A-modules, so is b • f , since if a r A(Z), then

. . . . a x f ux (m x b) = a x (b • f )u(m)

Now I have defined a B-module structure on ~A(M, N). The correspondence N "HA(M, N) is moreover functorial in N: if 0 is a morphism of A-modules from N to N', determined by morphisms 0z from N(Y) to N'(Y) for any G-set Y, then I define the morphism ~A(M, O) from ~HA(M, N) to ~A(M, N') in the following way: if Y is a G-set, if f is an element of HA(M, N) (Y) = I-IomA(M, Nz), and if U is a G-set, then

It is then clear that 7-~A(M, 0)z( f ) is a morphism of Mackey functors from M to N{., and since f and 0 are morphisms of A-modules, it is actually a morphism of A-modules. So there is a map 7-LA(M, O)y from 7-~A(M, N)(Y) to (HA(M, N')(Y), which defines a morphism of Mackey functors from 7{A(M , N) to 7-~A(M, N'). In other words, the Mackey functor 7-LA(M, N) is a sub-functor of ~ ( M , N), which is moreover invariant under 7-t(M, 0) if 0 is a morphism of A-modules.

So the bimodnle M defines a functor from A - M o d to B - M o d . A natural question is then to look for a left adjoint, as in proposition proposition 1.10.1. This question leads to the following definition:

Def in i t ion : Let A and B be Green [unctors for the group G, and M be an A-module- B. If N is a B-module, and X is a G-set, I set

6.6. THE FUNCTORS M | N 147

where J is the R-submodule generated by the dements

M.(f)(rn) @ n' - m | N*(f)(n') for f : (Y, ~5) ~ (Y', qS'), m E M(Y), n' E N(Y')

M*(f)(m') | n' - rn | N.(f)(n) for f : (I/, 4') --+ (Y', 4/), n / E M(Y'), n E N(Y)

r - | M(Y) , b B(Y), C N(Y)

L e m m a 6.6.1: T h e projec t ion M~)N(X) --+ M@BN(X) turns M@BN into a M a c k e y functor , and t h e m o r p h i s m A -+ A~)B turns it into an A - m o d u l e , quot i ent of M@N. M o r e o v e r , t he c o r r e s p o n d e n c e N ~-~ M@sN is a functor f r o m B - M o d to A - M o d .

Proof: To say" that the structure of Mackey functor of M@N is compatible with the projection is equivalent to say that the elements

generate a sub-Mackey functor of M@N. But if f : X ~ X ' is a morphism of G-sets, then

(m~N) . ( f ) ( [m.b | hi(Y#) - [m | b.n](y,~)) = [rn.b| n](y,f~) - [m @ b.n](y,/~)

Similarly, if now f is a morphism fl'om X ' to X, and if the square

a

y ' , Y

X ' I X f

is cartesian, then

. . . . [M*(a)(ra.b) | N*(a)(n)](v,,+,) -[M*(a)(m)|

But the proof of lemma 5.2.2 shows more generally that

m*(a)(m.b) = m*(a)(,n).B*(a)(b) X*(a)(b.,~) = B*(a)(b).N*(a)(n)

which proves that M Q s N is a quotient Mackey functor of M~N. To prove that MQBN is a quotient A-module of M~N, I must check similarly

that the elements (E) generate an A-submodule of MQN. But the product of the element a E A(Z) by the element [m| hi(y,,) of M ~ N ( X ) is given by

Since

148

I have

In part icular

, , ] CHAPTER 6. CONSTR[ CTION OF GREEN FUNCTORS

c~ x [m | n](Y,r = [(c~ x rr~) @ N* ( zyY) (n)](ZxY, ldxr

a x [m | b'n](v'r = [(a x rn) (3 N* ( zY)

= o, x . - ~ x (b . . . .

zyzy

It follows that

• ([~.b e ~l(Y,~)- In | b.~](y,~ 0 . . . .

= [((Ol X n Z).B. (~)(b,)@?Z](ZY, id;3)--[(o~ • , @B* (?)(b,.N* (~)(~)](ZY, Id4)) which proves that the A-module structure of M@eN passes down to its quotient M@BN.

It remains to observe that the construction N ~ M~)BN is functorial in N: if f : N --+ N ' is a morphism of B-modules, I can set for a G-set (Y, r over X

The map Mc~sf is well defined, because

(M@Bf) ([m.b|162174 = [m.b| fv(n)l(y,~, ) - [m| fv(b.n)l(v,r . . . .

. . . . [re.b| fy(n)](y,r - [m | b.,fv(n)](y,~)

The lemma follows. ,,

Moreover

Now the proof of lemma 5.2.2 shows that


Propos i t ion 6.6.2: Let A and 13 Green functors for the group G. If M is a n A-module-/3, if N is a /3-module and P an A-module , then there are i somorphisms of Mackey functors

~A(M+,,N, P) ~_ n,, (N, ~A(M, P))

which are moreover natura l in M,N and P.

In par t icular , the functor N ~-+ M@BN is left adjoint to the func tor P ~-+ n (M, P).

Proof : I have the following diagram of Mackey functors:

UA(M6.N,P)

1 1 (M6N, P) - ,

(7

where the bottom isomorphism ~r comes from proposition 1.10.1. I need to show that this isomorphism maps the left column to the right one. Let X be a G-set, and f be a morphism of Mackey functors from MQN to Px. It follows from proposition 1.8.3 that f is determined by bilinear maps

]y : M(Y) • N(Y) + P,:(Y) = P(YX)

More precisely, the image under f of the element [m | n](u,~) of MQN(Y) is given by

To say that f passes down to the quotient Mc~BN(Y) is equivalent to say that

for any G-set (U, r over Y and b E/3(U). This is also equivo,lent to the special case (U, r = (Is, Id), i . e .

= j (m, b.n) (6.5)

On the other hand, the element corresponding to f under (7 in

~(N, 7-t(M,P))(X) = HomM=&(a)(N, 7-t(M,P)x) = HomM~&(a)(N,'H(M, Px))

can also be defined using f : if Y is a G-set, and if n E N(Y), then (r(f)(n) is the morphism from M to (Px)z = Prx defined for a G-set Z and m E M(Z) by


Now cr(f)(n) is a morphism of B-modules if and only if for any G-set V and any b r B(V), I have

~(f)(b x n) = b • ~r(f)(n)

But

whereas

(6.7) Moreover, the expression of the product • from the product " . ' , and lemma 5.2.2 show that

Now equality of (6.6) and (6.7) becomes

f~v~ (M. (7)(~,, ~ (7)~, .~-(7)(~,) . . . .

It is now clear that (6.5) implies (6.8). Conversely, if I set

f z ,g(m,n) = cr(f)(n)z(m) �9 P ( Z Y X )

I know that f is a bilinear morphism from M, N to Px, and that I can recover f by the formula

Equality of (6.6) and (6.7) can also be written

Then i f Z = V = Y , Ihave

Since f is bifunctoriat, this is also

By a similar computation

= = N* Y ( b x n ) . . .

YY YY YY

which proves the equivalence of (6.5) and (6.8), and the isomorphism

7~(M6BN, P) -~ ~ . (X, ~(M, P))

It remains to prove that this isomorphism maps 7-gA(M~BN, P) into ~ e (N, 7-/a (M, P)) .

Let X be a G-set, and f be a morphism from MQBN to Px, determined by

Then f is a morphism of A-modules if and only if for any G-set Z and any a C A(Z) I have

s y(o x | <(u,+,)=a• sy(i | which can also be written

(

The right hand side is equal to

P* \ zr ]

and f is a morphism of A-modules if and only if

fzu(a • m, CB,Z • n) = a • ]u(m, n) (6.9)

It corresponds to f under c~ an element of HomB(N,~(M, Px)) , which maps n E N(Y) on the morphism from M to (Px)Y = PYx defined by

This morphism is a morphism of A-modules if and only if for any G-set U and any a E A(U) I have

. . . . fuzY ( M * ( u z Y ~ ( a • . . . .

. . . . f u z y ( a x M * ( ? ) ( m ) , N * ( u g Y ) ( n ) ) (6.10,

It is then clear that (6.9) implies (6.10), since

Conversely, the image of the left hand side of equation (6.10) in the case Y = Z under map P* (~;~) gives the

"" = a > < P * ( Y ) f Y 2 ( M * ( Y l Y 2 ) \ YI \ Y2 f

which, thanks to condition iii) of proposition 1.8.3, can also be written

The image of the right hand side is

. . . . • •

and equation (6.10) is equivalent to equation (6.9), which proves the isomorphism

nA(M+sN, P) ~- 7-@ (N, hA(M, P))

Those isomorphisms are deduced from those of proposition 1.10.1. So they are natural in M, N and P. Now evaluation at * gives the claimed adjunction property, and completes the proof of the proposition. �9

Chapter 7

A Morita theory

7.1 C o n s t r u c t i o n of b i m o d u l e s

Let A, B and C be Green flmctors for the group G. If M is an A-module-B and N is a B-module-C, then M is in part icular an A-module, and N is a C~ so M Q N is an A~C~ that is an A-module-C. It is easy to check that this s tructure passes down to the quotient and turns M ~ B N into an A-module-C. Moreover

P r o p o s i t i o n 7.1.1: L e t A, B, C b e G r e e n f u n c t o r s for t h e g r o u p G.

1. I f M is an A - m o d u l e - B , i f N is a B - m o d u l e - C , a n d i f P is a C - m o d u l e - D, t h e n t h e r e a r e i s o m o r p h i s m s of A - m o d u l e s - D

( M ~ B N ) ~ c P ~- M ~ B ( N ~ c P )

w h i c h a r e m o r e o v e r natural in M, N a n d P .

2. If M is an A - m o d u l e - B , then t h e r e a r e i s o m o r p h i s m s of A - m o d u l e s - B

A d A M ~-- M M Q B B ~ M

w h i c h a r e n a t u r a l in M.

P r o o f : The first assertion follows from proposition 5.3.2 and from the fact that the isomorphisms

( M 6 N ) ~ P ~_ M ~ ( N ~ P )

are compat ible with taking quotient, and are natural in M, N, and P.

The second assertion follows easily by adjunction fl'om the fact that

HomA(A, Mx) = M x ( . ) = M ( X )

which clearly implies 7-~A( A , M) ~_ M. �9

Similarly, if P is an A-module-C, then ~A(M, P) is a B-module-C: if X, Y and Z are G-sets, if b E B(X) , i f f E ~A(M, Pz), and c ~ C(Z), I can define b • f • c on the G-set U by

(b • f • c)u(m) = f u x ( m • b) • c E P ( U X Y Z ) = Pxyz(U)

154 CHAPTER 7. A MORITA THEORY

7 . 2 M o r i t a c o n t e x t s

Those constructions of bimodules over Green functors lead to try to generalize the notion of a Morita context(see Curtis-Reiner [4] 3.53), in the following way:

Defini t ion: Let A and B be Green functors for the group G. A Morita context (M, N, O, @) for A and B consists of an A-module-B M and a B-module-A N, and morphisms of bimodules ~ : M@sN ~ A and r : N@AM ~ B, which are balanced in the sense that the bilinear morphisms ~P and o2 associated to them are such that for any G-sets X , Y and Z

m • (gv, z(n,m') = ~x ,y(m,n) x m' Vm C M(X) , n E N(Y) , m 'E M(Z)

n • ~y,z(m,n') = ~x ,y(n ,m) x n' Vn E N(X) , m c M(Y) , n' E N(Z)

I will say that (M, N, ~, q2) is a surjective Morita context if ~ and �9 are surjective.

L e m m a 7.2.1: Let (M, N, O, ~) be a Mori ta contex t for A and B. Let (U, r be a G-set over X, and (V, ~b) be a G-set over Y. Then if m C M(U), n C N(U), p e M(V) and q e N(V)

[m@n](v,r ~Y([p@q](v,r = r162 x [p|162 in (M@BN)(X x Y)

Proof: The equation relating ~ and ~ is

Setting P = [m|162 x Oy([P|162

I have

P : [ (m• CB,Y)@ ( n • A.(~)A* (v;)(~g'v(P'q))](UxY,r

On the other hand

n x A . ( ~ ) A * vvV ~v 'v(P'q)=N* ur uvv nxr y(p,q) . . . .

. . . . N, ~ ( v ) ~ w

As (r x Id) o (Id • r = r • r I have also

~ ( ~ ) u~v (v•

Moreover

7.2. MORITA CONTEXTS 155

whereas expressing the product x using the product ".", I have

As the map N* (u;;) is compatible with the product ".", it is also

s o

Similarly, if I set O : ~x ([m | n](v,~/) x [p | q](v,~,)

I have

SO

Moreover

A.(r uuU ~v,u(m,n)• A. r uuv '

. . . . M. r uuv m x (gv, v(n,p)

As M * ( U V ) . . . .

. . . . M * ( Z ) (M'(U':12V)(m).B'(UluU~vV)~u,v(n,p)) . . . .

I have also

A.(r (u u ) Spu,U(m,n) x p : M. r (M* (uv) (m) .~u ,v(n ,p ) )

Finally, I have

Q = M* (m).~g,y(n,p)) | r (SA,X x q) (vxy,r215162

and as gA.X xq=N*(~:,~,)(q),Ihaveaiso

so P = Q in MQBN(X • Y), and the Iemma follows. �9

156 C H A P T E R 7. A M O R I T A T H E O R Y

P r o p o s i t i o n 7.2.2: Le t A and B be G r e e n f u n c t o r s for t he g r o u p G, and let (M, N, ~, qJ) be a M o r i t a c o n t e x t for A and B.

�9 I f ~o is su r j ec t ive , t h e n ~ is an i s o m o r p h i s m .

�9 I f ~P. and ~o are su r j ec t ive , t h e n �9 and q2 are i s o m o r p h i s m s . M o r e o v e r in t h a t case

1. T h e m o d u l e Ma is a p r o g e n e r a t o r for A - M o d and for M o d - B , and t h e m o d u l e Na is a p r o g e n e r a t o r for B - M o d and for M o d - A .

2. T h e r e are i s o m o r p h i s m s of b i m o d u l e s

N ~" 7-(A(M,A) ~_ ~Bop(M,B) M ~_'HB(N,B) ~_ "HAop(N,A)

3. T h e r e are i s o m o r p h i s m s of G r e e n f u n c t o r s

A ~- ~ B o , ( M , M ) ~- (']-{B(N,N)) ~ B ~-- 7-{Aop(N,N) "~ (7-~A(M,M)) ~

4. T h e f u n c t o r s P ~ N | P and Q ~ M ~ B Q are m u t u a l inverse equ iva l ences of ca t egor i e s b e t w e e n A - M o d and B - M o d .

P r o o f : As q5 is a morphism of A-modules-A, and as A is generated as a bimodule by cA, if ~ , is surjective, then eA is in its image, and then ap is surjective. To prove the first assertion, it suffices then to prove that q~ is injective.

Let X be a G-set. Let moreover [rei @ ni](u,,r for 1 < i < re be elements of M Q s N ( X ) such that

) ~X rei @ ni](U,r = 0

By hypothesis, the map ~. is surjective: let [pj | qJ](5,r for 1 < j <_ p be elements of M@N(e) such that

(the map r is then the unique map from Vj to *). Setting

m

v = F_.[ .~ | ~d(v,,~,) i=1

it follows from the previous lemma that

m p

v = v X eA = E ~-~[rniOni](ui,r x r ([pjO qj](v,r . . . . i= l j = l

i=1 j = l

j = l

7.2. MORITA CONTEXTS 157

which proves that r is injective, hence that it is an isomorphism.

Now if ~o and q*o are surjective, then ~ and �9 are isomorphisms (indeed the 4-tuple (N, M, ~, (I)) is a Morita context for B and A, and ~ and �9 play symmetric roles). Assertion 4) follows then fi'om proposition 7.1.1, since

N@A(M@BP) ~-- (N@aM)~BP ~- B@BP ~- P

L e m m a 7.2.3: Let M and N be Mackey functors. If X and Y are G-sets, then there are i somorphisms of Mackey functors

(M+N)xy ~- M.v+Ny

which are natural in M, N, X and Y.

P r o o f : This follows from the fact that

7~(Mx, N) ~_ 7-t(M, Nx )

naturally in M, N, and X. Then

(Mx Ny, P)) Px)y) Moreover, for any G-set Z

~(M, Px )y(Z) = ~(M, Px )(ZY) = HomM~k(o)(M, (Px)zz) . . . .

. . . . HomM~k(a)(M, PzYx) = HomM~ck(a)(M, (PYx)z) = "H(M, Pyx)(Z)

This gives the isomorphism 7-{(M, Px)Y ~- ~(M, PYX), and then

^ / r . . . .

U(MGN, Pxy) -

and the lemlna follows. "

R e m a r k s : 1) The isomorphisms of the lemma can be stated precisely. If Z is a ( o ) G-set, then a G-set over Z X Y is determined by a map ~(~)~(~)~(~) from U to ZXY.

I define then a map from (M@N)xr(Z) = M@N(ZXY) to (Mx@Ny)(Z) by

(u,-~)

which makes sense because M, ~(,~) (m) r M(UX) = Mx(U), and N, ~(~) (n) E

N(UY) = Xv(U). The inverse isomorphism is defined by

2) The case X = �9 gives in particular the isomorphism

(M(~N)x ~- M+(Xx) ~- (Mx)+N

and the naturality of this isomorphism shows that if M is an A-module-B, then for any B-module N

(M+BN)x ~- M+B(Nx) ~- (Mx)+~X P r o o f of p r o p o s i t i o n 7".2.2 (pa r t 2): Assertion 1)follows from the previous lemma, because an equivalence of categories preserves modules of finite type, and because Ma is the image of Be under the functor P ~ M@sP, since

m~ ~- (M@BB)a ~- M@B(Ba)

Concerning assertion 2), I observe first that evaluating at the set �9 the isomorphism

~ , (N, 7~A(M, A)) ~- 7%(M@uN, A)

gives HomB (N, 7-ta(M, A)) ~- HomA(M@BN, A)

Then the morphism (I) gives by adjunction a morphism 0 from N to 7-{a(m, A), very easy to describe: if n E N(X), then the image of n in ttomA(M, Ax) is the morphism defined by

m E M(Y) ~ 3Pv, x (m,n) E A(YX) = Ax(Y)

The morphism O is injective: the element n E N(X) is in the kernel of Ox if and only if for any G-set Y and any element m E M(Y), I have

~ v , x ( ' ~ , ~ ) = 0

But as �9 is surjective, there exists G-sets ~ and elements p~ E N(Y~) and q~ E M(YQ, for 1 < i < n, such that

i YlYl

Then

The expression inside hooks is also

Pi • ~.y,,x(qi, r~)

So it is zero for all i, which proves that n = 0 and that O is injective. To prove that O is also surjective, I must prove that if f E HomA(M, Ax), then

there exists an element n E N(X) such that for any G-set Y and any m E M(Y), I have

fy(m) -= ~y,x(rn, n)

If such an element exists, keeping the previous notations, I have

= pi x ~y,,x (qi, n) . . . . i \ yiYiX ]

7.2. MORITA CONTEXTS

i kyiyix/

Let then

i \ y i y i x /

Then for any Y and any rn E M(Y) , I have

' i ' \yiYiX/

As (~ is a bilinear morphism, it is also

{ \ yx I \yy;y,:z eY'h~x(rn'pi x fY,(qi))

As ~5 is a morphism of modules-A, I have

(~<<~x(rn,pi • fy'~(qi)) = ~y,y,(rn,pi) x fY,(qi)

and as f is a morphism of A-modules, it is also

o r

Finally

m;2 •

As f is a morphism of Mackey functors, it is also

Expression inside hooks is equal to

r n x B * ( ~ ) B * ( Y i )

so I have

159

which proves that O is an isomorphism. It is easy to see that it is an isomorphism of bimodules. The other isomorphisms of assertion 2) now follow, by switching the roles of A and B, or replacing them by their opposite.

Assertion 3) is proved by observing that

"]'{A(A, A) ~_ A ~


Moreover, if I denote by F the fimctor N~A-- from A - M o d to B - M o d , the isomorphisms

(N+aP)x show that F(Px) ~- F(P)x. As F is an equivalence of categories, then

HomB ( F ( P ) , F(Q) = HomA(P, Q)

for any P and Q. It follows that

. . . . HomA(P, Qx) = HA(P, Q)(X)

and it is easy to see that those isomorphisms induce isomorphisms of Mackey functors

"HB (F(P), F(Q)) "Ha(P, Q)

which are moreover compatible with the product 5 . Thus for any A-module, the Green functors "HB(F(P), iv(P)) and "HA(P, P) are isomorphic. Now for P : A, this gives

"HB(N, N) ~_ A ~

and assertion 3) follows. This completes the proof of the proposition. �9

7.3 C o n v e r s e

The previous proposition has a converse:

P r o p o s i t i o n 7.3.1: L e t A b e a G r e e n f u n e t o r for t h e g r o u p G, and M b e an A - m o d u l e such t h a t Mu is a p r o g e n e r a t o r of A - M o d . L e t N = "HA(M, A) and B = ('HA(M,M)) ~ T h e n t h e r e e x i s t s a s u r j e c t i v e M o r i t a c o n t e x t

(M, N, ~, ~ ) for A and B.

P r o o f : Let X and Y be G-sets. If rn r M(X) and f E B(Y) = Homa(M, M r ) , let

m • f = fx(m) E My(X) = M(XY)

It is easy to see that this definition turns M into an A-module-B. Then N = "HA(M, A) is a B-module-A. As moreover

"HA(M+sN, A) ~_ ~ s (N, "HA(M, A)) = "HB(N, N)

the ident i ty map of N gives a morphism of A-modules from M~BN to A, which can be described as follows: if X and Y are G-sets, let ~x,Y be the bilinear map from M(X) • N(Y) to A(XY) defined by

m E M(X), r E N(Y) = HomA(M, Av) ~ e x ( m ) E At(X) = A(XY)

It is easy to check that those maps define a bilinear morphism from M, N to A, associated to a morphism ~ from M+BN to A, which is a morphism of A-modules-A.

7.3. CONVERSE 161

Conversely, there is a bilinear morphism t) from N, M to B defined as follows: if X and Y are G-sets, if f E N(X) = HomA(M, Ax) and m r M(Y) , then ~ x , y ( f , m ) is the morphism from M to Mxv defined for a G-set Z and m E M(Z) by

~ x . v ( f , m) : 7Y~l ~ M(Z) ~ f z (m, ) • ro E M ( Z X Y ) = Mxv (Z )

The morphism from N@M to B associated to it passes down to the quotient, and defines a morphism from N@AM to B, which is a morphism of B-modules-B.

Before proving the proposition, I will give an equivalent formulation of the hypothesis on M, independent of fh

L e m m a 7.3.2: L e t A b e a G r e e n f u n c t o r for G, and M b e an A - m o d u l e . T h e f o l l o w i n g c o n d i t i o n s are equ iva l en t :

1. T h e m o d u l e Mn is a p r o g e n e r a t o r .

2. T h e m o d u l e M is a f in i t e ly g e n e r a t e d p r o j e c t i v e m o d u l e , and t h e r e e x i s t s a G-se t X such t h a t A is a d irec t s u m m a n d of Mx.

3. T h e r e e x i s t s G-se t s X and Y such tha t A is a d i rec t s u m m a n d of Mix and M is a d i r e c t s u m m a n d of Az .

Proo f i First I observe that if X and Y are G-sets such that X divides Y in CA, then Mx is a direct summand of M r : indeed, say that X divides Y in dA is equivalent to say that Ax is a direct summand of Av. But Mx is isomorphic to "HA(Ax, M), which is a direct summand of "HA(Ay, M) ~- Mz.

So if Ma is a progenerator, as �9 divides f~ in CA, it follows that M is a direct summand of Ma, so M is a finitely generated projective module. Moreover, there exists a set I and a surjective morphism

p : @ M# ) ~ A iE1

from a sum of copies of Ma to A. In part icular, there is a finite subset ,1 C_ I and /I/t(J) ' o" elements rnj E - - a ( ) , for j E J , such that.

~A = P.(~'~ mj) j~_J

Then the image B of the restriction of p to | j) is an A-submodule of A, and B(*) contains CA. This proves that there exists an integer n such that A is a quotient, hence a direct summand, of nMfl ~- M,~fl. Thus 1) implies 2).

If now hypothesis 2) holds, then as An is a progenerator, and as M is a finitely generated project ive module, there exists an integer n such that M is a direct summand

of nAG ~- A,~a. So 2) implies 3). Final ly if 3) holds, then M is a finitely generated projective module, because it

is a direct summand of the projective module Az. Then Mn is a direct summand of (Az)a ~- Aay, and Ma is a finitely generated projective module. Moreover, the module An is a direct summand of (Mx)a ~- Max. As f /X divides a mult iple of f~ in CA, it follows that Aa is a direct summand of a direct sum of copies of Ma. As Aa is a progenerator, so is Ma, and this proves the lemma. �9


P r o o f o f p r o p o s i t i o n 7.3.1: I must prove that if Mn is a progenerator, then 02 and q~ are surjective. It is equivalent to say that CA lies in the image of 4p,, and eB in the image of ~o.

I will show that if there exists X such that A is a direct summand of Mx, then 02 is surjective, and that if there exists Y such that M is a direct summand of A r (which is equivalent to say that M is projective and of finite type), then �9 is surjective.

So let X be such that A is a direct summand of Mx. Then there exists an element ct E HomA(A, Mx) and an element fl E HomA(Mx,A) such that

flo a = IdA

But I have seen that HomA(A, Mx) ~- M(X). Let m E M(X) be the image of c~ under that isomorphism. Then c~ is defined on the G set Z by

a E A(Z) ~ a x m E M ( Z X ) = Mx(Z)

On the other hand

HomA(Mx, A) ~_ HomA(M, Ax)

Under this isomorphism, the element fl maps to 5 E Homm(M, Ax) = N(X) , and I know that for any G-set Z and any m' E Mx(Z) = M(ZX) , I have

Now say that fl o c~ = IdA is equivalent to say that for any Z and any a E A(Z)

which can be writ ten as

This is equivalent to

a = a • X )

Then let

= [m | ? ] (x , ( : ) )E M~cBN(.)

The image of ~- under 02~ is precisely

= = 7 x ( m ) = c A X X X32

and this proves that 02~ is surjeetive.

Similarly, if M is a direct summand of My, then there exists a E HomA(Ay, M) and fl E HomA(M, Ay) such that c~ o/~ : IdM. But

HomA(Ay, M) "~ M(Y)

7.4. A REMARK ON BIMODULES 163

and c~ is determined by an element m E M(Y) , such that for a ff Ay(Z) = A (Z Y )

Now say that c~ o ~ = IdM means that for any G-set Z and any m' E M(Z) , I have

Let r ' = [3 | m](v,(~)) E N+AM( . )

The image of r ' under q2 is the element of HA(M, M)( . ) = EndA(M) defined by

which can be evaluated at a G-set Z by

It follows that ~(r')(m') = m', so q2(r') is the identity of M, that is the unit e~a(M,M). This proves that ffJ is surjective, and completes the proof of the proposition. �9

7.4 A r e m a r k o n b i m o d u l e s

If A is a Green functor for G, theorem 4.3.1 states that evaluation at fl is an equivalence of categories from A - M o d to A(g/2)-Mod. If B is another Green functor, and M is an A-module-B, then M ( a ) is an A(a2)-module, and a B~ As B~ s) ~_ B(~s) ~ the module M(~ s) is a module-B(fF). However, the module M ( a ) is not in general an A(aS)-module-Btas): indeed, if a E A(aS), if m E A(f/), then

\ CO 1 / \COlt~20.JS

Then if b E B(a 2)

\ cos / \colco:cos/


A. (colco2co3co4~ A* (co ico2 tM3cd4) (~x . l . x b ) \ colco3co4 / \t~lco2co20.)3t.04

~1&2~3

CO1~2~1~3/

)

\COlcolco2/

As the square



and it follows that

(a oa ~ ) oa b . . . .

\ /.,3 3 \ OJ 1 r 2/.,U 2 (.O 1 t.O3 /

A similar computat ion gives


As the square

C2&) f~3 , fp

\r /

is ca.rtesian, I have

\CUI(.U2CU2 \ 0.)10J2CJ 4 /

SO

a Of~ ( m 0 a b) ....

\ CU1CU2 / \O-~ICO2CU3~2 /

\ ~,g 1 \ Cgl0g 2 / \~,Ul~g20.)3Cd 2 / kt.U1CU20J3t~)3~.U 4 /

There is no obvious reason for which this expression should be equal to (E), so M ( 9 ) is not a bimodule in general. This is because (AQB~ e) is the tensor product of A(fF) and B(fF) ~ over the non-commutative algebra b(92) (i.e. Mackey algebra) as shown in the next proposition:

P r o p o s i t i o n 7.4.1: Let A and B be G r e e n f u n c t o r s for t h e g r o u p G. I f fl is a G - s e t , t h e n t h e m o r p h i s m f r o m b to A (resp . f r o m b to B) t u r n s A(f~ 2) ( r e sp . B ( f~ ) ) in to a module-b(Pt e) ( resp . a b(f~Z)-module), a n d

(A+B)(f~ 2) _~ A(f~2)(~b(a2)B(f~ z)

as R - m o d u l e s .

7.4. A REMARK ON BIMODULES 165

Proof : Lemma 7.2.3 shows that

(A@B)a2 -~ Aa@Ba

as Mackey fnnctors. Evaluation at the trivial G-set gives

(m~B)a2(,) = (m~B)(a 2) = (da6Ba)(~

and by definition of tensor product of Mackey functors

(do@Sn)(*) = da(f~) | B~(f~) = d(f~ 2) | B(f~ 2)

so proposition follows. �9

The algebra structure of A(f~ 2) | B(~ 2) can be recovered using the isomorphism

r B(f~2) | A(f~2) ~- (B@A)(f~2) ~- (A@B)(~22) -~ A(f~2) @b(n~)B(f12)

If a and a' are in A(f~), and b, b' in B(f~2), and if

r @ a') = ~ ai" @ bi" i

then (a | b)(a' @ b') = y~(aai") | (bi"b')

i

Chapter 8

Composition

8.1 B i s e t s

In [2], I have studied the following problem: if G and H are groups, what kind of functors F from G-se t to H - s e t induce by composition a functor from Mack(H) to Mack(G)? It seems natural to ask that the functor F transforms a disjoint union into a disjoint union, and a cartesian square into a cartesian square. The functors from G-se t to H - s e t having those two properties can be completely classified up to isomorphism by the (isomorphism classes of) H-sets-G:

D e f i n i t i o n s : Let G and H be groups. An H-set-G is a set X with a left H-action and a right G-action, which commute, i.e. such that if g E G, h C H and x E X

h (x g) -- (h.x) g

f iG, H, and K are groups, if X is an H-set-G, and Y a K-set-H, I denote by Y OH X the K-set-G de/~ned by

Y O H X = { ( y , x ) E Y x X I V h E H , y h = y ~ 3gE G, h . x = x . g } / H

where the action of H is given by (y, x).h = (y.h, h-ix) . The action of K and G on Y OH X is given

k.(y, x)g = (k.y,x4)

If X is an H-set-G, and if G and H are clear from context, I will also say that X is a set with a double action, or biset for short.

Wi th those notations, if U is an H-set-G, and X is a G-set, or G-se t - ( l ) , then U o a X is an H-se t - ( i ) , that is an H-set . This construction gives a functor from G-se t to H - s e t , that I denote by U oG - . The precise s ta tement proved in [2] is then

T h e o r e m 8.1.1: L e t G a n d H b e f in i t e g r o u p s .

* I f F is a f u n c t o r f r o m G-se t to H - s e t w h i c h t r a n s f o r m s d i s j o i n t u n i o n s i n to d i s j o i n t u n i o n s a n d c a r t e s i a n s q u a r e s in to c a r t e s i a n s q u a r e s , t h e n t h e r e e x i s t s an H - s e t - G U, u n i q u e u p to i s o m o r p h i s m of H - s e t s - G , such t h a t F is i s o m o r p h i c to t h e f u n c t o r U oa - .

168 CHAPTER 8. COMPOSITION

�9 C o n v e r s e l y , if U is an H - s e t - G , then the functor U o a - t r a n s f o r m s dis- jo int unions into disjoint unions and cartes ian squares into cartes ian s q u a r e s .

I also proved that in these conditions, the set U induces a functor from Mack(H) to Mack(G), defined by composition, and denoted by

M ~ M o U

If M is a Mackey functor for H, the Mackey functor M o U is defined over the G-set X by

(M o U)(X) = M(U o~ x )

If f : X --+ Y is a morphism of H-sets, then U oG f : U oa X --+ U oa Y is defined by

x): s(x)) and then

( m o U),( f ) = M.(U oa f) (M o U)*(f) = M*(U oa f)

E x a m p l e s : 1) If H is a subgroup of G, and if U is the set G, viewed as an H-se t -G by mult ipl icat ion, then the functor U oa - is tile restriction functor [rom G-se t to H-se t . If V is the set G viewed as a G-set-H, then V OH -- is the induction functor from H - s e t to G-se t . The functor M H M o U is then the induction functor for Mackey functors, and the functor N ~ N o V is the restriction functor for Mackey functors. 2) If N is a normal subgroup of the group G, if H = G/N, let U be the set H, viewed as an H-set-G, the group G acting by the projection G ~ G/N, and let V be the same set viewed as a G-set-H. Then the functor U oa - is the "fixed points by N" functor. It is easy to identify the functor M H M o U as the inflation functor from Mack(H) to Mack(G), defined by Th6venaz and Webb (see [14], [15]). The functor V oa - is the inflation functor from H - s e t to G-set . The functor N ~ N o V is the "coinflation" functor for Mackey functors (that Th6venaz and Webb denote by/3 ! in [15] Lemma 5.4, and I denote by Pg/N in [2]). 3) If U is an H-set-G, then U oa * ~- U/G. 4) If U is a G-set-G, and X a G-set, then

U o a X = {(u,x) E U • u . g = u ~ g.x=x}/G

8.2 Composi t ion and tensor product

Def in i t ion: Let G and H be groups, and U be an H-set-G. If M and N are Mackey functors for H, and if X and Y are G-sets, I denote by r U the map X,Y

YxY : M(U o~ X) • W(U oG Y) -+ (M+N)(U o~ (X • Y))

defined by

8.2. COMPOSITION AND TENSOR PRODUCT 169

L e m m a 8.2 .1: T h e m a p s TU, y f o r m a b i l i n e a r m o r p h i s m f r o m M o U, N o U

t o ( M @ N ) o U.

P r o o f : The maps TUg being clearly bilinear, all I have to check is their bifunctoriality. So let f : X --+ X ' and g : Y --+ Y' be morphisms of H-sets. If m E M(U oa X) and n E N(U oa X~), setting M' = M o U and N' = N o U, I have

t, x' ) M~*(f)(m) | m'* m~(g)(n) \ yt ) (U~

As the square f x l d

X x Y' ) X ~ x Y'

(x' G l,x X ) X ~

f is cartesian, its image under U oc - is also cartesian, and then

M* (U oa (x'Y'~

It follows that

~x,,~, (M:( f ) (~) , N:(g)(~)) . . . .

. . . . M . ( U o a ( f x I d ) ) M ' * ( m ) | Y' ]N'.(g)(n) . . . . (Uo~(X'xZ'),H)

= N'(g)(,,) . . . .

\ Y' } (UodX•215

\ y' ] t, y' ] J (UodXxY,),Uo~(S• But for the same reason, I have

t ,y')

This gives

TX, y, ( M : ( f ) ( m ) , N ' (g)(n)) . . . .

[ ( ? ....

\ Y' ) ) J(Uoa(XxZ),Uo~(Sxg))


proving finally that

so T is covariant. Now if rn' r M'(X') and n' E N'(Y'), then

r v {M'*tr'(rn '' N"(g) (n ' ) )= X , Y k ~ J J k ]~ �9 ""

On the other hand, as the square

U oa ( f • g) U o a ( X x Y ) , U o c ( X ' •

,d I l id u oo ( x • Y) , u oa (x ' • Y')

u oG (f • 9) is t r ivial ly cartesian, I have

. . . . M'*(f x g)M'* \ x' J (rn') | N'*(f • g)N'* \ Y' ,] (n') (Uoa(Xxz),/d)

As moreover

f o = f (x) \ x' ] o ( f x g) go = g(Y) \ Y' J o ( f x g)

I have "rUz(M'*(f)(m'),N*(g)(n')) = ( (M~N) o U)*'rx, z,(rn',n')

and r is also contravariant. The lemma follows. "

8.3 C o m p o s i t i o n and Green functors

When A is a Green functor for the group H, the product on A gives a morphism from A ~ A to A, hence a morphism from (A+A) o U to A o U. Composing this morphism with the morphism

(A o U)~(A o U) ~ (AQA) o U

deduced from T gives a morphism

(A o U)6(A o U) -+ A o U

8.8. COMPOSITION AND GREEN FUNCTORS 171

I view it as a product on A o U, that I denote by • It is natural to ask if A o U is a Green functor. To see this, I will first describe precisely the product: if X and Y are a-sets, if a C (A o U)(X) and b C (A o U)(Y) , the product a x v b is equal to

where the product "." on the right hand side is the product of A. Then

(~,x,y)(u,~,y) \ (~,x) ] (a) • \ (u,y) ] (~) . . . .

= A * ( ( u , x , y ) ) A . ( ( u l , x l , y l ) ( u ~ , x x , y 2 ) ~ ( a x b ) " (~, x, ~)(~, z, y) ~ (~l,x,)(~,y~) ]

whence finally

a• (~,x,y) ~(a•

This leads to the following definitions:

Def in i t ions : Let G and H be finite groups, and U be an H-set-G. I f X1, . . . X~ are G- sets, I denote by fu 1,...,x. the map from U oa (Xl,., Xn ) to ( U oGX1) X . . . • ( U Og Xn ) ,

defined by

f f A is a Green functor for the group H, I set

r = A.(pv/a)(eA) e (A o U)( .) = A(U/G)

where pv/a is the unique map U o �9 ~_ U/G to . .


a • b = A*($uz)(a x b)

where the product x on the right hand side is the product of A. The following remark will be useful:

L e m m a 8.3.1: T h e m a p (~u is in ject ive . X,Y

P r o o f : Indeed, as the square

(?) X • -~ X (x:) j l l:l

Y ~ �9

(:)

172 C H A P T E R 8. C O M P O S I T I O N

is cartesian, so is its image under U oa -

U o c ( X x Y) , U o a X

U o a Y ~ U o a .

which proves that U oa (X • Y) maps into (U oG X) • (U oa Y), precisely by the map 5 U So 5 U is injective. .. X,Y" X , Y

P r o p o s i t i o n 8.3.2: Le t G and H be finite g roups , and U be an H-se t -G. I f A is a G r e e n f u n c t o r for t he g r o u p H, t h e n A o U is a G r e e n f u n c t o r for G, for t he p r o d u c t x U a nd t he un i t CAoU.

P r o o f : The product x u is bifunctorial by construction. So I must check that it is associative and unitary.

Let X, Y, and Z be G-sets. If a E ( A o U ) ( X ) , if b E ( A o U ) ( Y ) , and if c E (A o U) (Z ) , then

. u �9 u b) x c ) = ( a x % ) • = A * ( 4 • 2 1 5 = A (~xx,.-z)(A (Sx.,,-)(a • . . .

�9 v �9 u IdvoGz)(a • b x c) A*(aUr, z ) (a x b x c) . . . . A (~Sxxy, z ) A (~SX, Y • =

5u Iduooz) o g 5 g since for (u ,x ,y , z ) E U o G ( X x Y • because ( x,Y x 5x• = x,Y,Z,

u o e u I dvocz ) ( (u , x, ( % r • Zd~o~Z) ~x x~,Z(~, x, ~, ~) = ( % ~ • y), (~, ~)) = . . .

. . . . ((~, x), (~, v), (~, ~)) = 4,~ ,~(~ , x, v, z)

On the other hand

�9 U * . . . . A (Sx , z • ( Iduocx • 5u, z ) (a • b • c) = A*(Su,z,z)(a • b • c)

since ( Iduo~x x 6~z ) o 5x,yx = 5Ux,y,z by a similar computation. Finally

a • (b x U c) = A*(Sgv,y,z)(U,x,y,z ) = (a • b) x U c

and the product • is associative. Moreover, if a �9 (A o U ) ( X ) , then

�9 v * a ) * v * I d x ) ( a ) eAoU • a A (5o,x) (A (pu/~)(e) • = = A (5 . , x )A (Pu/a •

But i f ( u , x ) � 9

(Pu/G X I d x ) o , 5 [ x ( U , X ) = pu/a • I d x ( ( u , . ) , ( u , x ) ) = (. , (u,x))

and with the usual identifications, this gives

~AoU • a = a

A similar computation shows that eAoU is a right unit for A o U, and the proposition follows. ..

8.4. COMPOSITION AND ASSOCIATED CATEGORIES 173

8 . 4 C o m p o s i t i o n a n d a s s o c i a t e d c a t e g o r i e s

Let G and H be finite groups. If U is an H-set-G, and A is a Green functor for H, then A o U is a Green functor for G. Each of these Green functors has an associated category, and the functor U oc, - defines actually a functor from Caou to CA:

P r o p o s i t i o n 8.4.1: Let G and H be f ini te groups , let U be an H - s e t - G , and A b e a G r e e n f u n c t o r for H. T h e n t h e c o r r e s p o n d e n c e w h i c h m a p s t h e G-set X to t h e H - s e t U[X] = U oa X, and t h e m o r p h i s m c~ E (A o U ) ( Y • X )

to U[a] = A.(5~,x)(C~ ) + A((U oc X) x (U oa Y)) is an R - a d d i t i v e f u n c t o r f rom CAoU to C.

Proof : I must check that if X, Y, and Z are G-sets, if a C A ( Y X ) and t3 E A ( Z Y ) , then

where o};7 is the product o for tile category CAo U, Tills equality can also be written as

A.( ~,~. )(~) O~ooy A.(~,~)(~) = A.( e~,x )(~ o~ <~) (s.~)

Let X ' = U[X], Y' = U[Y], and Z' = U[Z]. The left hand side of (8.1) is equal to

( A. \ z'x' ) \z 'y 'y 'x ' ) A*(5~'r • fu':")(/3 x a)

Z ' Y ' X ' ( z'y'y'x' X' ~ Z ' y ' 2x '

The square

(c)

U2 ~ ~18 8Y2 = Yl

Thus

The element (ul ,z , y l , sx) is in U oG (ZYX) : indeed, if t E G is such that uit = ul, then as (u l , z , y l ) E Y oa (ZY) , I have tz = z and ty 1 = Yl" As m o r e o v e r u2.s- l t . s =

ults = UlS = u> and as (u2, y~, s) E U oa (YX), I have also 8-1tsx = x, or t.sx = 8x.

o r ~ Uoc,(YX) and(z ' ,y ,x ' ) is c a r t e s i a n : i n d e e d , if ( ~ l , Z , y l ) ~ U o ( Z } )~ ( ~ 2 , y 2 , x )

Z 'Y 'X ' are such that

(=', v', v', .') = ((,,,, z), (~,, >) , (,,2, y2), (~, x))

then there exists a E G such that

174 C H A P T E R 8. C O M P O S I T I O N

Moreover

(~,z,y)(~,y,x)

and (721, yl, .$3:) = (~/1 8, 8-1yl , X) : (~2, Y2, 5C). Similarly (u, , sx ) = (u ls , x) = (u2, x) = x ~, and then

, --( )_- ~,~) ~zVy, x ( ~ ~ , ~ , , ~ ) ( ~ , , ~ ) , ( ~ , ~ l ) , ( ~ , ~ x ) (~', ' '

The map 5 U is the product of two injective maps. So it. is injective, and (C) is Z,Y,X cartesian.

It follows that the left hand side of (8.1) is equal to

A.\ z'~' ) \ (~,~,~)(~,~,x)

. . . . A . (u , z ) (u , :c ) \ ( u , z , y ) ( u , y , x )

On the other hand

zx / \ z y y x /

. . . . c ( ,, z x \ z y y x / /

which gives

(~,z,x) ] \ (~ ,z ,y)(~,y ,x) (9 • 4)

The right hand side of (8.1) is then equal to

\(~, z)(~, x)) t (~,z ,y)(~,y,x) (~ • ~)

which is the right hand side of (8.2). Moreover for any G-set X, I have

�9 . - , ~ , ~ , x ~ j ~ , . ~ j r . . . .

. . . . A. A* 'o (cA) = 1A(l~o~X)~

So I have defined a functor from CAoU to CA. It is clear that this functor is R-additive, and this completes the proof of the proposition. �9

8.5. COMPOSITION AND MODULES 175

8.5 C o m p o s i t i o n and m o d u l e s

Let G and H be finite groups, let U be an H-set-G, and A be a Green fnnctor for H. The above functor from CAoU to CA induces by composition a functor between the associated categories of representations: if F is an R-addit ive functor on CA, then the functor F o U is an R-addit ive fimctor on CAoU: if X is a G-set, then (F o U)(X) = F(U oa X). As the category of R-addit ive functors on CA is equivalent to the category of A-modules, the functor t" corresponds to an A-module N, and then the functor F o U corresponds of course to the module N o U:

P r o p o s i t i o n 8.5.1: Let G and H be f in i te groups , let U be an H - s e t - G , and A be a G r e e n f u n e t o r for H. If N is an A - m o d u l e , t h e n N o U is an A o U- m o d u l e , for t h e p r o d u c t x u de f ined by

( a , n ) E (A e U)(X) • (N o U)(Y) ~ c~ x U n = N*(5~,z)(a • n) E (X o U)(X x Y)

T h e c o r r e s p o n d e n c e N ~ N o U is a f u n c t o r f r o m A - M o d to A o U - M o d .

Proof: With the notations of theorem 3.3.5, I have to find the A o U-module N ~ = MFNoU, and to prove that it coincides with N o U as Mackey functor.

If X is a G-set, I have

X'(X) = (FN o U) (X) = r N ( ( : oa X) = N(U o~ X)

If f : X --+ Y is a morphism of G-sets, then by definition

N'.(f) = ( F N o U)( f~. r) = FN( U[ff.'])

where f.u is the element of (A o U) (YX) associated to f by lemma 3.2.3, i.e.

(x) (:/ f~. = (Ao U). f ( x )x (Ao U)* (CAoU)

In other words

((,,,x)) f a = A. \ ( u , f ( x ) , x )

then

A* (~, o)] ,4" ( (cA) . . . .

. . . . A. ( (u ,x)

= *( Y,x)(f. ) = A. 'o (E) = (U oc f ) . (~, f (x ) ) (~ , x)

SO

N' ( f ) = FN((U oG f ) . ) = N.(U oa f ) = (N o U) . ( f )

A similar argument shows that if fu,. is the element of (A o U) (XY) corresponding to f by lemma 3.2.3, then fu,. = (U oG f)*, and then

N'*(f) = FN(f u'*) = FN((U oa f)*) = N*(U oa f ) = (N o U)*(f)


which proves tha t N ' coincides as Mackey functor wi th N o U. The s t ruc ture of A o U-module of N o U is then defined as follows: if X and Y

are G-sets, if o~ E (A o g ) (x ) and n E (N o U)(Y) , then the product c~ x U n is the

e lement of (N o U)(X x Y) given by

o~• (AoU). xyy

Let

then

\ xyT/ /

. . . . A. ( ~ , , ~ , , y ) ( . ~ , y ) ] \ ( ~ , x ) ] ( , )

x ~ ~ = F N ( f l ) ( ~ ) = fl Ouo~y ,, . . . . .

Moreover

/~ X T/,= N. ( (ttltX'yl)(~2'~/2) ~ N* ((t~l'*t Ya)(?~2'y2)~ (dgx I%)

(~,=,y~)(~2,y2) '1 and g = ( (*~,=,y,)(~2,y2) ~ As the square Let f = (~l,=,y,)(~l,y=)(~2,y2)/~ \(~l,x,~l)(~2,y2)(~>y2)]"

(,,,x, ~)(~, y) u o~(xY) , (u o~ (xY)) x (u o~ r )

g

is car tes ian, I have

N* ( (ul'x'yl)(U2'Y2) I N, ( (*q'x'Yl)(u2'Y2) l . . . \(tz1,X,~ll)(tz2,Y2)(ll,2,Y2)/ ~(tzi -- , x, >)(u,, >)(u2, Y:)/

. . . . N. (~,.T, y)(~, y) (~ ,x ,y)(~,y)

and finally c~ x U n is equal to the image of c~ x n under the map

( ~ , x , y , ) X. \ ( ~ , x , y ) ( ~ , y ) o . . .

8.6. FUNCTORIALITY 177

oN'*( (~,x,y) )N.((Ul,X, yl)('U,2, Y2))_~ "'" (72, X, y)(t t , y) k (721, X)(2s Y2) ) ' ' "

which proves the claimed formula

8.6 Functoriality

If G and H are finite groups, and U is an H-set-G, then I have a functorial construction X ~ U oa X from G-se t to H-se t . This construction is not quite functorial in U: if f : U --+ V is a morphism of H-sets-G, there is in general no associated morphism U oa X --+ V oa X: this is because if (u, x) E U oa X, i.e. if the right stabilizer of u is contained in the left stabilizer of x, generally, the right stabilizer of f(u) is not.

I have studied this question in [2], and showed that it is natural to ask moreover that f is injective when restr icted to each orbit of G (or equivalently to ask that the right stabilizer of f(u) is equal to the right stabilizer of u, for any u C U). Then, there is a morphism of functors f oa - from U oa - to V oa - defined on the G-set

X by

In those conditions (see [2] prop. 10 and 11), if M is a Mackey functor for H, then M o U and M o V are Mackey functors for G, and the morphism f induces two morphisms of Mackey functors (denoted by jr. and f* in [2], but differently here to avoid confusion): a morphism Mf from M o U to M o V, and a morphism M / from M o V to M o U. Those morphisms are defined for a G-set X by

Mf,x = M . ( f oa X) : (M o U)(X) ~ (M o V)(X)

M~ = M*(f oc X ) : (M o V)(X) ~ (M o U)(X)

With those notations:

P r o p o s i t i o n 8.6.1: L e t G a n d H b e f in i t e g r o u p s , a n d l e t A b e a G r e e n f u n c t o r for H. L e t m o r e o v e r U a n d V b e H - s e t s - G , a n d f : U --* V b e a m o r p h i s m of H - s e t s - G , w h i c h is i n j e c t i v e on each r i g h t o r b i t .

�9 I f A is a G r e e n f u n c t o r for H, t h e n A f is a u n i t a r y m o r p h i s m of G r e e n f u n c t o r s f r o m A o V to A o U.

�9 I f M is an A - m o d u l e , t h e n r e s t r i c t i o n a l o n g A f g ives M o V a n d M o U s t r u c t u r e s of A o V - m o d u l e s , a n d t h e m o r p h i s m s M i a n d M f a r e m o r p h i s m s of A o V - m o d u l e s .

P r o o f : For the first assertion, I must show that if X and Y are G-sets, if a E (A o V)(X) and b E (A o V)(Y), then

A]x(a) • A~(b) = A]x• • b)


The left hand side is equal to

U A ( 6xy ) (A* ( f oG X)(a) • A*Cf oc Y)(b)) = A'(6~#y)A*((f oo X) x (fom Y))(a • b)

and the right hand side to

A*( f oc~ (X x Y) )A*(6~y)(a x b)

Now equality follows from

since for (u ,x ,y ) 6 U oG (X x Y)

((lo~x) x (,focy))o4,y(~,x,~)= ((,-ocx) xisoc,,-)) ((~,=), (~,~)) . . . .

. . . . ((s(=), =), (s(~), ~)) : ~.L.- (,f(,,),.~,,~) = 4,~o(so~(x • Y))(.,,,,,~) Moreover

A{@AoV) = A*(f oc ")A*(py/a)(Cm) = A*(pu/c)@A) = CAoU

since PU/G = Pv/c; o ( f o G .) . So the morphism A I is a unitary morphism of Green funetors.

For the second assertion, I must show that if a e (A o V)(X) and m 6 (M o U)(Y), then

a x v M !y (m ) = My,xy(Af(a) x U m) (8.3)

and that if m ' C (M o V)(Y), then

MY(a x y m') : AY(a) x u MY(m ') (8.4)

But

axVMyy(m)=M'(6~:'Y)(axM*(X~ ' (v,x)(f(u),y)](v'x)(u'Y) ] ( ax rn . )

The square

U oG (XY) O~,x,y)

V oa ( X Y )

(u,x,y) (f(~,), z)(,,, v))

)

( ("'~'Y) / (~,x)(~,y))

(V oo X) x (U oc Y)

\(~, x)( f (u) , y)] (v oa x) • (v oc Y)

(c)

is cartesian: if ((v,x), (u, y)) E (V oG X) x (U oo Y)and (v', x', y') �9 I/oG (XY) a r e

such that (v ' ,x ' ) = (v,x) (v ' ,y ' ) = ( f (u) ,y )

8.6. F U N C T O R I A L I T Y

then there exists s and t in G such that

t ?2 7--- ~)S SX t z X

179

v ' = f ( u ) t t y ' = y

In those conditions, the element ( u , t x ' , y ) is in U oa (XY): indeed, if r E G is such that ur = u, then as (u,y) E U oa Y, I have ry = y. Moreover

v ' t - l r t = f ( u ) r t = f ( u r t ) -- f ( u t ) = f ( u ) t - v'

and as (v', x/) E V oo X, I have t - % t x ' = ix' , or r.tx' = tx'. Moreover

( f ( u ) , tx', y) = (v't -1, tx', y) = (v', x', t - l y ) = (v', x', y')

and

( ( f ( u ) , ~ X t ) , ( ' U , , , y ) ) ~- ( ( V t ~ - i , t x t ) , ( ? j , ~ ] ) ) : ((Vt, x t ) , ( u , y ) ) . . . .

Conversely, if (Ul, Xl , Yl) E U oa ( X Y ) is such that

: :

then the last equali ty proves that replacing (Ul, Xl, Yl) by (uls -1 , SXl, syl) for a suitable s E G, I can suppose Ul = u and y~ = y. Then ( f (u) ,Xm) = (v , x ) = ( f ( u ) , t x l ) . So there exists r E G such that

f ( u ) r = f ( u ) r - ix1 = tx'

As f is injective on the right orbits, the first equali ty shows that ur = u, and since (u, Xl, y) E U oa ( X Y ) , I have rxl = Xl = tx', and

(u, , x , , y, ) : (u, tx', y)

which proves that (C) is cartesian. It follows that

\ ( f ( u ) , x , y ) ] \ ( f ( u ) , x ) ( u , y ) ]

But the right hand side of (8.3) is equal to

M f . x y ( A f ( a ) xU m) : M . ( f oa ( X Y ) ) M ' ( 5 ~ ; , y ) ( A * ( f o a X ) ( a ) • m) . . . .

" '" ( f ( ~ l ) , X), ('U2, y)

..=M,((u,~-,y) ~ M" ( (u,x,y) ) (a • ~) �9 ( f ( u ) , x , y ) ] \ ( f ( u ) , x ) ( u , y )

which proves equality (8.3).

Similarly, the left hand side of (8.4) is

M/(axVm')=M*(f~ z)(f(~), y) ( ~ ' ; ' Y ) ) (a •

whereas the right hand side is

* ~ " " r ) (~ ' ) ) A/(a) xVM/(m ') = M (hx,y)(A (foaX)(a) xUM*(foa . . . .

.. : i*((hU,y)M * ( ( ~ l ' X ) ( ~ 2 ' ~ / ) ~ ((/x;'7"/): ]~/./" ( (7.l,x,y) ) (a)<mt) �9 \ ( f (ul ) , z)(f(u2), y)] \ (f(u), x)(f(u), y)

This proves equality (8.4), and the proposition. "

8 . 7 E x a m p l e : i n d u c t i o n a n d r e s t r i c t i o n

Let O be a finite group, and H be a subgroup of G. Let U be the set G, viewed as a G-set-H, and V be the set G, viewed as ai1 H-set-G. If X is an H-set, then U oa X identifies with I n d e X , functorially in X. It follows that if M is a Mackey functor for G, then M o U is isomorphic to Res~M.

If Y is a G-set, then V oG Y identifies with Res~Y, functorially in Y, by the map (g, Y) ~ gY. Thus if N is a Mackey functor for H, then N o V identifies with Ind~N.

So if A is a Green functor for G, then Res~A is a Green functor for H, and this is not surprising. The case of induction is less clear, but corresponds to what Th6venaz calls coinduction (see [13]):

Propos i t ion 8.7.1: Le t H be a s u b g r o u p of t he g r o u p G, and B be a G r e e n f u n c t o r for H. then:

�9 T h e f u n c t o r Ind~B is a G r e e n f u n c t o r for G. I f K is a s u b g r o u p of G, then there is an i somorphism of rings (with unit)

( In4B)( I , ' ) = [ I B(H n ~I~) xEH\GII(

�9 T h e f u n c t o r B H IndCH B, f r o m the c a t e g o r y Green(H) of G r e e n fune- t o r s for H, to Green(G), is r igh t ad jo in t to the f u n c t o r A ~ Res~A f r o m Green(G) to Green(H).

P r o o f : The formula giving ( Ind~B)(K) is well known for Mackey functors. point is that is is still true for Green functors.

By definition of induction

But

( Ind~B)(h ' ) = B (Res~(G/K))

Res~(G/K) ~_ I_I H/(H n ~tr ~eHW/I<

The

8.7. EXAMPLE: INDUCTION AND RESTRICTION 181

the isomorphism (from right to left) mapping h(H N ~:I'() E H / ( H N ~K) to hxK E Thus if a and a' are in A(K) = A(G/K) = B ( V oa (G/K)) , setting

F = G/K , and denoting by , v , , the product "." on A, I have

~ , \ ( ~ , ~ ) ( ~ , ~ ) /

But I know that the ring (B(V oa F), .) is isomorphic to the direct product of rings (B(a3), .) for the various orbits w of H on V oG P. The first assertion follows.

Similarly, the second assertion is clear for Mackey functors. I must show that the adjunction passes down to Green functors. If A is a Green functor for G, if B is a Green functor for H, and (I) is a morphism from A to Ind/~B, then for any subgroup K of G, I have a morphism

~,~-: A(K) , @ B ( H N X K ) x E H \ G / K

Say that r is a morphism of Green functors means exactly by the previous remarks that qSK,x is a morphism of rings (with unit) fi'om (A(K), .v) to (B(H n xK), .), for

any x. But if L is a subgroup of H, the morphism q~L from A(L) = Res t (L ) to B(L) associated to (I) by adjunction is (I)L,1. So q? is a unitary morphism of Green functors.

Conversely, if �9 is a unitary morphism of Green functors from ResgA to B, then for any subgroup L of G, I have a morphism q'L of rings (with unit) from A(L) to B(L). The morphism ~ associated to �9 by adjunction is then defined on the subgroup K of G by

x K a ~K,::(a) = q~Hn~h'( rH,~nK( ))

It is the product of three morphisms of rings (with unit). So it is a morphism of rings with unit, and gP is a unitary morphism of Green functors. This proves the proposition. �9

The adjunction property shows in particular that if A is a Green functor for G, then there is a unitary morphism ~a of Green functors, adjoint to the identity of Res~A

r/A :A --+ IndGHReSaHA

But a G U) o IndHReSHA = (A o V = A o (U O H V )

It is easy to see that U OH V ~_ (G XH G), which is the quotient of G x G by the right action of H given by (gx,g2)h = (gxh, h-~g2). On the other hand, if I denote by I the set G, viewed as a "identity " G-set-G, the functor A is equal to A o I , and the morphism rlA comes from the morphism

f : U OH V ~ I (gl,g2)H ~-~ 9t-92 E I

This morphism is injective on the right orbits: indeed, if

f((gl,g2)g.g), ,, .~ f ( ( g l , g 2 g ) ~ = g l g 2 g ~- f ( ( g l , g 2 ) g ) = g ig2

182 C H A P T E R 8. C O M P O S f T I O N

then

91.92.9 = 91 .g2

which proves that g = 1. Now if M is an A-module, I know that Res~M is a Res~A module. Conversely, if

N is a Res~A-module, then Ind~N is a Ind~Res~A-module, so an A-module by the morphism flA. The proposition 8.6.1 shows that those correspondences are functorial. Moreover

P ropos i t i on 8.7.2: Let A be a Green func tor for the group G, and H be a subgroup of G. The func to r M H ReSaH M f rom A-Mod to ResaHA-Mod is left and r ight adjoint to the func tor N H Ind~N from Res~A-Mod to A-Mod.

Proof : Here again, this property is well known for Mackey functors. All I have to do is to keep track of the adjunction procedure: if M is an A-module, if N is a Res~A- module, and f is a morphism of nes A modules from R e s ~ M to N, then Ind~f is a morphism of Ind~Res/~A-modules from Ind~Res~M to Ind~N. Thus it is also a morphism of A modules from Ind~Res~M to Ind~N. The adjoint of f is obtained by composing this morphism with the morphism from M to IndGHResaM, which is a morphism of A-modules. So the adjunction at the level of Mackey functors maps morphisms of Res~A modules to morphisms of A-modules.

Conversely, if g is a morphism of A-modules from M to Ind~N, then ResaHg is a morphism of Res~A-modules fi'om Res~M to Res~Ind~N. The adjoint of g is obtained by composing this morphism with the morphism p from Res/~IndaH N to N, which is a morphism of ResaHA-modules: it follows indeed from the morphism of H-sets-H

@ : h E H H ( h , 1 ) E V o G U

which is injective: indeed, if hi and h2 in H are such that (hi, 1) = (h2, 1) in V oa U, then there exists g r G such that h2 = h~g and 1 = g.1. Sog = 1 and hi = h~. If B = Res~A, the morphism p is then a morphism of Res~Ind~B-modules from Res~Ind~N to N. As

Res~Ind~B = Res~Ind~/Res~A

I have the morphism ResaT1/from Res~A to Resalnd~Res~A, and the composite

aes,~A aes~qA Res~Ind~Res~A B e ) R e s ~ A

is the identity of B, because it expresses the adjunction of Res~ and Ind~ at the level of Mackey functors. So through Res~rlA I recover the same Res~A-module structure on N, and this proves that p is a morphism of Res~A-modules. Now the adjunction at the level of Mackey functors maps morphisms of A-modules to morphisms of Res~A-modules, and this proves that the functor Rest/is left adjoint to the functor Ind, . A similar argument. shows that it is also right adjoint. This completes the proof of the proposition. �9

Chapter 9

Adjoint constructions

N o t a t i o n s : If G and H are groups, and L is a subgroup of H • G, I denote by p~(L) the projection from L to H, and p2(L) its projection onto G. I denote by kl(L) (resp. k2( L ) ) the normal subgroup of p~ ( L ) (resp. of p2( L ) ) formed of elements h E H (11esp. elements9 E G) such that (h, 1) E L (resp, such that (1,9) C L). The groups pl(L)/kl (L) and p2(L)/I~2(L) are canonically isomorphic, and 1 denote by q(L) this quotient.

I denote by (H x G)/L the set of left classes of L in H • G~ viewed as an H-set-G by

h.(~,v)L.g : (h~,g-~v)L

With those notations, I have shown ([2] Lemme 2) that if U : (H x G)/L, and if X is a G-set, then

_ l ~ d u �9 ~ ( C / ,~, a , ~ , ~ ( L ) U oa X ~ . p~(L)mIpdL)/k~(L)rL[lteSp2(5)A )

denoting by TL the transport by the canonical isomorphismp~(L)/kl(L) ~_ p~(L)/k2(L) of a p~(L)/k~(L)-set.

It follows (see [2] 4.I.2) that if M is a Mackey functor for H, then

M o U = Ind~(L)Inf~:(L{ ka L OLj: (L) k. rL, ResH,L' M ( . / ( ) ; ( ) / . . . . .

where OL denotes the transport by isomorphism of Mackey %nctors for p~(L)/kl(L) to Mackey functors for p2(L)/k2(L).

As any H-se t -G U is a disjoint union of transitive H-sets-G, i.e. of the form (H • G)/L, it follows that the functor M o U is a direct sum of fnnctors composed of restriction, inflation, coinflation, and induction.

I already know that the functors of restriction and inflation are mutual left and right adjoints. Thhvenaz and Webb (see [14] 5.1, [15] 2.) have built left and right adjoints for inflation, and they also mention ([15] 5.4 and 5.6) that coinflation has a left adjoint (and a right one if the ground ring is a field). It follows that the functors M ~-~ M o U always have a left adjoint. I will show that they always have a left and right adjoint, and describe their structure in terms of G-sets.

9.1 A le f t a d j o i n t to t h e f u n c t o r Z ~-~ U OH Z

N o t a t i o n : Let U be a fixed G-set-H. [ denote by u ~ uH the projection from U to U/ H. If (Y, f ) is a G-set over U/ H, I denote by U.Y the pull-back product of U and

184 C H A P T E R 9. A D J O I N T C O N S T R U C T I O N S

Y over U / H , defined by

U.Y = {(u,y) E U • Y I f (Y ) = uH}

I view U.Y as a G-set-H: i fg 6 G and h 6 H, then

g.(u,y) .h = (guh,gy)

If (u, y) E U.Y, I denote by G(u, y) its orbit under G. I denote by G \ U . Y the set of orbits of G on U.Y, viewed as an H-set by

h.a(u, y) = G(uh- ' , y)

This construction is functorial in Y: if a : (Y, f ) ---, (Y', f ' ) is a morphism of G-sets over U/H, then the map

a \ U . ~ : G i u . Y -~ G\U.Y '

defined by

is a morphism of H-sets. Conversely, if Z is an H-set , and Pz is the unique morphism from Z to *, then

U o H p z is a morphism from UOHZ to UOH. = U/H. Thus Z ~ U o H Z is a functor from H - s e t to G-set~u/H. Actually:

P r o p o s i t i o n 9.1.1: T h e f u n c t o r ( Y , f ) ~ G \ U . Y f r o m G-se t lu /H t o H - s e t is l e f t a d j o i n t to t h e f u n e t o r Z ~-~ U OH Z.

P r o o f : Let (Y, f ) be a G-set over U/H, and Z be an H-set. If r is a morphism of G- sets over U / H from (Y, f ) to (U OH Z, U OH PZ), and if (u, y) E U.Y, then f ( y ) = u Y . Moreover, as

(U o g Pz)r = f (Y) = u g

there exists h E H and z E Z such that r = (uh, z). If h' E H and z' C Z are such that r = (uh', z'), then there exists h" E H such that

uh = uh'h" z' = h" z

In those conditions uh = 4h .h- lh 'h '', and as (4h, z) E U OH Z, it follows tha t h - l h ' h ' z = z, hence that hz = h'h"z = h'z'. In part icular the element hz C Z is well defined by the condition r = (uh, z) = (u, hz).

I can then define O(u, y) C Z by the condition

r = (u ,0 (u ,y ) )

Then as g(u, y) = (gu, gy) for g E G, if z = O(u, y), I have

r = gr = g(4, z) = (g4, z)

which proves that O(gu, gy) = O(u,y). So 0 is a map from G \ U . Y to Z. As moreover h.G(~ , y) = (~h-~ , y) , and as

r = ( ~ , z ) = ( 4 h - l , ~ z )

9.1. A L E F T A D J O I N T TO THE F U N C T O R Z ~ U OH Z 185

I have O(h.G(u,y)) = hz = hO(u,y). So 0 is a morphism of H-sets from G\U.Y to Z. Conversely, if 0 is a morphism of H-sets from G\U.Y to Z, and if y C Y, let u E U

such that f ( y ) = uH. Then (u,y) E U.Y. In those conditions, if z = O(G(u,y)) , the element (u, z) is in U OH Z: indeed if h C H is such that uh = u, then

h -1.u(u, y) = G(m~, y) = G(u, y)

thus z = O(G(u,y)) = hO(G(u,y)) = hz. The element (u , z )does not depend on the

choice of u such that f ( y ) = uH: if I change u to uh, for h E H, then

O(a(uh, y)) - - h - 1 0 ( G ( ~ , y ) ) : h-lz

and (u, z) = (uh, h- lz) . I can then set r = (4, z). As f (gy) = gI(Y) = gul l , and

as O(G(gu,gy)) : OiG(u,y)) : z, I have

r = (r z) = .q(u, z) = g r

Thus r is a morphism of G-sets from Y to U OH Z. The correspondences r ~ 0 and 0 ~ r are inverse to each other: indeed, the

equality

defines 0 from r and r from 0. The proposition will then follow, if I know that the unit and co-unit of this adjunction are functorial:

N o t a t i o n : If (Y, f ) is a G-set over U/H, I will denote by ~(y,S) (or ~,y if f is clear from context) the unit of the previous adjunction: it is the morphism

t,(yd) : Y ~ U OH (G\U.Y)

defined by ~(~,S)(Y)= (~, U(~, y)), if / (y)= ~H. If Z is an H-set, I will denote by 71z the co-unit of this adjunction: it is the

morphism rlz : G\U.(U oH Z) ~ Z

defined by 7jz(G(u' , (u,z))) = h - l z i f h �9 H is such that u ' = uh.

T h e n , is functorial: if c~: (Y, f ) ~ (Z, g) is a morphism of G-sets over U/H, and

if (u, y) �9 U.Y, then "(yd)(Y) = (u, G(u, y)). Thus

But gc~(y) = f (y ) = uH, so t,(zm)c~(y) = (u, G(u, c~(y))), which shows that

.(~,.)~ : (u o . (GXU..))~(~,j)

Simi lar ly , is functorial: if f : Z ~ T is a morphism of H-sets, and if (u', (u, z)) E

U.(UoHZ) , then thereex is t sh �9 H s u c h t h a t u ' = u h . T h e n , z ( G ( u ' , ( u , z ) ) ) = h - l z . As

o , , =

186 CHAPTER 9. ADJOINT CONSTRUCTIONS

I have

SO

[G\u (u o . i ) ( G ( < : i

,T (G\U.(U OH f ) ) = f~lz

9.2 The categories ~Du(X)

Defini t ion: If X is a G-set, let 39u(X) be the following category:

�9 The objects of Du(X) are the G-sets over X • (U/H). If (Y , I ) is an object of 2)u(X), I denote by fu the map from Y to U/H obtained by composing f with the projection from X • (U/H) on U/H, and f x the composition product of f with the projection on X. Then (Y, fu) is a G-set over U/H, to which I associate the pull-back product U.Y.

�9 A morphism ce from (Y,f) to (Z,g) is a mol'phism of G-sets from Y to Z such that g o c~ = f , and such that the morphism U.o from U.Y to U.Z associated to a is injective on each left orbit under G, i.e. such that

(u, y) ~ u.Y, g ~ G, v~ = ~, ~(gy) = ~(y) ~ gy = y

R e m a r k : In particular, if a is injective, then U.c~ is injective on each left orbit.

It is clear that ~Du(X) is a category: the product of two morphisms which are injective on the left orbits is injective on the left orbits, and the identity morphism is injective on the left orbits.

If r : X --+ X ' is a morphism of G-sets, let r be the morphism r x Idu/H fi'om X • (U/H) to X' • (U/H). There is an obvious functor from 7)u(X) to D u(X') , which maps (Y, f ) to (Y, e l ) , and the morphism o~: (II, f ) + (Z,g) to a : (Y, C f ) -+ (Z, q~g). I will denote this functor by 7)u,,(r Similarly, there is a functor from Z)u(X') to :Du(X), defined by inverse image along r if (Y', f ' ) is a G-set over X ' • (U/H), let Y, f and a be such that the square

a

Y , I,"

1 x • (U/H) , x'• (U/H)

is cartesian. If a ' : (Y ' , f ' ) ---+ (Z',g') is a morphism of G-sets over X' • (U/H), let Z, g, and b such that the square

b Z ~ Z'

X x (U/H) , X' x (U/H)

9.2. T H E C A T E G O R I E S D u ( X ) 187

is cartesian. Then as

g%'a = f ' a = C f

there exists a unique morphism ~ : Y --, Z such that the diagram

y a y ,

Z b �9 Z '

/ X x ( U / H ) ~) , X ' •

is commutat ive. Moreover, in those conditions, the square

a

Y , y '

1 l O~ Ol

Z ~ Z ' b

is cartesian. Indeed, if z E Z and y' E Y' are such that b(z) = ~ ' (y ' ) , then

g ' b ( z ) = Cg(z) = g ' a ' ( y ' ) = f ' ( y ' )

Then by definition of Y, there exists a unique y E Y such that

g(z) = f ( y ) y ' = a(y)

Moreover

ha(y) = c /a(y) = ce'(y') = b(z) gc~(y) = f ( y ) = g(z)

Then z and a (y) have the same image under b and g, so z = c~(y). And if another element yl E Y is such that c~(yl) = a(y) and a(yl) = a(y), then

goe(yl) = gc~(y) = f ( y l ) = f ( y )

and unicity of y implies yl = Y. Moreover, the morphism U.a is injective on the left orbits:

L e m m a 9.2.1: Let a

Y , y '

ol 1 Z ~ Z '

b

be a c a r t e s i a n s q u a r e of G-sets over U/H. If U.a' is i n j e c t i v e on each left orbi t o f G on U.Y', t h e n U.a is i n j e c t i v e on each left orb i t o f G on U.Y


P r o o f i Indeed, if (u , y ) r U.Y and (gu ,gy) have the same image under U.a, then gu = u and a(gy) = a(y) . Taking the image under b of this relation gives

ba(gy) = a 'a(gy) = ba(y) = a 'a(y)

so a ' ( g a ( y ) ) = a ' ( a ( y ) ) . As U.a' is injective on the left orbits, it follows that ga(y) = a(gy) = a(y). As moreover a(gy) = a(y) , and as the square Y, Y', Z, Z ' is cartesian, I have gy = y. ..

It follows that if I set D~(r f ' ) = (Y, f ) and D~( r = a , I get a functor from 7)u (X ' ) to Du(X) .

9.3 T h e f u n c t o r s Qu(M) D e f i n i t i o n s : I will say that an H-set ( Z, a) over G \ U.Y is v(vj)-disjoint (or v-disjoint for short) i f the square

0 , Y

1 1 U O H Z ~ U o H ( G \ U . Y )

U O H a

is cartesian, or equivalently i f

(U oH a)(U OH Z) n uv(Y) =

I f M is a Mackey functor for H, and (Y, f ) is a G-set over U / H , I set

Q u ( M ) ( Y , f ) = M ( G \ U . Y ) / ~ M . ( a ) M ( Z ) (z,~)

where the sum runs on the H-sets ( Z, a) over G k U . Y which are v-disjoint.

R e m a r k : Say that (Z, a) is not v-disjoint means that there exists z C Z and y' ff Y, such that if u ' H = fu (y ' ) and a(z) = G(u ,y ) , then there exists v C U such that (v ,z) G U o H Z a n d

(v, This means that there exists h E H such that v = u~h and

G(u, y) = h - lG(~ ', v') = G(u'h, y')

So there exists g E G such that

uth = gu y' = gy

Finally, there exists g E G and h E H such that

v = gu u' = guh -1 y~ = gy

9.3. THE FUNCTORS Qu(M) 189

Now (v, z) : (g u, z) belongs to U OH Z if and only if (u, z) E U OH Z does. Conversely, if a(z) = G(u, y), and if (u, z) E U OH Z, then (u, y) E U.Y, and

vy(y) = (u, G(u,y)) = (U OH a)(u,z)

Thus (Z, a) is v-disjoint if and only if

g e e Z, a(z) = G(u,y) ~ (u,z) ~ U oH Z

If X is a G-set, and Y is an object of Du(X), I set

Qu(M)(k] f ) = Qu(M)( fu)

L e m m a 9 .3 .1 : Let X be a G-set . T h e c o r r e s p o n d e n c e

(Y, f ) H Qu(M)(Y, f )

a : (Y , f ) + (Z,g) ~ M.(G\U.a): M(G\U.Y) ~ M(G\U.Z)

i nduces a f u n c t o r f r o m ~ u ( X ) to R - M o d .

P r o o f : I must prove that if a is a morphism in ?)u(X) from (Y , f ) to (g,g), if (T,a) is a v-disjoint H-set over G\U.Y, and if m E M(T) , then the image of M.(G\g.cOM.(a)(m ) in Qu(M)(Z, gu) is zero. This will be the case in particular if, denoting by /3 the morphism G\U.e, the H-set (T,/3a) is v-disjoint over G\U.Z. This is equivalent to say that the square

0 , Z

1 1 u oH T , u ou ( o \ u . z )

U oH (3~)

is cartesian. But this square is composed of the two squares

O~

(c)

0 Y ~ Z

1 1 1 U OH T U OH (G\U.Y) ~ U on (G\U.Z)

U OH a U OH

The left square is cartesian if (T,c~) is v-disjoint. Thus (C) is cartesian, since the right square is cartesian by the following lemma:

L e m m a 9.3.2: Le t a : (Y, fu) --+ (Z, gu) be a m o r p h i s m of G-sets over U/H, such t h a t U.a is in jee t ive on each left o rb i t of G on U.Y. T h e n t he s q u a r e

C~

y , Z

v " l I vz

U OH (G\U.~") , U OH (G\U.Z) U oH (G\U.~)

is c a r t e s i a n .


P r o o f : Let ('u,, G(u2, y)) E U OH (G\U.Y) and z E Z such that

If v �9 U is such that vH = gu(z), then uz(z) = (v,G(v,z)l . The above equali ty means that there exists h �9 H such that

k /

ulh = v h-lG(u2, e(y)) = c(,, ~) = c (~h ,~ (y ) )

So there exists x �9 G such that

v = xu~h z = x a ( y ) = a ( x y )

The image of the second equality under gu gives

g e ( Z ) = v H = g u o ~ ( x g ) = fu(xy)

Then

Conversely, let y l and Y2 be elements of Y having the same image under a and uy. Then i f u C U is such that fu(Yx) = u H , I have also

fu(Y2) = guc~(y2) = gua(y,) = fu(Y,) = uH

Then uy(yl)= (u,G(U, yl)) and ur(y2)= (u,G(u, ye)). These elements are equal if there exists h E H such that

tt ~-- tth h - lG ( t t , y l ) = G(tt, y2) = G(tth, y l ) = G(tt, y l )

Then there exists x E G such that u = my and Y2 = my1. As m o r e o v e r c~(y2) = c~(y l ) ,

and as U.c~ is injective on the left orbits of G on U.Y, it follows that Y2 = Yl, which proves the lemma. �9

Let r : X --+ X ' be a morphism of G-sets. If (K f ) is an object of 19u(X), I denote

by (Y', f ' ) = 2)u,.(r f ) = (Y, Cf) its image in Du(X'). As (r = fu, the sets U.Y and U.Y' are equal, and M(G\U.Y) identifies with M(G\U.Y'). Similarly, the u-disjoint H-sets over G\U.Y identify with u-disjoint H-sets over G\U.Y'. It follows that Qu(M)(Y,I) is natural ly isomorphic to Qu(M)(Y' , I ' ) , i.e.

Qu(M) ~ Qu(M) o Du,.(r

Conversely, if (Y', f ' ) is an object of Z)u(X'), I set (Y, f ) = :D~j(Y' , f ' ) . I have a cartesian square

a

Y , y '

x • (U /H) , X' • (U /H )

9.3. THE FUNCTORS Qu(M) 191

If m E M(G\U.Y'), then M*(G\U.a)(m) E M(G\U.Y). If (Z ' , a ' ) is a u-disjoint H-set over G\U.Y', I can fill with Z, ~, a the cartesian square

Ct

Z , G\U.Y

Z' , G\U.Y' I

Then (Z, (~) is a u-disjoint H-set over G\U.Y: indeed, let T, 7, and 6 be such that the square

T Y

1 U OH Z U OH (G\U.Y)

U oH ol

is cartesian. I have then a commutative diagram

T 7 ~}"

U oH Z U oH (

U o H 3 ~

U OH Z'

y t

U y ,

�9 u o. (a \uY ' ) U OH a'

As the bottom square is cartesian, I can fill this diagram by a morphism from T to 0, so T is empty, and (Z, c~) is u-disjoint.

As M*(G\U.a)M.(c~') = M.(cjM*(fl), the image of M*(G\U.a)M.(~') in the quotient Qu(M)(Y, f) is zero. So I have built a morphism T:y, S') from Qu(M)(Y', f') to Qu(M)(Y, f). This construction is moreover functoraal in (Y, f ' ) : if c~': (Y', f ' ) --* (Z',g') is a morphism in 7?u(X'), then the square

Qu(M)(Y', f')

Qu(M)(a') 1 Qv(M)(Z',g')

Ti~,,y, ) )

)

T r (Z,,g,)

Qtr(M) o D~(r f')

1 Qv(M) o ~b(r Qu(M) o 73b(C)(Z', ~')

(c)


is commutative: indeed, I have a diagram

y a },.-/

X • (U/H) ~ , X ' •

and to show that (C) is commutative, it suffices to show that

M.(G\U.o~)M*(G\U.a) = M*(G\U.b)M.(G\U.c~')

which will follow from the fact that, the square

g . a U.Y i U.Y'

1 U.Z , U.Z'

U.b

is cartesian, as the pull-back of a cartesian square. I will apply to this square the following lemma:

L e m m a 9.3.3: Let a

Y ~ y '

1 1 C~ C~

Z ) Z' b

be a c a r t e s i a n square of G-sets . If ~' is i n j e c t i v e on each orbi t o f G on Y', t h e n t h e square

c\Y , G\Y'

G\,~ 1 I G\,~' ) G\z r G\Z'

is c a r t e s i a n .

Proof: Let Gz r G \ Z and Gy' E G \ Y ' be such that

( G \ b ) ( C z ) : ( C \ ~ ' ) ( O y ' ) : Cb(z) : C ~ ' ( y ' )

Then there exists an element g E G such that 9b(z) = b(gz) = c?(y'). So there exists a unique element y E Y such that

9.4.

In those conditions

and moreover

THE F U N C T O R S L u ( M )

( a \ a ) ( G y ) : G~(~) : @ '

( G \ ~ ) ( @ ) = O~(~) = Ggz = Gz

193

9 . 4 T h e f u n c t o r s s

The previous lemma leads to the following definition:

D e f i n i t i o n : I f X is a G-set, and M is a Mackey functor for H, I set

s = lira Qu(M)(M, f ) (v,f)Ov(x)

If ~ : X ~ X ' i s a morphism of G-sets, then the isomorphism T~ induces a morphism

lira Q ~ d M ) ( Y , f ) ~ lim Qu(M)o~gu , . (~ ) (Y , f ) --* lim Q u ( M ) ( Y ' , f ' ) -------+ (v,l )e~,( x ) (v J )~u( x ) (Y',f ')~u( x')

Now if Gyl and Gy2 are elements of G \ Y having the same image under G\a and G \ a , I have

Ga(yl) -- Ga(y2) Ga(yl) = Ga(y2)

So there exists elements g and g' of G such that

ga(yl) = a(y2) g 'a(yl) = a(y2)

The image under b of this relation gives

a(yl))

Then as ~(y~) = a-~(y2), I have

As a ' is injective on the left orbits, I have g'g-la(y2) = a(y2), or

g-la(y2) = a(g-ly2) = a(yl) = 9'-la(y2)

As c~(y~) = a(g'-ly2), it follows that YI = g'-iY2, and then Gy~ = Gy2, completing the proof of the lemma. "

I have finally built a natural transformation T e from Qu(M) to Q u ( M ) o D~(r so I have proved the following lemma:

L e m m a 9.3 .4: I f ~p : X ~ X ' is a m o r p h i s m of G-se t s , t h e n q~ i n d u c e s an i s o m o r p h i s m of f u n e t o r s

T, : Qu(M) --* Qu(M) o Z)u,,(r

and a na tura l t r a n s f o r m a t i o n

T~ : Qu(M) ~ Qu(M) o ~gb(~) )


i.e. a morphism s from s to E u ( M ) ( X ' ) . Conversely, the trans[ormat~ion T ~ induces a morphism

M Y' ' lira c-2u(M)oDu(r , f ' ) ---+ ~ Q u ( M ) ( Y , f ) lim Qu( )( , f ) - ~ ~ c * " lim (v',Y')e~)u( x ') (Y',f')evu( x') (v,De~v( x )

i.e. a morphism t;u(M)*(~) r,-om &~(M)(X') to & ; ( M ) ( X )

Proposition 9.4.1: The previous definitions turn s into a Maekey func- tot.

Proof: It is clear that if 4) : X --+ X' and 0' : X' --+ X" are morphisms of G-sets, then

s162 = s163 s162162 = 12u(M)*(r163

It is also clear that s and s are the identity morphisms. It suffices then to check axioms (M1) and (M2) for s I will start with (M2):

let a

Z - - - , T d

be a cartesian square of G-sets.

Notations: Let (E ,e ) be a G-set over Y • (U/H) , and m E M(G\U.E) . I denote by m.(E,~) the image of rn under the composite morphism

M(G\U.~) --+ Qv(M)(E, ~) --+ &~(M)(Y)

Furthermore, I denote by X ~ X the fimctor X ~-~ X • (U/H) from G-set to G-set.

To compute Eu(M)*(a)(rn(z,~)), I fill the cartesian square

C~ F , E

X ------+ Y

The image of m(E,~) by s is then

C ~ ( M ) * ( a ) ( . ~ ( ~ , ~ ) ) : M*(a\U.~)( , . )(Fj)

The image of this element under s is

s163 = M*(G\U.a)irn)(FSf)

On the other hand, the image of m(E,e) under ]~u(M).(c) is

Z;v(M)~(c)(-~(u,~)) = m(u,~)

9.4. THE FUNCTORS Lu(M) 195

As the square o~

F , E

Z , T J

is cartesian, because it is composed of the previous one and of the cartesian square

a )

l Z , T

g

the image under s of m(E,Te) is then

s163 = M*(G\U.~)(~)(F,;s) which proves that

s163 = s

So s satisfies (M2).

To check axiom (M1) , I consider two G-sets X and Y, and their disjoint union Z = X LI Y. As 2 = X H,Y, it is clear that a G-set (E, e) over 2 is the disjoint union of a G-set (El, el) over X and a G-set (E2, ~2) over r . Then G\U.E is the disjoint union of G\U.E1 and G\U.E2. Let il and i2 be the respective injections from E1 and E2 into E. If rrz E M(G\U.E), let m~ = M*(G\g.il)(rn) and m 2 = M*(O\U.i2)(m). Then

m = M.(G\U.i,)(ml) + M.(G\U.i2)(m2)

I can view this equality in s as

m(u,r : M.(G\U.it)(rnj(u,~) + M.(a\u.ij(m~)(u,~)

But as i1 is injective, it is a morphism in 2)u(Z) from (E,, eh) to (E, e), and in s I have

a , ( G \ U , il)(rrq)(E,e ) : (TI%I)(EI,eil)

Now if ix is the injection from X to Z, I have

( r f t l ) ( E l , e h ) : s

The same argument for is shows then that, denoting by iv the injection from Y to Z, I have

/ %

s M).(iy )((mj(E~,~))

In particular, the map (s Cu(M).(,r)) from Cu(M)(X) * C~(M)(Y) to s is surjective. The following lemma now shows that axiom (M1) holds for s and this completes the proof of the proposition. �9


L e m m a 9 .4 .2: Let L b e a b i f l m c t o r on G-se t , sa t i s fy ing a x i o m ( M 2 ) , and such t h a t L({3) = {0}. I f for any G-se t s X and Y, t h e m a p

(L.(ix), L.(iy)) : L(X) �9 L(Y) ---+ L(X I_I Y)

is s u r j e c t i v e , or t h e m a p

L*(ix) | L*(iy) : L(X H Y) ~ L(X) | L(Y)

i n j e c t i v e , t h e n L is a M a c k e y f u n c t o r for G.

Proof : Let 0 be the map (L.(ix),n.(iy)). As L satisfies (M2), as L(0) = {0}, and as the square

1 1 Y , X H Y

iy

is cartesian, the products L*(ix)L.(Q) and L*(ir)L.(ix) are zero. As ix is injective, the square

Id X , X

x , X L I Y ix

is cartesian. It follows that L*(ix)L.(ix) is the identity of L(X), and that L*(iz)L.(iv) is the identity of L(Y). Now the map r = L*(ix) | L*(iy) is such that r = IdL(x)eL(Y). Then 0r = 0, and r162 = 0- If 0 is surjective (resp. if r is injective), the first (resp. the second) of these equalities implies 0r = IdL(xHz), so (M1) holds. This proves the lemma. �9

9.5 Left adjunction

N o t a t i o n : If 0 : M ~ M' is a morphism of Mackey functors for the group H, and if X is a G-set, I de~ne a map s from s to s by setting', for an object, (Y, f ) of:Du(X), and a~ Cement m E M(G\U.Y)

s = Oc\uY(~) (r j )

L e m m a 9.5 .1: T h i s de f in i t i on t u r n s s in to a m o r p h i s m of M a c k e y func- t o r s f r o m s to s Proof : First, the map s is well defined: if a : Z ~ G\U.Y is u-disjoint, then

s (M.(G\U.a)(m)(z,f)) = OG,\uyM.(a)(m)(zd) = (M.(a)Oz(m))(v,/) = 0

Moreover, if a : (Y , f ) ~ (Z,g) is a morphism in Du(X), then

c . ( o ) ~ M.(~)(.~(~,j~) = (O~w.Y . (G \U .~ ) ( .~ ) ) (a~ ) . . . .

9.5. LEFT ADJUNCTION 197

Now let r : X ~ X ~ be a morphism of G-sets. Then

Z_u(M).(r = s162 . . . .

. . . . O G \ U . y ( m ) ( y , ~ f ) = . ~ u ( O ) x , ( m ( y ' ~ . f ) ) . . . .

. . . . z:u (0)~,~:~ (M).(r

If moreover (Y', f ' ) is an object of :Du(X'), and if m' ~ M(G\U.Y'), then let Y, f , and a filling the cartesian square

a

Y , y '

X ~ X I


�9 ' ' ) s (r163 = s Oa\u.v,(m )(y,,f,) . . . .

. . . . s (M*(G\V.a)(m')(yj)) = s

which proves the lemma. .,

T h e o r e m 9 . 5 . 2 : L e t G a n d H he f ini te g roups , and U b e a G-se t -H . T h e c o r r e s p o n d e n c e

V ~ Cu(M)

0 E Homu~,ck(u)(M, M') ~ s C Homuack(a)(s s

is a f u n c t o r f r o m Mack(H) to Mack(G), which is left ad jo in t to t h e f u n c t o r N H N o U .

Proof i It is clear that the correspondence M ~-~ 12u(M) is functorial in M. The main point is the adjunction property.

So let M be a Mackeyfunctor for H, and N be a Mackeyfunctor for G. I f 0 i s a morphism of Mackey functors from M to N o U, then for any H-set Z, I have a morphism

Oz : M(Z) --, (W o U)(Z) = N(U OH Z)

In particular, if (Y, f ) is a G-set over x • (U/H), I have a morphism

o~\~.y : M(a\U.V) ~ N(u o,~ (G\U.V))

Composing this morphism with N*(ug) gives

N*(uy)Oa\v.v : M(G\U.Y) ~ N(U cu (G\U.Y)) ~ N(Y)


If (Z, a) is a u-disjoint H-set over G\U.Y, I have the diagram

OZ M(Z) , N(U OH Z) , N(O)

) M(a\U.Y) N(U oH (G\U.Y)) N(Y)

This diagram is commutative: the left square is because 0 is a morphism of Mackey functors, and the right square is because (Z, a) is r-disjoint. It follows a morphism from Qu(M)(Y,f) to N(Y). Now I can compose this morphism with the morphism N.(fx) : N(Y) ---, N(X) deduced from f .

If c~: (Y,f) ~ (Z,g) is a morphism in 7?u(X), I have the following diagram

oG\v.y N*(.y) N.(fx) M(G\U.Y) , N(UoH(G\U.Y)) , N(Y) N(X)

) M(G\U.Z) Oa\u.z' N(UoH(G\U.Z)) N*(uz) N(Z) X.(gx)' N(X)

The left square is commutative because 0 is a morphism of Mackey functors. The middle square is commutative by lemma 9.3.2 because N is a Mackey functor. The right square is commutative because gc~ = f .

It follows a morphism ~bx from s to N(X). Now if r : X ~ X ' is a morphism of G-sets, if (I/, f ) is a G-set over X • (U/H), and if m E M(G\U.Y), then by definition of ~bx

On the other hand

Cx(m(y,j)) = N.( fx )N*(~'r )Oa\v.y(m)

s M).( r )(m(y,])) = m(g,-~])

SO

Cx,s162 = N.(r = N.(r

and then r163162 = N.(C)r Conversely, if (Y', f ' ) is a G-set over X ' x (U/H), and if m' C M(G\U.Y'), then

the image of mlg,,], ) under L:u(M)*(r is obtained by filling the cartesian square

a

Y , y '

si ? X , X'

More precisely, I have then

s162 = M*(G\U.a)(m')(y,])

9.5. LEFT ADJUNCTION 199

The image under ~bx of this element is by definition

Cx Cu( M)*( r )(m{y,,f,)) = N.( f x )N*( uy )Oa\u.z M*( G\ U.a )(rn')

As 0 is a morphism of Mackey functors, I have

Oa\u.zV*(G\U.a) = X* (U OH (G\U.a))Oa\u.z,

Moreover, as the square

Y

u o , (a \u . z )

is commutative, I have

a

+ U oi~ (G\U.a)

y!

l l/y1

U oh, (G\U.Y')

N*(uy)N*(U OH (G\U.a)) = N*(a)N*(uz,)

Furthermore, it is easy to see that the square

a

Y ~ y '

X ~ X '

(9.1)

is cartesian, and then N.(fx)N*(a) = N*(r ). Finally, I can write equation (9.1) as

N (d~)N.(f'x,)N (uy,)Oa\u.z,(m) ~xs162 = " ' * '

On the other hand

Cx,(m(y,j,)) = W.(,f'~,)N'(~y,)Oawy,(,~')

It follows that Oxf-.u(M)*(r = N*(r and this shows that the morphisms Cx define a morphism r of Mackey functors from Eu(M) to N.

Conversely, let ~b be a morphism of Mackey functors from s to N. Then for any G-set X, I have a morphism ~bx from f-.u(M)(X) to N(X). In particular, if Z is an H-set, I have a morphism ~uoHz from f-.u(M)(U OH Z) to N(U OH Z ) =

(N o U)(Z). But if m c M(Z), then M*(vz)(m) e M(G\U.(U on Z)). Moreover,

setting rrz(u,z) = ((u,z) ,uH), I define a morphism of G-sets fez from U OH Z to

(U OH Z) • (U/H), and then U OH Z is an object of Du(U OH Z). In particular, I can consider the element

Az(m) = M*(~z)(m)(vo.z,,~z) E f-.u(M)(U OH Z)

Finally, I have the composite morphism

Oz : M(Z) )~z , s on Z) CUo.Z N(U OH Z)


If r is a morphism of H-sets from Z to Z', and if m' C M(Z'), then

OzM*(r = CUo.zAzM*(r

Moreover AzM'(r = M*(rlz)M*(r

It is clear moreover that the square

~z G\u.(u o . z) - ~ z

G\U.(U ~ r 1 [ r G\U.(U o. z') , z '

~Z'

is commutative. So I have

The square

~M*(r : M" (O\U.(U o . r

UOHr UoHZ , UoHZ'

UOHZ UOH; UoHZ'

is trivially cartesian, so the square

U o u r UoHZ , UOHZ'

UOHZ ~ UoHZ' UoHr

is also cartesian. It follows that

M*(C\U.(Vo.~))M*(,7~,)(.;)(~o.Z,.~) : ~u(M)*(Uo.~)(M*(,~,)(-~')(~o.~, ~,)) : .

. . . . s on r

Finally, I have

OzM* ( r = r163 M)* ( U oH r )A z,(,~')

and as r is a morphism of Mackey functors, it is also

OzM*( r )(m') = ( N o U)*( r AZ,(m') = ( N o U)*( r )Oz,(m')

Thus OzM*(r = (N o U)*(r The previous proof shows that

AzM*(8) = s OH r

9.5. L E F T A D J U N C T I O N 201

I will now prove that

Az, M , ( r = s oH r

so that A will be a morphism of Mackey functors from M to s e U. Then 0 will be composed of A and of the morphism r o U from s o U to N o U. It will also be a morphism of Mackey functors.

First I fill the cartesian square

(2 YI ~ Z

a \ u . ( u oH z') , z ' ~lZ'

and I denote by i the morphism from G \ U . ( U OH Z) to II filling the commutat ive diagram

c \ u . ( u o . z )

o a \ u . ( u o . z ') rlz,

Z

Z'

L e m m a 9.5.3: I n t h e p r e v i o u s d i a g r a m :

�9 T h e m o r p h i s m i is i n j e c t i v e .

�9 I f m o r e o v e r II = Im(i)]_I H', t h e n IY is a u - d i s j o i n t H - s e t o v e r t h e se t G\U.(U oH z').

Proo~ Let G(~', (~,~))and ~ (~ , (~,,z,))be two elements having the same image under i. If u' = uh and u~ = u lh l , this means that

G(ut,(?~,r : G(~i , (ul, r ) h - l z : h l l z l

Then Zl = h l h - l z , and there exists g ~ G such that ( g u , r = ( u l , r and

gu' = u' t. So there exists h' C H such that gu = u ,h ' and r = h ' - l r In those conditions

a(~'l, (~l,Zl)) = a(g,,', (~,,,,h-l~/) . . . .

But t --1 u l h ' h h l I = gUhhl 1 = gU'hl ~ = u lh 1 = ul


As ('gl, Zl) ~ U O H Z, it follows that h'hh~lZl = z1, i.e. h'z = h l h - l z . Then

G('u/1, (Ul,Zl)) ---- ~'(lt", (~]),"-1, htz)) ---- G(/s (/s

and this proves that i is injective. To prove the second assertion of the lemma, I must prove that the square

0 ~ U OH Z'

1 [VUo~Z, / \

U o . n' , u o . ( a \ u . ( v o . z ' ) ) U o H b

\ - - -

is cartesian. This means that if (u", 7r) 6 U OH II and (ul, z~) 6 U OH Z' are such that

(U O H b)(u ' , 71-) = //UoHZ,('U.1, Z'I) (9.2)

then ~r is in the image of i. The element ~r is of the form

with (u,z ' ) e U O H Z ' , and ho l z ' = r if ho e H and who = u'. Now b0r ) = G(u',(u,z')), and then

(u o .

On the other hand [ ( '))] 1]UoHZ,('ts = ~ I ,G "t/1,(Ul,Z 1

Now equality (9.2) means that there exists h 6 H such that

~" = ?.tlh h-IG(ul,(tll,Z'I)) = G ( u i h , ( U l , Z ; ) ) : G(?-t", (l/,,zt))

Then there exists g 6 G such that (gu~,z'l) = (u , z ' ) and gu lh = u'. Finally, there exists g 6 G, and h, h' 6 H such that

gulh = u' gu lh ' = u h'-lz ' l = z'

71"= [G(ut,(?~,Z")),Z] = [G(gulh,(gll, lh",hor ] . . . .

. . . . [G('u.lh,(ulht, ho~(z))),z] : [G(u' ,(u'h-lht , ho~(z))),z] . . . .

. . . .

Furthermore, as (u",Tr) E U OH II, it follows that if hi 6 H is such that u"hl = u", then hl~r = % which implies in particular that hlz = z. So (u", z) E U o H Z . Moreover

u " h - l h ' h o = ulh'ho = g - luho = g - l u ' = u lh = u"

'a" : 'U. 1 ]I

In those conditions

9.5. LEFT ADJUNCTION

It follows that h-~h'hor = r = r and then


To prove the equali ty

Az, M.(~b) = s o . ~)Az

I choose rn E M(Z). Then

203

Then U oH r is a morphism of G-sets over U oH Z' from (U OH Z , (U oH r to (U OH Z' , ~rz,). Moreover U.(U OH 5) is injective on the left orbits of G on U.(U OH Z):

indeed, if (u,(u',z)) has the same image than g.(u,(u',z)) = (gu,(gu',z)), then

gu = u and there exists h E H such that gu' = u'h and hS(z) = r But there

exists h0 such that u = u'ho. As gu = u, I have also gu' = u', and then (gu, (gu', z)) =

As Uo/4~ is a morphism in Z)u(UoHZ'), I have the following equality in f~u(M)(Uos Z')

M*(~z)(m)(uo.z,(uT-s = M.(U oH r

Moreover

M.(U oH r = M.(b)M.(i)M*(i)M*(a) . . . .

. . . . M.(b)M*(a) + M.(b) (1 - M.(i)M*(i))M*(a)

If j denotes the injection from II ' into l-I, I have 1 - M.( i ) i* ( i ) = M.(j)M*(j), and

M.(b) (1 - M.(i)M*(i))M'(a) = M.(b j ) i*(a j )

But ( I I ' , b j ) is a v-disjoint G-set over G\U.(U OH Z'). The image of M.(bj) in Qu(M)(U OH Z') is then zero. It follows that

M.(U OH ~)M*(~z)(m)(uo.z,,~z,) = M.(b)M*(a)(m)(VosZ',~z,) . . . .

. . . . M*(~z,)M.(d))(m)(UonZ,,.z,) = Az, M. ( r

proving that A z , / . ( r = s oH ~b)Az

so A is a morphism of Mackey functors from M to f~u(M) o U.

Now I have a correspondence A : 0 ~

HomM~ck(H)(M, N o U) ~ HOmM~ck(a)(s N)


and a correspondence B in the other direction. It is easy to see that those constructions are functorial in M and N. The theorem will now follow, if I prove that they are inverse to each other. It is equivalent to check the relations on unit and co-unit (see Mac-Lane [10] chp. IV): if 0 is the identi ty of N o U, I must check that B o A(O) = O. Similarly, if ~b is the ident i ty of s I must check that A o B(~b) = ~b.

But A(O) is the morphism from s o U) to N defined on a G-set X by

m(yd) E s o U)(X) ~ N.(fx)N*(uv)Oa\u.v(m) E N(X)

- ( ) where (!/, f ) is a G - s e t over x , and m is an element of M G \ U . Y . In other words

A(O)x is defined by

m(v,S) C s o U)(X) ~-+ X, ( fx )N*(uz) (m) �9 N (X)

Then B o A(O) is the endomorphism of N o U defined on an H-set Z by

,~ �9 (N o U)(Z) H A(O)uo,,Z~z(,j

Moreover I have here

Az(n) = (N o U)*(~lz)(n)(uo,z.,z) = N*(U OH •z)(n)(Uo,Z.,z)

So A(O)UogAz(n) : N.((~Z)Uonz) N*(uUo,z)N*(U OH ,z)(n)

But (lrZ)UonZ is the identi ty of U OH Z. And as u and ~ are the unit and co-unit of the adjunction of Z ~-~ U OH Z and Y H G\U.Y, I have

(U O H TIz)IIUoHZ = IduoHz

So B o A(O) is the identi ty of N o U. Conversely, if ~b is the identi ty of s then B(~b) is the morphism from M to

s o U defined on the H-set Z by

m �9 M(Z) ~ Iz(m)

The endomorphism A o B(~b) of s is then defined for a G set X, a set (Y, f ) over X, and an element ra �9 M(G\U.Y) by

re(v,]) C s ~ s

But

Aa\v.v(ra) = M*(Tlc\v.y)(m)(uon(a\u.z).,aw.r)

Let e be the map from Y to Y defined by e(y) -- (y, tidy)]. It is clear that the square \

l/y Y U , oH t ~ \ u . Y )

e ! ~TrG\U. Y

z , u o , (a \U.V) Yy

9.6. THE F U N C T O R S Su(M)

is cartesian, and then

s = M*(G\U.uv)M*(~a\u.v)(m)(z,e)

Thus

205

s )s M)*(uz )Aa\v.y(m) : M*(G\U.uz )M*(rla\u.z)(m)(z, Txc )

But

so f x e = f . Moreover rlatu.z(G\U.uy ) is the identity map, because u and r/ are the unit and co-unit of the adjunction of Z H U OH Z and Y ~ G\U.Y. Finally

s M ) , ( f x )s M )*( uy )AakU.Z(m ) = m(z,S)

and AoB('~b) is the identity morphism of s This completes the proof of theorem 9.5.2. �9

Remark: Let X be a G-set. The expression of the colimit over Du(X ) shows that if M is a Mackey functor for G, then

s : ( | M ( G \ U . Y ) ) / f l YLXx(UIH)

where J is the submodule generated by the submodule

M.(<,)(M(z)) whenever (Z, a) is a u-disjoint H-set over G\U.Y, and by the elements

m - M, (G\U.a) (m)

whenever .~ ~ M ( G \ U . Y ) and ~ : (Y, f) ~ (Y', f') is a morphism of G-sets over X x (U/H) which is injective on each G-orbit.

9 . 6 T h e f u n c t o r s S u ( M )

To build right adjoints, I need the following dual definition:

Def in i t ion : If M is a Mackey functor for H, and (Y, f) is a G-set over Y / H , I set

S u ( M ) ( Y , f ) = ~ KerM*(a) (Z,a)

where the intersection runs over u-disjoint H-sets ( Z, a) over G\ U.Y .

L e m m a 9.6.1: The correspondence

(Y, f ) ~ Su (M) (Y , fu)

c~ : ( Y , f ) ~ (Z,g) H M*(G\U.a ) : M(G\U.Z ) ~ M ( G \ U . Y )

i nduces a c o n t r a v a r i a n t f u n c t o r f r o m Du(X) to R - M o d .

206 CHAPTER 9, ADJOINT CONSTRUCTIONS

Proof i [ have seen that if a is a morphism in /Pu(X) from (K.f) to (Z,g), and if (T, a) is a u-disjoint //-set over C\U.Y, then setting/3 = G\U.c~, the H-set (T,/3a) is a u-disjoint set over G\UZ. Thus if m. E $u(M)(Z,g), then M'(fla)(rn.) = 0 = M*(a)M*(/3)(m), and this proves that M*(~3)(n~)is in 8u(M)(Y, f) . m

L e m m a 9.6 .2: If 4) : X ~ X ~ is a m o r p h i s m o f G-sets , t h e n b i n d u c e s an i s o m o r p h i s m

S~ : 8u( M ) o ~u, . ( iJ) --+ 8u( M)

and a n a t u r a l t r a n s f o r m a t i o n

s ~ : & , ( M ) o z~.(+) -+ $~,(~.v~)

Proof: The first assertion is clear: let (}, f ) be an object of ID0-(X), and let (Y', f ' ) = Du,.(4)(Y,f) = (Y, 4f) . As (~f)u = fu, the sets U.Y and U.Y' coincide, and the u-disjoint sets over G\U.Y and G\U.Y' are the same. Thus Su(M)(Y, f ) = S~,(M)(Y',y).

For the second assertion, let (Y', ]") be an object of lPu(X'). I fill the cartesian square

g y - - - - , y '

X , X'

Then if (Z',c~') is a u-disjoint H-set over G\UY', and if I fill the cartesian square

cr

Z , G\U.Y

,2 1 1 6 ' \ U.a

Z' --, G\U.Y' I

I have seen that (Z, a) is u-disjoint over G\U.Y. Then if rn 6 $u(M)(Y, f) , I have

M'(~')M.(C;\U.~)(,~) : M.(#)M'(~)(~.) = o

and M.( G\ U.a ) induces a morphism S~,d, ) from 8u( M)(}s f) =$u( lVl)oD~r( ~ )( Y', f ') to Su(M)(Y' , f'). It remains to see that this construction is functorial in (Y', J"): but if a ' : (Y ' , f ' ) ---+ (Z',g') is a morphism in Z)u(X'), then the diagram

b(y,.f,) Su(M) o Db(~)(Y' , f ' ) ------, Su(M)(Y ' , f ' )

S~(M) o ~(r 1 I $v(M)(~') $u(M) o lPg.(c~)(Z',g') -~ Su(M)(Z',9')

,5'[~x,,9,)

is commutative: indeed, this is equivalent to say that if m 6 8u(M)(Y, f ) , then

M*(G\U.a')M.(GiU.a) = M.(G\U.b)M*(G\U.a)

and this equality follows from lemma 9.3.3. This proves the second assertion. ,.

9.7. THE FUNCTORS Ru(M) 207

9 . 7 T h e f u n c t o r s g u ( M )

The previous section leads to the following definition, which is dual to the definition of L;u(M):

Def in i t ion : If X is a G-set, and M a Machey functor for H, I set

Ru(M)(X) = lim s f) +______ (Y,I)~vu(x) ~

If r : X --+ X ~ is a morphism of G-sets, then the isomorphism Sr induces a morphism

lim Su(M)(Y' , f ' ) -+ lim Su(M)oDu.(~)(Y,f) --+ lira 8u(M)(Y,I) <-------- ( ' < . - - - -_

(Y- ' , f ' )eDu(X') ~ (Y , f )EDu(X) ~ (Y , f ) eDu(X) ~

i.e. a morphism Tgu(M)*(~) from "R.u(M)(X') to Tgu(M)(X). Conversely, the natural transformation S r induces a morphism

lira Su(M)(Y,f)--+ lim Su(M)oD~(r lim Su(M)(Y',I ') ( + - - - - - (

(Y,f )eDu( X ) ~ (Y ' , J ' )eDu( X ' ) ~ (Y ' , f ' ) eDo ( X') ~

i.e. a morphism ~u(M).(c~) from 7r to 7r

Propos i t ion 9.7.1: The above definitions turn T @ ( M ) into a Mackey functor.

Proof: It is clear that T~u(M) is a bifunctor on G-set. It suffices then to check axioms (M1) and (M2) of Mackey funetors. For (M2), let

a

X , Y

1 Z , T

d

be a cartesian square of G-sets. The module T@(M)(Z) identifies with the set of sequences rn(s,~/, indexed by the objects of Du(Z), such that m(E,~l E M(G\U.E), and such that

: 0

whenever (T, a) is a u-disjoint H-set over G\U.E, and such that

* ? M (G\C.~)('~(F,~/) = m(E,~)

if a is a morphism from (E, e) to (F, f ) in the category Dr (X) . The image under gv(M)*(b) of the sequence m(u,~) is the sequence m}F,i ) indexed

by the objects of De(X) , and defined by

!

~(F,f ) = m(F,~i)

The image under "R.u(M).(a) of this sequence is then the sequence ra{~,,y,) indexed by the objects of 2)u(Y), and defined by filling the cartesian square

and setting

O �84

F , F '

X ~ Y E

I! ~(~,,:,) = M.(a\U.~)( .<. , : ) ) = M.(G\U.~)(m(F,~:))

On the other hand, the image under T@(M).(d) of the sequence m(E,~) is the sequence nlE,,r ) indexed by the objects of Z)u(T), defined by filling the cartesian square

and setting

E , E '

1 l' C e

Z , T

"{~',~'/= M.(O\U.5)(-~(E,~))

The image under T@(M)*(c) of this sequence is the sequence hi'r, y, ) indexed by the objects of ~Du(Y), defined by

l] ! n ( F , , f , ) = r ~ ( F , , c f , )

As the squares C%

F , F '

l Z , T

g

are cartesian, because they are composed of two cartesian squares, I have

/ f '~(F,,~s,) = M*( O\ U~ )(~(F,~:))

so I have 7r = T@(M).(a)TCu(M)*(b), and (M2) holds.

To check (M1), I observe as before that an object (E,e) in 7)u(XIjY) is the disjoint union of an object (El, el) of Igu(X) and an object (E2, e2) of / )u(Y) . Let i~ and i2 be the respective injections from E1 and E2 into E. If m E g u ( X L[ Y), and if (E, e) is a.n object of Du(X II Y), let r~ = m(Ed), and

Tt 1 = M*(G\U.il)(n) 7~2 = M*(G\U.i2)(n)

so that n = M.(G\U.il)(n~) + M.(G\U.i2)(n2).

9.8. R I G t t T A D J U N C T I O N 209

As il is injective, it is a morph ism in ~)u(XI_[Y) from (E~ , f i l ) to ( E , f ) . Thus rt 1 = m ( E x , f i l ) . But I have a cartesian square

il E1 ~ E

X ~ X I I Y 7 v

Z X

which proves tha t f i l = ix f l , and that

R v ( M)*( ix )(m )(El,f,) = m(e,,i~ fl ) = m(E~,:i~) = n~

Similarly, for a sui table map f2, I have

TCu(M)*(iy)(m)(E2,:~) = n2

Then if 7r = 0 and 7r = O, I have n l = n2 = O, and then n = 0 for any (E , e ) , so m = 0. Then the map

n u ( M)*( ix ) | 7r M)*( iz )

is injective, and l e m m a 9.4.2 shows that ~ u ( M ) is a Mackey functor.

9.8 R i g h t a d j u n c t i o n

N o t a t i o n : If 0 : M ~ M ~ is a morphism of Mackey functors for the group H, and i f X is a G-set, [ de/ine a ma p T~u(O)x from ~ u ( M ) ( X ) to "Fgu(M')(X) by setting, for m C T @ ( M ) ( X ) , and fox" an object ( Y , f ) of~?u(X)

~v(O)x (m)( r , s ) = Oa\v.r (m(r , s ) )

L e m m a 9.8.1: This def init ion turns T~v(O) into a m o r p h i s m of M a c k e y functors f rom ~v(M) to 7~u(M').

Proof: First , the map Tgu(O) is well defined: if a : Z ~ G \ U . Y is u-disjoint , then

M*(a) T@(O)x(rrz)(rd) = M*(a)Oa\uy(m(rd)) = Oa\u.vM (a)( (rd)) = 0

Moreover, if c~ : ( Y , I ) ~ (Z,g) is a morph i sm in D u ( X ) , and if m E M ( G \ U . Z ) , then

M*(~) (~(0)~(.~)(~,~)) : M*(~)O~\~.~(.~(~,~)) . . . .

. . . . Oa\v.r M*( ~ )( m(z,~) ) = Oa\v.r(m(Y,S)) = ~r O)x(m)(yj)

Now let ~ : X ~ X ' be a morph i sm of G-sets. Then , if m' E Tgu(M)(X' )

~u(M)*(g))~u(O)x,(~')(r,f) = 7r )(r,~?) = Oa\v,r( (r,;'s)) . . . .


If moreover (Y', f ' ) is an object of Du(X'), and if m q M(G\U.Y') , then let Y, f , and a fill the cartesian square

a y , }i,

X , X '


7 ~ ( M ) . ( ~ ) Z e ~ ( O ) x ( , . ) ( ~ , ~,) -- M.(G\U.~)(r~u(O).~(~,~)(y,s)) . . . .

. . . . M.(G\U.a)Oc\u.y(m(y,f)) = Oa\v,M.(G\U.a)(m(y,f)) . . . .

. . . . o~\~, (7~:(M)( .~)(~, :,)) = r~u(o)x,r~.(M)(.~)o:,,:,)

which proves the lemma. "

T h e o r e m 9.8.2: Let G and H be f in i te groups , and U be a G-se t -H. T h e f u n e t o r M ~ Tgu(M) is r ight adjo int to the fune tor N ~ N o U.

Proof: The proof is dual of the proof of theorem 9.5.2. Let 0 be a morphism of Mackey functors from M o U to N. Then for any H-set Z, I have a morphism

Oz : (M o U)(Z) = M(U o . Z) ~ X(Z)

In particular, if (Y, f ) is a G-set over X • (U/H), I have a morphism

Oc\u.y : M(U oH G\U.Y) --+ N(G\U.Y)

Composing this morphism with M.(vy), I have a morphism

Oa\u.yM.(vv) : M(Y) + M(U oH G\U.Y) + N(G\U.Y)

If (Z, a) is a u-disjoint H-set over G\U.Y, I have the diagram

M.(uy) 0cW.r M(~' ) , M ( U o , G\u .}" ) , N(C, \U.Y)

1 M*(U ~ [ N'(~) M(O) , M(U OH Z) , N(Z)

Oz

This diagram is commutative: the left square is because (Z, a) is u-disjoint, and the right square is because 0 is a morphism of Mackey functors. So this gives a morphism from M(Y) to Su(N) (Y , f ) , that can be composed with the morphism M*(fx) from re(x) to M(Y).

9.8. RIGHT ADJUNCTION 211

If oe: (Y, f) -+ (Z,g) is a morphism in ~9~(X), I have the diagram

M*(gx) M.(uz) Oa\g.z M(X) , M(Z) , M(UoHG\U.Z) , N(G\U.Z)

> M(Y) , M(UoHG\U.Y) , N(G\U.Y) M(X) M*(fx) M.(uy) Oa\u.y

The left square is commutative because gc~ = f. The middle square is commutative by lemma 9.3.2. The right square is commutative because 0 is a morphism of Mackey fimctors.

So I have a morphism e x from M(X) to Tgu(N)(X). If 4) : X + X ' is a morphism of G-sets, and if rn C M(X), then ~x(rn) is the sequence indexed by the objects (Y, f) o f / ? u ( X ) defined by

~bx(m)(y,f) = Oa\urM.(ur)M*(fx)(m)

Then 7~v(a).(4))Ox(m) is the sequence rn{r,,y, ) indexed by the objects of Du(X' ) , and defined by filling the cartesian square

g

Y , y '

X , X '

More precisely

m}y,,f,) = M.(G\U.a)(~bx (re)if, i)) = N.(G\U.a)OG\uyM.(py)M*(fx)(m)

On the other hand, the element r is the sequence n(y,],) defined by

As the square


* !

n(y,,f,) = Oa\u.y,M.(uy,)M (fx,)M.(4))

a

Y ~ y '

X , X' r

M*(f~,)M.(r = M.(a)M*(fx)

Moreover as ~ , o a ; (U o~ ( G W . ~ ) ) o ~Y, Ihave

M . ( ~ , ) M . ( a ) = M.(U o . ( C \ U . a ) ) M . ( ~ )

Finally as 0 is a morphism of Mackey functors, I have

Oa\v.y,M. (U OH (G\U.a)) = N.(G\U.a)Oa\u.y

and finally

n(v,,l,) = N.(G\U.a)Oa\u.vM.(uv)M*(fx)(m) = m}v,4, )

This equality proves that

7~v(M).(r = ~ ,x ,M. ( r

Now if rn' E M(X') , then ~'x,(m') is the sequence indexed by the objects (Y', f ' ) of Du(X'), defined by

~x,(rn')(v, f,) = Oa\v.y,M.(uy,)M*(f~x,)(rrz ')

Its image under 7r162 is the sequence indexed by the objects (Y, f ) of Du(X), defined by

, =

But (~f)x , = e fx , and then

(T@( M)*( r (y,f) = Oaxu.y M.(uy )M*(fx )M'( r = ~x (V*(r

which proves that 7r162 = $xM*(r

Thus ~b is a morphism of Mackey functors from M to ~u(M). Conversely, if ~b is a morphism of Mackey functors from M to 7~u(N), I have for

any G-set X a morphism %bx from M(X) to "fCu(N)(X). In particular, if Z is an H-set, I hame a morphism

~ v o . z : M(U o . Z) -+ Tau(N)(U o . Z)

An element of TQs(N)(U OH Z) is a sequence n(r,f) indexed by the objects of 7)u(U OH Z), with n(v,f) E N(G\U.Y). I can then consider the element

and its image under N.OTz), which is an element of N(Z). I get a morphism Oz from M(U oH Z) = (M o U)(Z) to N(Z), defined by

To prove that 0 is a morphism of Mackey functors, it suffices to observe that 0 is composed of the morphism ~/J o U from M o U to ~u(M) o U deduced from ~b, and of the morphism O from 7r o U to N, defined on the set Z by

,~ ~ (T~u(X) o V)(Z) = 7r o . Z) ~ Oz(~) = X.(~z)(n(uo.a .~) )

It suffices then to prove that this is a morphism of Mackey functors. So let r : Z --* Z ~ be a morphism of H-sets. The element

( ~ ( N) o u) . ( r )( ,, )(Vo,<Z,,~,~)

9.8. RIGHT ADJUNCTION 213

is obtained by filling (trivially) the cartesian square

Then:

so that

UOHr U OH Z ) U OH Zt

U O H Z ) UOH Z! U O H r

o ~ , ( ~ ( N ) o v).(r = x.( ,~,)N.(G\U (U o. ~))(~(~o.~,.z>)

As rlz, (G\U.(U OH r : CzIz, this is also

thus Oz, (7~u(N) o U),(r = N,(r Conversely, if n' E Tgu(N)(U OH Z'), then

= N . ( r

(, ) N*(r = N (r n(uouZ,,~,z)

On the other hand (TZu(N) o U)*(r is the sequence indexed by the objects (Y, f) of Du(U OH Z), defined by

* t

(Y,(Uonr It follows that

* ! !

But I have already observed in the proof of theorem 9.5.2 that the morphism U OH r

is a morphism in 7?u(U OH Z') from (U OH Z, ((U OH r to (U OH Z',Tcz,). Then in Tgu(N)(U OH zr), I have

n' ~ : N.(U oH r (Uo.Z,(UoHr

SO that * l r

I have again the commutative diagram

a \ u . ( u o . z )

o G\U.(U oH Z') Vz, ' Z'


and lemma 9.5.3 shows that i is injective, and that if II = Ira(i) [I 11', denoting by j the injection from 11' into II, the set (II', bj) is v-disjoint. I have then

N*(r Sz,) = N.(a)N*(b)(nluouz,,r . . . .

. . . . ) . . . .

As n' E Tgu(N)(U oH Z'), and as (II',bj) is v-disjoint, the second term is zero, and this gives

N*(r = N.(ai)N*(bi)(nluo,Z,fz)) . . . .

. . . . N . ( . . ) N ' ( G W ( V o .

Finally, I have proved that

and O is a morphism of Mackey functors.

To complete the proof of the theorem, it remains to state that the correspondences A : 0 F-+ ~b and B : ~b ~-+ 0, which are clearly functorial in M and N, are inverse to each other. It suffices to check that if 0 is the identity, then so is (B o A)(O), and that if ~b is the identity, so is (A o B)(r

Let 0 be the identity endomorphism of M o U, and ~b = A(O). If X is a G-set, the morphism ~bx from M(X) to 7r o U)(X) maps the element m C M(X) to the sequence '~bx (rn)(rj) defined by

~x(rn)(r,]) = M.(vy)M*(fx)(m) e M(U OH (G\U.Y)) = (M o U)(GkU.Y) \

Then if O' = B(r I have for any H-set Z and any m E M(U oH Z)

O'z(m ) = ( M o U).(7?z ) (ZbUo.z(m )(yo.z,,~z) )

which gives e'z(m) : M.(U oH 7lz)M.(vUouz)M* ((~rz)x)(m)

But I have seen that (U OH rlz)VUo.Z is the identity map, as well as (Trz)x. Thus O} is the identity, and so is 0'.

Now if ~b is the identity endomorphism of TCu(M), and if 0 = B(~b), then for any H-set Z and any

m' C (Tiu(M) o U)(Z) = ~u(M)(U OH Z)

I have

If r = A(O), if m E R u ( M ) ( X ) , then r is the sequence indexed by the objects (I/, f ) of Du(X) defined by

r = Oa\u.yT~u(M).(vy)TCu(M)*(fx)(m)

9.9. EXAMPLES 215

Sett ing m ' = T~u(M).(uz)Tiu(M)*(fx)(m), I have then

But nu(M)*(fx)(m) is the sequence n(E,e)indexed by the objects of T)u(Y), defined by

n(E,e ) = m(E,~xe ) Moreover rn' is the image of n under Tiu(M).(uy). The component m}uo~(a\u.z),,~a\uy ) is then obtained by observing that if e is the morphism from Y to Y defined by e(y) = ((y, fu(Y)), then the square

is cartesian. Then

fly Y U , o . ( u \ u . ~ )

e ! ~rCaiu.Y

Y , U OH (a\U.Y) Py

, )( ) m(uoH(G\U.y),rCc\uy ) = M.(G\U.uy m(y,~xe )

As moreover fxe = f , I have then

r = M.(rla\u.y )M.( G\ U.uy )(m(zj) )

As finally rla\u.z(G\U.uz ) is the identi ty map, I have r = re(y j ) . Thus ~ ' is the identity, which completes the proof of the theorem. �9

R e m a r k : Let X be a G-set. The expression of the l imit over :Du(X) shows that if M is a Mackey functor for G, then Tiu(M)(X) is the set of sequences m(gj), indexed by G-sets (Y, f ) over X x (U/H), such that rn(y,f) E M(G\U.Y), and

M*(a)( -~(y , s ) ) = 0

whenever (Z, a) is a r-disjoint H-set over G\U.Y, and moreover

m(z j ) = M*(G\U.a)(m(y,j,))

whenever a : (Y, f ) --+ (Y', f ' ) is a rnorphism of G-sets over X • (U/H) which is injective on each G-orbit.

9 .9 E x a m p l e s

9.9,1 Induction and restriction

Let G be a group, and H be a subgroup of G. If U is the set G, viewed as a G-set-H, then the fimctor N ~ N o U is the restriction functor for Mackey functors from G to H.


As U/H = G/H, an object (Y, f ) over U/H is of the form Ind~Z, with Z = f - l ( H ) (see lemma 2.4.1). An object of 7?u(X) is then an H-set Z, with a morphism from Ind/~Z to X, i.e. a morphism from Z to Res/~X. Moreover, the group G acts freely on U, so on U.Y. Thus if c~ is a morphism of G-sets over X x (U/H), then U.c~ is injective on each left orbit. In other words, the category 7?u(X) identifies with

H-set~aes~X. Moreover, the set G\U.Y identifies with Z, by the map

G(u,y) E G\U.Y H u-ly E g

This map is indeed surjective, because if z E Z, then z is the image of G(1,z) . Conversely, if u-ly = u'-ly ', then

G(u, y) = G ( u , u u ' - l y ') = G(1, u '-~y ') = G(u' , y')

It follows that U OH (G\U.Y) ~- Ind~(G\U.Y) ~_ Ind/~Z ~_ Y

and that vg is an isomorphism. Then if the H-set (T, a) over G\U.Y is v-disjoint, the image of U OH a is disjoint of the image of yr . As vy is surjective, I have U OH T ---- 0. But U OH T = Ind~rT, and then T = 0.

In part icular , I see that

Qu(M)(Y, f ) = M(G\U.Y) = M(Z) = Sv(M)(Y, f )

As ReSaH X is a final object of 7?u(X), I see that

Cu(M)(X) = lim M(Z) = M ( R e s ~ X ) ze~u(x)

As ReSaH X is an initial object of :Du(X) ~ I have also

T@(M)(X) = M(ReSUH X)

and the following isomorphisms follow easily:

s ~- I n d , ( M ) _~ Tiu(M)

I recover that way the adjunction properties of induction and restriction. Now switching the roles of H and G, I consider V = G as an H-set-G. The functor

N o V is then the induction functor for Mackey functors from H to G. As V/G = . , an H-set (]I, f ) over V/G is just an H-set , and V.Y = V • Y. The group H acts freely on V, so if a is a morphism of H-sets from Y to Z, then V.c~ is injective on each left orbit of H on U.Y (because h.u = u implies h = 1). The category :Dv(X) identifies then with H-setJ , x , and has a final object X.

If Y is an H-set , then H \ V . Y identifies with Ind~Y. Let (T ,a ) be a v-disjoint G-set over H\V.Y . If t E T and a(t) = H(v, y), then as G acts freely on V, I have (v, t) C V oG T, which contradicts the hypothesis on (T, a). So T = 13.

In those conditions, it is clear that

Lv(M) ~ Res~,(M) _~ 7~v(M)

and I recover once again the adjunction properties of induction and restriction.

9.9. EXAMPLES 217

9 .9 .2 I n f l a t i o n

Let N be a normal subgroup of the group G, and H be the quotient G/N. If U is the set H, viewed as an H-set-G, then the functor M ~ M o U is the inflation functor for Mackey functors: indeed, if X is a G-set, then U oa X = X N.

As U/G = *, an object of 7?u(X) is just an H-set over X. If (!/, f ) is such a set, then U.Y = U • Y, and H\U.Y identifies with I n k Y . As H acts regularly on U, if ~: (Y, f) ~ (Z,g) is a morphism of sets over X, then U.a is injective on each orbit of H on U.Y. So X is a final object in 7Pu(X), and then for any X, I have

s = Qu(M)(X) ~u(M)(X) = Su(M)(X)

Furthermore as U oc (G\U.Y) = ( I n ~ Y ) N _~ Y, the morphism uy is an isomorphism for any Y. So an object (T,a) over H\U.Y = In~HY is , -dis joint if and only if U oa T = T N = O. Finally:

P r o p o s i t i o n 9.9.1: L e t N b e a n o r m a l s u b g r o u p of t h e g r o u p G, a n d H = GIN. I f M is a M a c k e y f u n c t o r for G, a n d X is an H - s e t , I se t

MN(x) = M(In~HX)/ ~ M.(a)M(T) (T,a)

Mar(X) = ~ KerM*(a) (T.~)

w h e r e t h e s u m a n d i n t e r s e c t i o n r u n ove r t h e G- se t s (T ,a) ove r InfaH X such t h a t T N = O.

I f ~ : X --~ X ' is a m o r p h i s m of H - s e t s , t h e n t h e m a p s M . ( I n ~ r a n d M * ( I n ~ r i n d u c e m o r p h i s m s b e t w e e n MN(X) a n d MN(x') , a n d b e t w e e n Mar(X) a n d Mar(X'), w h i c h t u r n M N a n d MN in to M a c k e y f u n c t o r s for H. T h e f u n c t o r M ~-* M N is lef t a d j o i n t to t h e f u n c t o r L ~-~ I n , L, a n d t h e f u n c t o r M ~ MN is r i g h t a d j o i n t to i t .

R e m a r k : Wi th the notations of Th~venaz and Webb (see [14], [15]), it is easy to identify M x with M +, and MN with M - : any set T such that Tar = 0 is indeed isomorphic to a disjoint union of sets of the form G/K, for N g K, and it follows easily that if L/N is a subgroup of H = G/N

MN(L/N) = M ( L ) / ~ tL-M(I() MN(L/N) = A Ker r i K

where the sum and intersection run on the subgroups K of L (which give morphisms from G/K to G/L) not containing N.

9 .9 .3 C o i n f l a t i o n

Let N be a normal subgroup of G, and H = G/N. Let V be the set H, viewed as a G-set-H. Then if Z is an H-set , the set V on Z identifies with I n , Z, and the functor M H M o V is the functor that I have denoted by p~ (and denoted by fl! by Th6venaz and Webb see[15].5).

Here again, the set V/H is trivial, so if X is a G-set, an object (Y,f) of Dr(X) is just a G-set over X. The set V.Y is the product 1/ x Y. Let c~ : (Y,f) ---, (Z,g) be a

morphism of sets over X. Then V.a is injective on the left orbits of G on V.Y if and only if the hypothesis

imply gy = y. But the stabilizer in G of any point v of V is equal to N. Thus V.a is injective on the left orbits of G on V.Y if and only if c~ is injective on the orbits of N on Y. a morphism in l ) v (X) from (Y , f ) to (Z,g) is then a morphism a of sets over X, which is moreover injective on each orbit of N.

Now the set G\V.Y identifies with N \ Y . Then V OH (G\V.Y) identifies with InfaH(N\Y). The morphism uy maps y to its orbit Ny by N. In part icular , it is surjective. So an object (T ,a) over G\V.Y is ~,-disjoint if and only if V OH T = In~HT = ~, i.e. if T = ~. It follows that

Qv(M)(Y, f) = M ( N \ Y ) = Sv(M)(Y, f )

Finally:

P r o p o s i t i o n 9 .9 .2: Let N b e a n o r m a l s u b g r o u p of G, and H = G/N. I f V is t h e s e t H, v i e w e d as a G - s e t - H , a n d if X is a G - s e t , t h e n :Dv(X) is i s o m o r p h i c to t h e c a t e g o r y w h i c h o b j e c t s are t h e G-se t s over X , and t h e m o r p h i s m s are t h e m o r p h i s m s of s e t s over X w h i c h are m o r e o v e r i n j e c t i v e on each o r b i t of N .

I f M is a M a c k e y f u n c t o r for H, t h e n

s = lira M ( N \ Y ) Tiv(M)(X) : lim M ( N \ Y ) (vJ )~Vv( X ) (v,f )~Vv( X ) ~

N o t a t i o n s : K, ordered by the following relation

L Z L ' ~ {

If M is a Mackey functor for the group

lim L6wN(K)

Let K be a subgroup of G. I denote by wN(K) the set of subgroups of

L C L ' L A N = L'NN

H = G/N, I denote by

M ( L N / N )

the quotient of OLc_KM(LN/N) by the submodule generated by the elements of the .L'N/N form ~ L N / N m - - m , for L ~_ L' and m 6 M(LN/N) .

The group K acts on l im M(LN/N) , and I denote by LEwN(K)

< ~ ( M ) ( K ) = ( lira M(LN/N))~ . L6wN(K)

the biggest quotient on which K acts trivially. If L is a subgroup of K, and if m is an element of M( L N/N) , I denote by m f the image of ,~ in ~ ( M ) ( A ' ) .

ff K C K', I denote by t K' (resp. K' _ r K ) the map from L~(M)(K) to f i ( M ) ( K ' ) (resp. from L~4(M)(K' ) to f i ( M ) ( K ) ) defined by

" K' ~K'z IK', {x L'N/N ,~K = ~ 7"(K~nLON/Nrn )KnxL,

xEK\K'/L'

9.9. EXAMPLES 219

[ f x ~ G, I s e t ~ ( m f ) = ~x v K

Dually I denote by lira M(LN/N) (_._._

L6wN(K) ~

the set of sequences m[ indexed by the subgroups of K, such that m~ E M(LN/N) , L'N/NI--n~ = mhn " whenever L -< L'. The group K acts on this set, and such that I'LN/N k,,~L,]

and I denote by f i ( M ) ( K ) = ( ~im M(LN/N) ) K

LEwN(K) ~

the set of its fixed points. K' r~/) fiH(M)(K) to If I< C_ K', I denote by t K (resp. the map from ~ ( M ) ( K ' )

(resp. from ~ ( M ) ( K ' ) to Z~(M)(K)) defined by

K' K' -L'N/N , x K ~ z K' ,\K' tK (tK m)u = E t (L ,AXK)N/N( m,L ,~nK ) ( r h. m )L : m L

x E K \ K ' / L '

Finally, if x E G, and if m is an element of fiH(M)(K), I denote by ~m the dement of ~a(M)(~K) defined by (~m)L K = ~(mn~

Proposition 9.9.3: Those definitions turn t~(M) and i~4(M) into Mackey f u n c t o r s for G. T h e f u n c t o r M ~-~ t ~ ( M ) is lef t a d j o i n t to t h e f u n c t o r L ~-~ p(L) , and the functor M ~-~ i~(M) is right adjoint to it.

Proof: I have to identify s and Tiv(M)(X) in the case X = G/K. If (Y, f ) is a G-set over G/K, I can choose a system of representatives S of G \ Y contained in f - l ( K ) . If s E S, then its stabilizer G, is contained in K , and I have a natural map #~ : gG, ~-~ g.s from G/G~ to Y. The union [ I ,es #~ is an isomorphism

: 11 a/a - - , Y

sES

If m E M ( N \ Y ) , then M*(N\#~) (m) E and

N\(G/G~) ~_ (G/N)/(G~N/N)

The map Av defined by

Ay(m) = ~ M*(N\#~)(m)~ E ~a(M)(K) sES

does not depend on the choice of the system S inside f - l ( K ) : indeed, changing s in g~s, so that f(g~s) = g~f(s) E K forces g~ E K. Then M*(N\#~)(m) is replaced by its conjugate under g~, which has the same image in f i (M) (K) .

This map passes down to the quotient s if a : (Y, f ) ~ (Z,g) is a morphism of G-sets over G/K which is injective on each N-orbi t , I choose a system of representatives T of G\Z contained in g-l(K). Then for any s E S, there exists a unique t~ 6 T such that a(s) E Gt~, i.e. a(s) = x~t~, for x~ E G. As ga = f , I have f (s) = x~g(t~), and then x, E I f .

Moreover, I have G~ C_ ~'Gt,, and if n E ~Gt , ( / N , then

= = = = =


If a is injective on each N-orbi t , I have ns = s, which proves that N n ~:' Gt, = N M G~, s o

Then let 7r~ be the map gG~ ~ gx~Gt, from G/G~ to G/Gt,, and 7r be the map

II ~s: II c / c ~ ~ I I c / c , sES sES tET

As

The square # Y

YLes G/G~ , Y

1 o I ~ t e T G / G t - , Z

#z

is commutat ive. As #y and #z are isomorphisms, it is cartesian. As moreover a is injective on N-orbits , by lemma 9.3.3 the square

N\#y N\I_I,~C/C~ - , N\Y

, N \ Z N\LIt~TG/G~ N\#z

is also cartesian. Then I have

M*(N\#z)M.(N\a) = M.(N\~r)M*(N\#z)

But if m E M(N\Y) , I have

M*(N\fy)(m) = (~ M*(f=)(m) sES

SO r / xs G * \ x s n i (

M.(N\~r)M*(N\fy) = @ [( tc , *"M (N\#v)(m)) Jc,, sES

As xs E I f , this is also ~ t ~ " M*( N\#y )(m )~:a,.. sES

and as G~ ~ ='Or,, this is (~ M*( N \ f y )(m)~ sES

i.e. Ay(m). Furthermore

M * ( N \ # z ) i . ( N \ a ) ( m ) = ( ~ M*(ft)) M.(N\a)(m) = AzM.(Nka)(m) tET

which proves tha t A passes down to the quotient.

9.9. EXAMPLES 221

Conversely, to the element ,rz~ E ~C~(M)(K) I associate the image A'(rn) of rn in s corresponding to the G-set G/L --+ G'/K over G/K. This map is welt defined, because if L ~ L' then the projection G/L --+ G/L' is injective on N-orbits: indeed, if gL' = ngL' for n E N, then n g E L' N N = L Cl N, so gL = ngL. Similarly~ if k E K, then the conjugation G/L ~ G/kL is bijective, so it is injective on the N-orbits.

If L C K, I can take {L} as a system of representatives of G\(G/L) for the computation of Ac/c, and then it is clear that AA' is the identity.

Conversely, the isomorphism #y : I]l~es G/G~ --+ Y is an isomorphism of sets over G/K, and it follows that A'A is the identity. This states the isomorphism

s ~- ,~(M)(K)

Tile formulae giving the transfers and restrictions from c~(M)(K) to ~a(M)(K') follow easily from isomorphisms k and A'. The isomorphism

and the corresponding formulae can be proved by an argument dual to the previous o n e . �9

R e m a r k s : 1) The set aaN(K) is seldom connected: indeed if L and L' are in the same connected component, then L N N = L' N N. For a subgroup J of K FI N, if I denote by co~,(J, K) the set of subgroups L of K such that L f3 N = J, then coN(J, K) is the connected component of coN(K) containing J , and I have the isomorphism

JC_KNN LEwN(J,K) J rnod. K

and ~ similar isomorphism for ;(M)~(K). 2) If K N N = {1}, then wN(K) has a biggest element K, and then

4 ( M ) ( K ) = M(KN/ N) =

It follows in particular that ~ ( M ) ( 1 ) = M(N/N). 3) It is easy to see that for any Mackey functor L for H

This follows from the fact that,

By adjunction, it follows that

G G puInitt(L) _~ L

a.t the level of H-sets

(Inf~Z) 'v~" Z

~C(L) ~ ' f t ~- L

That can be checked directly: let K be a subgroup of G containing N. If L is a subgroup of I f not containing N, and if rn r M(LN/N), the element m~" of z.~(M)(K) is equal to t~(m~) . And if H _D N, then H < Ix', and then m.~ = tH/NUn)I"''~/N" ,~ So if rr

is the projection from ~(M)(K) onto its quotient ~v(M)~V(K/N), tile morphism

c M(K/ N) c 4 ( M ) N ( K / N )


is surjective. Conversely, it is easy to see that the morphism

mr[ E ~ ( M ) ( K ) ~ { 0K/Nif N ~ L tn /N(m ) otherwise

passes down to the quotient t~v(M)N(K/N) , and induces an isomorphism inverse of the previous one. 4) The associativity of the product OH and the adjunction show that if G, H, and K are groups, if U is a G-set-H, and V is an H-set-K, then

s o s ~- s 7{u o Tgv ~- T~uouv

5) The Burnside functor b for the group H is such that

HomM~ck(u)(b, Nz) ~- N(Z)

if N is a Mackey functor for H, and Z is an H-set. Then if U is a G-set-H, and M a Mackey functor for G

HomM~k(,) (b, ( M o U)z) ~- M ( U OH Z) ~_ HomM~ck(a)(b, Muo.~Z) ~-- . . .

. . . ~-- HomM~k(G)(buo,Z, M) ~ HomM~k(H)(bz, M o U) ~- . . .

"~HomM:ck(a)( f -u(bz) ,M)

As this isomorphism is natural in M, it follows that

s ~- bUo.Z

where b in the left hand side is the Burnside functor for H, and in the right hand side the Burnside functor for G.

Chapter 10

Adjunct ion and Green functors

10.1 Frobenius morphisms

Let G and H be groups, and U be a G-set-H. If X and Y are H-sets, I have defined the maps

6 v (X x X , Y : O o H Y ) --+ ( g OH X ) x (U OH Y )

by 4,~(~,x,y) = ((~,,.), (~,y))

L e m m a 10.1.1: I f f : X --* X ' and g : Y --~ Y' are m o r p h i s m s of H-se t s , t h e n t he s q u a r e

U oH (X x Y) /

0 H ( f X g) / U

U oH (X' x Y')

~U X,Y )

)

~Tu, y,

(U oH X) x (g oH Y)

I (U OH f ) • (U OH g)

(u o . x ' ) • (u o . y')

(c)

is cartes ian .

Proof: Indeed, if

(~X,,y,(~,(xl, y')) = (([.70 H f)X (Uo H g)) ((UI,X),(u2, y))

then in U o H X ' , I h a v e (u , z ' ) = ( u l , f ( x ) ) , and in U o H Y ' I have (u,y') = (u2 ,g (y ) ) . Then there exists elements s and t of H such that

' f ( ) ' ( ) u . s = u l x = s . x u . t = u 2 y = t . g y

In those conditions, the element (u, (sx , ty ) ) is in U OH (X x Y): indeed, if r C H is

such that u.r = r, then U l s - l r s = u> As (u,~,x) E U oH X, I have s - i r s . x = x, so r .sx = sx. Similarly, since t t 2 ~ - l v t = U2, and since (u2, t) C U o H Y , I have t - l r t . y = Y, or r . ty = ty , which proves that r . (sx , ty) = (sx , ty) if u.r = u.

Moreover

u o , , (s • = (,,,

224

and

CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS

The injectivity of 6 u (see lemma 8.3.1) now shows that the square (C) is cartesian. X,Y The lemma follows. �9

Observing that if P is a Mackey functor for G, then

P(U Olt ( x • YI) = (P o u ) I x • Y) = (P o u ) y t x )

P ( U oH x x U oH Y ) = Puo, ,r (U o . x ) : (PuoHZ o U ) ( X )

lemma 10.1.1 shows that the maps

P . ( ~ y ) : (P o U)r (X) ~ (P,o,~. o U)(X)

and p, u u)(x) (PoU)~(x) (~x,Y) : (Puo~,r o ---,

induce for any Y morphisms from (P o U)y to PUoHr o U. Those morphisms are U moreover functorial in Y and P, in an obvious sense. Moreover, the injectivity of f x y

shows that

so (P o U)r is a direct summand of PUo,V o U. Then if M is a Mackey functor for H, there are morphisms

HOmM~k(H) (M, (P o U),~) ~ HOmM~k(H)(M, P u o j o U) (10.1)

The left hand side is also

HomM~k(H)(Mr, P o U) ~- UOmMack(G) ( ~-.u( My ), P)


HomM~k(a)(s ~- HomM~k(a)(s P)

Thus the functor HomM~fG) ( s

is a direct summand of the functor

Now Yoneda's lemma shows that s is a direct summand of s The left hand side of (10.1) is also equal to ?-f(M, P o U)(Y), and the right hand

side is equal to

HomM~k(H)(M, PUoHY o U) ~_ Homf~k(c) (s PUo.Y) . . . .

10.1. FROBENIUS MORPHISMS 225

The functoriality in Y shows that the above morphisms induce morphisms of Mackey functors

a ,p: (M,P o u) P) o U

and

such that VM,pAM,p = [d. Those morphisms are moreover functorial in M and P. If N is a Mackey functor for H, then by composition I have morphisms

By proposition 1.10.1, the left hand side is

HomM~r P o U) ~- HomM~k(a)(s P)


HomM~ok(a) (s 7-/(s P)) ~- HomM~c~(a)(s163 P)

Thus the functor HomM~k(~)(s

is a direct summand of the functor

HomM~k(c) (s M)+s N),-)

Now Yoneda's lemma shows that the functor s is a direct summand of the functor s M)+s

Furthermore I have morphisms

HomM~k(.) ((P o U)v, M) ~ HOmM~k(H)(PUo.r o U, M) (10.2)

The left hand side is also

HomM~k(H)(P o U, My) -~ HomM~k(c)(P, Teu(My))


HomM~(G)(Pcro/~Z,7~u(M)) -~ HOmM~,a)(P,~u(M)uoj)

It follows that Teu(My) is a direct summand of 7~u(M)UoHY. The left hand side of (10.2) can be written as

Homi~k(u)(P o U, Mz ) = 7-t( P o U, M)(Y)

and the right hand side as

HOmMadc(G')(Puo,V, T'gu(M)) "" HOmMa~k(G)(P, ~g(M)uong) . . . .

226 CHAPTER JO. ADJUNCTION AND GREEN FUNCTORS

So 7Y(P oU, M)is a direct summand of [TY(P,T@(M))] oU. Then

is a direct summand of

HOmM~k(H)(N, 7-l iP, T~u(M))oU) ~_ HomM~(a)(L;u(N), N(P, T~v(M)) ) ~- . . .

It follows that (P o U)@N is a direct summand of [P@s o g. Then denoting by M I N the relation "M is a direct summand of N", I have also

N@( P o U) I [s N)+P] o U

So for any M

which can be written as

HomM~k(H)(P o U, ~( N, M)) I HomM~k(a,(/2u(N)@P, TQr(M))

or

it foHow that M)) a direct s mma d of In the case U/H = �9 (i.e. when H is transitive on U), as the square

U o H (X X Y) ' V OH X

UOHY , U o H , ~ _ , U~ (Y.)


U oH (X • Y) ~_ (U oH x ) • (U oH Y)

so 5 U is an isomorphism. Then all the previous split monomorphisms are isomor- X,Y phisms.

Finally, I have proved the

P r o p o s i t i o n 10.1.2: Let G and H be groups , and U be a G - s e t - H . I f P is a M a e k e y f u n e t o r for G, if M and N are M a e k e y f u n e t o r s for H, and if Y is an H - s e t , t h e n

(P o U)y t Puo.r o U

s Is ~u(Mr) l~v(M)uoHr

10.2. LEFT ADJOINTS AND TENSOR PRODUCT 227

n(M, P o U) I n(C~(M), P) o V

s

M+(P o U)

~(P o U, M) I "H(P, Tiu(M)) o U

Zu(M)+s

(C~(M)+P) oV

"H(s Tiu(M))

Moreover , if U/H = *, all these split m o n o m o r p h i s m s are i s o m o r p h i s m s .

Example : In the case when H is a subgroup of G, if U is the set G, viewed as an H- set-G, I have U/G = �9 (the roles of H and G have to be switched in the proposition). The functor - o U is the induction functor for Mackey functors, and the functors Z;u and 7~u are equal to the restriction fnnctor. The previous isomorphisms give

(Ind~.P)v ~ Ind~(Ma~gy)

G G ~(M, Ind~P) ~_ IndH~(ResHM , P) 7-{(Ind~P, M) -~ I n d ~ ( P , Res~M)

aes~(M~N) ~_ ReSaHM~)aes~H

M~Ind~P -~ Ind~(Res~(M)~N)

R e s ~ ( N , M) ~ H(Res~N, ResiN)

The last but one of these relations explains the name of "Frobenius morphisms" for this section.

10.2 Left adjoints and tensor product

Let G and H be groups, and U be a G-set-H. If M and N are Mackey functors for H, proposition 10.1.2 shows that there are natural morphisms

I will describe those morphisms explicit]y. Let P be a Mackey functor for G. I have the following diagram

HomM~c~(H)(N, 7-t(M,P oU))

HomMa~k(H)( M @ H, P o U)

-___+

0

HomM=~k(H)[N, 7-I(s oU]

Hom.ock( , P)]

i HomM~k(a) ( s M)@s N), P)

where ~ = HomM~ck(H)(N,--/-/-~M,P), and O is the map obtained by composition of (I) with the isomorphisms of the diagram. Then the morphism I am looking for is

228 CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS

obtained by taking for P the functor s it is the image under @ of the identity map of s

By adjunction, I have the morphism

~M6N : M[~N ~ s o U

that I already used in the proof of theorem theorem 9.5.2. It is defined for an H-set V by

(M| (rlv)(t)(vouV,~v)

The associated bilinear morphism is then defined for H-sets V and W by

m E M(V), n E N ( W ) ~ (M+N)*(,lv• w(m,n))(Uo.(V•215 )

where I set

\ W / J (VxW, Id)

Now this gives through q) a morphism fl'om N to ~(s P)oU: it maps n E N(W) to the element

p E [7-{(f~u(M),P) o U] ( W ) = HOmM~k(c)(s

obtained by adjunction from the element

which maps the element m E M(V) to the element rv, w(m, n) of s OH V x U OH W) defined by

U Now the definition of P*(@,w) for P = f~u(M@N) shows that

Then if X is a G-set, and (S, f ) is an object of 7?u(X), the morphism p maps the element m E M(G\U.S) to

P.(fx x Iduo.w)P*(v(s,]) • Iduo.w)ra\u.s,w(m,n) E PUo.w(X)

This morphism fl'om N to 7-g(f-.u(M),P) oU gives by adjunction a morphism from

s to ~(s P ) , o r a bilinear morphism fl'om s s to P, defined

as follows: if Y is a G-set, if (T,g) is an object of Du(Y), and if n E N(G\U.T), then the image of m(s,f), n(T,g) is the element

P. ( Idx• ) P* ( Idx• P. (fxxlduo~(C\u.r))P* (U(S,s)Xlduohia\U.T) )ra\u.s,a\u.r(m, n )

I0.2. LEFT ADJOINTS AND TENSOR PRODUCT 229

of P ( X x Y). As the square

S x T i

f x x IdT )

X x T

Ids x P(T,g) )

Idx • u(T,g)

S • U o~ (G\U.T)

l fx x Iduo~(a\U.T)

X • U o , (G\U.T)

is cartesian, it is also

P,(Idx • gz )P,(.fx x IdT)P*( Ids x U(T,9))P*(u(Sj) X IduoNG\U.T))rG\U.S,G\U.T(m, n)

o r

* / / P. ( fx • g~,)P ((s,s) • . (~,~))~a\~,~w~( m, ~)

To compute this expression, I must fill the cartesian square

a

r , u o , [ (a \U.S) • (G\U.T)]

s • T ~ U o , [(G\U.S) ~ (a \U.T)] I / (S , I ) • P (T ,9 )

where I denote by d the map u ,, (Sa\U.S,a\U.T)~ra\u.s,G\U.T. Let (s, t, u H) E S~-T , and u, u' E U such that fu(s) = uH and gu(t) = u'H. Then

(.(s,s)-~(~,.))(~, t, u"H) = ((~, a (~ , . ) ) , (~', a(~ ' , t))

Furthermore, let (u,, G(u2, so), a(u3,to)) e g oH [(G\U.S) x (G\U.T)]. Its image under d is equal to

It is equal to (u(Sj)~'U(r,g))(s, t, u"H) if and only if

This is equivalent to say that there exists elements h and h' in H such that

4, = uh a (u2 , So) = a ( ~ h , ~) ~, = ~'h' a ( ~ 3 , t o ) = a ( ~ ' h , t ) u " H = ~ H

In those conditions, I have

uH = ulH = u'H = u"H

and then fu(s) = gu(t). Thus (s, t) is in the pull-back of S and T over U/H.

N o t a t i o n s : If (S, f ) and (T,g) are G-sets over U/H, I denote by S.T their pull- back over U/H. If s E S and t E T are such that (s, t) ~ S.T, I denote by s.t the couple (s, t) , and G.s.t its orbit by G. Is X and Y are G-sets, if ( S, f ) is a G-set over

230 C H A P T E R 10. ADJUNCTION AND GREEN F U N C T O R S

Z = X x (U/H) , and (T, g) is a G-set over ? = Y x (U/H), I denote by f .g the map from S .T to X x Y x (U/H) defined by

With these notations, I can define a map

~_ : S . T --* U OH [(GkU.S) • ( G \ U . T ) ]

by setting a(~.t) = (u, Cm.s ,G.u . t ) if fu(s) = gu(t) = uH

Indeed, if h 6 H is such that uh = u, then

h( a.~,.s, a .u. 0 = (a.~,h -~.s, 6'.~,h -~.t) = (a .u.s , G.u. 0

Thus a(s. t) 6 U OH [(G\U.S) x (G\U.T)]. And if I replace u by uh, then

(uh, G.uh.s, C.uh.t) = (uh, h- lC.u.s , h - l G . u . t ) = (tt, O.tt.8, C.tl.t)

I have also a map/3 from S.T to S ' ~ T , defined by

It is clear that the square

(3: S .T U oH [(G\U.S) • (C\U.T)]

s • T , U o , [ (a \U.S) • (G\U.T)] I/(S,]) • 1/(T,9)

is commutative. The previous argument shows that the associated map i from S.T to F is surjective. As/3 is injective, and factors through i, it follows that i is injective. Thus F is isomorphic to S.T, and the above square is cartesian.

In those conditions, I have

P * ( I/(S,] ) • I](T,g))FG\U.S,G\U.T( KO., • ) . . . .

^ . ^ .

. . . . ( M E N ) (G \U .a ) (M| 07a\U.S•

Furthermore, if G.u.s.t 6 G\U.S.T, then

Notation: If ( S , f ) and (T,g) are G-sets over U/H, I denote by x u the map S,T

US, T : GkU.(S.T) --* (G\U.S) x (G\U.T)

defi, ed br ,~,~( a.~.~.t ) = ( C.~,.~, a.~,.t ). So I have

P*(V(S,I) X V(T,g))rG\u.s,G\U.T(m, n) = (M@N)*(~UT)WGkU.S,G\U.T(m, n)

10.3. THE GREEN FUNCTORS Lu(A) 231

As

[ O) G\U.S,G\U.T(~7~,t L M* k G.u.s ] \ G.u.t ] (aXU.Sxa\u.r,Ia)

I have finally

^ * U (M| ( ~s,r )a~aXU.S,CXU.r(m, n) . . . .

[M.( UT)M. (G.u.s G.u.t'~ (G.u.s G.u.t'~ ' : L ~, \ G.~.~ / ( '~) | N * ( ~ g , ~ ) N * (?%)] \ G.u.t / J (a\u.s.T,1d)

It is also the element

[ o(m, ?%) t M* k G.u.s / \ G.u.t / J (GXU.S.T,Id)

Finally, I see that

P*(u(sj) • U(T,g))raxu.s,aXU.r(rn,?%) -= O(m, 7?,)(S.Tfl )

and the image of m(s,l), n(T,g) is equal to

P.( fx x gy)(O(m, n)(s.:<z)) = O(m, n)(s.T,(i~xgr)O)

Since

I have proved the following lemma:

L e m m a 10.2.1: Let X and Y be G-sets. Let (S , f ) be an object of 19u(X) and (T,g) be an object of 7)u(Y). Then the map which to m E M(G\U.S) and ?% 6 N(G\U.T) associates

G.u.s.t'~ N* G.u.s.t~ (n)] E (M~)N)(G\U.S.T)

induces a bilinear morph i sm from s s to s defined by

(m(s,l), n(r,g)) ~ O(m, n)(S.T,l.g)

1 0 . a The Green funetors s

When A is a Green functor for H, and M is an A-module, I can compose the above bilinear morphism O

s s ~ s

with the morphism from s to s induced by the product A@M ~ M. I obtain a product that I denote by Ux

s163 -~ s


L e m m a 10.3.1: L e t X a n d Y b e G-se t s . I f (Z , f ) is a G-se t ove r X , i f (T,g) is a G-set over ? , i f a E A(G\U.Z) a n d m C M(G\U.T), t h e n

a(z,f) U• m ( T , g ) = M.(~cUr)(a • m)(z.TJ.g)

P r o o f : Indeed, by definition

O(a(z,f), m(T,g)) = O(a, m)(Z.T,y.g )

and

[ (a.~.z.t) (a) o M* (a.~ z.t) (.~)] ~ (A6M)(G\U.Z.T) O(a, m) A* \ G.u.z ] \ G.u.t ] (a\U.Z.rdd)

By composition with the product, I obtain the element

(G.~.z.t~ (a).M* (a.~.z.~) (m)] M.(Id) [A* \ G.u.z ] \ G.u.t ]

But

A* (G'u'z't ' l (a).M* (G'u'z't 'l \ G.u.z ] \ G.u.t ] (m) . . . .

: M* ( G. .z.t • M* (m)) : "'" \G.u.z.t G.u.z.t/ (A* \ G.u.z ] (a) \ G.u.t ] "'"

..=M*( a.~.z.t ~M.(a.~l.~l.tla.u2.z2.t~(axm) . . . . �9 \G.u.z.t G.u.z.t] \ G.Ul .Z 1 G. t t2 . t 2 ,]

M * ( G.u.z.t ~ ( a • �9 u m) . . . . . M (~;Z,T)(a • \ G.u.z G.u.t /

and the lemma follows. �9

N o t a t i o n : Let Pa\u be the unique morphism from G\U to .. Then (U/H, Id) is a G-set over �9 x (U/H), that is an object of / : )v( . ) . Moreover U.(U/H) ~_ U. I denote by

Cs ) = A*(pG\u)(CA)(U/H,Id )

the image of the element A*(pa\u)(eA) of A(G\U) in s

P r o p o s i t i o n 10.3.2: L e t G a n d H b e g r o u p s , a n d U b e a G - s e t - H .

�9 I f A is a G r e e n f u n c t o r for H, t h e n s is a G r e e n f u n c t o r for G for t h e p r o d u c t u • w i t h u n i t r T h e c o r r e s p o n d e n c e A ~-+ s is a f u n c t o r f r o m Green(H) t o Green(G).

�9 I f M is an A - m o d u l e , t h e n s is an s a n d t h e co r - r e s p o n d e n c e M ~-+ s is a f u n c t o r f r o m A - M o d to s

10.3. THE GREEN FUNCTORS Lu(A) 233

Proof." The product e• is bifunctorial by construction. I must check that it is associative and unitary.

To check associativity, I consider a-sets X, X' and X ' . Let (Y, f), (Y', f') and (Y",f") be G-sets over X • (U/H), X ' • (U/H) and X" • (U/H) respectively. Let a E A(G\U.Y), a'C A(G\U.Y'), and m" E M(G\U.Y"). Then

a' "• ,~" : : ~ I * ( , ~ , ~ , , ) ( ~ ' • ~")

So

a Ux (a' Ux r n " ) = * g M.(~v,z,,)(a,x M(~y,z,.y,,)(a• m")) . . . .

. . . . M*(tC~,y,.y,,)M*(Id x t~u,,v,,)(a x a' x rn")

On the other hand a u • a'= A*(nuy,)(a • a')

which gives

�9 U * U W t ' ) . . . (a ~• a') ~x ,~" : M (~yy, y,,)(A (~y,~.,)(a • a') • :

�9 u * u I d ) ( a • m " ) . . . . M (l.Cy.y,,y,,)M (ny, y, • •

But after identification of (Y.Y').Y" with Y.(Y'Y"), I have

U U

which proves that a ux (a' ux r n " ) = (a u• a') u• m".

Moreover, if (Y,f) is an object of Du(X), and if m E M(G\U.Y), then

�9 u �9 m) = M*(nU/H,y)M*(p)(m) eCu(A) v• m = M (aU/H,y)(A (pa\u)(e) •

where p is the projection from (G\U) x (G\U.Y) onto G\U.Y. But identifying

(U/H).Y with Y, and G\U.((U/H).Y) with G\U.Y, and U.(U/H) with U, the map U ~U/H,Z is the map

G\U.Y-~ (G\U) x (G\U.Y)

which maps G(u, y) to (Gu, G(u, y)). Then p o tcU/H,y is the identity, and it follows that

5 s ) UX m = rZt

The previous arguments, in the case M = A, show that Z;u(A) is a Green functor (it remains to check by a similar argument that eCv(A) is also a right unit). This proves the first part of the first assertion. The case of an arbitrary A-module M proves the first part of the second.

If 0 : A ~ B is a unitary morphism of Green functors from A to B, then 0 induces a morphism of Mackey functors s from s to Z:u(B). If X and Y are G-sets, if (Z,f) is an object of ~u(X) and (T,g) an object of s if a E A(G\U.Z) and a' E A(G\U.T), then

c~(o)x• Ux ~,~)) : O~\~.~.~(A*(~,~)(~ x ~)(~.~,s.~))


As 0 is a morphism of Mackey functors, this is also

B*(tcU, T)OG\u.ZxG\U.T(a • at)(Z.T,.f.g)

and as 0 is a morphism of Green functors, it is

so that

~u(O)x• u• a~T,g))= s U• s

Moreover

s = s (A*(PG\U)(r = OG\uA*(pc\u)(eA)(U/H,rd) . . . .

. . . . B*(PG\u )O,(c A )(U/H,Id) ---- B*(PG\U )(r )(U/H,Id) = gs

Thus s is a unitary morphism of Green functors, if 0 is. This proves the first assertion.

And if 0 : M ~ N is a morphism of A-modules, then 0 induces in particular a morphism of Mackey functors s from s to s defined for an object (Y, f ) of 7?u(X) and a element m of M(G\U.Y) by

s il) = Oaw.y(m)(vj)

Then if (Y', f ' ) is an object of Du(X'), and if a' E A(G\U.Y'), I have

' v~ )( /) . . . . a(y,,],) s = N*(nu,,y a' • Oa\u.y(m (y,.y,f,.y)

* U t . . . . N (xz,,y)O(a\u.y,)•215 • m)(y,.y,y,.y) . . . . * U I t . . . . OG\U.(y,.y)M (t~y,,v)(a • = OG\u.(y,.y)(a U• m)(y,.y,f,.y) . . . .

. . . . z~u(O)x'• ~• "~(Y,s)) which proves that the correspondence M H s is functorial in M. �9

10.4 s and adjunction

Let G and H be groups, and U be a G-set-H. Let moreover A be a Green functor for the group H. In the proof of theorem 9.5.2, I have considered the morphism deduced by adjunction from the identity morphism of s It is the morphism

AA : A ~ s o U

defined as follows: if Z is an H-set, then (U OH Z, ~rz) is an object of 7)u(U OH Z). If a 6 A(Z), then

A*(qz)(a) e A(G\U.(U OH Z))

and I set

,~A,Z(a) = A*(rlz)(a)(vo~z,~z) Similarly, if M is an A-module, I denote by /~M the morphism of Mackey funct-or-s- M ~ s o V. The module s is an s so the module s o U is an s o U-module. In those conditions:

10.4. L v ( A ) - M O D U L E S AND A D J U N C T I O N 235

L e m m a 10.4.1: Let Z and Z' be H-sets. I f a C A(Z) and m E M(Z ' ) , then

~ M , Z x Z , ( a • m ) = )~A,Z(a) • /~M,Z,(rrt )

where the p r oduc t in the right hand side is the p roduc t of s o U as s A) o U-module.

Proofi With the notations of the lemma, I have

A M , Z x z , ( a • rrt) = M*(rlZxZ,)(a x m)(uo~(zxz'),,~z•

On the other hand, I have AA,z(a) E ( s o U) (Z) , and AM,z,(m) E ( s o

U)(Z ' ) . Their product for the functor s o U is equal to

AA,z(a) x u AM,Z,(m)= s z,)(AA,z(a) ux AM,Z,(m))

where the product ux in the right hand side is computed inside s i.e.

)~A,Z(a) UX )~M,Z,(ra) = A*@z)(a)(voj ,~z) vx M*(~z,)(m)(Vo.Z,,,~z, ) . . . .

v*( u ) ( A * ( ) ( ) *( ) ( ) ) M r/z, m ((Uo.Z).(uo.z'),.z.,~z,) �9 . . z I, ZUoHZ,UoHZ , ~]Z a X ~ . . .

�9 g *

. . . . M (auouz,uouz,)M (71z x ~Tz,)(a • m)((uo~z).(uouz'),,~z.~z,)

Finally

/~A,Z(a) • /~M,Z , (m) . . . .

. . . . s x ~?z,)(a x m)((UonZ).(UoHZ'),,~z.,~z,)) (10.3)

To compute this expression, I note that the square

U oH (Z x Z') , U oH Z (zz"~

U o H Z ' ~ U o u ' ~ - U / H

is cartesian, as the image under U oH - of a cartesian square. Thus 6~, z, induces an isomorphism

O~,z, : U OH (Z • Z') ~_ (U oH Z) .(U o . Z')

Then the square

OU, z , u o . ( z • z ' ) , ( u o~ z ) . ( u o . z ' )

7rZ• ~ I 7rZ'TrZP

U oH ( Z x Z') , U oH Z x U oH Z' ~SU, z ,


is cartesian, and expression (10.3) can be written

AA,z(a) • AM,Z,(m) . . . .

* U * U * . . . . M (G\U.Oz,z,)M (~Uo,Z,Uo,z,)M (~z x ~lz,)(a • m)(zxz',,,z•

Let G(u',(u,(z,z'))) be an element of G\U.(U oH(Z • Z')) . Then there exists h E H such that u ~ = uh. Moreover

[G(u',(u, (z,z')))] :

Hence

=

and finally

. . . . (h - l z , h - l z t) z h - l ( z , z ,) : ,ZxZt[G(t t t , (U,(Z, Zt)))]

Now it follows that

)~A,Z(a) • )~M,Z,(m) = M*(rlzxz,)(a x m)(z•215 = AM,Zxz,(a x m)

which proves the lemma. �9

It follows in particular that the morphism AA is a morphism of Green functors from A to s o A, non unitary in general: indeed, by definition

AA,.(CA) = A*(rl.)(cA)(uo, .... )

Identifying U OH �9 with U/H, the map ~r. is the diagonal injection from U/H into U/H • U/H, and r/. is the (unique) morphism from G\U.(U/H) "~ G\U t o . , that I denote by pa\u. Thus

AA,.(eA) = A*(pa\u)(U/H,~.) On the other hand, the unit of s o U is by definition

f--u(A)*(PU/H)(eCv(A) = f~u(A)*(PU/H)(A*(pa\u)(CA)(U/H,H))

I compute this element with the cartesian square

So

ulH u2H'~ u2H )

U/H x U/H , U/H

I d ~ l i d

U/H , i PU/H

(ulH u2H'~,A . . . . . ~s ~- A*(G\U. ~ u2H )) (PG\U)(eA)(U/HxU/H,Id)

10.4. Lu(A)-MODULES AND ADJUNCTION 237

o r

eCv(A)oV = A*(pa\u.(V/H)2)(e A)(V/H•

which is not equal to )~A,.(CA) in general.

Notat ion: If N is an s the module N o U is an s o U module, which gives an A-module by restriction along AA, that I will denote by xog lA , defined on the H-set Z by

(N o UtA)(Z ) = ~..(eA) • (N o U)(Z) C_ (N o U)(Z)

where the product in the right hand side is the product of N o U as an s o U- module.

Propos i t ion 10.4.2: Let G and H be groups, and U be a G-set-H. Let A be a Green f u n c t o r for H. Then the correspondence

N ~ NOUIA

is a f u n c t o r f r o m s t o A - M o d , which is right adjoint to the funetor M ~ Co-(M).

Proof: It is clear that the correspondence N ~ N o UIA is a functor, because it is composed of the functor N ~ N o U and of the functor of restriction to A along tA-

I must then prove the adjunction property. Let ~ be a morphism of s modules from s to N. Then ~ o U is a morphism of s o U-modules from s o U to N o U. Restricting along I , I obtain a morphism ~b o UIA of A-modules from s o UiA to N o UIA. But lemma 10.4.1 shows that the morphism from M to s o UIA is a morphism of A-modules. Composing with ga o UIA , I obtain a morphism 0 of A-modules from M to N o UIA.

The square

Homcv(A)(s N)

1 HomA(M, N o U]A)

c___+

c_+

HomM~&(a)( s M ), N)

HomMack(H)(M, N o U)

is commutat ive: indeed, any homomorphism 0 from M to N o U which is compat ible with the product of A is a morphism of A-modules from M to N o UIA , since if Z is an H-set , and if m 6 M(Z) , then

OZ(~Tt ) = OZ( s A X TrZ) = /~A, . (gA) x U OZ(fO') C7_ ( N o UIA)( Z )

It follows that the correspondence ~ H 0 defined above is injective. Conversely, if 6 is a morphism of A-modules from M to NOUIA , then I can compose

0 with the inclusion N o UIA ~ N o U, and obtain by duali ty a morphism of Mackey functors ~b from s to N. If I know that ~b is a morphism of s the proposit ion will follow.

Let X be a G-set. If (I/, f ) is an object of 2)u(X), and if m 6 M(G\U.Y) , then by construction of ~b, I have

~x(rn(y,f)) = N , ( f x )N*(y(gj))Oa\u.y(m) (10.4)

23s CHAPTER I0 ADJVNCT ON AND GREEN FVNCTORS

L e m m a 10.4 .3: Let dy : Y --+ Y • (U/H) be the map (v]~(v)). Th en if a C A(G\U.Y) and m E M(G\U.Y), and if 0 E HomA(M, N o UIA), I have

N*(u(y,f))OG\u.y(a.m) = ay, dy.N*(u(yj))OGku.y(m)

where the p r oduc t in the right hand side is the p roduc t "." for the s m o d u l e N .

Proof: I denote by Z the set G\U.Y. If 0 E HomA(M,N o UjA), then

Oz(a..~) : aYOz(.~)

where the product ,.u,, is the product of the A-module N o U. By definition, I have

Oz(m)) z z /

�9 = N* U OH N (hz,z)(AA,Z(a) X z z

where the product in the right hand side is the product of the s N. As

AA,z(a) = A*(,z)(a)(~o.Z,~z)

I have finally

( ) N*(u(y,f))Oz(a.m) =X'(u(y,f))X* U OH N*(hu, z) A'(qz)(a)(go,Z,,~z)xOz(m) z z

As moreover, if y �9 Y and fu(Y) = uH, I have

5Uz,z U O H ZZ U ( y j ) ( y ) Z,Z ZZ U, =

. . . 5 g = =

this gives

N*(u(Y"o)Oz(a'm) = N* ( Y ) (A*(rlz)(a)(U~ x uu(y) . . . .

= ( ) * a Oz(m)) .. . N" Y N*(uy x uy) (A (Tlz)( )(Uo.Z,~,z) • YY

and as N is an s it is also

N*(u(yj))Oz(a.m) : N * ( yyy)[s . . . .

. . . . s (A)*(uy)(A" (Vz)(a)(uo.z,~,z)).S'(uY)Oz(m)

Now the lemma follows from the next lemma:

L e m m a 10.4 .4: Let Z = G \ U . Y . T h e n

s A)*(uy ) (A*(qz)(a)(vozZ,.z)) = a(y, dv)

10.4.

P r o o f : Indeed the square

Lu(A)-MODULES AND ADJUNCTION

//y Y ~ U o H Z

Y ~ U O H Z 1]y

is cartesian. Then

239

= (A ' (G\U dy)

and the lemma follows, since rIz o (G\U.uy) is the identity. �9

Let then X ' be a G-set. If (Y', f ' ) is an object of Du(X') , and if a E A(G\U.Y ' ) , then the product of equality (10.4) by a(y,j,) gives

a(z,j,) x ~bx(rn(z,f)) = a(z,,s,) x N.(fx)N*(utz,]))Oa\u.z(m) (10.5)

Moreover, setting dz, = y' Sy,(y') ' I have

a(g,j,) = s

As N is an s equality (10.5) becomes

ao,., f, ) x ~x(rn(yj)) = N . ( S X, • fx ) (a(y,,dy,) x X*(u(z,f))Oa\u.y(rn))

Let e be the unit of the ring (A(G\U.Y) , . ) , equal to

e ---- A*(pG\u.y)(eA)

denoting by PG\u.Y the unique morphism from G\U.Y to . . Then by lemma 10.4.3

N*( , (Y , s ) )Oaxv .y ( ,~ ) = N*(~' (y , i ) )Oaxv.y(~.m) = ~(y,e~).N*(~,(y, j))Oaxv.Y(m)

and then

a(y,j,) • g,x(rn(gj)) = N . ( f } , x f x ) [a(z,,dy,) x e(z, dv).N*(u(zj))OGxu.z(m)] (10.6)

L e m m a 10.4 .5: Let A be a G r e e n f u n c t o r for G. I f X and Y are G-se t s , if a E A(X) and b,c E A(Y), t h e n

Proof." It suffices to compute

a x (b'c) = a x A* ( Y ) (b x c) = A* ( xY ) (a x b x xyy . . . .

. . . . A * ( x Y ) ( \ xyl ] \ Y2 ] . . . .


Thus in equality (10.6), the expression inside hooks is equal to

a(Y'~dY~ • e~Y~1Y).N~(~Y~f~)0~.Y(m) ~ (a~Y'~1~.~ • e~d~) .N* ( y;y) N~(~Yd) )0~ .Y (m)

I set

Now equality (10.6) becomes

~(Y,j,) • ~ ( ~ ( , . , j ) ) = x. tf . i . , • .f~.)((~(Y,,~,) • ~(Ys,-)).T)

By definition, I have

a(y',dw) • e(z, dy) = A*(t~:,,y)(a • e)(Z'.Z.dy,.d,)

I set P = A*(~,u,z)(a • e). I denote by i the injection from Y ' .Y into Y' • Y, and J the complement of Y ' . Y in Y' • Y, If j is the injection from J to Y' • Y, and dy,.y is the map from Y'.Y to (Y'.Y) • (U/l t) defined by

, (, )) ( ' , ,)) dy, y (~ ,~ ) = y , y , f~(y = Y,Y, fb(Y

then

As

and as

s dy,) x e(y, dy)) = s y, dy, y) ) = 0

since the images of i and j are disjoint, I have

But the identity of N(Y ' • Y) is equal to N.(i)N*(i) + N.( j )N*(j ) . It follows that

a(y,,f,) • = N . ( f ) , x fx)(N.(i)N*(i)+N.(j)N*(j))((a(y, ,dy,)• = . .

. . . . N . ( f ' X, • Jx)N.(i)X*(i)((a(y,,dy,) • e(z, dr)).T)

As moreover by lemma 10.4.5

(a(y',dr,) • eO1,dy)).T = a(y,,dr, ) • e(y, dy).N*(~(y,y))OG\u.y(m) . . . .

. . . . a(y,,dr, ) • N*(~(y,f))Oa\u.y(m)

10.4. L u ( A ) - M O D U L E S A N D A D J U N C T I O N 241

I get finally

a(g,j,) • = N.(f~x , x f x ) N . ( i ) N * ( i ) ( a ( y , , d y , ) x N * ( v ( z j ) ) O c \ u . z ( m ) ) (10.7)

On the other hand

a(y , j , / x m(v , s /= M'(~:~, y) (a • rn)cy,.y,s,.s/

Thus

~x ,xx (a ( z ,d , ) x m(yd) ) N . ( ( f . f ) x ' • * . u m) = ' N ( vy , . y )OG\u . ( y , . y )M ( ~ y , , y ) ( a •


�9 U Wt) U) (Ny,,y)O(G\U.Y,)x(G\U.y)(a • OG\U.(y,y)M (gy, ,y) = ( N o * u

As 0 is a morphism of A-modules, I have

O(a\v.z,)• x m) = a x u Oa\u.z(m)

where the product x u is the product of the A-module N o U. Then

• 0G\u.y( ) . u • , = N (~G\v.Y,,cxv.Y) Oaxu.Y(~))

It follows that ~bx,• :,) • re(r,:)) . . . .

But the composite map

U 8~\u.y,,c\u.~,(U oH ~y, y )~,y,.y

has the following effect on (y ' , y ) E Y ' . Y , if fu (Y) : f~'(Y') = u H

. . . .

So it is the restriction to Y ' . Y of ~,y, • ~:),, or (~:z, • z:z) o i. Then

@X'xX (a(y,,y,) • rn(y,f)) . . . .

. . . . N . ( ( f ' . f ) x , • 2 1 5 2 1 5 . . . .

It remains to observe that setting Z' = G\U.Y ' , lemma 10.4.4 shows that

s163 : s : a(z,,dy,)

Finally

As moreover ( f ' . f ) x ' • = (fix, • f x ) o i, this expression is equal to the right hand side of (10.7), and then

which proves that ~ is a morphism of s and completes the proof of the proposition. �9


10.5 R i g h t a d j o i n t s a n d t e n s o r p r o d u c t

Let G and H be groups, and U be an H-set G. If M is a Mackey functor for H, and P is a Mackey functor for G, I have built morphisms

(s M)&2P) o U ~ M~( P o U)

In the case P = Tiu(N), for a Mackey functor N for the group H, I have in par t icular a morphism

Composing this morphism with the morphism

M + ( R u ( N ) o U) ---+ M ~ N

deduced from the co-unit gu(N) o U --* N, I obtain a morphism

o u M ~ N

By adjunction, I have a morphism

A tedious computat ion shows that this morphism can be described as follows:

Notat ion: Let G and H be groups, and U be an H-set G. If X is a G-set, if (Z , f ) and (T, g) are objects of Du(X), [ denote by Z~T their pullback over X

Z~T= {(z, t ) �9 a • T I f (z) = g(t)}

I denote by z~t the couple (z, t) of Z~T. I denote by f~g the map from Z~T to X defined by

(f~g)(z~t) = f(z) = g(t)

Definition: [[ M and N are Maekey functors for H, if X is a G-set, if (Z, f ) is an object of Du(X) , if m �9 M(G\U.Z), and if n E TCu(N)(X), I define a sequence Ox(a(y,f),n)(w,g ) indexed by the objects (T,g) of Du(X), such that Ox(a(y,l),n)(r,g) C (M~N)(G\U.T), by setting

Ox(a(rd),n)(r,g) = [M* ( G'u'(z~t) ~ (m) | n(zbrd~g)] "~ ( ~,). \ GAt.z ] J(G\U.(ZbT),( ~:~, ))

L e m m a 10.5.1: The previous equality defines a bilinear map

Ox : • r u(N)(X) --,

Proof: First I must check that Ox is well defined. So let (S, a) be a u-disjoint H-set over G\U.Z, and let m �9 M(S). Then

\ G.u.z ] M.(a)(m) | n(z~r,/~g) (G\U(Z~r)'(G2:(,:~O))

10.5. R I G H T A D J O I N T S AND TENSOR P R O D U C T 243

Let S', ct' and/3 filling the cartesian square

! O~

s' , G\U.(&T)

\ a.~.~ ] S , G\U.Z

O'

Then M* (a.,,.(z~t)'~ M,,(cz) = M,,(a')M*(/3), and < G.u.z ]

. . . . [M* (/3)(m)| N*(a')(n(zbT,]~))] (S',(G2:[:~'))o~')

But (S', a ' ) is a u-disjoint H-set over G\U.(Z~T): indeed, if c~'(s') = G.u.(z~t) and if (u, s') E U CH S', then

( G.~.(z~t) a.~.z (J) : ~9(~') = a.~,.z

and (u,]3(s ')) 6 U OH S, which is impossible if (S,c~) is . - d i s j o i n t . T h e n as n

S u ( N ) ( (G\U.(Z~T)) , I have

N*(a')(n(z~rj~)) = 0

so Ox(M. (a ) (m) ( z , j ) , n ) = O. Now if c~: ( Z , f ) --+ (Z ' , f ' ) is a morphism in "Du(X), then

ox (M.(a\ U.~)(.~)(~, s,), n)(~,~) . . . .

. . . . M* ( a.u.(z'~O

It is clear that the squares

~T U.(~T) Z~T , ZgT U.(Z~T) ,

(z:') 1 1 , Z , Z' U.Z )

O: U . ( ~

U.(Z'~T)

\ ~ . z ' ] U.Z'

are cartesian, and as U.c~ is injective on the orbits of G, lemma 9.3.3 shows that the square

G\U.(Z~T) , O\U.(Z'~T)

I a \ u . z a \u .~ a \ u . z '

244 C H A P T E R 10. A D J U N C T I O N A N D G R E E N F U N C T O R S

is cartesian. Then

ox (M.(a\u.~)(,~)(~,j,), ~)(~. ) . . . .

But the morphism

\ a.~.~(~) )

is a morphism in Du(X) : indeed, the morphism U . ( ~ T ) is injective on the orbits of

G on U.(ZhT): if x E G, and if

then xu = u, xt = t and c~(xz) = c~(z). Thus xz = z, since U.c~ is injective on the

orbits of G. It follows that

and then

ox ( M.( G\ v.~ )(,~)(z,,s,), ~) (~,~) . . . .

: [M" (a.~.(z~)) (,~) | Ox(m(z, f) ,n) "'" \ G.~z.z ] r*(ZbT'Ibg).I(G\U.(Zi~T),(~:(:~,)))

Finally, the map Ox is well defined.

Now if (S, c~) is a u-disjoint H-set over G\U.T, and if m E M ( S ) , then I fill the cartesian square

so that

/ O~

s' , a\U.(Z~T) ~1 [/G'~(z~')'/ ' \ a .~ . t )

S ~ G\U.T

�9 \ a.,~.z ] ('~) | N ( ) ( (z~r, i~)) (s,,~)

But (S' , a ' ) is as before a u-disjoint H-set over G\U.(Z~T) , and then N*(c~')(n(z~Tj~g)) = O~ so

(M+N)*(~) (0x(~(z,j), ~)(T,g)) = o

I0.5. RIGHT ADJOINTS AND TENSOR PRODUCT 245

Finally if c~: (T', g') --~ (T, g) is a morphism in Z)u(X), then as the square

G\U,(Z~.)

and then

and moreover

* n

{ = M" ( M*(r t G.uz ] \ C.u.t ]

Fir~a]Iy (M+N)*tG\U.a) (Oxtm(zj), n)(T,g)) = Ox(m(z,l), n)(T,,,,)

which proves that the image of Ox is contained in T@(M~N)(X), and the lemma follows. �9

L e m m a 10.5.2: T h e m a p s Ox define a bi l inear m o r p h i s m from s Tiu(N) to Tgu(M@N).

Proof: I must check the three conditions of the proposition 1.8.3. Let then r : X --* X' be a morphism of G-sets. If (Z', f ' ) is an object of l)u(X'), if m 6 M(G\U.Z'), and if n e 7"{u(M)(X'), then to compute s162 I fill the cartesian square

a

Z , Z'

X ~ X'

s162 = M*(G\U.a)(m)(z,])

In those conditions

Ox ( s ( M )*( r )(m(z,j,} ), 7~u(N)* (q~)(r0)(:r,g) . . . .

= M* \{G'u(z~t)G.u.z ]1 M*(G\U.a)(m) | 7~u(M)*(r162

O\U.(Z~T') O\U.(Z~T)

) G\ U.T' O\ U.o~ G\ U.T

is cartesian, setting r = G\U.(Z~oO, I have

( M Q N )"( G\ U.a) (Ox(m(z,.f}, n )(?,~) ) . . . .

Now the morphism U.(Z~a) is injective on the orbits of G, thus

246 CHAPTER 10. ADJUNCTfON AND GREEN FUNCTORS

. . . . [M* (G.u.(zOt)~ (m)@n(ZbT,.$(/~g)) ] \ C.,,.a(z) ] (aW.(zm,(~:t:~ ') ))

On the other hand

(nu(M@N)*(~)Ox,(m(z,j,), n))(r,9) = Ox,(rn(z,j,), n)(T,gg ) . . . .

= [M* (G.u.(z'~t) rt(Z'~T'f'~("~g)) ] ' \ a . u . / ) ( , ~ ) e

But the diagram

a

Z[T , Z , Z'

'l l" T ~ X > X'

g

is composed of two cartesian squares. Thus the square

(lo.s)

Z[T -+ Z'

T , X '

is cartesian. It follows that in equality (10.8), I have Z~T = Z'~T, and with this (c.~.(zb0 (G.~.(z'b0'~ becomes the map t,a.~.a(z))" Those remarks prove identification, the map \ a.~.z, j

that

( �9 . ) ) ~ . Ox s (~)(rn(z,./,)),T@(N) (r = 7~u(M| (r

which is condition iii) of proposition 1.8.3. Now if m E M(G\U.Z), then

s M).( (o)(m(z,/) ) = m(z.'gs )

so that for an object (T,g) of Du(X'), and n E nu(N)(X') , I have

Ox, ( s ( M). ( ~ )(m(zj) ), n)(T,g) . . . .

' " G.~.~ ] (c\u.(ar),(~:(:.~') )~

10.5. R I G H T ADJOINTS AND TENSOR PRODUCT 247

where the pull-back product Z~T fills the cartesian square

(zl ) Z t T , T

Z , X '

I must compare this element with

~ f

To compute this element, I fill the cartesian square

so that

O /

Z' > T

X > X '

( c )

( z t z ' ~

ZI ] Og Z~Z' , Z' , T

Z , X , X ' f

/ . . \

Moreover

0x (~(=,s), nu(Nl'(r = ..

..--/M- ( G.u.(zhz') ~ ] �9 [ \ G.u.z ](m)|162 , , ,

J (GW.(=b='),( ~ ~:V~ ) ) )

On the other hand

This gives

a ~ z t z ' . . . M* ' ( ) m =

' G.~.~ / ( '~) |

Then the following diagram is composed of two cartesian squares


It follows that that in (C), I can identify Z[T with Z[Z'. With this identification, I }lave

(r = r k G.u.t ] \ C.u.a(z') so that I have

which is condition i) of proposition 1.8.3. Finally for condition ii) I must compare

E ' = Ox, (m(z,~), ~ u ( N ) . ( r

and E" = Tiu(M@N).(r (s162 n)

when (Z, f ) is an object of Du(X') and m E M(G\U.Z), and r~ E Ru(N)(X). For an object (T,g) of IDu(X'), I have

E~T,g)= [ M* {G'u'(z~t)~ ] \ C.u.z ] (m) | T@(N).(r (a\u(Z~T,(C2:t:~O))

To compute this expression, I fill the cartesian square

so that

Thus

Ol

S , Z~T

X , X'

T@( N).( r = N.( G\U.a)(n(s,~))

E~T'g) = [ M* ( Gu(zlt)'~ 1

which is also

On the other hand, to compute s162 I fill the cartesian square

a

S' , Z

X I X '

(c,)

(c~)

10.5. R I G H T A D J O I N T S A N D T E N S O R P R O D U C T 249

so that

Let

Then

s162 = M*(G\U.a)(m)(s , ,b)

,, ( )) E (T ,g)= I ~ u ( M @ N ) , ( r (T,9)

To compute this expression, I fill the cartesian square

Then

! a

S" T

X X '

(c~)

E"(T,a) = (M+N) . (G\U.a ' ) (F(s , , ,b , ) )

Moreover

G.~.s, ] (G\u.(s, ts,,), (* ;~:.?,~")))

Hence

a.~.s, J (~\u.(s,bs,,),( S:!:,:;:'; ) )

In those conditions, the set 5" identifies with the pull-back SPaS " of S ' and S" over X: indeed, if s'[ls" E S'[IS", then b(s') = b'(s"), so

Cb(s') = f a ( s ' ) = ga'(s") = Cb'(s ' )

and then I have the element a(s ')~a'(s") of Z~T. Moreover

( f ~ ) (a(~,)~,(~,,)) = fa(~,) = ~b(~ p)

Then as (C1) is cartesian, there exists a unique s E S such that

~(s') = ~(~) ~(~ ' )~ ' (~") = ~(~)

So I have a map s'~s" ~-~ s from S'~S" to S. Conversely, if s E S, and if a(s ) = z~t, then

( f~g)(z~t) = f ( z ) = ( f~g)a ( s ) = r

As (C2) is cartesian, there exists a unique s' E S' such that

b(~') = ~ ( ~ ) a (~ ' ) = z

Similarly, a.s ( f~g)(z~t) = g(t) = ( f~g)a ( s ) = r


and as (C3) is cartesian, there exists a unique s" C S" such that

b'(s") = 9(s) a ' (s") = t

Moreover b(s') = b'(s"), so I have the element s'~s" of S'~S". This correspondence s ~ s'~s" is clearly inverse of the previous one. As they are morphisms of G-sets, I have S ~- S'~S".

With this identification, I have

Ct(,S'~S" ) = a( s')~a' ( s" )

and I can write E~T,g ) as

EIT'g) = [ M*(G\U'a)M* (G'u'(z~t))\ G.u.z (m)

Moreover

a.u.z ] =

Then

Z ( ~ ' ~ " ) = b(~') = b'(~")

=, ( ))

(G.u.(z~t)~ (G.u.(a(s')~a'(s"))) . . . . G.u.z ]

. . . . C.~,.~(~') = (G\U.~) \ C.u.~' ]

(G.u.(z~t)~ M* (G.u.(s'~s")~ M*(G\U.a)M* \ G.u.z ] (m) = \ G.u.s' ] M*(G\U.a)(m)

whence E~T,e ) = E"(T,g), SO E t = E ' , and condition ii) of proposit ion 1.8.3 holds. This proves lemma 10.5.2. �9

1 0 . 6 n u ( M ) as /2u(A) -module

Let G and H be groups, and U be a G-set-H. If A is a Green functor for H, then s is a Green functor for G. If M is an A-module, then 7~u(M) is a Mackey functor for G. I have moreover a bilinear morphism

s x nu(M) -~ nu(A@M)

that I can compose with the morphism

n~j(A@M) -, n~(M)

deduced of the product A@M -* M. Now I obtain a product

f-,u( A )@Tiu( M) ~ nu( M)

10.6. R u ( M ) AS Lu (A) -MODULE 251

L e m m a 10.6.1: Let X be a G-set, let (Z, f ) be an ob jec t of Du(X), and a C A(G\U.Z) . I f m E Tgu(M)(X) , the above p r o d u c t is such t h a t for any ob jec t (T, g) of 7?u(X)

(a(zj).m)(r,g) = M* {G.u.(z~')] {G.u.(z~t)]

where the p r o d u c t in the r ight hand side is the p r o d u c t "." for the A- m o d u l e M.

Proof : This follows from the definition, and from the formula

G.u.z ] (a) | J (G\U.(Z~T),( ~:~uzt ))

L e m m a 10.6.2: Let (Z, f ) and (Z', f ' ) be ob jec t s of T~u(X). I f a E A(G\U.Z) and a 'E A(G\U.Z ' ) , t hen in s I have

, '

a(zj) 'a(z"s') \ G.u.z ] \ G.u.z' ] (a)(z~z,,S~I,)

Proof : Indeed, by definition

() ' = s x (a(z,]) x a(z,j,)) . . . . a(z,s).a(z,,s,) xx

. . . . L;u(A)* (xxx) (A*(~u'Z')(ax a')(z.z,,s.S')) To compute this expression, I fill the cartesian square

i P ~ Z.Z'

X ~ X x X

(') Z X

It is then clear that P is isomorphic to the pull-back of Z and Z' over X, with

i(z~z') = z.z' j(z~z') = f ( z ) = f ' (z ' )

In other words, the map j identifies with f~f ' . Then

, : A . / , a(zj) 'a(z 'J ') \ G.u.z.z' A* \G .u . z G.u.z '] (a x a )(z~z',I~I') . . . .

= A* ( C . u . ( z ~ z ' ) ~ (a x a')(z~z'jby') " ' " \a.u.zG.u.z']

252

Moreover

CHAPTER 10. ADJUNCTION AND GREEN FUNCTORS

\ G.u,.z ] (a).m* \ G.u2.z' ] (a')

(c.~.(z~z') ~ I have and then denoting by ~b the map \G .... c.~.z,),

a(z,]).a~z,j, ) . . . .

. . . . A*(r G.u,.z ] \ G.u2.z' ] (a') . . . .

= A. (G.u.(z~z')~ (a).A.(G'u'(z~z')) , "'" G.u.z ] \ G.u.z' (a )(zbz',]~]')


Proposition 10.6.3: T h e a b o v e p r o d u c t

s A )~Tiv( M) ~ T~u( M)

t u r n s T~u(M) in to an s

P r o o f : The product "." satisfies by construction the three conditions of proposition 1.8.3. Then the associated product "x" is bifunctorial. It suffices then to check that it is associative and unitary, which is equivalent to check that the product "." itself is associative and unitary.

Let (Z, f ) and (Z' , f ' ) be objects of l)u(X), and let a E A(G\U.Z) and a ' E A(G\U.Z'). If m E Tiu(M)(X), then for any object (T, g) of I)u(X), I must compute

v = ( a ~ z , , ~ . r . . . .

. . . . J A" \ G.u.z ](a)'(a(z"f')'m)(zhT,f~gl

Moreover, identifying Z'~(Z~T) and (Z'~Z)~T with Z'~Z~T

,

(a(z,j,).m)(Z~Tjbg) = M. \ G.u.(z~t) ) A* G.u.z' ] (a ).m(z'~zbT,f'b$~g)

I set E = A * " " " ""(C'u'~z'~z~t)~ ,

\ G.u.z' ] (a).m(z'bzbT,f'bfbg)

and

\ G.u.z ] (a)

Then

= (a.~.(z'~z~t)~ A* (G.u.(z~t)~ (a).(a~z,,l,).m))(Z~T,lbg) F.M. G.u.(z~t) ) (E) . . . . k C.~.z )

10.6. Ru(M) AS Lu(A)-MODULE 253

.... M.\ G.,,.(~b,)J [A" (~).~] \ c.~.(zhO ] As

\ a.u.(z~t) ] (F) = A* \ C.u.z ] (a) I have

A. (G.u.(z'~z~t) "] .(G.u.(z'Oz~t))(a).A.(G.u.(z'~z~t)' ] �9 \ C.u.(z~t) ] (F).E:A G.u.z \ G.u.z' ](a').m(z,bz~T,f,~f~g)

Moreover

\ G.u.z ] (a).A* (a') . . . . \ C.~.~' )

. . . .

: ~ c.~.(~z,) ~ \ c.~.~ )t~.~, t c.~.(~,) )~' \ c.~,.~, ](~) . . . .

" ' \ G.u.(z[z') ] [A" (a).A* Setting

F' = A* { G.u.(z'~z~t) ( G.u.(z~z')'~ ( G.u.(z~z')'] (a')) .m(z,~z~T,],~]~s) \ G.u.(z~z,) ) (A* (a).A* \ c.~,.~ ] \ c.,~.z, ]

this gives

P= M. \ C.u.t ] M* \ G.u.(z~t) ] (F') o1"

P = M. \ c.~.t / ( F')

Setting

I have

= \ G.u.(z[lz,) ) (A* )(a].A* \ G.~.~ \ G.~.z' ]

(c.~.(~ '~t)~

P = M. c.~.~ j [E ' .m~,~,~ ,~ ,~ , ] . . . .

. . . . M. \ C.~,.~ ] M. \a.u.(~'~t)]

Moreover

(G.u.(z'bz~t)'~ [E'.m(z,~z~Tj,~ybg)] : M* I G'u'(z~zgt)'~ [E'.m(z,~Z~T,I,W~g)] . . . .

Finally

\ c.~,.(z,~z~),] (E') . . . .

" ' " = \ G.u,(z~z') ] (A* (a) .A" \ c.~,.z ] \ c.~,.z, ] ]

(zbz'~Q is bijective, hence a morphism in 7)u(X), As m 6 gu ( M) ( X) , and as the map \ , ' ~ t / I have also

M* (c.~.(~b~'~t)) \ O.u.(zgzbt) (m(z'~zbr,1'~jbg)) = m(ZbZ'br.S~'bg)

Finally, setting

this gives

and since

\ C.~,.z (,~).A'\ C.~.~' }

\ c.~,.(z~') ] (a).mi~z,~r,i~i,~g )

a(zd).a(z, ~,) = a (z~z'd~]')

I have p : ((a(z,~.alz, g , )) .q~,g )

which proves associativity of the product. Moreover the unit ex of (s is equal to

As the square

_ (:) X , U/H

X , i

(:) is trivially cartesian, I have also

~ _- ~-~.(x))~,/>~)~<~,,~, _- ~, ( % % ~<~.,~)

Then if m 6 T@(M)(X), I have, for an object (T,g) of :Du(X)

10.7. Lu(A)-MODULES AND RIGHT ADJOINTS 255

As the square Id

T , T

X , X Id

is trivially cartesian, I have X~T = T, and then

since A* (a: . t ) (CA) is the unit of the ring (A(G\U.T), .) . So the product is unitary, k /

and this completes the proof of the proposition. ..

10.7 s and right adjoints Let G and H be groups, and U be a G-set-H. If A is a Green functor for H, I have defined s which is a Green functor for the group G, and the functor N ~ NoUIA from s to A-Mod: if N is an s and Z is an H-set, then

(:v o u ,A)(z ) = :,~,.(~) • (N o u ) ( z )

where the product in the right hand side is the product of the s o U-module N o U .

L e m m a 10.7.1: Let n E (N o U)(Z). Then denot ing by P~\u.(uoHz) the unique m o r p h i s m from G\U.(U OH Z) to *, I have

where the p r oduc t in the right hand side is the p roduc t "." for the s module N.

Proof : Indeed by definition

xUTt * ( ~( ( ~ U n) ~A,.(CA) = X ,~ . ,Z ,~A, . ,CA, •

where the product in the right hand side is is the product of the s N. Let X = U OH Z. Then 5A,.(CA) E s and n E N(X) . Thus

' \ u H ]

So

\ u g ] -IA'~ (~)

But if x -= ('u, z) E X, then

256

and then

Moreover

CHAPTER 10 ADJUNCTION AND GREEN FUNCTORS

N*~5 U ~N* ( u H z )

uH / '

U o H Z , U/H

U O H Z ) U / H

Z~IA)'/e[z)e~IA)" (~H x~ (~../cA)) = ' \ u H /

... = . O H oHZ,Tr Z


P r o p o s i t i o n 10.7.2: Le t G and H be g roups , and U be a G-se t -H. I f A is a G r e e n f u n c t o r for H, t h e n t he c o r r e s p o n d e n c e M ~ 7r is a f u n c t o r f r o m A - M o d to s which is r igh t ad jo in t to t he f u n c t o r N ~ N o UIA.

P r o o f : I already know that the correspondence M ~ g u ( M ) is a functor between the categories of Mackey functors. Moreover, if r : M --+ M' is a morphism of A-modules, then the morphism Ru( r is given for a G-set X, an element m E 7r and an object (T,g) of Du(X) by

nu(r = Ca\v.r(~(r,~))

Then if (Z, f ) is an object of Du(X) and if m E M(G\U.Z), I have

n~(r = ~>~w.~ ((~(~,s).~)(~,~)) . . . .

and then

Hence

~ ( ) / . ,S~(A) ~ ~H ) (~A,./cA)) : ~(A)"

To compute this expression, I use the cartesian square


As r is a morphism of Mackey functors, it is also

G.u.t ] r [A" \ G.u.z ) (a).m(Z~T,lbg)]

Finally as r is a morphism of A-modules, I have

TgU( r )( a(zj).m )(r,g) . . . .

�9 \ c .u . z ] . . . .

. . . . (a(z,j).T~u(r

which proves that T@(r is a morphism of s

Let 0 be a m o r p h i s m o f A-modules from NOUIA to M. As NOUIA is a direct summand of N o U as Mackey fnnctor, I have a morphism O' of Mackey functors from N o U to M, defined on the H-set Z by

0 (m) = • m)

By adjunction, it correspond to 0' a morphism r of Mackey functors from N to Tiu(M), defined as follows: if X is a C-set and (r,g) an object of 7)u(X), and if n E N(X), then

r = O~\u.TN.(vT)N*(gx)(n) 6 M ( G \ U . T )

Thus ~(n)(T,~) = 0a\U.T(AA,.(CA) • N. ( .~)N*(gx)(n) )

The previous lemma shows that setting T' = G\U.T

AA,.(eA) X U N.(vT)N*(gx)(n) = A*(pc\u.(Uo.T,))(CA)(Uo.T,,~r,).N.(~'T)N*(gx)(n)

But as N is an s

A*(pa\u.(Uo,T,))(C a)Wo,T,,~,).N.(,T)N*(gx )(n) . . . .

Let dr be the map from T to 2F defined by dT(t) = (t,gg(t)). The square

is cartesian. So

l/T T , UoHT ~

T , UoHT'

PT

. . . . A*( G\ U.~'T )A*(pG\u.(UoHT') )(C A )(T,dr) . . . .

. . . . A*(PG\U.T)(eA)(T,dT)

This gives finally

~(n)(T,g) = Oa\u.TN,(vT)(A (PG\U.T)(~A)(T,dT).N (~X ) ( / 7 ) )

I must show that ~b is a morphism of s from N to g u ( M ) . So let (Z, f ) be an object of �9 and a E A(G\U.Z). Then by definition of the product on Tgu(M), I have

(o<z,,,.,o<,.,,: M . , , ~.,,, j[A. , ~.,,.~ j <~/+~,,~<~..,~,]

Moreover

I set

so that

= * sV*

(~{~"~~)(~,g/: M, < c.~.t / L "4. < o.~.z J

As 0 is a morphism of A-modules, I have

A* {a.u.(z~t)

where the product .t~ is the product of the A-module NOUIA. Setting Z ' = G\U.(Z~T), and

a': ,X~,~,A* (C.~,.(~t)~C;..<,.z S (a) I have

Then by definition

(z,) (z,) d Y E : (W o v)* _-'~' (a' • E) : (N o U)" z'~' X*(~, ,z , ) (a ' • E)

where the product in the right hand side is the product of the s N. But if (u, z') C U OH Z', then

ZIZ !

It follows that

dYE=N ~( (~' ~') ) \ (~ , ~')0,, ~') (~' • E) : a ' .B

where the product in the right hand side is the product "." of the s N. Now setting

E' = A*(pa\u.(ZbT))(C A )(Z~T,dz~r).N* ((f~g )x ) (n)

I have E = N,(~'ZbT)(E'). But as N is an s

= N . (~ 'ZbT) (s aI.E

Moreover

To compute s I use the cartesian square

lYZbT Z[T ~ U OH Z'

< , Z~T ~ U OH Z'

which shows that

{C.,,.(z~t)) s = A*(G\U.uz~T)A*(~z,)A* \ G.u.z ] (a)(ZbT,dz~T)

As A (G\[ UZ~T)A (r;z) is the identity (by an adjnnction argument), it follows that

Cu(A)*(uz~r)(a') = A* (G.u.(z~t)~

Then

s : . . .

, ,) = A. (G.~L.(z~) "'" \ G.u.z (a)(Z~T,dz~r).A*(pa\cr(ZbT))(SA)(ZbT,dz~r).N*((f~g)X)(n)

Moreover, the square Id

Z[T , Z~T

Id I ? z ~ T

Z~T ) zbr dzbT

is cartesian, because dz~T is injective. Then by lemma 10.6.2

\ G.u.z / (a)(ZhT'dz~r)'A (Pa\u.(Z~T))(~-A)(Z~T,dz~r)= A ' l (7.u.z J

SO

As moreover the square


Then

Z~T , Z

dz~r ~_ ! dz

Z~T , Z

A* (G.<z~))

Moreover (S~glx = ~x o U )' so

. . . . N* a(z,dz).N*(fx)(n)

Finally, I have

(a(z,/).n)(T,s)= M. G.u.t ] OG\U.(ZIT)N.(L'ZtT)N" [a(z,dz).N'(fx)(n)]

As 0 is a morphism of Mackey functors, I have also

M . \ G.u.t ] OG\U'(Z~T) = OG\U'TN* U ~ \ G.u.t

Moreover ( N. U ott \ a.u.t / ] N.(z/Z~T) = N.(UT)N. I t

This gives finally

The square

ZiT . . . . . . -~ Z

T - - ~ X g

10.7. L u ( A ) - M O D U L E S AND R I G H T A D J O I N T S

being cartesian by definition, I have

and then

(a(zd).n)(r,g) = Oa\u.TN.(uT)N*(g)N.( f)[a(z ,az) .N*(fx)(n)]

Let p be the projection from X onto X. Then f x = f o p, and

* Tt . . . . s (P)( )

Moreover

N*(g) [s dz)).N*(p)(n)] . . . .

�9 .. = Eu(A)*(g)Eu(A) , (f) (a(z,az)). N*(g)N*(p)(n) = .. .

. . . . Eu(A)*(g)gu(A) . ( f ) (a(z ,dz) ) .N*(gx) (n)

Finally s = af~ z, and the square

<) Z~T ~ Z

dT(t)] I . , Y

is cartesian, Then

whence

gu(A)*(g)(a~dz) = A* (G.u.(z~t)~

\ a.~.z ] (a)(zh~,,~;%;X

I must compare this element with ~(a(z,]).n)(T,g). By definition of ~b, I have

r = 0o\u.TN.(~T)(A*(pcw.~)(~A)(~,~).N*(g~)(a(~j).~))

But N*(gz)(a(zd) .n) = s

261


Let P , a and fl fill the cartesian square

O/

P , Z

T , X gx

Then

and I can write

s = A*(G\U.a)(a)(p,z)

r = Oa\u.TN.(vT) [A*(pc\u.r)(cA)(T,dr).A*(G\U.a)(a)(p,z).N*(gx)(n)]

To compute this product by lemma i0.6.2, I fill the cartesian square

O, !

T~P , P

T , T dz

so that

A*(Pa\v.T )(e A )(T,dT).A*( G\ U.a)( a )(P,Z) . . . .

"'" \ G.u.t )(eA).A* a*(G\U.a)(a)(The, dr~)

Then the square OLO~ t

ThP , Z

T , X g-xdT

is composed of the two previous cartesian squares. It is then cartesian, which proves that T~P identifies with Z~T. With this identification, I have fl'(z~t) = t, and a'a(z~t) = z. Then

[ = Oa\u.TN.(vT) A* \(G'u'(z~t)]G.u.t J (a)(Z~T,(~;}'O)) "N ( g x ) ( ) r

which proves finally that

~(a(z,~).~) = a(z,1).r

and ~b is a morphism of s

Conversely, being given a morphism of L~u(A)-modules ~b from N to T~u(M), I have a morphism of s o U-modules ~b o U from N o U to TZu(M) o U. Composing

10.7. Lu (A) -MODULES AND R I G H T ADJOINTS 263

this morphism with the morphism of Mackey functors from 7~u(M) o U to M, I obtain a morphism of Mackey functors from N o U to M. If I know that this morphism is compatible with the action of A, then it will be entirely determined by its restriction to N o UIA: indeed, the image of an element is then equal to the image of its product by AA,.(eA). In other words, the proposition will be proved if I know that the co-unit morphism

rb : 7 r o U --* M

is compatible with the action of A. But q) is defined for an H-set Z and an element rn E ( ~ u ( M ) o U)(Z) = "gu(M)(U oH Z) by

(~z(rn) = M,(~z)(rn(uo.z,~z)) E M ( Z )

Let a r A(Z) . Then a acts on (T@(M) o U)(Z) through the morphism A : A s A ) o U. In other words

a.rn = •A,a(a).Um

I already observed that AA,z(a).Um = AA,z(a).rn, this product being the product of n u ( M ) an Cu(A)-module . Then

=

But ,~A,Z(a) = A*(r?z)(a)(uo,Z,~z), so setting T = U oil Z, I have

(/kA,Z(a).m)(uOHZ,~z) = (A*(rlz)(a)(T,,z).rn)(T,,z)

To compute this product, I must build TIT: but the square

Id T ~ T

T , T 7C Z

is cartesian since 7rz is injective. Thus T~T = T, and then

(A*(~z)(a)(T,,~z).rn)(T,,~z) : M. ( Id ) [A*(Id)A*(rlz)(a).m(T~T,~z~,~z)] . . . .

In those conditions

. . . .

42z(a.rn) = M,(rlz ) (A*(rlz)(a).rn(T,,~z)) = a.M.(rlz)(m(T,,~z) ) = a.•z(rn)

which proves that r is compatible with the action of A, and completes the proof of the proposition. �9

264 C H A P T E R I0. A D J U N C T I O N AND GREEN F U N C T O R S

10.8 Examples and applications

1 0 . 8 . 1 I n d u c t i o n a n d r e s t r i c t i o n

Let G be a group, and H be a subgroup of G. Let U be the set G, viewed as a G-set-H, and V be the set G, viewed as an H-set-G. Then I know that if A is a Green functor for G, the functors A o U and s are both equal to the restriction of A to H. It is not clear a priori that the products defined on A o U and s are the same.

But they are: the isomorphism

s ~- Res~A(X)

comes from the fact that the category D r ( X ) identifies with H - s e t S x , which has a final object (X, Id). Moreover, the set H \ V . X identifies with I n d e X . If a and a ' are elements of A( Ind~X) , then

But obviously X ~ X = X , so

a(X,ld).a~X,ld) = a,a{x,id)

and the product of L;v(A) ~s the product of Res~A. A similar argument shows that if B is a Green functor for H, then the products

defined on s ~- Ind~B ~- B o V

are the same.

1 0 . 8 . 2 T h e c a s e U / H = .

P r o p o s i t i o n 10.8.1: L e t G a n d H b e g r o u p s , a n d U b e a G - s e t - H . fo l l owing c o n d i t i o n s a r e e q u i v a l e n t :

T h e

1. T h e g r o u p H is t r a n s i t i v e on U, i .e . U / H = . .

2. F o r a n y G r e e n f u n c t o r A for H, t h e m o r p h i s m

/~A : A ~ s o U

is u n i t a r y .

3. T h e f u n c t o r A H Cu(A) f r o m Green(H) t o Green(G) is le f t a d j o i n t to t h e f u n c t o r B H B o U.

P r o o f : I recall the formulae

AA,.(r = A*(Pa\u)(r r = A*(pa\u.(U/H)2)(eA)((U/H)2,H)

10.8. EXAMPLES AND APPLICATIONS 265

Moreover, as 7to is the diagonal injection from U/H to (U/H) 2, it is a morphism from

(U/H, rr.) to ((U/H) 2, Id) in the category Z)u(U/H). So in s I have also

AA,.(~A) = A.(rc~

If 1) holds, then U/H ~_ (U/H) ~ _~., and in this case

~A,.(e A) = ec~(a)o~

so 2) holds. Conversely if 2) holds, then for any Green functor A for H and any s N, I have

NoUIA = N o U

Then

HomA(A, N o UFA ) ~_ (N o U)(.) = N(U/H) "z_ ttomLv(A)(s N) ~ N ( . )

Taking for A the Burnside functor b, and for N the module f~g(b) ~- bv/u, this gives

b(U/H) ~_ b((V/H) ~)

But (U/H) 2 is the disjoint union of its diagonal, isomorphic to U/H, and of its complement C in (U/H) 2. Then the above isomorphism shows that b(C) = 0. But the only set X such that b(X) = 0 is the empty set. Then U/H ~- (U/H) 2, and U/H ~ . , so 1) holds.

Now if 3) holds, then as s is the unit of the adjunction, it is a unitary morphism of Green functors. So 2) holds.

And if 2) holds, then to prove 3), I must prove that the unitary morphisms of Green functors from s to B are sent by adjunction to unitary morphisms of Green flmctors from A to B o U, and conversely. But if g5 is a unitary morphism of Green functors from s to B, then 05 o U is a unitary morphism of Green functors from s o U to B o U. Composing this morphism with hA, which is a unitary morphism of Green functors by hypothesis, I obtain the morphism from A to B o U which is associated to r hence this is a unitary morphism of Green functors.

To prove the inverse correspondence, it suffices similarly to prove that the co-unit of the adjunction

0 : s o U) - , B

is a unitary morphism of Green functors. But if X is a G-set, as U/H = . , the objects of � 9 are just the G-sets over X. Let (Z, f ) be such a set, and b e (BoU)(G\U.Z) (furthermore the set U.Z is equal to U x Z in this case). Then

Ox(b(z,f)) = B.(f)B*(l,z)(b)

The unit of s o U) is by definition

scv(BoU) = (BoU)*(pa\u)(S,oC~)(u/udd) = B*(UoHpa\u)B*(p~s/H)(SB)(.,Id) . . . .

. . . . B*(PUoMa\U))(SB)(.,Id)

So o . ( c ~ ( s o v ) ) = B . ( I d ) B * ( , . ) B * ( p V o , ( G \ v ) ) ( ~ , ) = B ' ( p . ) ( c , ) = ~

266 C H A P T E R 10. A D J U N C T I O N AND GREEN F U N C T O R S

Thus O is unitary. It remains to see that it is a morphism of Green functors. So let Y be a G-set, let (T, g) be a set over Y, and let b' E B(G\U.T) . Then

i : B . [ U b(z,]) x b(T,g ) t~z,r)(b x b )(Z.TJ.g)

But Z.T = Z • T and f .g = f x g. Then

! * * U Ox• • b(T,g)) = B . ( f • g)B ( . z • (az, T)(b x b')

Let moreover u be any element of H. Then

SO

Final ly

Ox• • b~T,g)) = B . ( f x g)B'(v 'z x .T)(b x b') . . . .

. . . . B.( f )B*(v 'z)(b) x B.(g)B*(V'T)(b') = @x(b(z,y)) x Oy(b(T,g))

Thus O is a uni tary morphism of Green functors. The proposition follows.

10.8 .3 A d j u n c t i o n and M o r i t a c o n t e x t s

Let G and H be groups, and U be a G-set-H. Let A be a Green functor for H, and M be a module-A. If N is a Mackey functor for H, then /-/(M, N) has a natural s t ructure of A-module (see proposition 6.4.2), defined as follows: if X and Y are H-sets , if a E A ( X ) and

r E TI(M, N ) ( Y ) = HomM~k(H)(M, Ny)

then a x r is the morphism from M to Nx• defined for an H-set Z and m E M ( Z ) by

(a • r = Czxx(m x a) e N ( Z • X • Y) = N x •

The Mackey functor s has also a structure of module- / :v(A) . Then if P is

a Mackey functor for G, the functor ~ ( s is an s So the

module " H ( s is an s By restriction along the morphism

AA : A ~ s o U, I obtain an A-module 7Y(s P) o UIA. On the other hand, the functor 7-t(M, P o U) is also an A-module. I have built

morphisms of Mackey functors from ~ ( M , P o U) to 7-l(s P) o U. It is na tura l to ask if these morphisms are compatible with the product of A.

Proposition 10.8.2: L e t G and H b e g r o u p s , a n d U b e a G - s e t - H . L e t A b e a G r e e n functor for H, and M b e a m o d u l e - A . I f P is a M a c k e y f u n c t o r , t h e n t h e m o r p h i s m s 1-t(M,P o U) ~ " H ( s o U i n d u c e isomorphisms of A - m o d u l e s

TI( M, P o U) ~- ) "H(s M), P) oUla

Proof." Let O be the morphism from "H(M, P o U) to ~ ( s P) o U. If I prove that O is compatible with the product of A, then as the module in the left hand side is unitary, the image of O will be contained in ~ ( s P) o UIA.

So let X and Y be H-sets. Let a E A(X) and

r E I~(M, P o U)(Y) = gomMack(H)(M, (P o U)y)

Then let

= O(c~)E [ ~ ( s ( Y ) = HomM=ck(a)(s

If S is a G-set, and (T,g) is an object of Du(S), and if rn E M(G\U.T) , then the image |162 is given by the following diagram

. . ~ + , ~ > • M~o< P(r • (Cro. Y)) '~ • ~ ~ P((x • (Vo. Y))

In other words, denoting by 7 the functor from H-se t to G-set$u/H defined by -y(X) = U OH X, and co its left adjoint, defined by co(T) = G\U.T, I have

U ~s(rn(T,g)) = P.(gs • Id~(z))P*(u(r.g) x Id .~(y ) )P . (5~(r ) , y ) r )

To simplify this expression, I filI the cartesian square

C~ r , ~(co(r) • r )

r • 7(Z) , ~co(r) • ~(Y) U(r,g) • Ida(y)

Let (t, u, y) E T x 3'(Y). If u' E U is such that gu(t) = u'H, then

Let (ul ,G.u2. t , ,y , ) e 7(w(T) x Y). Then

6wU(T),y(ttl, a'U2"tl, ~]1) = ((?/1, a'tt2"tl), (~1, Yl))

Those images are equal if and only if

(~t', ~ . ~ " . t ) : (Ul, a.tt2.tl) (tt, y) : (ttl,Yl)

This is equivalent to say that there exists h, h ~ C H such that

ul = u~h G.u~.tl = G.u~h.t uh ~ = Ul htyl = y

It follows in particular that u~H = ulH = uH. Moreover uh' = u'h, thus

(ul, ~.t~2.tl, Yl) : (uh', G.ulh.t, yl) = (tth', G.uhl.t, yl) = (u, G.u.t, h'yl) = (tt, G.u.t, y)

268 CHAPTER 10. ADJUNCTION AND GREEN F U N C T O R S

Then the element (t, u, y) is in T.7(Y). If C~T,y is the map from T."/(Y) to 3' (w(T) x Y) defined by

~ r , r ( t , ~ , y ) = (~, a .~ . t ,y )

I have C~T,y(t, u, y) = (ul, G.u>tl, yt). If moreover/~r,y is the inclusion from T.~/(Y) into T x "~(Y), then the square

sT,y • Y)

T x "~(Z) , ")a(T) x ~f(Y) l/(T,g ) X Id~(z)

is commutat ive. The above argument shows that the associated morphism s from T."/(Y) to F is surjective. As 13z,y factors through s, the morphism s is also injective. Thus F is isomorphic to T."/(Y).

In those conditions, I can write

~Ss(m(T,g)) = P.(gs • Id-r(Y))P.(•T,Y)P*(aT,Y)r

As the composite map (gs x Id,(y)) o/3T,y

is nothing but (g.rry)sx~(y), I have finally

On the other hand, if Z is an H-set , and if m' C M(Z) , then

(a x r = r215 • a)

It follows that

O(a x r = P.((g.lrxxy)s•215215162215 x a) (10.9)

The image of a under An is the element

b= AA,x(a)= A'(~x)(a)(~(X)#x, C (f~u(A) o U ) ( X ) = s

Then, by definition of the product on ? f ( s I have

P) • a • ~ = 7 t ~ s

where the product in the right hand side is the product of ~ ( s s 5S

module. Then

(a x r = P*(Ids x 5U,y)~bSx.r(x)(m(T,.q) X b) (10.10)

Moreover

: * u (rn A*(~x)(a))(T.~(x),g.~x) : rn(T,g) x b M (~r,~(x)) x . . .

10.8. E X A M P L E S A ND A P P LICA TIONS 269

M*/nu ~M*~r-* . . . . [ T,7(X)) [ l c ~ ( T ) x ~x) (m x a)(r.,(x),g.,~) = M*(r ) (m x a)(T..y(X),g.r~x) where I denote by r the composite map

U r = (Ida(T) x fiX) o XT,,~(X)

Finally,

(a • ~b)s(mlr,~)) . . . .

�9 .. x y ) P . g.rrx.rcy)Sx.,(x)x,(y) P (C~T..,(X)y)r ( r ) (m X a)

As q~ is a morphism of Mackey functors, I have

O~(f.-r(x))M*(r) = (P o U)*(r x Idz)(o~(r)•

SO

(a x ~)~('~(r,~/) . . . .

U * * = P*(z~• ((g.,-~.,-Y)~•165 (,~.~(x),y)(PoU) (,.•162215

I must compare this expression with (10.9). I have already observed that

~(x) . -~ (Y) ~_ ~ ( x • Y)

Indeed, the map p : 7 (X x Y) ~ 3'(X).7(Y) defined by

p ( , , , x , v ) = ( ~ , x ) . ( ~ , v )

is an isomorphism. Now it follows from the injectivity of 5xV,g that the square

T."/(X x Y )

(g'TrXxY)Sx~r(XxY) 1 S x ",/(X x Y )

is cartesian. Thus

P )

)

Ids x 5 u X,Y

T.-~(X).-~(Y)

[ (g.;r x.;ry )sx,,/(X)x-ffy) s • -r(x) • -y(y)

and then

(~ • ~)s( '~( r ,~ / ) . . . .

. . . . P, ((g.~x •215 P*(p)P*(<~.;(x),~)(P o v)*(~ • I~)r215 • a)

Let t . (u , x , y ) E T."/(X x Y), Then

Thus

270

Then

C H A P T E R 10. A D J U N C T I O N A N D G R E E N F U N C T O R S

U

As r (Ida(T) • ~x) u and as ( )) z, = o tCT,~(X) , r/X G . u . ( u , z = this gives

Finally

(a • r = P . ( (g .~rxxy)s •215162 x a)

The comparison with (10.9) shows that

a • ~ = O ( a • r = . • O ( r

hence O is compatible with the product of A. In particular

Conversely, let r U ( e u ( M ) , P) o U(Y)

I can apply fornmla (10.10) in the case X = . , and a = CA: this gives

b = A*(rl.)(ea)(~l.),~.l

SO

Moreover

(CA X @)S(m(T,g)) = P*( Ids x ~~176 x b)

m(T,g) • b = M * ( r ) ( m • a)(T.~(.),g.~.)

Here m • a = m, and T.'y(.) = T . ( U / H ) ~_ T. The map g.~r. identifies then with the map

t E T H (g( t ) ,gu( t ) ) r S x ( U / H ) 2

Now the map r is r = (Ida(T) x rlo ) o ~u T,-~(.)

Then identifying w (T.7( .)) with w(T), the map r becomes the identity of w(T) , and then

(CA • O)s(rn(y,g)) = P*( Ids • 5.,v)~s•

Moreover

m(T,g.,.) = s

where dr is the map from T to T x ( U / H ) de~ned by dT(t) = ( t , gu ( t ) ) . As ~ is a morphism of Mackey functors, I have

~ s • 1 6 3 = P . ( f • Id , (y ) )~T

and then

(CA z r = P*( Ids x 5 . y ) P . ( f z Id~(v))r

10.8.

It is easy to see that the square

EXAMPLES AND APPLICATIONS

/~T,Y T.7(Y) ~ T • ~,(Y)

(9.r~y)sx~(z) I [ f • Id.~(y)

T • .y(Z) ~ T • ( U / H ) • -~(Y) IdT • ~Uy

271

is cartesian. Finally, this gives

The question is now to know if there exists ~b E ~ ( M , P o U)(Y) such that

(~A • ~ ) s ( .~ ,~ ) ) = P, ( ( g . ~ ) ~ •

This relation must hold for any T and g. This forces the equality

P ' ( 9 ~ , ~ ) ~ ( ~ , ~ ) : P*(~r,Y)~{r)(~) (10.11)

In the case T = ~/(X) for an H-set X, and rn = M*(Tix)(m'), this relation gives

P*(/L+x.),r)~.~(x)( M*(qx )(m')c~cxl,~x) ) = P*(a.y(x)y )~.~(x)M*(rlx )(rn')


r ) = P*(U oH (71x • Idg))Ox

Thus

P*(13.y(x),y)~b.,(x)(M*(rlx)(m')(,(x),,,x)) = P*(a.~(xly)P*(e OH (fix x Idr))(ox(m')

The map a.~(x),y is a map fi'om ~/(X).'~(Y) to "~(w(7(X)) x Y). I can compose the

previous equality with P*(p), where p is the isomorphism from 7(X x Y) to "/(X).7(Y). Then, as

this gives c)x(m') = P*(p)P*(/3~(x)y )'g,~(x)( M*(rlx )(m')(~(x),~))

Moreover, the map fl~.(x),Y o p is the inclusion from 3'(X • Y) into 7(X) x "),(Y), that is (~u Thus for any X and any m E M(X), I have X,Y"

which proves at least unicity of & (I know that | is a split monomorphism). I observe that this eqnMity can be written as

~x(m) = P*(sU,y)~UoHX/~M,X(m)

In this form, It is easy to check that this equality defines a morphism of Mackey functors f iom M to Po U. Moreover, relation (10.11) holds: indeed, with this definition of qS, I have

P*(C~Ty)6~(T)(m ) = P*(C~T,y)IP*((5~(T),y)t',..,~(T)(:}I*(rl~(T))(rn)(W(T),~:~(T)))

Moreover, if t.(u, y) C T.7(Y), then

Thus 5~iT)y o a t ,y is the restriction to T.7(Y) of u(r,g) • Ida0, ), that is

5wU(T),y 00~T, r = (1/(T,9) X /d,(v)) o flT,Y

In those conditions,

P*(aT,Y)O~(T)(m) = P*(flT,y)P*(u(T,g) • [d.yo'))g'.).~(T)(M*(Tlw(T))(m)(.-r~(r),r~(T)))

As ~b is by hypothesis a morphism of Mackey t\mctors from s to PUoY, I have

P* ( U(T,g) • I G(r) )g'.~,(T) = I~'T s ]~ )* ( IIT )

So

P*(aXy )CD~(x)(m) = P*(flX,y )~f.,fs M)*(UT)( M'(rl~(r))(ra )(~(T),~(rl))

As the square


I/T T , . ~ ( T )

dT I ~ ( T )

T ,-yco(T) YT

f2u(M)*(ur)( m*(rl,,,(r))(m)(~,(T),,~(r))) = M*(G\uT)M*(rl~o('r))(m)(T, av) = m(T,dr)

Then

n * ( ~ r , y ) r = P*(flr,~')~r(.~(r,<~))

which proves (10.11), hence that c x ~b = | So the image of 0 is exactly "H(s o UIA , and the proposition follows. ,,

P r o p o s i t i o n 10.8.3: Let G and H be groups , and U he a G-se t -H. If A is a G r e e n f u n c t o r for H, i f M is a m o d u l e - A and N an A - m o d u l e , t h e n

s ~- ~U(~I)@s163 )

Proof: This result follows from the previous one by adjunction: if N is an A-module, then by proposition 6.6.2, I have

HomA(N,~f(M, PoU)) : HomM~ok(H)(M~aN, P o U ) : HomM=~k(~)(s


On the other hand

HomA (N, ~(f-.u(M), P)o UIA ) = Hom&;(A)(.s ~ ( s P)) . . . .

. . . . nomM.~(c) (s163 P)

The comparison of these two equalities now gives the claimed isomorphism.

R e m a r k : Similar tedious computations show that if N is an A-module, then

~ ( P , ~ u ( N ) ) o U~A --~ ~ ( P o U, N 1

s o QA ~- M ~ ( P o U)

as A-modules.

P r o p o s i t i o n 10.8.4: Let G and H be groups, and U be a G-set -H. If A and B are Green functors for H, if (P,(2, q~, ~P) is a surject ive Mor i ta c o n t e x t for A and B, if @ (resp. 0') is the i s o m o r p h i s m from s163 to s (resp. f rom s163 to Z:u((2~BP)), then the 4- tuple (s163163163 is a surject ive Mor i ta contex t for s

and s

Proof: This is clear, since if P~AQ ~- B, then

Moreover, the isomorphisms of proposition 10.8.3 are natural enough to give here isomorphisms of bimodules. �9

Taking for U the set H, viewed as (l)-set-H, then the fnnctor s is the restriction functor to the trivial subgroup, that is the ewlua.tion functor at the trivial subgroup.

N o t a t i o n : If M is a Mackey functor for the group G, and H is a subgroup of G, I se t

M(H) = (!tesa~(H)M)H(H/H) = M(H)/ ~ t~V(L) LcH

LcH

IrA is a Green functoz3 then A.( H) is a unitary., ring for the product " ,, of (Res~.G(H)A and A(H) is a two-sided ideal of (A(H), .).

c; A H Now the rings .4(H) are evaluations at the trivial group of functors (ResNc(H) ) which are the functors L:u(A) for suitable sets U: indeed, let U = H \ G , viewed as a

t ' N(~(H)/H-se -G by ~.Hx.g = Hnxg

where n ~-+ r7 is the projection from Nc~(H) to Na(H)/H. Then for any G-set X, I t l a v e

U oG X = X H


Indeed, in (H\G) oe X, I have (Hy, x ) = (H, gx), and (H,x) 6 (H\G) oa X if and only if z E X H. Moreover (H,x) = (H,x') if and only if x = z'.

Thus if M is a Mackey functor for G, I have

s c, H = (ResNc(HtM)

This gives in particular the following lemma:

L e m m a 10.8.5: Let H be a subg roup of G, and A, B and C be Green functors for G. I f M is a B-module -A, and N is an A-module -C, t hen

~- (Reslv~(H)M) "@{F{esNCG(H}A)H(Res%G(H)N) t/

as (Res~va(H)B)H-modules-(Res~c(H)C)".

Now evaluating this isomorphism at the trivial Nc;(H)/H-set, I have the following consequence:

P r o p o s i t i o n 10.8.6: Let G and H be groups. I f A and B are Green func tors for G, if (P,(2,~, tit) is a sur jec t ive Mor i t a contex t for A and B, t hen for any subg roup H of G, the 4- tuple (P(H), ~)(H), ~, ~) is a sur jec t ive M o r i t a con tex t for A(H) and B(H).

Chapter 11

The simple modules

11.1 Genera l i t i e s

Let A be a Green functor for the group G. Theorem 3.3.5 states the equivalence between the category of A-modules and the category of representations of C.4. I can in particular apply to A - M o d the generalities on representations of categories proved in [3]. In particular, the simple A-modules can be described as follows:

P r o p o s i t i o n 11.1 .1: Let A be a G r e e n f u n c t o r for t h e g r o u p G.

1. I f X is a G-set , and V is a s imple m o d u l e for t he a lgeb ra EndeA(X) = A(X2), t h e n t h e m o d u l e Lx,v defined by

Lx,v(Y) = A(YX) •A(x2) V

has a u n i q u e m a x i m a l s u b m o d u l e Jx,v, def ined by

Jx,v(Y) = {~i ai| 'Vr A(XY), ~-~.(r a,).vi =O } i

T h e q u o t i e n t Sx,y = Lx,y/Jx,v is t h e n a s imple m o d u l e .

2. Converse ly , if S is a s imple A - m o d u l e , and if X is a G-set such t h a t S(X) r O, t h e n t h e m o d u l e V = ,q'(X) is a s imp le A(X2) -modu le , and S is i s o m o r p h i c to Sx,v.

Proof : The first assertion follows from [3]. The second is not stated there explicitly, but it is a consequence of the following argument: if W is a non-zero A(X2)-submodule of S(X), then by adjunction there is a non-zero morphism from Lx,w to S, which is onto since S is simple. Then as Lx,w(X) = W, it follows that l/V = S(X), thus S(X) is a simple A(X2)-module. Then ,5" is a simple quotient of Lx,s(x), which has a unique simple quotient Sx,s(x). Thus S _~ ,5'x,s(x). "

11.2 Class i f i cat ion of the s i m p l e m o d u l e s

In the special case of a functor on CA, I have moreover the notion of minimal subgroup: a subgroup H of G is called minhna/for the functor F on CA if F(G/H) r O, but

276 CItAPTER 11. THE SIMPLE MODULES

if F(G/K) = 0 for any subgroup K of C of strictly smaller order. Such a subgroup always exists if F is non-zero, since F is additive.

Let S be a simple A-module, and II be a minimal subgroup for S. Let K a subgroup of G such that 1Is < iH[, and ~) be an element of A((G/H) 2) which factors

through G/K in Ca, i.e. such that.

: oG/K/7 for C A ( r x (r /9 • IC,/H))

Then setting P = G/H, and V = S(F), I have the following equality in Lr,v(F) = V for v E V

1A(r~) | r = r | v = (c~ %./~,- ~3) q) v = Sr.v(a)(/7 @ v)

But as H is minimal for S, the element/7 @ v of S(K) is zero. So the element 4.v is zero.

Thus the module V is annihilated by any endomorphism of F which factors by a

set G/Is such that IKI < IHI.

N o t a t i o n : Let A be a Green [unctor, and H be a subg~'oup of G. I denote by ]~(H) the quotient of A((G/H) 2) by the two-,sided ideal generated by the elements of the fo~m ~ oa/s, 17, fo~ IKI < IHI.

With this definition, I have the following classification

P r o p o s i t i o n 11.2.1: Let A be a G r e e n f u n c t o r for t h e g r o u p G.

1. I f S is a s i m p l e A - m o d u l e , and H is a m i n i m a l s u b g r o u p for S, t h e n V = S(H) is a s imp le A ( H ) - m o d u l e , and S is i s o m o r p h i c to Sc./Hy.

2. Converse ly , if H is a s u b g r o u p of G, and V is a s i m p l e / i ( H ) - m o d u l e , t h e n Sa/~,v is a s imp le A - m o d u l e , t he g r o u p H is m i n i m a l for Sc,,/u,v, and m o r e o v e r Sa/H,v(H) ~- V.

3. Le t H be a s u b g r o u p of G, and V be a s i m p l e / i ( H ) - m o d u l e . I f X is a G-se t such t h a t Sa/H,v(X) r 0, t h e n X u r ~. In p a r t i c u l a r , t h e m i n i m a l s u b g r o u p s of Sc,/H.v are t h e c o n j u g a t e s of H in G.

4. Le t H and K b e s u b g r o u p s of G. If V is a s imple A ( H ) - m o d u l e , and if W is a s i m p l e / t ( K ) - m o d u l e , t h e n the m o d u l e s Sc,/s,v and ,5'c,/r,,w are i s o m o r p h i c if and o n l y if t h e r e e x i s t s x E G such t h a t K = XH and W ~ V .

R e m a r k s : a) I have identified a A(H)-module V with the associated . 4 ( ( g / H ) 2 ) -

module. b) The notation W _~ ~V means that V maps to W by the isomorphism c:~,. fi'om G/H to G/K in CA, deduced from the conjugation c~: : G/H --+ G/K defined by

c~(g.H ) = 9z-I.K.

Proof : I have already proved assertion 1). For assertion 2), I already know that if V is a simple / t(H)-modnle, viewed as an A((G/H)2)-module, then ScUH,v is a

i1.2. CLASSIFICATION OF THE SIMPLE MODULES 277

simple A-module such that SH,v(H) "z_ V. It remains to show that H is minimal for Sa/u.v. So let K be a subgroup of G such that IKI < Igl. Let a | v be an

element of La/n,v(K). Then a | v E Ja/n,v(K): indeed, if r C A((G/H) x (G/K)) ,

then r oa/K a factors through G/K, so annihilates V. Thus a | v E Ja/H,v(K), and Sc/I~,V(K) = O.

Under the hypothesis of assertion 3), there exists a G A ( X x (G/H)) and v C V such that a | v is non-zero in Sa/n,v(X). In particular, there exists an element

E A( (G/H) x X ) such that 6 ox a has non-zero action on V, so does not factor

through sets G / K such that IK[ < IH[. But

L e m m a 11.2.2: Le t X and Y be G-sets , a nd a C A ( X Y ) . Let

b = A. xy (a) C A ( Y X Y ) = Hornc~(Y, Y X ) c = E Hornc~(YX, X) yxy x .

T h e n a = c ovx b in Ca.

Admitting this lemma for a while, I see in particular that a factors in CA through (G/H) x X , and so does r a. As

( a / H ) • x II n xEG\X

g~H\G/Gx I see that there exists x and g E G such that [H N G~I = Igl, and then gx C X H, which proves that X H r O. Then if K is another minimal subgroup of Sa/H,v, and if W = Sc/~,v(K), I know that Sa/H,V ~-- Sc/K,w. The group K has fixed points on G/H, and H has fixed points on G/K, so K is conjugate to H. This proves assertion 3).

Under the hypothesis of assertion 4), the group K is minimal for SC/H,V, so it is conjugate to H. I can then suppose H = K, and the proposition follows then from the fact that V ~ Sa/H,v(H) ~- SG/H,w(H) ~ W .

P r o o f of l e m m a 11.2.2: With the notations of the temma, I have

c o y x b = A . ( X l y l x 2 Y 2 1 A , ( Xlylx2y2 / (cxb) \ xly2 / \XlYlX2YlX2y2/

c x b = A .

But

As the square

. . . . A . ( ylZlX2y2 IA*(yl~lX2Y21(a) kxlylxly2x2y2/ k x2y2 /

(&) X Y ~ Y X ~ Y

k xlylxl y2x2y2 /

(Xy)2 ( XlYlX2y2 i (XY)3

\xWlx2ylx2Y2

278 CHAPTER 11. THE SIMPLE MODULES


A * ( xlylx2y2 I A . ( ylx~x2y2 / = A . ( xy ) A * ( xy ) \XlYlX2YlX2y2] XlYlXlY2X2~]2 ] xyxy yxxy

and then

e ~ x l y 2 ] xl]xy ~xxy \ x2y 2 / . . . .

. . . . A. xy xy


11.3 The structure of algebras A(H)

Having classified the simple A-modules, 1 have still to describe the structure of the algebras /i(H):

Nota t ions : Let K be a group acting by automorphisms on ~n R-algebra A. I denote by A | K the tensor product A | RK, with the following multiplication

(a | k).(a' | k') = ak(a') Q kk'

If H is a subgroup of G, [ denote by NG(H) the quotient No(H)/H.

Propos i t ion 11.3.1: Let A be a Green functor for the group G. If H is a subgroup of G, then A(H) identifies with 5.(H) | No(H), by the map

c~ E A((G/H) 2) ~ ~ A*(iH,~,.)(c 0 | n ~ ( H )

where iH,n,H : G/H -'~ (G/H) 2 maps gH to (gH, gnH).

Proof: By definition, I have

A((G/HT)) ~- 0 A(H n ' H ) xEH\G/H

the isomorphism being obtained by the maps

c~ E A((G/H) ~) ~ ~ A'(iH,~,H)(a) xEH\G/H

iH,x,H : G/(H ~ ~H) ~ (G/H) 2

defined by ig,~,H(g(H M ~H)) = (gH, gxg)

The inverse isomorphism maps the element fl E A(H N ~H) to A.(ig,.,g)(fl). But if x • No(H), then m.(iH,x,u)(fl) factors through G/(H N ~H):

where i l l ,z , H is the map

11.3. THE STRUCTURE OF ALGEBRAS A(H) 279

L e m m a 11.3.2: L e t X a n d Y b e G-se t s . I f f a n d g a r e m o r p h i s m s of G - s e t s f r o m X t o Y, a n d if a a n d b a r e e l e m e n t s of A(X), t h e n

(x) A. g(x)x (a) oxA. xf(x) (b) z A . g(x)f(x)

P r o o f : It suffices to compute

(x) A. g(x)x (a) ox A.

. . . z A .

The cartesian square

x )) (b) . . . . xf(x

y, xY2) A.(y,xy2) [A. x,x2 ~(axb)] YlY2 \ y l x x y 2 \g (x l )X lX2f (x2) ]

(x) XX X ~ X 2

\g(xl)xlx2f(z2)] Y X Y ( ylxy2 } YX~Y

\ylxxy2

shows that

A.(ylxy2 IA.( ZlX 2 ~ ( X )A.(Z) kY,XXy2/ \g(x,)x,x2f(x2)] = A. g(x)xf(x) xx

so that

A. g(x)x (a)oxA. xf(x) ( b ) = A . A. = . . \ YlY2 / g(x)xf(x) xx

. . . . A. (a.b) g(x)f(x


Taking Y = G/H and X = G/(H N XH), and

f : G/(H r~ XH) --~ G/H

g: G/(H A XH) ~ G/H

f(u(HN~:H)) =uH

a = 9 ~ A ( H n ~H)

this l emma proves that A.(iH,~:,H)(fl) factors through G/(H A xH) in CA. Its image in A.(H) is then zero if x ~ Na(H). If 7r denotes the canonical project ion from

A((G/H) 2) o n t o / i ( H ) , there is a surjective morphism 0

O: (~ A(H) ---+ f4(H) ne~Va(H)


which maps the element/3 of the component n of the left hand side to ~rA.(iH,n,H)(/3). If K is a proper subgroup of H, and if ~ = tH(y), for 7 ~ A(I() , then denoting by

7r H the projection from G / K to G/H, I have

A.(iH,~,H)(/~) z A.(iH,,~,H o ~rH)(?)

Lemma 11.2.2 shows that this element factors through G / K , and its image under ~r is zero. Final ly I have a surjective morphism 0

0: �9 X(H)+A(H) ~Na(H)

Denoting by ~ | n the element/~ of the n component, I have

0(8 | n) = 7r A.( iH,~,H )(/3)

If n and n' are elements of Na(H) , and if fl and/3 ' are elements of A(H), then

A.(iH,~,H)(/~) O~/H A.( i . ,~, , . ) (~ ') = A.(ig,n~,,H)(/~.~l~ ')

Indeed, sett ing X = G/H, and denoting by x ~ xn the map from X to X which sends gH to gnH, I have

d.(ig,,~,H)(/~) Oa/H d.(iu,n, U)(/~') . . . .

\ X lX2 / \ Z l X 2 X 2 X 3 / X l X l r t X2 Z2 n! (/~ X /~1)

As the square (x) X Xft

X - - --~ X 2

x 1 [ xlx2

X 3 ~ X 4 Z l X 2 X 3 I

X l X 2 X 2 X 3 /


A xlx2x3 xlx ) ( x )A,( x ) \ X l X 2 X 2 X 3 ] X l X l ~ X2 X2 rtt z A. x z n xnn' x x n

whence

The right hand side can also be writ ten as

A. (/~ x •') . . . . X Xn?'l I XX \ X 1 X 2 r t / '

. . . " * x (•' = A.(iH,,~n,,H)(/~.~/~ ') XX x n

11.3. THE STRUCTURE OF ALGEBRAS A(H) 281

It follows that 0 is a morphism of algebras from A.(H) | Na(H) to /~(H). This morphism is moreover unitary, since the unit of A(H) | Na(H) is e~-]-ff N 1, which is mapped to

Conversely, if a E A((G/H)e) , and if n e -No(H) I can consider the element

d*(ig,,~,u)(a) of 7 . (g) . If a factors through Y = G/I<, then setting X = G/H, there exists b E A ( X Y ) and c E A (YX ) such that a = boy c. Then for n E No(H), I have

A*(iH,n,H)(a) = A* ( x ) A. (xlyx2~ A* ( x~yx2 ~ ( b • c) xn '. xlx2 / \ x l yyx~ /

As the square

) X Y X Y X

<11 l X ~ X 2


As X Y = (G/H) x (G/K) ~- I I G/(H N gK)

geH\GIK

then for any c~ r A(XY) , the element A. ( 7 ) ( c ~ ) i s a linear combination of elements

of the form tHHn~Kag. If the order of K is strictly smaller than the order of H, the

groups YNgI ( are proper subgroups of H, and it follows that the image of A. ( 7 ) (c0

in J . (H) is zero. So I have a morphism r f rom/~(H) to A(H) | -Na(H), defined by

n6No(H)

Moreover r | n) = ~ A*(iH,n',H)A.(iH,n,H)(/3) | n'

n'6NG(H)

The only non-zero product A*(iH,n,H)A.(iH,~.H) corresponds to n' = n, and it is equal to the identity of A(H). It follows that r is the identity, so that 0 is injective, hence an isomorphism. Then r is the inverse isomorphism, which proves the proposition. �9

11.4 T h e s t r u c t u r e of s i m p l e m o d u l e s

I know that the isomorphism classes of simple modules can be indexed by the con- jugacy classes of couples (H,V), where H is a subgroup of G, and V is a simple ft.(H) | Na(H) module.

G N o t a t i o n : I will denote by SHy or SHy (instead of SO/H,V) the module associated to the couple (H, V).

The module V can be viewed as 7VG(H)-module, by the homomorphism n eC/H | n from No(H) to A(H) | NG(H). Then denoting by [Y] the permutation module associated to a set Y, with the set Y as a basis, I have the

P r o p o s i t i o n 11.4.1: Le t H be a s u b g r o u p of G, and V be a s imple A(H) @ ~Tg(H)-module . I f X is a G-set , t h e n IX H] is a JVG(H)-module, and

S~,v(X) ~- Horn(IX'], V)~ G<m

I t is t h e set of m o r p h i s m s of N u ( H ) - m o d u l e s f r o m [X H] to V which f ac to r t h r o u g h a p r o j e c t i v e m o d u l e .

I f f : X -+ Y is a m o r p h i s m of G-sets , t h e n for c~ r SH,v(X) and y E yS , I have

SH,v,(f)(cO(y ) = ~ c~(x) z E X H ](x)=y

I f g : Y --+ X is a m o r p h i s m of G-sets , t h e n for a E SH,v(X) and y E yH, I have

=

Fina l l y if a C A(X) a nd f E SHy(Y), t h e n a • f is t he m o r p h i s m f r o m [(X • y)H] = IX u • yH] to V def ined by

(a x f ) (x ,y)= (A*(m~)(a)| l ) . f (y )

w h e r e mr is t h e m o r p h i s m of G-sets f r o m G/H to X def ined by m~(uH) = ux.

P r o o f : I will first prove the isomorphism

&,v(x) Horn(IX'l, ~

and the other formulae will follow.

m

11.4.1 The isomorphism S H y ( X ) ~ Hom([XH],v)~ ~(H) First I observe that X H identifies with the set of morphisms of G-sets from G/H to X, by the map sending z to m~. Let a | v be an element of LG/H,v(X). Then

a 6 A(X • (G/H)). If x 6 X H, then m• C A((G/H) • X) , and I can consider the

-o ( ) product m r x a, which is an element of A (G/H) 2 Its image rr(m* ox a) is then in

11.4. THE STRUCTURE OF SIMPLE MODULES 283

A(H). This is identified with A(H) | Na(H), which acts on V. Now I can consider the map

),~,v : x ~ X ~ ~ 7~(m; ox a).v

Proposition 11.3.1 shows that this is also

7r(rn*~ ox a).v = E [A*(iH,~,H)(m* Ox a)| n] .v nE/~G(H)

But 1 can view a as an element of AG/H(X), and then

* A'a~ H ( d)( m x ox a = (m~)(a) = A* m~ • I a)

Thus

Aa,v(X) = E nEI~G(H)

[A*(iH,.,H)A*(m~ x Id)(a) | n] .v . . . .

ux unH ) . . . .

neNG(H)

Now let g~,~ be the linear map from IX H] to V defined by

ux uH (a) | l .v E V

TrN~ ~(H)(g. , . )(z) = Then

This can also be written as

or

neNG(H)

TrNla(H)(g~,.)(x) =- ~ [~A* (un_lxuH)uH (a) | nE~r a(H)

TrNa(H)(ga,.)(X) = ~ A* (unH) A* (un_lx uH) (a)Qn .v ne~Va(H)

whence finally

TrNl C(H)(ga,.)(x) = ~ A* (ux unH) (a) | .v

This formula proves that A~,v = Tr~C(H)(ga,v), hence that A~,. is a morphism of JVa(H)-modules from [X H] to V, which factors by a projective module. I obtain that way a morphism

~: A(x • (G/H)) | V -~ ~om([X"l , V )~ G(") a | v H Ao,~

Moreover, i f bEA( (G/H)2) , then

/~aOG/Hb,v z /~aw(b). v

284

Indeed, for x E X H

CHAPTER 11. THE SIMPLE MOD ULES

(" ) m~ o x a o a / . b .v = ~ ( ~ ; o x a )~ (b ) .~

The morphism ~ is then a morphism from LG/H,v(X) to Hom([XH], v)N1 G(H), which passes down to the quotient SH,v(X): indeed, if ~ a~ | v~ is an element of Ja/H,v(X), then for any ~b E A((G/H) x X ) , I have

~ ( r ox a i ) . v i = ~ 7r(r o x ai).vi = 0 i i

It follows that for any x E X H

E 7r(m*z ~ ai).vi = 0 i

so that ~ i ~a,,~, = 0. Thus I have a morphism, that I still denote by ~, from SHy(X) to Hom([XH], V)~ c(H).

Conversely, let f E Hom([XH], V)~ a(u), expressed as

f = TrNIG(H)(g) (11.1)

where g E Hom([XH], V). If x E X H, then m~,. E A ( X • (G/H)), and I set

#(f) = ~ m~,. | g(x) E SHy(X) (11.2) x 6 X n

Then # ( f ) does not depend on the choice of the element g in (11.1): this is equivalent

to say that if TrNa(H)(g) = 0, then the expression (11.2) is zero. In SH,v(X), this

means that for any r E A((G/H) • X) , I have

(r ox , , ,~ . . ) .g (x) = 0 x 6 X H

Let g be any element of Homn([XH], V). Then

x E X H ~EX H

But setting Y = G/H, I have

' \ YlY2 / \ylXlXly2/

Moreover

r • rex,. = r • A. m~(y)y (e) = A. ~kylxlmx(Y2)Y2] \ ylXl


As the square

YlY2

y2 > Y X Y ( YlY2 ~ ( ylxly2

\ylzlm=(y2)y2 )

\ y l X l X l y ~ ]


A*( Y~xlY2 I \ylxlxly2 /

and this gives

~boxm~,.= A,(y lx ly2) \ YlY2 /

Finally

Z ( r zEX H

( ylXlY2 ~ ~ YlY2 ~ A * ( YlY2

\YlI= \ylm=(y=)/ (~) = A*(Id x m=)(@)

[A*(iH,.,H)A*(Id • m=)(@) | n] .g(x) . . . . xEX H

~eR~(H)

' =~x- ~ H ~ x @) | ~ . s . . . .

�9 ( ~ ) @ Tt . g ( r t - l x ) . . . .

.~Ra(H)

xEX H

Thus if TrlNG(H)(g) = 0, then the right hand side of (11.2) is zero�9 It follows that # ( f ) is well defined by equations (11�9 and (11.2). So I have a morphism from

Hom([XH], V)~ G(H) to SHy(X). Moreover, if a | v E SHy(X), then as ~ = Tr~G(H)(9~,~), I have

~(Ao,~) = ~ .~=,. e go,v(=) x~.k " H

If ~b �9 A((G/H) • X) , the previous computation shows that I have

286 C H A P T E R 11. THE SIMPLE MODULES

But

�9 , � 9 z

E xs H

nENG(H)

[ ""1 ] A* ( u H ux (~b) | 1 .Aa,v(X) . . . . xEX H

? (,x%),~ ....

= E xEX H

neNG(H)

u H uH

"uH uH "uH

"'" uH u g u lH UlX U2X u2nH.] (~ x a) . . . .

"'" ux ux "unH) (r x a) = A* ( u H uH

This gives

E (r OX lT~x,*)*ga,v(X) : E xEX H xEX H

n6/~G(H)

A* (~ x a ) | .v uH ux ux unH

On the other hand

(~ ox a).v = E [A*(i.,.,.)(r o~ a) | ~] .v

Moreover

A ( ' . , . , g ) (~ OX a) = A (ZH,.,H)A. \ u ,H "u,H / \ u , H x x u , H / ( r x a)

As the square

uH x "~

u H x "unH ) G / H x X ) (G /H) x X x ( G / H )

uH x \ u lH u2H /

G I H , (GIH) 2 uH

( 'uH un, H )


A . ( i H n H ) A . ('ul H X "u2gx~ : A . ( ' u S x~ A* ( u g x ) ' ' \ u l H u2H / \ uH / \ u H x u n H /

and then

( u H X ' ] A . ( u I I x ) A * ( ~lHX'u2g ) A*(iH,n,H)(r = A. \ uH ] \ uH x unH] 2u~H x x usH ( r . . . .

(uH x ) A* ( uH x "~ . . . . A. \ uH ] \ uH x x unH] (~b x a)

This gives

(r o x a ) . ~ : A, 2(UHuHX'~] A* (uH x x unH ] (r x a) | n .v E ,ae~Va(H)

But the map sending (uH, x) E (G/H) x X to g-~x 6 H \ X induces a bijection

G\((G/H) • X) ~_ H \ X

Then the map c~ defined by

~ : u ( H n G ~ ) e I I a / ( H n a ~ ) ~ ( u ( H n C ~ ) ) =(gH, gx) e ( O / H ) • zeH\X

is an isomorphism of G-sets. In particular,

IdA((a/H)• = Y~ x~H\X

Then

( r = Z xEH\X

(u(HnG~) But A. \ ~H

A.(cr~)A*(~)

A . u H x �9 �9 u H x ( u" )A.(ax)A (a~)A (uHxxunH)(~b x a ) | . . . .

E neNc(H) xEH\X

= t H HnG~"

[ {u(HNG~,)]A. (u(HNG~) ~ ] A . \ uH " ] \ u H u x u x u n a ] (r • 1 7 4 .v

Thus i f H ~: G~, then A ,k ~H ] ( a ) = 0 for a n y a . Now I can restrict the summation to x E H \ X H = X H, and then

(~20X a).tl : E zEX H

[ ,,. ) ] A, uH u H u x u x u n H (r • 1 7 4 .v . . . .

[-< ) ] . . . . ~ A* uH ,~Zc(H) u H u x u x u n H (~ x a ) | .v

xEX H

Finedly, for any ~, e A((G/H) x X) , I have

xEX ~

288 C H A P T E R 11. THE SIMPLE MODULES

which proves that in S H y ( X ) , I have

:zE X H

or that # o A is the identity.

Conversely, if f E Hom([XH], V )~ G(H), i.e. if f is of the form

f = Tr~c(H)(g)

then

so that for y E X H, I have

A o # ( f ) ( y ) =

# ( f ) = ~ m~ , .Qg(x ) x E X ~

neJVc(/4) zEX H

M o r e o v e r

uy unH (m~,,) = A* \ u y unH A. \ u x u H /

For g ivenn , x a n d y , let Q ..... 9, a .... ~ and b .. . . y be such that the square

ar~,m ,xj Q .... ~ > G / H

C / H , x • ( G / H )

uy unH

is cartesian. Then if Q~,y is non-empty, there exists u and u' in G such that ux = u'y and uH = u 'nH. These equalities imply that

uH.x = {ux} = {u'y} ~ u'nH.x = {u'nx}

u H u H s o y : nx. Thus if y ~ nx, then Q . . . . y : 0, and the product A* ( ~ u ~ - , ) A. (=~,H)

is zero. And if y = nx, then Q .... y identifies with G/H, the map b,~,,:,y being the identity, and the map a .... y being right multiplication by n. So I can write

. . . .

-e~a(H) t \ - - ~ j

.e~o(H) ~eRa(H)

. . . . T~lVC(")(g)(~) = f ( y )

which proves that I o # is also the identity, hence that )~ and # are mutual inverse isomorphisms.


1 1 . 4 . 2 T h e A - m o d u l e s t r u c t u r e o f SH,V

Let f : X ~ Y be a morphism of G-sets, and h E Hom([ZH], V)~ ~(H), expressed as

h = Tr~G(H)(k). Then

# ( f ) = ~ m~,. | x 6 X H

so that SH,V,.(I)@(f)) = ~ m.( f x Id)(m~,.) | k(x)

x 6 X H

Under the isomorphism A, this element is sent to the map which sends y 6 yH to the element

~ex" uy uH A . ( f • Id)(m~,.) | .k(x)

neNG(H)

Moreover

A . ( f • Id)(m~,.) : A , ( f • Id)A, (r : A. (Ca~H)

Let Q~,~,y, a . . . . y and b .. . . y such that the square

a n , x , y

G/H , g • (G/H)

uy unH

is cartesian. If Q .... y r 0, then there exists u and u' in G such that uy = u'f(x) and unH = u'H. These equalities imply that

unH.f(x) = {unf(x)} = u'H.f(x) = {u'f(x)} = {uy}

uH u H Thus if y r nf(x) , then Q .. . . y---- O, and the product A" (~yuH)A. (,d(~)uH)is zero.

If y = nf(x) , then Q,:,y ~- G/H, the map b~,~,y is the identi ty and the map a . . . . y is right mult ipl icat ion n. This gives

e ---- E n E N G ( H )

x e X H , f ( n z ) = y

IA* uH n I .k(x) = ( u n H ) (Ea/H| ""

So I have proved that

H (~7-~ | ~).k(~-'x) = ~ h(x) xEX H nENc(H) x 6 x H ](~)=y ](x)=y

SH,y,.(f)(h)(y) = H h(x) x E X H

f(~)=y

290 C H A P T E R 11. T H E S I M P L E M O D U L E S

E xE X J4

~Va(H)

Now if g is a morphism of G-sets from Y to X, then

S;/,y~(h) = ~ A*(g • Id)(m~,.) 0 k(x) x ~ X H

The image under A of this element is the map sending y E yH to

uy u H A*(g • I d ) ( m x , , ) @ n .k(x) . . . .

So I have

~e_x H ug(y) u H n~Nc(H)

= ~ o , ( h ) ( g ( y ) ) = ~ g ( y )

s;~,v(g)(h)(y) = hg(:y) Finally, let X and Y he G-sets, and

h = Tr~G(H)(k) E Hom([rH], V)

Then #(f) = ~ my, . |

yEY H

Thus, if a E A ( X ) , I have

a • # ( f ) --- (a • .~ , . ) 0 k(y) yEY H

The image under A of this element is tile map sending z C Z H

element

y 6 Y H

Moreover

. . . . A*( " " x, ,H ) \ u z u n H / x uy u H )

Let Q ..... y, a .... ~ and bn,z,~ such that, the square

= X H X y H to the

an,z ,y Q .... ~ - - ~ X •

" uy u H ] G I N , z • ( a / H )

uz u n H

11.5. THE SIMPLE GREEN FUNCTORS 291

is cartesian. If z = (x0, Y0), and if Q~,z,y ~r 0, then there exists u and u' in G such that uz = (UXo, uyo) = (x, u'y) and unH = u'H. In those conditions

e = E ~e~r a(H)

u'H.y = {u'y} = {uyo} = unH.y = {uny}

Thus if y0 r ny, the set Q,~,z,y is empty, and the product A* ( . . . . H) A. \x ~ ~HJ zero�9 and if Y0 = ny, then Q,~,z,~ ~- G/H, the map b~,~,y is the identity, and the map an,z,y is defined by

a,~,~,y(uH) = (UXo, uH)

It follows that

[A* (uxUoHuH) A* (x:H)(a)@rt] . / r . . . .

nC/VG(H) L \ / j u x 0

. . . . (A*(rn~:o)(a) | 1) • (CG/H@rt).~(l't-lyo) . . . . ~Na(H)

�9 , . z . . .

. . . . | l ) . h ( > )

which proves that

(a • h)(xo, yo) = (A*(mxo)(a) @ 1).h(yo)

and completes the proof of the proposition.

11.5 The s imple Green functors

The previous results give a new proof of a theorem of Th6venaz (see[13]) on simple Green functors. The notion of simple Green functor relies on the notion of functorial ideal, that Th~venaz defines as follows (see [13] 1.8): a functorial ideal I of the Green functor A for the group G is a sub-Mackey functor such that for any subgroup H, the module I (H) is moreover a two-sided ideal of A(H). It is equivalent to say that the Green functor structure of A passes down to the quotient A/ I .

Translating this definition in terms of the product x, I see that for any G-sets X and Y, I must have

I ( X ) • A(Y) C_ I (X x Y) a ( x ) • I (Y) g I ( X • Y)

Conversely, if I is a sub-Mackey functor of A such that this condition holds for any X and Y, then the product • of A passes down to the quotient A/ I .

In other words, a functorial ideal is nothing but an A-submodule-A of A. Thus a Green functor A is a simple Green functor if and only if A is simple as an A-module-A.

But the A-modules-A are also the A@A~ Thus if A is a simple Green functor, then there exists a subgroup H of G and an AQA~ | JVa(H)-module V such that ~,A~A ~ A "~ ~H,V


as AQA~ (the exponent A(~A ~ recalls that the simple module SHy is a simple module for this Green functor). Then H is the unique minimal subgroup of A

up to conjugation, and V = A(H) = ~.(H) is a simple A~A~ | Na(H)-module . Now

(A~A~ g ~ AH~(A~ H ~ AH~)(AH) ~

This follows from lemma 10.8.5, since moreover for U = H\G as a .Nc(H)-set-G, I have U/G = . , and so

(Res~va(H)b) H = s = bu/G = b

This isomorphism is now clearly an isomorphism of Green functors, and it follows from evaluation at H the isomorphism of algebras

A@A~ ~_ A(H) | A(H) ~ = A(H) | A(H) ~

Now say that A(H) is a simple A(H) | A(H) ~ | JVa(H)-module, is equivalent to say that d ( Y ) is a _Na(H)-algebra, o1" an algebra with an action of -Na(g) , having no proper two-sided ideal invariant by Na(H). Moreover, if A is non-zero, then A(*) # 0. As

SH@g4~ ~ Hom({.l, v)N1 o(H)

I see that Tr~(H)(A(H)) has to be non-zero. As it is a two-sided invariant ideal of % /

A(H), I must then have

A ( H ) :

which proves that A(H) is moreover a projective/Va(H)-algebra. Conversely, if B is a Nc(H)-algebra, then I have a Green functor FPB, such that

for any subgroup K of NG(H), the ring FPB(H) is equal to B H, the restrictions being inclusions, and the transfers being relative traces. In those conditions, for any /Va(H)-set Z, I have FPB(Z) ~-- Hom~(H)( [Z ], B).

If B is a projective _NG(H)-algebra, having no proper two-sided invariant ideal, then let

G NG(H) A = IndNG(H)InfK%(H)FPB

Then for any G-set X, I have

A(X) = Hom~G(HI([X"], B)

and as B is projective, it is also

A(X) = Hom([XH], B)~ c(H)

In those conditions, the group H is a minimal subgroup for A, and A(H) = A(H) = B is a simple A(H) | A(H) ~ | Na(H)-module , so that

A ~- S A~gA~ H,B

as A-modules-A. Thus A is a simple Green functor.

So I must find, for a group K, which are the algebras B with a K-action which are projective and have no proper two-sided invariant ideal. Let ] be any maximal

11.5. THE SIMPLE GREEN FUNCTORS 293

(proper) two-sided ideal of B as an algebra.. Such an ideal exists by Zorn's lemma. Let L be the stabil izer of [ in K. Then for any k 5 K, the image k(I) of I by k is also a maximal two sided ideal of B. In part icular

f-) k(t l = 0 kEK

since it is a two-sided invariant ideal. Let P be a subset of K of maximal cardinali ty such that

J = (~ k(i) # o kEP

Then P :~ K, and there exists ko C K - P. Replacing P by kolP, I can suppose k0 = 1 ~ P , and then

I N J =O

Moreover, the maximal i ty of I implies I + J = / 3 , thus

B = I |

Then J identifies with B/I , which is a simple B-module-B. Thus J is a minimal two-sided ideal of B. Then JI is contained in J V/I = 0. So I is contained in the right annihilator of J , which is a two-sided ideal. As J # 0, the annihilator of J is

equal to I . On the other hand the sum

keKIL

is a non-zero two-sided invariant ideal of B, so it is equal to B. As J.k(J) is contained in g A k(J ) , which is a sub-two-sided ideal of J , I have J.k(g) = 0 if k ~ L. Thus if (jk)keK/n is a family of elements of J such that

Z k(jk) = 0 kEK/L

then for any kl 6 K , I have

o = kl(J) E k, ( E = 1(J.5 1) keK/L \ k e K / L ]

Then Jkl is in the annihilator of J , hence in I , hence in I C~ J . Thus Jkl = 0, and

B = (9 k(J) keK/L

In part icular , the unit of B decomposes as

IB : E ek ke1,/L

for some ek E k(J). If kl E K/L, I have

keh'lL

294 CHAPTER 11. THE SIMPLE MOD ULES

and unicity of the decomposition implies ek = k(el). Moreover

el ~--- el Z k(el) z e~ kEK/L

and for any b E B, I have

elb = e l b

e l = Y~ kl EK kEK/L klk6L

E k ( C l ) : Z elbk(el)=elbel kEK/L kEK/L

because elbk(el) G J.k(J) = 0 if k ~ L. The same argument shows then that be1 = etbel, so that el is central in B. Then J = elB is a simple algebra with an action of L, and B is isomorphic to the induced K-algebra

B _~ I n d 2 J = RK | J

with product defined by

( k | 1 7 4 0 if k - l k ' ~ L k @ j.(k-lk')(j ') otherwise

Now say that B is projective means that there exists elements jk, for k E K/L, such that / \

kl ( ~ k(jk)) =1t3= ~_, k(el) kl EK \kEK/L / kEK/L

The product by el gives

~l]C(Jk) z E l ( jk) : Zrf ( E jk) kEh'/L \keK/L I

IEL

It follows that J is projective as L-algebra. Conversely, if

el = TrL(j) = ~ l(j) IEL

then T@'(j) = TrI~-TrL(j) = Tr~L'(el) = 1B

and B is projective. Similarly, if J is a simple algebra on which L acts, then IndLKJ is a K-algebra

without proper two-sided invariant ideals: if U is such an ideal, and if el is the unit of J , then V is the direct sum of k(el)V, for k E h'/L, and k(el)U is isomorphic to elU, which is a two-sided ideal of J. If elU = 0, then U = 0, and if elU = J, then U = B .

Thus the algebra B is isomorphic to I n d , J , where L is a subgroup of K, and J is a simple algebra with an action of L, the algebra J being moreover projective as L-Mgebra.

Finally, I observe that the group L and the L-algebra J are entirely determined if B is known as K-algebra: indeed, if/1 is a maximal two-sided ideal of B, then k(el)I1

11.6. SIMPLE FUNCTORS AND ENDOMORPHISMS 295

is isomorphic to elk-l(/1), which is a two-sided ideal of J , hence equal to zero or J , and

I , = @ k(el)J kEK/L

k(e~ )11 r

As I1 is maximal, there exists a unique ko C K/L such that

h = @ k( l)J keK/L-{ko}

The maximal two-sided ideals of B are then conjugate by K. The group L is the stabilizer of a maximal two-sided ideal of B, and the algebra J is the unique simple quotient of B: the couple (L, J) is then unique up to conjugation by K.

Going back to the case of a simple Green functor A, and observing that induction and inflation of algebras commute with the functor FP_, it follows that there exists a subgroup M of Na(H), containing H (such that M/H is the above group n), and a simple algebra S (equal to J) on which M/H acts projectively, such that

a M p A ~- IndMInfM/HF S

So the previous discussion gives a new proof, under slightly weaker hypothesis, of the following theorem:

Propos i t ion 11.5.1: 'Th6venaz [13] T h e o r e m 12.11)

1. Let A be a s imple Green func tor for G. T h e n t he r e exis ts a subg roup M of G, a n o r m a l subg roup H of M, and a s imple a lgebra S on which M/H acts pro jec t ive ly , such t h a t

A G M ~_ IndMInfM/HFPs

The t r ip le (M,H,S) is unique up to conjuga t ion by G (and up to i somorphism of M/H-algebras for S).

2. Converse ly , if H_~ M are subgroups of G, if S is a s imple a lgeb ra on which M/H acts pro jec t ive ly , t hen IndaMInfM/HFPs is a s imple G r e e n func tor .

11.6 S i m p l e f u n c t o r s a n d e n d o m o r p h i s m s

Let G be a group and A be a Green functor for G. In this section I will study the G relations between the simple A-modules SH, v and the functors - 8 - and 7 - / ( - , - ) .

First I observe that proposition 11.4.1 gives a definition of S c for any J.(H) | H,V Na(H)-module V (non-necessarily simple). Moreover, the structure of Mackey functor

G of SH, y depends only on the restriction of V to NG(H) ~- b(H) | Na(H) : in other words

aesbS ,v a = SH,Res~rO(H)V

Similarly, I can define:


Definit ion: Let H be a subgroup of G, and V be an A(H)| [denote by FP~, V = FPH,v the A-module defined like SHy, but with any homomorphisms, and not only those which factor through a projective module: if X is a G-set, then

FPg, v (X ) = HomK~G(H)([xH], V)

ff f : X ~ Y is a morphism of G-sets, then for a 6 FPI~,v(X) and y 6 yH, [have

FP~,v.(f)(cO(y ) = ~ a(x) x 6 X H f(~)=y

If g : Y ~ X is a morphism of G-sets, then for c~ E FpGH,v(X) and y E yH, I have

FP~,~*(g)(9)(y) = ~g(y)

Finally, ira E A(X) and f E FP~,v(Y ), then a x f is the morphism from [(X • y)H] = [X H x yH] to V defined by

(a x m)(x, y) = (A*(m~)(a) | 1).f(y)

where mx is the morphism of G-sets from G/H to X defined by m~(uH) = ux. I define duMly FQaH,V for a G-set X by

F a QH,v(X) = [x'] | v

If f : X ---+ Y is a morphism of G-sets, then

FQ~,v . ( f ) (x | v) = f (x) | v

If g : Y --+ X is a morphism of G-sets, then

G * FQ~,v (f)(x | vl = E Y | v y 6 Y H g(y)=x

Finally, ira E A(X) , and y | v C FQ~,v(Y ), then

a x (y | = ~ (x ,y) | x E X H

It is easy to see that these definitions turn FPI~,V and FQ~,v into A-modules. The notation comes from the fact that if A is the Burnside functor, then FP~,v and FQH, VG identify respectively with the functors denoted by

Ind~%(H)InfN:((HH))F Pv IndGxc(H)Inf~7:((SH))FQv

by Th6venaz and Webb. The following lemma is a way to recover this isomorphism:

Lemma 11.6.1: Let M be an A-module . Then, for any subgroup H of G, the modules _~r(H) and M__(H) are A(H) | Moreover

HomA(FQGH,v, M) = Hom~(H)| ) (V, M(H))

HomA(M, FPg, v) = Hom~(H)| ) (M(H), V)


P r o o f : It is clear that the action of A(H) | Nc(H) on M(H) induces an action of A(H) | N o ( H ) on M(H) , since

H H (tH(a) | n).m = tH(a).n.rn = th-(a.rK(n.m))

Similarly, if m E M(H) , and if K C H, then

H /-/ | = tK (a = 0

because = n. /,.~(m) : O. So M(H) is also an J,(H) | .Nc(H)-module. Moreover if r is a morphism from a FQH,V to M, then r is a morphism of

A(H) | Nc(H)-modules from F Q ~ , v ( H ) to M(H). But it is clear that FQH,v(H) is isomorphic to V, and that FQ~,v(K ) = 0 if K is a proper subgroup of H. Then the image of r is contained in M(H) , and r is a morphism of A(H) | Na(H)- modules from V to M_M_(H).

Conversely, if ~b is a morphism of A(H) | NG(H)-modules from V to M ( H ) , and if X is a a-set, I define a morphism ~bx from FQ~,v(X ) to M(X) by setting

Cx(x o =

It is easy to see that this defines a morphism of A-modules from a FQH,v to M. The first adjunction of the lemma follows.

The second one follows from a dual proof. �9

L e m m a 11.6.2: Le t H be a s u b g r o u p of G, and V be a ATc(H)-module . I f K is a s u b g r o u p of G, t h e n

S~,v(K) : 0 = SaH,v(K) if It* r H

S~,v(H) : V : S~,v(H)

P r o o f : By definition, as (G/H) H = Na(H)/H, I have

S~4,v(G/H) = Hom([(G/H)"], V)~ a(H) = Horn(R-No(H), V)i ~G(H) -~ V

Then v E V is associated to the morphism

from RNc(H) to V. Moreover, if X is a G-set, if x0 E X, and if 5~0,v is the morphism from IX H] to V defined by

f 0 if # X Xo

' l , v i f x = x o

then

nWVc(H) n . v o : x

Let rn~ 0 the morphism of G-sets from C/H to X defined by rn~:o(GH ) = gxo. follows from proposition 11.4.1 that

~elV~(H) nxo:x

thus Tr[a(s4)(5~o,, ) = Sa,v.(m~o). If I( is a subgroup of G, then

S~4,v(I<) = Hom([(G/Ii)U], V)~ c(H)

is zero if H is not contained in K modulo G, because then (G/K) s argument proves moreover that

S~,v(K ) ~ ,, a = tH.Su,v(H ) ~:EG

H ~ C A -

It follows that

It

= 0. The previous

= 0

if Ix" #G H. It is clear moreover that S~,v(H ) = S~I,v(H) = V. A dual argument proves the assertions on S~,v(K), and the lemma follows. "

P r o p o s i t i o n 11.6.3: Le t G be a g roup , and M and N be M a e k e y f u n c t o r s for G.

* I f fo r any s u b g r o u p H of G, one of t h e m o d u l e s ~r(H) or N ( H ) is zero , t h e n M@N = O.

. I f for any s u b g r o u p H of G, one of t he m o d u l e s M(H) or N ( H ) is zero , t h e n N(M, N) = 0

P r o o f : Indeed, I have already seen that if M and N are Mackey functors for G, then

M~N(H) " M(H) | N(H)

If one of the modules ~r(H) or iV(H) is zero for any H, then M@N(H) = 0 for any H. If M@N # 0, there is a minimal subgroup H for MQN. Then

0 # (M~N)(H) ~- (M~N)(H) = 0

This contradiction proves the first assertion of the proposition. The second one follows similarly from the equality

n u n ( M , N) = n(s ~u(N))

for U = H\G, viewed as a IVa(H)-set-G (see proposition 10.1.2), which proves that

7{( i , N)(H) = H o m . (_~(H), N ( H ) )

Then if 7{(M, N) ~ 0, for any minimal subgroup H of 7-/(M, N), I have

0 r 7{( i , N)(H) = 7t(M, N)(H) = Hom.(-~r(H), N__(H))

so the two modules M(H) and _N(H) are non-zero. ,,

Proposition 11.6.4: L e t H a n d K b e s u b g r o u p s of G. I f V is a Na(H)- m o d u l e , a n d W a Na(K) -modu le , a n d if H a n d K a r e n o t c o n j u g a t e in G, t h e n

G ~ G G G S H , V e S K , W ~- 0 ~ t ~ ( S H , V , SK,w) ~--- 0

M o r e o v e r , i f K = H, t h e n

G ~ G F G G G G SH,VeSH,W ~-- QH, v . . w H(X~,v, SK,W) ~-- FP~,Hom.(V,W)

as M a c k e y f u n c t o r s .

P r o o f : The first assertion follows from proposition 11.6.3 and from lemma 11.6.2. As moreover the module S~,v(K ) is zero for any proper subgroup of H, it is clear that

the same is true for 6' ^ a - (SH,vQS~,w)(Ii), and that moreover

~- SH, v(H)eSH,w(H ) ~-- ViW"

Then (S~4,v+S~,w)(H) = a ^ a (SH,V| and lemma 11.6.1 shows that there exists

a a ^ a G-set X b y a morphism <I) from FQH,v~w to SH.V| defined for a

r : x @ (v | w) E FQH,v+w(X) ~

Conversely, if X and Y are G-sets, it is easy to check that the correspondence which maps

r = TrFa(H)(r E S~,v(X) = Hom([XH], V) Na(H)

r E S~I,w(Y) = Hom([yH], W)F a(H)

to Ox,y(r162 = ~ (x ,y) | (r | r E FQ~,v(X x Y)

x E X H yEY H

is well defined, and bifunctoriah indeed, if r = TrF~(H)(r I have also

OX,Y( r r = E iX, y) @ (r | ~)O(n--lx)) . . . .

i

x E X H yEY H

~eRG(H)

xEX ~/ yEY H

nE-NG(H)

E zE X .4 yEy tt

nERo(H)

E . . . . x E X H yE y ti

nEI~G(H)

(x,y)e(n-lr174162 (x,v)|162174162 xE X ~I yEY H

It follows that Oxy(r r is independent of the choice of r such that r = Tr N~ If moreover f : X --+ X' and g : Y -+ Y' are morphisms of G-sets, then for x' E X',

I have r

x E X H f(=)=~'


Setting

r = ~ Co(X) x 6 X H f(~):~'

I have SHa,v.(f)(r = TrNa{H)(r , and then

G G , , i X l s G = Z )) . . . . x ' 6 x ' H y'6y 'H

.... ~ (x',y')|174 = FQaH,v.(f xg)Ox,Y(C,r (x,y)e(x •

(: xg)(~,y)=(:~',r

Now if f is a morphism from X' to X and g is a morphism from Y' to Y, then for x' C X'

G * SH,V (f)(C)(x') = r

Setting C~ = C0 o f , I have S~,v*(f)(C) = Trfc(H)(C'o), so

X / @ I E (x',y')| . . . . ( x ' , y ' ) 6 x ' H • 'H

. . . . ~ ~ (x',y')|174 = FQ~,v*(f xg)Oi,r(m,n) (~,y)e(x xY) ~ (~%')e(x'xY') H

(: xg)(~',r

Thus Ox,y is bifunctorial, and induces a bilinear morphism from SGH,v, S~,,w to G G ^ G FQH,v| This gives a morphism 0 from SH,V@SH, W to FQH,v6w , defined for

a G-set X by

e x : [r | r �9 (S~,y@S~,w)(x) ~ ~ f(y) | (r | r �9 FQaH,V| y 6 Y H

where Co is a morphism f r o m [yH] t o V such that

Then (I) and 0 are mutual inverse isomorphisms: if

G x | (v | w) �9 FQH,v|

then ~(x) = [5~ | 5.o](a/H,~)

Let 5o,~ be the map from Na(H) to V defined by

v if n = l 5o,.(n) = 0 otherwise

Then 5. = Tr~C(H)(5o,~), and

0 0 ( x | 1 7 4 = E rn~(n)|174174174174174 n6/VG(H)

i1.6. SIMPLE FUNCTORS AND ENDOMORPHLY'MS

So O~ is the identity,. Conversely, if (Y, f) is a G-set over X, then

ox([r o <(~,s)) : ~ ,r(:,j) e (~o(~j) o ~,(,,j)) y~yH

where r is such that Tr~G(s)(r = r Thus

y6Y H

Let r q,G = ~ e Z ~ ~ mv.(m~)(6r . Then if Y0 ~ y H I have

H

r : ~ Siv.(,,,,,)(ar : ~ ~,,0(.~,(,~) . . . . yEY ,qEY ' t

n6Nc;(tt) m v n ) = 9 0

. . . . Z ,,,o(y) = Z ,,,~o(,-b0) = r y6Y t!~ n6NG(H)

ny:yo

'H,V ( ',)( Co(y))" Then

- , G ~ /

yEY H

Moreover if n 6 NG(H)

&f,v (-,..~)(~')(,.): V,(,,v) & ,~( , J

So

301

(1~.3)

(S~,v)X ~-- '-" H,HomR([Xq.V)

yEY H

As moreover .fm~ = rn](,v), this expression is equal to the right hand side of (11.3), thus 0 0 is also the identity, proving that

~G 5 ~<; F G ~%~ H,VQg'gH,W "~ QH,V|

c;c; qG To prove that ~( , H,t .... H,W)) ~- FP[j.Hom~,(V,W)' I wi]] use the following lemma:

L e m m a 11.6.5: Le t H be a s u b g r o u p of G, and V be an A(H) | NG(H)- m o d u l e . I f X is a G-set , t h e n Homz,([Xn], V) is an 7,( t t )Q ,~G(H)- inodule by

: (,, r

and t h e r e is all i s o m o r p h i s m of A - m o d u l e s

302 CHAPTER 11. TIlE SIMPLE MODULES

Proof: Indeed, if 4) e S~v(X) and i fa@n and a'~Sn' are elements of A ( H ) Q , ~ o ( H ) , then

. . . .

thus HomR([xH], V ) i s an A(H)C ' !VG(lI)-module. Moreover. if Y is an H-set

: sg, O"x) = Hom([ 'H • x ' ] , v ) )

The structure of Na (H) -modu le of Hom1{([xU], V) is such that

Hom([Y H • xH], V)~ a(Ir ~_ Hom([yH], HomR([xHI, V))~Va(~r)

So ,G ,"

(S~,v)x(Y) ~ 5H,,om,~([XH3,V)(~ )

It is easy to see that these isomorphisms are compatible with the structures of A- modules of both sides, and the lemma follows. �9

It follows in particular that

= HomM ~k(a)( 5H,V, ~ H,HomR (IX H].W) )

Then I use the following lemma:

L e m m a 11.6.6: Let H be a subgroup of G, and V and W be iVc(H)-modules . T h e n

G ,G 5H,W) Homg~(H)(~/; W) HomM~k(o)( SH,v, ~_

SHy to ,SH,W, then OH Proof: Indeed, if 0 is a morphism of Mackey functors from ,a ,a is a morphism of Na(H)-modules from ,5'~,v(H ) = V to S~,w(H ) = W. Conversely, if ~/, is a morphism of .Na(H)-modules from V to W, and if X is a G-set, I define a. morphism ~bx from S~,v(X) to Ssa w(X) by

f x : ~ r Hom([XH], V)~ G(H) ~ ~ o

It is clear indeed that if ~ factors through a projective module, so does bb~. It is easy to see that these correspondences are inverse to each other, and this

proves the lemma. �9

In those conditions

?-t ( S~,v, Sg,w ) ( X ) ~_ I-Iom~vG(H)(V, HomR( [xH], W)) = Honwa(H) ( V | [ X H], W) . . . .

. . . . HomK%(H)([xH], MotoR(V, W ) ) = [~'PH,HomR(V,w)(X)

Those isomorphisms are moreover natural in X, and this gives an isomorphism of Mackey functors

7-((,fH,V , SH,W ) ~-- F PIt,Hom:a(v,w)

11.6. S I M P L E F U N C T O R S A N D E N D O M O R P H I S M S 303

and completes the proof of the proposition. �9

R e m a r k s : 1) In the case when V and W are A(H) @ NG(H)-modnles, then the module Hom g(H)(V, W) becomes a Nc;(H)-module, and then I have the isomorphism of Mackey functors

, l PH,Hom~(H)(V,W )

2) Let H be a subgroup of G, and V be a simple A(H) Q Nc,(H)-modnle. Then Iet

B ~C ~,C G

The Green [unct, or B admits H as a minimal subgroup, and B ( H ) ~- EndRV. This algebra is simple if and only if R is a. field k, and l/ is finite dimensional over k. Now

~ B | ~ the functor /~ is a simple Green functor if and only if it is isomorphic to ,. H,E=a~V This is equive~ient to say that for a.ny (-,'-set X, I ha.re

The case X = �9 shows that this condition holds if" and only if

(Endk l/)Nc.(Ju) = (EndkV)f G'(H)

i.e. if the kAWc,(H)-module V is projective. In those conditions, denoting by V ~ = Homk( V, k) the dual of V, I have V~ kV -~ Endk V if and only if V is finite dhnensional over k, and q,c; ~qc, ho Then '- t t ,V o ~ k'~ H , V ] �9

-,G o < ~,O c,G A q, G G S G ( b H , V ) r v ~ DH,Vo'JO'- t I ,V ~ FQHyo| ~ QH,EnakV

Moreover, it is clear that F(~)~ E,a v ( t f ) = End~.V = FPI~Endkv(H) . , k ' "

F a - , The morphism QH,E,dkV ---+ b P~',S,d~V which follows from lemma 11.6.1 factors - , C 1 ' through 5H,~:na~v, which is ~ quotient of ["Qu,s~a~v and a submodule of FPU,E,~d~V.

It is an isomorphism if and only if

G , q,G ~ F G FQH,Endk~ ~ ~ H.EndkV -- P~I,EndkV

which is equivalent to say that Endj,.V is projective as k/Vc,(H)-module. Finally, those considerations prove the following proposition:

P r o p o s i t i o n 11.6.7: Let R = k be a field. Let C be a group , and A b e a G r e e n f u n c t o r for G. Let m o r e o v e r It be a s u b g r o u p of G, and V be a s i m p l e A(H) 4) ~ c ; ( H ) - l n o d u l e . T h e fo l l owing c o n d i t i o u s are equ iva l en t :

,G q,C,' 1. T i l e G r e e n f u n c t o r ~(5HV, ~tr is s i m p l e .

2. A s M a c k e y f u n c t o r s

3. T h e m o d u l e V is f ini te d i m e n s i o n a l over k and p r o j e c t i v e as kNG(H)-

m o d u l e .

304 ( ' t fAPTER i1. THE SIMPLE MODULES

I will say then that S~, v is endosimple. The number of endosimple modules is related to Alperin's conjecture in the following way: let .4 = b~ be tile in-part of the Burnside functor, for a prime number in: if X is a C-set. then bp(X) is the Grothendieck group of the full subcategory of G-sets over X, formed of G-sets }" such tha.t for any .y E Y, the stabilizer of y in G is a in-group. Then [~(tt) is zero if H is not a in-group, and isomorphic to h otherwise.

Let k be an algebraically closed field of characteristic p. Then if ~~) is a in-subgroup of G, and V is a simple NG(Q) module (hence finite dimensional over k), the bp-

,G module ,SH, v is endosimple if and only if the module V is simple and projective. Thus Alperin's conjecture (in its global form) can be expressed by saying that the number of endosimple bp-modules is equal to the number of simple/,'G-modules.

Chapter 12

Centres

12.1 T h e c e n t r e o f a G r e e n f u n c t o r

D e f i n i t i o n : Let G be a group, and A be a Green fimctor for G. I call centre of A, and I denote by Z(A) , the commutant of A in A: if X is a G-set, then

Z ( A ) ( X ) = {o~ E A ( X ) I v Y , v/~ c A ( Y ) , c~ x /3 = e~ x ~ f l }

With this definit ion, the functor Z(A) is a, sul>Creen functor of A, which is clearly

commuta t ive .

D e f i n i t i o n s : If A and B are Green flmctors fox" the group G, then the direct sum

A | B of A and B is the direct sum of A and B as Mackey flmetors, with the product defined for O-sets X and Y, and elements a E A(X) , b E B ( X ) , c E A(}") and

(z ~ B(Y) bx

(,~ r b) • (~ ~ d) = (a • c) r (t, • d) c ( A ~ B ) ( X • Y )

The unit of A @ t3 is the element CA @ C B of ( A @ B )( . ). If M is an A-module, and if z E Z(A) ( . ) , I denote by z x M the A-module defined

for a G-set X by (~ • M)(X) = ~ • M ( X )

1Te is an idempotent of Z (A) ( . ) , I denote by e x A the subfunctor of A defined for a

O-set X by (~ • A ) ( X ) - ~: • A ( X ) c_ A ( X )

Then e x A is a sub-Green functor of A (the inclusion being not unitary in general), with e = e x CA C (e X A)( . ) as unit. Moreover, if M is an A-module, then ( x M is

a e x A-module

The module z • M is an A-module , because if X and Y are G-sets, if c~ E A ( X ) and rn E M ( Y ) , then

o~ x ~ x m = (<~ x ~ :~) • m ( z x ,~) x '.~

If e is an i dempo ten t of Z(A)(e) , the functor e • M is an e x A module , since

(e x ~) x (e x , ~ ) = e x ( . x ~ c) x m = ~ x ~ x a x b = ~ x <, x . ,

With these definitions:

306 ( 'HAPTER 12. CENTRES

L e m m a 12.1.1: Let A, B, and C be G r e e n f u n c t o r s for t h e grou p G. T h e n A is i s o m o r p h i c to B @ C if and o n l y if the re ex i s t s o r t h o g o n a l i d e m p o t e n t s e and f o f Z ( A ) ( . ) , such t h a t e + f = <.1, and B _~ e • A and C ~ f • A.

Proof: Indeed, the unit of B is in Z(B H)C)( �9 since for a G-set X and elements b r B ( X ) and c E C ( X ) , I have

cB x (b@c) =

and similarly

Then if A ~ B r C, the image idempotents , and e + f = CA. Z ( A ) ( . ) such that e + f = e_4,

a C A ( X ) ~ (e x

(es @ 0) x (b<"b c) = (CB x b) @ 0 = b

c) x eB = (b x ~ ) + : : ~ 0 = b

e of ~s and the linage f of go' in A(I ) a.re ort.hogonal Converse�9 it" c a~(t .f are ort.hogonaI idempotents of then it is clear that the maps

a) ~ ( . f x ~) r (~: x A ) ( X ) r (.f x A ) ( X )

define an isomorphism from A to (e x A) I~t (.f x A). The main interest of this lemma is that since Z(A) is a Green functor, there is

always a. (unique) uni tary morphism of Green functors from b to Z(A). Thus any decomposition of Cb as a sum of orthogona.1 idempotents of b(�9 hldttces a decomposit ion of A as a direct sum of Green fnnctors.

L e m m a 12.1.2: L e t G and H b e groups , and A be a G r e e n f u n c t o r for G. L e t i b e t h e i n j e c t i o n f r o m Z(A) i n to A.

�9 If U is a G-se t -H , t h e n

�9 If V is an H-se t -G , t h e n

Proof: Indeed, if X and Y are H-sets, if

z c (Z(A)o U)(X) -- Z(A)(U o. X)

and if a r (A o U)(Y), then theh' product for A o U is

Shni lar ly , in (A o U) ~

(}Sz) ,u z xV~ = ( A o U). k~:~/ A*(ay, x ) ( . x z)

But a • z = A* ( (,~,y)(.~,~,z)'~ ( z x a ), and moreover

o@, x o U o n

12.1. THE C E N T R E OF A GREEN FUNCTOR 307

Thus

z x U ~ = A. U OH my

and the first assertion follows. Similarly, for the second assertion, if X and Y are H-sets, and (Z, f ) and (T,g)

are H-sets respectively over X and Y, and if z C Z ( A ) ( H \ V . Z ) and a E A(H\ i<T) ,

then in Z . v (A) (X • Y) , ~ h~,~e

Z(ZJ) • a(Tm) = A*(t~; ,T)(Z • a)(z .TJm)

On the other hand

But a • z = A* ( a . , , . t a ..... ~ (z • a), and \ G.va.z G.vl.t )

k G.v2.z G.vx.t ) T,Z = Z,T 0

Thus

\ x Y )

(t~) is bijective, in s • Y), l h a v e As zt

~ / [ o r e o v e r

s xy ~ ,

o = ,ave Finally as ~y tz

z(zd) x~ a(Tm) = A*(nVT)( z x a)(z.T#mt = Z(zd) x a(r,~l

which proves the second assertion of the lemnm. �9

I have already used an application of these lemmas, when I have built for an s N the module N o UI, ~ = )<<.(CA) • (=V o U): indeed, the element

XA,.(SA) is in Z ( s o U)( . ) .

N o t a t i o n : I will denote by rc = ~rR(G) the set of 1)rime factors of the order of G

which are not invertible in 19.

I will suppose for simplicity that l{ is contained in A(o) (otherwise I can replace R by its image in A(- ) ) . Then the Burnside ring b(.) = bR(G) with coefficieui,s in

308 CHAPTER 12. (_:ENTRES

Fg has a family of idempotents f~ indexed by rr-perfect subgroups of G (see theorem (9.3) of Thavena~ and Webb [~5], or chapter 5.4 of Benson's book [1]): the primitive idempotents of bq(G) are indexed by the subgroups L of G, up to conjugation, and given by the following formulae of Gluck (see [7])

1 e ~ - INn(L) I Z I I " I~] I ; ,GG/I '~

h C L

The idempotent c~ is characterized by the fact that for any G-set X, I have

~'.x -- I X H I 4 gI1

and in particular I(ea)~'~ I is zero if H and Ix are not conjugate in G, and equal to 1 otherwise.

Then the idempotent .f~ corresponding to a ,-r-perfect subgroup H of G is given by

fD = E ~:2 O=(L)=II

L rood.Nee(H)

where O~(L .) is the smallest normal subgroup N of L such that the quotient L/N is a solvable ~r-group.

Now a theorem of Dress (see [1] 5.4.7 and 5.4.8) shows th&t the idempotents f~ , as H runs through a system of representatives of the conjngacy classes of rr-perfect subgroups of G (i.e. subgroups H such that H = O~(H), or equivalently subgroups having no non-trivial p-quotient, for p r 7r), are mutual orthogonal idempotents of

sum GIG in bR(G). It follows that for any Green functor A for G, I have

A ~_ @ ,l';i • A H

where the sum runs on 7r-perfect subgroups of G up to conjugation by G. In this expression I also denote by f~; the image of the element .f~ of b(G) in Z(A)(.).

I.~A o l 'Na(H) Now observe that f~ . . . . . Nc,(Htau , a.nd it, seems natural to compa,re the

Green fnnctors ,f~ x A and f ~ ( H ) x Reset H A It is actually easier to compare their ( ) �9 associated categories:

~,Nc,(H) G L e m m a 12.1.8: Le t B = f~ x A a n d C = .~H • ResNG(H)A. L e t m o r e o v e r

F = G/No(H). T h e n

1. F o r a n y G - s e t X

a a " A(V x X) (ReSNa(H),4)( [{esNc;( H).~ )

M o r e o v e r , if Y is a G-set , if a E A(I" x X) a n d fl E A(E x Y), t h e n t h e i r G p r o d u c t a x T fl f o r t h e f u n e t o r H.esN,:;(fi)A is given by

c~ x~ fl = A* ( ?x!/ l (c~ x d ) \ ?x"/y /

12.1. THE C E N T R E OF A GREEN F U N C T O R 309

2. For any G-set X

3. T h e c o r r e s p o n d e n c e 7~ m a p p i n g the G-set X to Res~G(H)X , and the

e l e m e n t u 6 B ( Y X ) to fNHC(H) X U is a fully fai thful func to r f r o m Cs to

Cc.

4. M o r e o v e r , a n y objec t of Cc is i somorph ic to a d i r e c t s u m m a n d of an

ob jec t in the image of 7~.

Proof : Set N = No(H) . The first assertion is clear, since for any G-set X

(Res~A)(Res~X) = A(Ind~vRes~X)

Moreover Ind~ResCNX ~- F x X, and assertions 1) and 2) follow from the definition of the product for the functor Res~A.

Now if u E A ( Y X ) , then A* ( ~ ) (u) E A(rYX), and

\ ? T y x \ yx / \ 7 7 y x \ ~yx ]

\ TYX l

This shows that Tt(u) 6 C(ReSCN Y x Res~vX ). Now if X, Y and Z are G-sets, if u 6 A ( F Y X ) and v E A ( F Z Y ) , then the

composition of ~(v) and 7r is given by

\ 7 zx I \ T z y y x ]

. . . . A . ( ? z Y X ] A * \ 7zx I

As f ~ is in the center, this is also

\ 7zx / \ ~ z y y x ]

( Tzyx ] (fZ • v • fZ • u) ? z y ~ y x /

• . . . .

\ y zx ] "77 \ z y y x ]

B u t A ' ( < ) ( S Z • N N = f~l.f~l = fN , SO

k "Tzx / zyyx

\ zx i k z y y x ]

310 C H A P T E R 12. CE NTRE S

M o r e o v e r

Ts x 1A(X2)) = fN • fr~ X 1A(X2 ) = f~ X fH • A. (s) = ZZ

Furthermore f ~ . A ( f~) N a C, = = fH .ResNfn

But a c ResN.fH is the sum of the f ~ , where K runs through the different conjugates of H in G which are contained in N. So fN • f~ = fN, and

which it the unit of C(Res~X2). So 7s is a flmctor from CB to Co. Note that the previous argument shows that

Now if X and g are G-sets, and if v 6 A(F • Y x X), then define

\ y X l

The equalities

show that Ts and 8 are mutual inverse bijections between B ( Y X ) and C ( /~(Y)R(X)) , so the functor R is fully faithful.

ResNIndNY , it is a direct summand in Finally, as any N-set Y is a sub-N-set of a c (Jm of a.n object in the image of Ts �9

Corol lary 12.1.4: The functors

c indic(H) L M ~ fN~(U) • ReSNa(H) M and L

are m u t u a l inverse equivalences of categories be tween B - M o d and C-Mod.

Proo f i If C is an R-additive category, define the category C in the following way: the objects of C are the couples (X,i), where X is an object of C, and i ~nd idempotent endoinorphism of X in C. A morphism in C fi'om (X , i ) to (Y , j ) is a morphism f from X to Y in C, such that f o i = f = j o f . In other words

Homd((X, i), (Y , j ) ) = j o Homc(X, Y) o i

The composition of morphisms is the composition in C. The category C is obtained by "adding direct summands" of objects of C. It is also an /~-additive category (the direct sum of (X , i ) and (Y , j ) is of course (X @ Y,i @j)).

12.1. THE CENTRE OF A GREEN FUNCTOR 311

There is a canonical functor X ~ (X, ldx) from C to C. So i f /~ is an additive functor from d to R - M o d , it gives by composition an additive functor fi'om C to R-Mod . Conversely, if F is an additive functor fi'om C to R - M o d , then F admits a unique extension/~ to C, defined by

i) = F ( , ) ( F ( x ) )

This is because if (X, i) is an object of C, then i : X --+ X defines two morphisms in d

i+ : (X,i) ~ (X, Idx) i- : (X, Idx) ---, (X, i)

Moreover i - o i + is equal to i, which is the identity morphism of (X, i) in C. So (X, i) is a direct summand of (X, Idx) in C.

This shows that the categories of representations of C and d are equivalent. Now if �9 : C - 4 / ) is a functor between additive categories, it induces a functor ~ : d -~ 75. It is clear moreover that if ~ is fully faithfld, then ~5 is fully faithfuh indeed, the bijection

F : Homc(X, Y) --* Horny (F(X) , F(Y) )

restricts to a bi ject ion/" from I-Iornd ((X, i), (IG j ) ) = j o Home(X, Y ) o i to

,om F(Y,j)) : F(j)o Homo(F(X), F(Y)) o

If moreover any object Y o f /9 is a direct summand of an object F(X) in the image of F, therl there are morphisms

~ : P ( X ) ~ Z fl: Z ~ F ( X )

such that c~ofl = Idv. I f i is any idempotent in Endv(Y), then / 3 o i o c ~ is an idempotent endomorphism of F(X) . So it is equal to F(7) , for a unique 3' E Endc(X). This implies that "y is an idempotent. Now the morphisms

are mutual inverse isomorphisms in D. So /~' is essentially surjective, hence it is an equivalence of categories.

Thus in the situation of the lemma, the functor ~ is an equivalence of categories: this implies in particular that 7-4 induces an equivalence of categories between the categories of representations of CB and Ct). It remains to check that this equivalence is as stated in the corollary, which is clear since if L is a C-module, then for any G-set X

fH c x ( Indaa(~jL)(X) = f~' x L(Res~a(H)X ) . . . .

. . . . f~ x fN~(H) • L(ResGva(H)X ) = fNG(H) X L(Res~%(H)X ) = L(Res~va(H)X)

which proves that f,~ x Ind~G(H)L = IndNc;(H)L. �9

L e m m a 12.1.5: Le t H b e a n o r m a l s u b g r o u p of G. I f A is a G r e e n f u n c t o r for G such t h a t f~/ x A = A, t h e n fl c'/h' • A H = AH~ and t h e c o r r e s p o n d e n c e X ~ X H i n d u c e s an e q u i v a l e n c e of c a t e g o r i e s f r o m Ca t o CAH.

312 CHAPTER 12. CENTRES

P r o o f : Let Y be a (G/H)-set. Then

AM(y) = A(In~/ ,~ ' ) / ~ A . ( f ) ( A ( Z ) ) (z,:)

where (Z, f ) runs through the G-sets over In~/HY such that Z H = O. Now since f ~ G G is the unit of A, and as ResKf ~, = 0 if H Z K, it follows that A(K) = 0 if H ~ K,

so A(Z) = 0 if Z H = O. Thus AH(y) = A(In~/HY ). Let ix denote the injection from In~/H(X H) into X. Then

AH(x") = A(In~/HX n)

The unit A(X) --* (In~/HAH)(x) = A H ( x " ) i s tlae map A'(ix). N o w as

X = (In~/HX H) ~l Z

with Z H = 0, it follows thai, A(Z) = 0, and the map A*(ix) induces an isomorphism A(X) ---+ AH(X~).

Now it is clear that the correspondence

X H X g f C A(YX) ~ A*(iyx)(f) r AH(yHx H)

is a fully faithful functor from CA to CAH. As any (G/H)-set Y is isomorphic to (Inf~a/HY) H, this functor is essentially surjective, hence it is an equivalence of cate-

gories. Finally, the unit of A H is f ~ / H because .f~ - In~/Hf]/H is a linear combination of G/If , for H ~ K, so it acts by zero on A. �9

C o r o l l a r y 12.1 .6: Let H be a n o r m a l s u b g r o u p of G. If A is a G r e e n f u n c t o r for G such t h a t fH a X A = A, t h e n t h e f u n c t o r s

M ~ M H and L ~ In~/NL

are m u t u a l in verse e q u i v a l e n c e s of c a t e g o r i e s b e t w e e n A - M o d and A H - M o d .

P r o o f : This follows from lemma 12.1.5, and from the definition

(In~/HL)(X) = L(X H)

So In~/g is an equivalence of categories, and its inverse is the left adjoint functor

M ~ M H. �9

L e m m a 12.1 .7: Le t H be a s u b g r o u p of G, and M be a M a e k e y f u n c t o r for N o ( H ) . I f Z is an e l e m e n t of b( . ) , t h e n

G Na(H) , ,C Inf N~(H) IzH M) Z • IndNc(y)InfNa(H)M ~-- tn(]NG(H ) ~Va(H) [ X

P r o o f : I f Z i s a G - s c t , I k n o w t h a t Z=b,(~.)b'(~.)(cb). Thus if X is a G set, aad

L is a Mackey functor for G, then for I 6 L(X) I have

L. (y) L, (y)

12.1, THE CENTRE OF A GREEN FUNCTOR 313

G N Set N = No(H) and N = Na(H). If L = tndNInfvM , I have L(X) = M(XH), and

setting Z ' = Z H and X ' = X H, the maps L*(":) arid L. (;~) are respectively equal z i g I

to M" ( ~ ' ) and M. ( ~ , ) . It follows that the image of L. ( ~ ) L* ( ~ ) i s the image zlm # z#x ! o f . . ), that ,n o her woras, the e ,ement on

M(X H) via Z H, for any Z 6 b(G). It follows that

(Z x Ind~InfNN-M)(X) = Z" • M ( X H) = indNinf~v( g a N H X M)(X)

This proves the lemma. �9

L e m m a 12.1.8: Le t H be a z , -perfect n o r m a l s u b g r o u p of G. T h e n in bR(G/H), I have

= fC,,/H , 1

{ ,cG~H (G/H P r o o f : Let /7 = K / H be a ~r-perfect subgroup of G/H. If va~s~ "a/</H is non-zero,

then there exists a subgroup L of G with O'S(L) = H and a subgroup U = L'/H of G/H with O'~(L ') = K such that

, a,H OIH a OlU h' I(In~/H U ) I (L ) = teg ) It Z,) 0 # teL) .eg = ( % . I n ~ l s ~ e s = e c l H L e a " " G/H'ClS-f"eC'"

Thus L/H is conjugate of ~7 in G/H. Then L is conjugate to L' in G, and then O'(L) = H is conjugate in G to O"(L ') = K. Thus H = K, and the lemma follows.�9

C o r o l l a r y 12.1.9: I f H is a rc,-perfect n o r m a l s u b g r o u p of G, t h e n for any A - m o d u l e M a n d a n y A N - m o d u l e L

(f~s • M) H ~- fl WH • MH f~ x InfaalHL ~- Ing/s_s(f~ C(") x L)

Finally, corollary 12.1.4, 12.1.6 and 12.1.9 show that if A is a Green functor for G, and H is a rr-perfect subgroup of G, then the functor

G NG(H) L ~ IndNc(H)Inf~c,(H)L

is an equivalence of categories from ,fl ~G(H) • (Resg-a(H)A)H-Mod to fH c' x A Mod . The inverse equivalence is the adjoint functor

(Res,%.(H)M)" In particular, the unit morphism

a N~(H) f~ • A ~ IndNa(H)In~a(H)(f~ X (Res~vc(H)A) H)

is an isomorphism of A-modules. As it is a morphism of Green functors, it is an isomorphism of Green functors.

The following lemma gives a characterization of the Green functors .f~ • A as the functors which are projective relative to the set E~(G) of solvable rr-subgroups of G (i.e. relative to the set I]H G/H, where H runs through E~(G)):

P r o p o s i t i o n 12.1.10: Le t G be a g roup , and A be a G r e e n f u n c t o r for G over R. Le t Ir t h e set of p r i m e fac to r s of IGI which are no t inve r t ib le in R. T h e fo l lowing cond i t i ons are equ iva len t :

1. T h e f u n c t o r A is p r o j e c t i v e re la t ive to t he set of solvable re-subgroups.

2. T h e i d e m p o t e n t fl a ac t s as t he i d e n t i t y on A, i.e. A = ,f~ x A.

Proof i It is clear that 2) implies 1), because if A = f a • A, then <4 is the image in A(.) of fla. This element is a sum of el , for L E E,(G), hence a linear combination of G / K , for K C_ L, thus t( E E~(G). Then the identity map of A(.) = A(G) is a

a E.(G) Thus A is projective relative to the set linear combination of tic-rio for K E of solvable rr-subgroups.

Conversely, if A is projective relative to the set of solvable ~r-suhgroups of G, then eA can be written as

~A ~ G = t ~,- (~ ,")

Then t~.:(rl,.f , .oeK)

IqEE~(G)

and it is easy to see that for any subgroup Ix" of G

G a = f [ Resr,,fl

and that f#" = K / K if I ( is re-solvable. Thus .f~ = r and 2) holds. .,

Finally, I have proved the following proposition:

P r o p o s i t i o n 12.1.11: Le t G be a g roup , and A be a G r e e n f u n c t o r for G over R. Let rr t h e set o f p r i m e factors of I01 w h i c h are n o t i n v e r t i b l e in /2. T h e n t h e r e is an i s o m o r p h i s m of G r e e n f u n c t o r s

A ~ - ( ~ S ~ x A H

w h e r e t h e s u m m a t i o n runs t h r o u g h a set of r e p r e s e n t a t i v e s of t he con ju - gacy classes of 7r-perfect s u b g r o u p s of G. I f H is such a s u b g r o u p , t h e n

, , a , ~Nc(H) { ~Fc(H) J • a • H)

T h e f u n e t o r fife(H) x (Res~o(H)A)n is p r o j e c t i v e re la t ive to t h e set of so lvable re-subgroups , and t he f u n c t o r s

G No(H) S~c,(H) G H M ~-+ IndNG(y)InfFrc(H)M L ~-~ x (ReSNa(H)L)

are m u t u a l inverse equ iva lences of ca t egor i e s b e t w e e n ( f~ x A ) - M o d and

/1No(H) X (ResaNa(s)A)U-Mod.

12.2. THE FUNCTORS (A 315

R e m a r k s : 1) In the case of Mackey functors (i.e. the case A = b), this is essentially theorem 10.1 of Th4venaz and Webb ([15]). 2) In the case 7r = 0, i.e. when the order of G is invertible in R, the solvable ~r- subgroups are trivial. In those conditions, I have fa c = ~a[G/1. Thus if f ~ • A -- A, then the maps

a 6 A(H) ~ rlHa 6 A(1) H b E A(1) H ~-~ 1-~--tHb 114[

are mutual inverse isomorphisms between A and FPA(1). Thus if the order of G is invertible in R, then the isomorphism of the proposition gives

A ~ ( ~ Indff'c(") I n f N o ( H ) ~ N~.~/F - G [ ) P~'H'( ) H

This is theorem 13.1 of Th&enaz (see [13]).

12.2 T h e f u n c t o r s (A

12.2.1 Another analogue of the centre

The definition of the centre of a Green functor given in the previous section is certainly natural, and mimics the definition of the centre of a ring. However, one can also define the centre of a ring from its category of modules, as the endomorphism ring of the identity functor ([1] prop. 2.2.7).

If A is a Green functor for the group G, then the endomorphisms ring of the identity functor of A - M o d is just a ring, isomorphic to the centre of A(ft2). I would like to have a similar construction, using the category A-Mod , and producing a Green functor (A- The idea is then to mimic the construction of the Green functors 7-/(M, M), setting for a subgroup H of G

CA(H) = Endf~nct(Res~)

where EndFu~,(Res~) denotes the endomorphisms of the restriction functor from A - M o d to Res~A-Mod: such an endomorphism is determined by giving, for any A-module M, a morphism OM of Res~A-modules from Res~M to itself, such that if ~b : M --+ N is a morphism of A-modules, the square

OM Resr~M , Res~M

t R e s i n , Res/~N

0~

is commutat ive. With this definition, it is clear that if A is a Green functor on R, then (A(H) has a natural R-algebra structure. It is natural to try to turn (A into a Green functor.

First I observe that if 2- denotes the identity functor of A-Mod , and 2-G/H the functor M H MO/H from A - M o d to A-Mod, then

~A(H) ~-- Homf~n~t(Z, Za/H)


Indeed, I know that 2-a/H(M) ~- Ind~Res~M, flmctorially in M, and that Ind/~ is right adjoint to Rest . The above isomorphism follows then from the next lemma:

L e m m a 12.2 .1: Let C and D be ca tegor i e s , and F be a f u n c t o r f r o m C to 7) h a v i n g a r igh t ad jo in t F ' . T h e n if g is a ca t egory , if A is a f u n c t o r f r o m $ to C, and B is a f u n c t o r f r o m g to 19, I have

HomF~n~t(F o A, B) _~ HomElier(A, F ' o B)

P roo f i Indeed, a morphism of functors from F o A to B is determined by giving, for any object X of g, a morphism Ox from F o A(X) to B(X), such that if q~ is a morphism in 5" from X to Y, the square

OX F o A(X) , B(X)

F o A(f) [ I B( f )

f o A(Y) , B(z) Oy

is commutative. Denoting by u ~-+ u ~ the bijection

Hom•(F(Z),T) ~ Homc(Z,F'(T))

deduced of the adjunction, I obtain for any X a morphism 0) from A(X) to F o B ( X ) . Moreover the square

0) A(X) , F' o B(X)

A(f) l [ F o B ( f )

A(Y) , F'o B(Y) O{

is commutative, because its diagonal h : A(X) --+ F' o B(Y) is both equal to (B( f ) o

Ox) ~ and to (0y o ( F o A)(f)) ~ Thus the maps 0~c define a morphism of functors from A to F' o B. There is an

obvious inverse construction, which gives the isomorphism of the lemma. �9

Now for any G-set X, I can define the endofunctor Zx of A - M o d by I x ( M ) = Mx. Then, for any G-set X, I define

(A(x) : Hom~.~,(Z, Zx)

If f : X ~ Y is a morphism of G-sets, then f determines a natural transformation from I x to 2-z: if M is an A-module, and Z is a G-set, then the maps

M,(Idz • Mx(Z) --~ My(Z)

define a natural transformation from Mx to Mz, which is itself natural in M. Similarly, the maps

M*(Idz x f ) : My(Z) --+ Mx(Z)

12.2. THE FUNCTORS (A 317

define a natural t ransformation from IT to 2- x. This gives by composition morphisms

( J , , ( f ) : (A(X) -~ (A(Y) r : r ---' (A(X)

Finally, if c~ r CA(X) and /9 r r then for any A-module M, I have a morphism aM : M ~ Mx, and a morphism ~M : M ~ My. I set then

(o, x ,3)M = <,,,1 x aM

the product in the right hand side being computed in 7-((M, M). In other words, if Z is a G-set, I have for nz E M(Z)

(a x 3)M,z(,,,) M. CZYz I \ zx!t /

Then if 0 : M ~ N is a morphism of Mackey functors, I have

(c~ x 3)~v,zOz = X . \ ~.W / o C~N,Zr o ~N,Z o Oz('~) . . . .

. . . . N. zx~ :~r9]

o a,~i ,n. o ~ M , z ( m ) = Ozxy o (c~ x ,J)M,z . . . . Oz.,x'y o AI. z.W

which proves that a x/3 r r x Y).

L e m m a 12.2.2: The above definit ions turn CA into a Green functor, such that

(A(*) = EndF,~,~o~(Z) ~- Z(A(fl2))

Proof: The verifications to make are not difficult, and similar to those made for the functors "H(M,M). The equality CA(e) ~-- Z(A(f i2)) foIlows from the fact that the

category A - M o d is equivalent to A ( f F ) - M o d . Thus r identifies to the endomorphisms of the identi ty functor in this category, hence to the centre of the algebra A ( W ) .

Let X be a G-set, and ~ E r Then for any A-module A'/, I have a morphism c~M : M --~ Mx, such that if q5 is a. morphism of A-modubs from !1,1 to N, the square

(~A,I M > Mx

N - - ~ N \ ' o N

(c)

is conmmtative.

318 C H A P T E R 12. C E N T R E S

In part icular , if Y is a G-set, I can take M = Ay. Then HomA(A},-, M) ~_ M ( Y ) (see proposition 3.1.3): the element m �9 M(Y) is associated to the morphism 4,~ : Av --* M, defined for any G-set Z by

a �9 A v ( Z ) = A ( Z x Y) ~ a %-,)) = M. M* zyy

It follows in part icular that the morphism aAy : Av ~ ( A v ) x = A ( X Y ) is determined by an element

ay �9 Axy(Y) = A ( Y X Y )

Let n �9 N ( Y ) . I must have a commutat ive square

OZ,4 y Ay ~ A x v

N , Nx CtN

Then for any G-set Z and any a E A v ( Z ) = A(ZY) , I must have

The case Z = Y and a = 1A(V~) gives then

Conversely, if this equali ty is used to define a n y for any N and any Y, then the commutat iv i ty of the square (C) is equivalent to

fo any Y and any m r M ( Y ) . But

As r is a morphism of Mackey functors, the right hand side is

N. ( y ' x Y 2 1 N * ( YlxY2 t r • m ) k ylx } kylXy2y2/

As r is a morphism of A-modules, it is also

N. (yIxy2 / N* ( ylxy2 ) ( a y x ~ y ( w i ) ) = ay oy @y(fil) k ylx / kylxy2y2

Thus the square (C) is commutative. The only condition on the elements a r comes then from the fact that the morphism

c~v is determined twice by the previous argument: I must have for any Z and any a �9 A v ( Z ) = A ( Z Y )

Cry(a) = a Oy ay = az OZ a

12.2. THE F U N C T O R S (a 319

Finally, if those conditions are satisfied, then the morphisms C~N,Z are morphisms of A-modules: indeed, if a E A(Z) and ~ E N(Y) , then

x o,:,,~,-(,~) = a x (,: ,~,-o~,-,~)

whereas

But

z~ A* (a) o y . . a x n = A. zyy

As A. (7) (,,) A( YY), have

( o,.,, . . . . azy ozy (a x n) = azy ozy A. zyy

( z Y ) A * ( ; Y ) ( a ) o y a y ) o y n = A . ( Z Y ) A ' ( : J ) ( a ) ~ . . . . (A. zyy zyy

. . . . a • (ay oy n)

So I have

g, yOy1q.) ~ . . .

If f : X ~ X ' is a morphism of G-sets, then the sequence (ay) determines a morphism from 2- to 2-x,, defined by composition for an A-module 3J and a G-set Z by

M . ( I d z x f ) M ( z ) , ~,~x(z) = M ( Z X ) , A , I ( Z X ' )

In part icular , if M = Az and Z = Y, the image of 1A(Z2) in A r ( Y X ' ) = A ( Y X ' Y ) is

Ar , . ( Idz x f ) (ay ) = A . ( l d z x f x Idy ) (ay)

In other words, if s is the sequence ( a r ) , | have

(A, . ( f ) ( s )r = A . ( l d v x f x Idr ) (ay )

It is clear similarly that if s' = (@) is a sequence defining an element of (A(X') , then

G ( f ) ( s ' ) r = A*(ldy x ,f x [ d r ) ( @ )

If s = ( a r ) defines the element a C (A (X), and t = (br) defines the element/3 E G ( Y ) , then for any A-module M, any G-set Z and any m E M ( Z ) , I have

\ zxy / \ zxy /

. . . . M. ( z Y x l (azz ozv b z o z ,n) \ z x y /

Moreover as bz E A ( Z Y Z ) , I have

a z y O z y b z = b z Oz a z

320 CHAPTER I2. CENTRES

Thus

( z Y : r ) ( b z o z a z o z m ) (~ • 3)M,Z('~) = M. \ : w

In par t icular , if M = Am and m = la(z_~), this equali ty shows that (~ x /~ is defined by the sequence .~ • t such that

( s x t ) y = A* ( z~itl:rz21(bz ~

Finally, I have proved the following proposition:

P r o p o s i t i o n 12.2.3: Let A be a Gr een f u n c t o r for t h e group G, and X be a G-se t . T h e n ~.4(X) i d e n t i f i e s w i t h t h e se t of s e q u e n c e s .s = (.~<), i n d e x e d b y t he G - s e t s , s u c h t h a t st. C A ( Y X Y ) a n d

a Oy ,gy = *~Z OZ (/

for any G - s e t s Y a n d Z a n d a n y a C A(ZY).

I f f : X ~ X ' is a m o r p h i s m of G- s e t s , i f s E CA(X) a n d s' C ( a ( X ' ) , t h e n

CA,.(f)(S)y = A.(ldy • f • Idy)(.sy) r = A*(Idy • f • Idz)(Jy)

I f s E CA(X) and t 6 (A(Y), t h e n for any Z

(s x t)z = A. (Z~:rz~) (tzoz ~z) \ zl ;r.7t z2

Let X and Y be G-sets, such that X divides a mul t ip le of Y in CA. This is equivalent to say tha t there exists e lements a{ E A(XY) and fl~. E A(YX) , for 1 < i < n, such that

1A(X~) : ~ ~i oy fl~ i

Then let Z be a G-set, and s E (A(Z). I cart write

8X ~- 3X OX 1A(X 2) = Z 3X OX O:i Oy /3 i = E Q" Oy 8 y Oy fli i i

This formula shows that sx is de te rmined by sy. The e lement sv is such tha t .syogtt = u oy sy for any u E A(Y2).

Conversely, if T is a G-set such that any G-set divides in CA a mnl t ip le of T, if I choose an e lement .s E A(TZT), such that s OT u = zL OT s for any u C A(T2), and if I choose for any X elements ai ,x E A ( X T ) and fli.x C A(TX) , for 1 < i < rzx, such that

7t X

1A(x~) = ~ c~{..< oT/3i,x 4=1

I can se~ ~z X

sx = ~ c~,x oT s OT 5<X C A ( X Z X ) i=1

i2.2. THE FUNCTORS (A 321

This element does not depend on the choices of c~i,x and di,x: indeed, if

!

1A(X~) = ~ C~j,X o7c fl' j,X d=l

then rz X

I / SX Ox O~j, X = E Cti,X OT S 0 T fli,X OX Ctj,X

i=1

As fl~,x ox ct}, x r A(T2), it is also

nX

SX OX O~j, X : ~ Ozi,X OT fli,X OX dtj,X ~ S : 1A(X2) Ox as OT S : ~ j , X OT 8

i=1

It follows that

and summing over j

1 ! / / SX Ox Ctj,X 01 / ' f l j ,X = (~j,X 0 T S 0 T flj ,X

~k X;' 9' 6 X = Ozj, X 0 T 8 0 T j,X j = l

Moreover, the sequence sx is an element of (A(Z): indeed, if X and Y are G-sets,

and if a E A ( X Y ) , then rzy

a = )__s a o r c,j,~,. O T / 3 j , r j=l

Thus ,S X 0 X a = v--,L ~ OT 5 o T fli ,x o x a Oy O~j,y 0 T f l j ,y

l<i<nx l~_j~_ny

As fli,x ox a %, cU,y C A(T2), it is also

8 x O x O , E 1 <i<n x

~ly

O : i , X O T ~ i , x O x ( I O Y d t j y O T SOY flj,Y = E a ~ 1 7 6 T S O Y flj,Y = (tOy S y j = l

Then I see that (A(Z) identifies with the set of elements s in A(TZT) such that

s OT u = 'u of s for any u E A(T2). But

and I have also A( T 2) ~- J'{ A ( AT, AT)(') . Moreover, the product of u r ?-{ A ( AT, AT)(') and s r "HA(AT, AT)(Z) for the functor J-(A(AT, AT) is precisely zt OTS. It follows that

CA is the commutant of J-((AT, AT)(.) in "}-{(AT, AT). Now say that any X divides a multiple of T in CA is equivalent to say that AT

is a progenerator of A - M o d . A natural question is then to know if, whenever P is a.

progenerator of A - M o d , I have


First it is clear tha t if a E (A (X) , then o'p is a element of Hom~ (P, Px" ) = ~.4( P, P ) ( X ) . I ob ta in tha~. wa.y a na tu ra l morph ism (.4(X) ~ 7~A(P, P ) ( X ) , which induces a morphism

0 : ( . ~ -~ ~.~(P, P)

If f E E n d a ( P ) = ~ a ( P , P) (*) , then the square

( rp P ~ Px

P ~ t~ 0 p

has to be commuta t ive . This expresses exactly the fact that the image of (9 is con-

ta ined in the c o m m u t a n t of 7-LA(P, P ) ( . ) . Let p(O he a direct sum of copies of P. It. is easy to see that the com m uta t i v i t y

of tile squares

) X

P ' Px (.'r p

where fi is the project ion on the component of index i E I , forces at ,!n = (c~p) (t). Then if a p = 0, I have an(~ 0 for any i. As any A-module M is a quot ient of some

p(I); for a sui table ( I) , the commuta t iv i ty of tile square

p(1) _C~P(1)> [)~I)

M ~ M x OC M

forces then C~M o ~r = 0, thus a M z 0 if (7 is surjective. Then ct = 0, which proves that

@ is injective. Convers~ e.lv,, if ct E HomA(P, Px) is such that all the squares

(~ P - ' Pv

,1 I fX P -- ) Pv

are commuta t ive , then set t ing ap(• = (c~) (/), it is easy to see tha t all the squares

,1 lsx p(d) p(3)

- - ) X O'p(J)

(c)

12.2. THE F U N C T O R S CA 323

obtained for index sets I and d and a morphism f : p(S) __+ p ( a ) are commutative. In particular, if L is the image of f , I see that ap(,l(L) C Lx. So if M is an arbitrary A-module, and ~r a surjective morphism from p(a) to M, then there exists a unique morphism C~M such that the square

P( J) _~ Rid)

O I ~. OX

M ~ Mx C* M

is commutative. The commutativity of the squares (C) and the projectivity of p(1) shows that C~M does not depend on (d) or on ~r. It is then clear that the c<vi define an element of ~A(X). Thus (9 is smjective. Finally, [ have proved the following

proposition:

P r o p o s i t i o n 12.2.4: Let A be a G r e e n f u n c t o r for the g r o u p G. I f P is a p r o g e n e r a t o r of A - M o d , t h e n

(A "" C~A(P,P)(EndA(P))

In p a r t i c u l a r , if T is a G-set such tha t any G-set d i v i d e s a m u l t i p l e of T in

CA, t h e n for any G-set Z

CA(Z) = {s ff A(TZT) IV. E A(T2), .s OT u = u or s}

12.2.2 E n d o m o r p h i s m s o f the restr ict ion functor

Let H be a subgroup of G. The first definition of ~A(H) is

where Reset is the restriction functor from A - M o d to Res,~A-Mod. Setting

A-CG KCH

the evaluation M ~-+ ]IJ(Q) is an equivalence of categories from A - M o d to A(Q 2) G 4 M o d , and the evaluation M' ~-+ M'(gt') is an equivalence of categories from iRes~, -

M o d to (Res~A)(f2'2)-Mod. Moreover

= G Qt2 (Res~a)(fY ~) A(IndH )

and the product of the elements c~ and fl of (Res~A)(~t '2) is defined by

e of~, fl (Res~A). w'w2c% c; * (c* x ' fl) ' = (ResHA) ~lw2w3 / / I / , / /

where the products o' and x ' are those of the Green functor Res~A. Thus

( ) ( , , , c'~o a , f l =A . In H \ c0~w~ ] \w,w2co2w3)]

324 C H A P T E R 12. CENTRES

G #2 2 where 5 is the map from Ind~FY 4 to (IndHf~) defined by

0 . . . . /) 031'022'023'024) = ,q'031'032)' (g ' ~'3'CJ'I

It follows that

# / , ! , ' ' ,' (g,~o~,02~,~) '~ ( ~ x / 3 ) , / l / I l # \ (~,~,02~) ) (~,02~,02~)(g,02~,02~)]

Let i be the map fi'om Inds~fl ~2 to f~2 defined by

i ( g , J ~ , J ~ ) - (,qJ1,g022)

where the expression g03 ~ is equal to 9hK if 03' = DK C fY. The map i is injective: indeed, if

i(g,03',,03;) = i(.q', 02'/, 03;')

then g02~ = g'03'[ and as 02~ and w'/are contained in H, I have g'tl = gH. Then there exists h E H such that

g # = gh 02'1 = ]702'1; 02; = ] 7 0 2 7

# #1 ## ! i Then (g ,02~,022)= (g'h,03~,022) = (g ,%,022) , and i is injective. It follows that A.(i) is an injective morphism of (Res~A)(fi '2) into A ( ~ ) . More-

over

( ' ~ , ,, , ( (~,~,03~,02~)",, ~ (o, • ;~) A.(i)(a o~,/3) = A . . (g'02''cc~'a~3). A* # # ~ , / / / ,1 \ ~02,,~02~ ) (g,02,,~)(~,02~,~,~))

On the ol, her hand

A. ( i ) (o~)o f~n , ( i ) ( f f ] ) : A . (021022~ A* ( 021032023 ~ (n . ( i ) (o~) x n.(i)(/_~)) : . , k 021023 / \0210220220"13 /

! ,t , ~/ l

' �9 �9 # # , / , ,1 k 031033 / \021022~'2023/ glCOlglO.:2g2~3g2",. /]

Let (C) be the square

(g,03~, 02,1,~;) (g, 02~, 02~)(j, 02~I, 021) J

1G , ' ~ ;3 n<~.~ , (Ind~fY~) 2

~ ( 02,022023 I ~ k 031C02C02cO3 /

This ~qu~re ~s cartesian: h~deed, if (~o1,~.,~,~:~) ~nd t(~,~.,,<~)(.<~,02~.~':,)) a,'e s n c h

that (w~, 022,02~, w3) ~ ~ ~ = ( 9 t % , .(k 022, g2a3, ff203~)

12.2, THE FUNCTORS (A 325

t ! then 91022 = g202;. As 02; and 023 are contained in H, this forces gl t t = g2H, and there = ' = ha;;, and exists h E H such that 91 g2h. Then %

I ! --1 / P / I t - - 1 ! \ ( g 2 ' c d 3 ' 0 2 4 ) = (g l h ' had2,024) = [g1 ,022 , " 024)

Let then u = (g, 02'1,02;, h-~o2~) E Ind/%'*. I have

( ( g ' < ' < ' < ) ) ( , , ) = ((gl, ' (9, ~ , 021)0,4, 02.~) ) 021,

and

= , , , =

I / I I /

( . . . . ) On the other hand, the map i = 0,~,~,%) is injective: indeed, if O,q,~;)Cg.%.~i) �9 I I l , ! I I I! I I

then there exists h E H such that

g' = gh -1 J ( = h02t1 o J; = h02; 0,23 = h02;

and then I I t I I I I l I f = (g, 021, % , % ) ( g , 0 2 1 , 0 2 2 , 0 2 3 ) ( g h - 1 , hcu1, h022, h023) : ' ' '

So the square (C) is cartesian, and then

A.(i)(~) oa A.(i)(fl) . . . .

" \ g021wi / \ (g ~ 4 ) 0 , 4 , 4 ) / % • . . . . ! . . . . i(~ o~, fi)

It follows that A. (i) is a morphism of algebras (non-unitary in general) from Res~rA(fY 2) to A(f~). In terms of generators, if K and L are subgroups of H, if x E H, and if iK,~-,L is the map from H/ (K N XL) to fy2 defined by

then i o Ind~iK#,L is the map

g(K n ~L) e a / ( I ( n ~L) ~ (gK, gxL) C a s

It follows that

d ~.,,,K . , xr~L ~ K L *~)(~ KNXL A a,Kn=L F[znL) -~ tKnXL~a,Kn~LXTK=nL

tn other words, the morphism A.(i) is just the inclusion

a " "fy2x (ResHA)/ ] = @ A(K rn ~L) ~-~ @ A(K n ~L) = A(a 2) K,LCH K,LC_G

xel i~H/L xeK\G/L

I have finally proved the


L e m m a 12.2.5: Le t H be a s u b g r o u p of G. T h e n the inc lus ion i f r o m Ind~fY 2 in to f~2 induces an in jec t ive m o r p h i s m of a lgebras

A,(i) = (Res~)(A)(fY 2) ~-~ A(f~ ~)

R e m a r k : This is the only point on which I disagree with Th6venaz and Webb (see [15] 5.3 to 5.4): this lemma shows in particular that the morphism b.(i) from the Mackey algebra, of H to the Mackey algebra of G is injective.

If M is an A-module, then

K C H - N C H

In other words, the restriction functor can be translated in terms of the algebras B = A(Q 2) and C = (Res~A)(fY 23 as

(Res~M)(W) = A,( i ) ( lc ) oa M(f~)

Let e be the idempotent A.(i)(1c) of B. Then e oa B is a C-module-B, and

aes~M(a') ~_ (~ o~ B) O~ M(n)

The restriction functor is then given by tensoring with a bimodule, and then I can apply the following lemma:

L e m m a 12.2.6: Le t A a nd B be R-a lgebras , and M be an A - m o d u l e - B . T h e n

E n d F ~ t ( M @B - ) ~-- EndA|

In p a r t i c u l a r , if f : A --~ B is a m o r p h i s m of a lgebras , if e = .f(1A), and M = eB, t h e n eM @B -- is t he f u n c t o r Resj of r e s t r i c t i on a long f , and

E n d f ~ M a e s s ) ~-- {5 ~ ~B~ IV(, ~ A, bf(a) = f(a)b)

Proof i An endomorphism r of the functor M | - is determined by giving, for any B-module L, a morphism of A-modules eL from M @B L to itself, such that for any morphism of B-modules ~b : L --+ L', the square

r M | , M |

M @ B ~ t I M| M @B L' M | L' )

eL'

~B M @ B B ~ _ M ~ M |

I

@t? #I ~ ~ M @B #t M 4.

M @ L 1 ' M | cs

(c)

is commutative. In particular CB is an endomorphism of M | B, which is isomorphic to M by the map 0 defined by O(m | b) = rnb. Then ~ = OCBO -1 is an endomorphism of M. Moreover HomB(B, L) _~ L for any L, the e lement /E L defining the morphism #l : b ~-+ bl from B to L. The commutativity of the square

12,2. THE F U N C T O R S s 327

forces then

eL(re| = (M|174 (M | (M@pz)((I ) (m)@l)= O(m) |

Thus ~ determines r Moreover, if a E A and b E B, I mnst have

The case L = B now implies that ~ is an endomorphism of M as A-module-B. Conversely, being given r I can define (I)L by qSc(rn| = (I)(m)| This definition

makes sense, because for b C B

Cc(,~b | l) = ~(mb) | Z = ~(m)b O l = ~(m) O bl = CL(-~ | bl)

If moreover �9 is a morphism of A-modules, then for any a E A, I have

CL(a,~ | l) = ~(am) | 1 = a~(n~) | l = aCL(,~ | l)

Thus qSc is a morphism of A-modules from M to Mx. With this definition of eL, it is clear that all squares (C) are commutative. This proves the first assertion of the ]emma.

For the second assertion, I must find the endomorphisms of eB as an A-module-B. Such an endomorphism r is entirely determined by the image of e E eB, since

r = r

Then I must have r ~) = ~5(e) = r and also

So r E eBe. Moreover, if a C A, then

f (a)e = f (a ) f (1A) = f(a) = f (1A) f (a) = ef(a)

Conversely, if p E | commutes with any element in the image of f , then setting

r = pb

I obtain an endomorphism of eB as A-module-B, since

r = f3(f(a)ebb') = r = pf(a)bb' = f(a)pbb' = a.(pb).b'

This proves the lemma. "

This lemma applies to the morphism A.(i) from (tles~/A)(ft a) to A(f~2), and this gives the

Propos i t ion 12.2.7: Let A be a Green functor for G, and H be a subgroup of G. T h e n s ident i f ies w i th t he set of e lements in

4A(n~)t~: I ( C H

which c o m m u t e to all the e lements

t~'~:nx rfi1,n* C,~Xr~xnC

for K,L C H, for a C A(K) and x E H. In particular, the ring (CA(I),.) is

i somorphic to the centralizer of ,4(1) in A(1) | G.

Proof: The first assertion is only a reformulation of the previous results: if

z E ( ~ t~[)A(fl2)( E t~[) KCH NCH

and if z commutes with all the t~:, for K _C H, then

Z E ~K .L . h . L Ix" I N . L = ~KZ~L ~ = Z ~I( z = E = LKZ~ h- ~lggL z

K,LCH h-,LC_H KC_H KCH

thus z C L, tK ~ ~ K, and the first assertion holds. The second one follows from the case H = {1}: the algebra t lA(n~)f i is formed

of the elements t] ~\l ,a:rr]

for a 6 A(1) and x r G. Hence it is isomorphic to A ( 1 ) @ G . The image of A(fl '2) corresponds to the elements such that x = 1. Thus it is A(1), and the proposition follows. "

~K ~i, Say that z commutes Let z E (A(H) . Then z = Z~,cH zK, where zK = ~Kz~K. with all the ~K ~ xr L is equivalent to say that I, Kn~: L KN~L,a K#nL

K L I~" L (12.1) ZKI~Kr3zL)~KcT~L,aXrI<znL = ~KV~:~L~KnZL,aXrIs

which gives in part icular , for L C K C_ H, for a C A ( K ) and x E H

z K t ~ h i4 h tL ZL rL ZK = Z L ? ~ L ZN.~K,a ~ /~IC,aZI.2 XZNTs -1 = Zx K

Conversely, these relations imply equalities (12.1). A computat ion similar to the one made to identify the functors H(M, N) (see

proposit ion 1.4.1) shows then the following proposition:

Proposit ion 12.2.8: Let A b e a G r e e n f u n c t o r for G. I f H is a s u b g r o u p of G, t h e n (A (H) i d e n t i f i e s w i t h t h e se t of s e q u e n c e s (zi,-) i n d e x e d b y t h e

K 2 N subgroups of H, such that z~- E ti,,A(f~ )t K, a n d

zKt~- K K I~ : t L ZL f L ZK = ZLT L Zl,[/~lq,a = )~B.',aZK X Z K X -1 = Zxlx"

for any subgroups K D L of H, a n d a n y a G A ( K ) a n d x G H.

I f H ' C H a n d z = (zh) G ( A ( H ) , t h e n

r.",(~)K = ~, W~" C_ H '

i f H C_ H ' and ~ = (~,,-) ~ G ( H ) , t h e n

H ~ (t H z)i , ~ .r~ -1 I, VK C H ' .= ~ K n ~ H X Z K x n H X r K n ~ H __ xC=K\H'/H

F i n a l l y if g C G a n d z = (zK) E ( A ( H ) , t h e n

(9,7.)12 = 9(Zh'g ) V J ; C 9 9

T h e p r o d u c t z . z ' of two e l e m e n t s of ~A(H) is g iven b y

= ' V I ( C H (~.~')~, ~ ~,,- _

12.2. THE F U N C T O R S CA 329

1 2 . 2 . 3 I n d u c t i o n a n d i n f l a t i o n

Let G and H be groups, and U be a G-set-H. If A is a Green functor for G, and X is an H-set , then an element a E CA(U OH X) is a sequence (ay), indexed by the G-sets, such that ay E A ( Y • (U OH X ) • Y) . Then if T is an H-set , I set

bT �9 u (A U ) ( T X T ) = A (@,X,T)(aUo,T) E o

L e m m a 12.2.9: If U/H = i, the above construct ion is a unitary morphism of Green functors f r o m ~A o U to ~AoU.

Proof: First , if U / H = o, and if X and Y are G-sets, then

U o . (X • Y) ~_ (u o . x ) ( u o . Y) = (U o . x ) • (U oH Y)

so that the maps FUxy are bijective. Then the maps

U ~u = ((SU, x x IdUoHT ) o @X,T T,X ,T

are also bijective. Let then S be an H-set, and ~ E (A o U)(TS) . Then denoting by o' and x ' the products o and x for the functor A o U, I have

bTO'~/3 = (AoU). (t,xt2s'~ (AoU)* ( t,xt2s ,. , , . . , . . . .

( . . . . (AoU). (AoU)* \ t l X t 2 t 2 3 J \ t l xs /

Moreover, set t ing/3 ' = A.(@U,s)(/3), I have

�9 U * U * U / * U A (@,X,T)(aUosT)• = A (@,X,T)(aUouT)X A (@, s ) ( / 3 )= A (@,X,T,T,S)(aUouT X/3')

Setting T' = U O H T X' = U OH X S ' = U o H S

it is then clear that

( '1x'28 "~ . u ( t'lx't'2s' "~ (A o U)* \ t , x t 2 t ~ s / A (@,X,'r,T,s) = A*(5~X,T,s)A* tlx,t,2t,2s, ]

Similarly

(A U). (l~lX*2s~ A*(~SUx TS) (A U). (~1x'28"~ u -1 o = o A . ( ~ T , X , T , s ) . . . . \ t l x t 2 / ' ' ' \ t l x s /

= A.@Uxs- ' )A. (t~x't'2s"~ �9 . . , , \ ~;~'~ ' /

so that finally bT OtT /3 * U = A (~ST,X,S)(a T, 0 T' /~')

As a E ~A(X'), and as /3' E A(T'S ' ) , I have also

t * O t / b r o T f l = A ( ~ S T , X , S ) ( / 3 0 S , a s )


The same argument applied to fl o~ bs shows that

!

bT O T /3 = ~ os bs

hence that the sequence bT defines an element of (AoU(X). Now the lemma follows from proposition 8.4.1, which says that the correspondence

X ~-* U on X a C A ( Y X ) ~-* A.(Su,y)(a) C A((U oH X ) x (U OH Y ) )

is a functor from CAoU to CA. �9

In some cases, the morphism from (A o U to (Aog is an isomorphism:

Propos i t ion 12.2.10: Let L~_K be subgroups of G, and A be a Green funetor for the group K / L . T h e n

ind a , a , - (indl~.inf~/LA ~-- lx.lnIi(/L(A

Proof: I will show that if G and H are groups, if U is a G-set -H such that U / H = �9 and such that G acts freely on U, then

(A o U ~-- (AoV

This assertion is equivalent to the proposition, because with the hypothesis on U, there exists a subgroup P of G x H such that

p:(P) = G k , (P) = {1} U -~ (G x H ) / P

In those conditions, if K = P2(P) and L = k2(P), I have an isomorphism 0 from K / L on G, and if Z is an H-set , I have

V o n X ~ - 0 ( ( a e d Z ) L)

Then if G = K / L and 0 = Id, and if A is a Green functor for G, I have

A o U H I, = Ind1(Inf1~/LA

With the notations of lemma 12.2.9, if br = 0 for any T, then ay = 0 for any Y: the hypothesis implies indeed that ay = 0 if Y is of the form U ou T, since A.(SU, x,T) is an isomorphism. But if Y is a G-set, I have the morphism

uy : Y --* g oH (G\U.Y)

defined in this case by . y ( y ) =

for an arbi t rary element u C U (indeed U.Y = U x Y if U/H = . ) . If moreover G acts freely on U, then vy is injective: indeed, if

12.2. THE F U N C T O R S (A 331

then there exists h E H and g ff G such that

Whence gy = y' and gu = uh = u. Then g = 1 and y~ = y. Thus any G-set is a subset of a set of the form U OH T. Then in CA, any object

divides an object of this type. But I know that if Y divides Y' in CA, and if a E (A(X), then ay, determines ay. In those conditions, I have then ag = 0 for any Y, and the morphism from (A o U to (AoU is injective.

It is also surjective: if b C (Aou(X), and if T is an H-set, I have an element

br E ( A o U ) ( T X T )

I have thus an element

b~ : A.(~T,X,T)(bT) r A((U OH T) • (U oH X) • (U oH T))

and it is easy to see by proposition 8.4.1 that if a E A((U OH T) x (U OH T')) , then

setting o / = A*(fU, T,)(v~) r (A o U)(TT')

I have

/ 0 U U / U t t bT UonTa = A.(@,X,T)(bT)OUoHT A*(~T,T,)(OI ) = A*(@,X,T')(bT~ ) . . . .

U t I U i U t �9 . . = o r , b T , ) = =

If Y is a G-set, then Y maps into U OH (G\U.Y) via uy, and I set for T = G\U.Y

�9 ! ay = uy OUoffT b T OUoHT Yy,.

If Y' is another G-set, if T' = G\U.Y ' , and if a r A ( Y Y ' ) , then

ely Oy OL ~ Y y OUolr T bT OUoI4T b'y,. Oy OZ Oy~ lYyr OUoHT~ IZyl,.

But uy,. oy a oy, @, r A((U OH T) x (U OH T')) . So this gives

a y Oy O~ = b'y OUoHT lYy,. Oy O~ Oy, l ly, OUoHT, b ! T' OU~ lYy, . = OL Oy, a y ,

So the sequence ay defines an element of (A(U OH X). If moreover Y = U OH T, then for T ' = G\U.Y , I have

a y : Uy OUoHT, bT, OUoHT, 12y..

But uy,, is an element of

A((u o . r') • y) : A((v o . r') • (u o . r))

Then a y = l]y OUoHT, Py. . OUoHT bT = b~T

and then �9 U * U !

A ((~T,X,T)(auo;_.:T) = A ( ~ T , X , T ) ( b T ) = bT

which proves the proposition. �9


1 2 . 3 E x a m p l e s

Let A be a Green functor for G. If M is an A module, if Y is a G-set, and if z C CA(Y), then z induces by definition a morphism zM of A-modules from M to My. It is clear that I obtain that way a morphism of Mackey functors from (A to ~ ( M , My). The construction of the product on (A, by the formulae

(~ • = o:M • Z .

where the product of the right hand side is the product of 7~(M, M), shows that this morphism is compatible with the product. It is moreover clearly unitary. Thus M is a (A-module, and the image of ~A in 7-{(M, M) is contained in 7~A(M , M) (since z M is a morphism of A-modules). Thus M is an A(~(a-module.

P r o p o s i t i o n 12.3.1: Let A be a G r e e n f u n c t o r for t he g r o u p G. I f M is an A - m o d u l e , and if X, Y and Z are G-sets , t h e n t he p r o d u c t

aEA(X), b ~ ( A ( Y ) ' r n E M ( Z ) H a x b x m = a x M * ( z Y ) ( b z ~

t u r n s M in to an A ~ ( A - m o d u l e .

In particular, the primitive idempotents of CA ( ' ) = Z (A(gt 2)) lead to a decomposition of A-modules in blocks: the block j is formed of the A-modules M such that j x M = M.

1 2 . 3 . 1 T h e f u n c t o r s FPB

Let B be a G-algebra, and A = FPB be the fixed points functor on B, defined for a G-set X by

FP~(X) = nom~([X], B)

Then if X and Y are G-sets, the module FPB(XY) identifies with the set of matrices re(x, y) indexed by X x Y, with coefficients in B, which are invariant by G, i.e. such that for any g C G

A similar computation as in proposition 4.5.2 shows that if Z is a G-set, if" p(y,z) is a matrix in FPB(YZ), then the product m oy p is obtained as a product of matrices, i.e.

(.~ oy p)(~, ~) : X~ "~(~, y)p(y, ~) yEY

where the expression in the right hand side is computed in the algebra B. Let X be a G-set, and rn E (A(X). Then for any G-set Y, I have an element

my C A(YXY), that is a "matrix" my(y,x, y') indexed by YXY, and invariant by G, i.e. such that for any g E G, I have

-~y(gy, g~, gy') = g.~y (y , . , y')

12.3. EXAMPLES 333

Say t ha t rn E ~A(X) iS equivalent to say tha t for any G-sets Y and Z, for any a E A (YZ) , and for any (a:,y,z) E X Y Z , I have

y'ffY #EZ

Fi rs t I will t~ke Z = G/1. I set

m( : r , z ) = ma/ l (1 ,x , z ) tor x E X, z E G = (;/1

so tha t mo/l(z ' , x, z) = ~ ' rn(z t - lx , z'-lz). Similarly, if & is a map from Y to B, and if

I set

I ob ta in an e lement of A ( Y • ( G / l ) ) , and any e lement of A ( Y • ( G / l ) ) i s of this

form. In par t i cu la r , if r = 1 for y = y0 and r = 0 otherwise, equa l i ty (12.2)

becomes ?Tty(Y'X'ZyO) = E zt'TL(zI-lx'~ZI-Iz)

z'EG z t-1Y=YO

or for y ' = zyo 7Tty(y,',C,y') : E z'~lYt(zt--lx;zt--lz) (12.3)

z'E(; zt--1 ~I=Z-- I 9 t

Then tak ing Y = G/1, this equal i ty forces

,~G/, (> ,~, v') = ~-~(v - 1 . , y - l , / ) = y~,-, ~ .~ (z - lv ,v - lx , z - , v , y - l z )

which can also be wr i t t en as

m(y -1 x, y-iv,) = ~'-' =.m(z-~v,y-lz, z-~y,y-lz)

This re la t ion must hold for any x C X, and any y, S , z and z' in G. Changing x to

yx and z to y'z, this gives

g,z( x, y- ly,) = z?,,.(z-lx, z- l y- , ytz )

or changing fu r the rmore y' to yy'

.~.(.~, ~') = - ' r e ( z - l . , ~ - l v '~ )

which gives finally, changing z to z -1 and y' to y

In those condi t ions , equa t ion (12.3) becomes

zJEG ~.l--I 9--Z--1 7:Ji


which gives (y y')

Using this relation in equation (12.2) gives

m(z,g) 9EG

9y=y'

y'EY z~CZ 9EG 9EG

gy=yt gzt=z

or

that is

Z ?Tt(X,g)(;t(g~], Z) = Z (/(Y'g-Iz)?T/(X' g) 9EG 9EG

9EG 9r (;

Let again Y = Z = G/1. Let b E /3, and r be the function from Y to /~' which is equal to b in Y0 and to 0 elsewhere. If a(y, z) = ~r equation (12.4) gives

TYt(X, zyoy-1).Zb = gy~ b.fft(x, zy0y -1 )

This equation gives for Y = Yo = 1

Conversely, it is clear that this relation implies equation (12.4). Thus CA(X) is isomorphic to the set of matrices re(z, z), indexed by X x G, and satisfying the following conditions

Zm(x, y) = ~(~z, ~y) V:~ ~ X, W, y ~ G

Then let r'n(z) = ~ e a re(z, z) | z �9 B | G. The second condition is equivalent to the fact that r'n(x) commutes with all the elements of B, I have then a map

7~ : X ---, CBoc(B)

The first condition shows that for z E G

(1 e z),a(x)(1 o z -1) = ~ z~(~ , y) | z~ = }2 - 4 % zy) e z~ = ,~,(~x) yEG yEG

In other words, if C = CB| with its natural structure of G-Mgebra (induced by the morphism 9 ~ 1 | g from G to B (x) G), I see that rh is just a morphism of RG-modules from IX] to C.

On the other hand, the product in CA of the element rn E CA(X) by the element 0 ' r CA(X') is defined by

tZZ2) I (o • , . ' )y = A. kzlx~'z~ (~y or my)

12.3. EXAMPLES 335

Then if m corresponds to the matrix re(x, z), and m' to the matrix m'(x' , z), I have

(,~• = ~ ~ ( y , ~ ' , v )-~(y , x , y ) = ~ m'(~',g).~(~,g') . . . .

y" E Y gY=Y" yy,,=yl

. . . . F~ ~ ~'(x',g).~(x,g') 9"C=0 9,g~EG

g,,y=yt gig=g,,

It follows that

But

,~(x).~,(x') =

~T.~.,(~:,s) = ~ ,./(~:',g).~(x,J) og'g g,g~EG

E g I / l = I I I

g,gSEG g,glEG

Then (A(X) ~-- Hom~([X], C), and it is clear that this isomorphism induces an isomorphism of Green functors from (a to FIX:.

Now let b C Z(B)(X). Then b is a morphism of RG-modnles from IX] to B, such that for any Y, and any morphism c of RG-modules from Y to B

The case Y = G/1 shows then that b(x) E Z(B), and it follows an isomorphism of Green functors Z(FPB) ~-- FPz(B). Finally, I have the following proposition:

P r o p o s i t i o n 12.3.2: Let B be a G-a lgebra , and C = CB| T h e n

Z(FPs) ~- FPz(B) @e. ~-- FPc

as Green functors . In part icular @p.(*) ~- Z(B | G) and

Thus the centre of Yosh ida algebra (see propos i t ion 4.5.2) is i s o m o r p h i c to the centre of J~G.

R e m a r k : If the algebra B is an interior algebra (see [11]), then the algebra C is isomorphic to Z(B)G: indeed, if 9 ~ P(g) is a morphism from G to the group of invertible elements of/5', and if g acts on B by conjugation by 0(9), then say that ~g % ~ g is in C is equivalent to say that for any g, the element %p(g) -~ is in Z(B), hence that Eg(%p(9)-l)g E Z(B)G. This correspondence is moreover an isomorphism

of algebras C ~- Z( B)G.

12.3.2 The blocks of Mackey algebra

Let R = k be a field of characteristic p > 0 and A = bp the p-part of the Burnside functor, with coefficients in k. Then A(1) ~ k, and (A(1) ~-- kG. On the other hand, (A(G) is the centre of the p-part #e(G) of the Mackey algebra (i.e. the piece of the Mackey algebra corresponding to the central idempotent fl a, or the subalgebra of the Mackey algebra formed of elements ~px'1t rKp~, for any subgroups H and K of G, any

336 C H A P T E R I2. C E N T R E S

element z of G, and any p-subgroup P of 1t N ~K). The lq(G)-modules are exactly the Mackey functors which are pr@ective relative to p-subgroups.

Then I have the map rl a : Z(#I((-/)) --4 (k.(;) c" = Z~:G. It is a morphism of algebras, it is unitary, and surjective: indeed, Za'(; is generated as ~'-module by the elements of the form

for x E G. Let for K C_ Ca(x)

It is easy to see that this defines an eleme,tt of (.a (Cc;(.~')). Moreover

G G - , , r 111 rd>.(~.)( r

Thus r~ induces a surjection f rom Z(/ll((")) to Z~'(;. The kernel of this surjection

is moreover nilpotent, because if e is an idempotent of the kernel, then 7'~;(e') = 0. The module e#l(G) is then a projective Mackey functor, which is projective relative to p-subgroups, and equal to zero at {1}. Hence it is zero (see [15[ corollary 12.2) and

It follows that rl a induces a bijection between the blocks of kG and those of #~(G) (see [15] Theorem 17.1).

Bibliography

[1] D. Benson. Representations and cohornology L volmne 30 of Cambridge studies in advanced mathematics. Cambridge University Press, 1991.

[2] S. Bouc. Construction de foncteurs entre categories de G-ensembles. J. of Algebra , 183(0239):737 825, 1996.

[3] S. Bouc. Foncteurs d'ensembles munis d'une double action. J. of Algebra, 183(0238):664-736, 1996.

[4] C. Curtis and I. Reiner. Methods of representation theory with applications to ,finite groups and orders, volume 1 of Wiley classics library. Wiley, 1990.

[5] A. Dress. Contributions to the theory o.f induced representations, volume 342 of Lecture Notes in Mathematics, pages 183 240. Springer-Verlag, 1973.

[6] L. G. Lewis, Jr. The theory of Green functors. Unpublished notes, 1981.

[7] D. Gluck. Idempotent formula for the Burnside ring with applications to the p subgroup simplicial complex, lllinois J. Math., 25:63--67, 1981.

[8] J. Green. Axiomatic representation theory for finite groups. J. Pure Appl. Alge- bra, 1:41-77, 1971.

[9] H. Lindner. A remark on Mackey functors. Manuscripta Math., 18:273-278, 1976.

[10] S. Mac Lane. Categories for the working m.athematician, volume 5 of Graduate tezts in Mathematics. Springer, 1971.

[11] L. Puig. Pointed groups and construction of characters. Math. Z., 176:265 292, 1981.

[12] H. Sasaki. Green correspondence atld transfer theorems of Wielandt type for G fimctors. J. Algebra, 79:98-120, March 1982.

[13] J. Th~venaz. Defect theory for maximal ideals and simple ftlnctors. J. of Algebra, 140(2):426-483, July 1991.

[14] J. Th~venaz and P. Webb. Simple Mackey functors. In Proceedings of the 2nd International group theory conference Bressanone 1989, volume 23 of l?end. Circ. Mat. Palertoo, pages 299 319, 1990. Serie II.

[15] J. Th~venaz and P. Webb. The structure of Mackey functors. Trans. Amer. Math. 5'oc., 347(6):1865-1961, June 1995.

338 BIBLIOGRAPHY

[16] P. Webb. A split exact sequence fin' Mackey functors. Commerzt. Math. Helv., 66:34 69, 1991.

[17] T. Yoshida. On G-functors (II): Hecke operators and G-functors. J.Math.Soc.Japan, 35:179-190, 1983.

I n d e x

.v, 181 A-Mod, 48 A(X2), 81 A ~ 127 A | K, 278 Av.j, 115 CM(L), 143 CM(O~), 143 FP~,v, 296 F a OH,v, 296 FM, 71 G.s.t, 229 G\U.Y, 18a G\U.c~, 183 GreenR(G), 47 H| 97 Lx,v, 275 M o U, 168 M@BN, 146 Ms-+My, 8 M o, 130 MF, 72 MackR(G), 7 N o UIA , 237, 256 S.T, 229 S r 206 S a H,v, 282 Sx,v, 275 Sr 206 T r 193 Tr 193 U.Y, 183 U[X], 173 Z(A), 305 Z~T, 242 [m @ n]K, 14 [m@ n](y,r 16 fl, 84 fQ, 112 E~(G), 313

(LI), 5o OH, 167 5 U 171 X1 ,...,Xn ~lz, 185 G-set~x, 53 A(H), 276 d, ao ~, 21s ~ , 218 ~u 230 S,T~ Az, 199 lq (G), 335 >disjoint, 188 u(yj), 185 ~N(K), 218 v-perfect, 308 rrR(G), 307 ~z, 199 M(H) , 273 M(H), 27a ~ ( H ) , 27s r U 168 X y x ~ 127

U • , 171 xH, 181 U• 231

CA, 134 5AoU, 171 s 232 X, 194 (A(H), 315 (A(X), 316 a.m, 49 a ox m, 72 a oz a', 65 a x b, 46 a x r n , 47,76 b, 52 b 5 a, 125 e~, 308

340 INDEX

f.,f*, 68 fH a, 308 kl(L), 183 k2(L), 183 rn(g.~), t94 pI(L), 183 p~(L), ls3 q(L), 183 Du(X), 186 ~(M, N), 9 ~A(M, N), 145 Zx, 316 ~:(M,,..., ,%; P), 29 s 193 s 196 Qv(M), 188 7~(M), 2o7 T~u(O), 209 8u(M), 205

adjunction between ~ and ~ , 38 between @B and "HA, 148 co-unit, 185 unit, 185

algebra associated to a Green functor, 84,

99, 164 Alperin's conjecture, 304

balanced, 99, 154 bifunctoriality, 46, 47 bimodule, 141

construction, 15a structure on 7-{(M, N), 142

biproducts, 68 biset, 167

composition with, 168 blocks of Mackey algebra, aa5 Burnside functor

as Green functor, ,55 as initial object, 57 as Mackey functor, 52 as unit, 59

cartesian product in CA, 103 in CA • CA, 109

category Du(X), 186 adding direct snmmands to a, 310 associated to a Green functor, 67 equivalence of, 79, 310 .312, 314 representation of, 71, 81

centre, 305 of Yoshida algebra, a35

coinduction. I80 coinfiation, 168, 217 commutant, l,t3 commutative Green functor, 110, 129,

305 commute, 136 composition

and associated categories, 173 and Green functors, 170 and modules, 175 and tensor product, 168 with a biset, 168

direct summand in CA, 82

divides in C.4, 82

dual of a module, 130

embedding of a.Lgebras, 112 endosimple A module, 304 equivalence

of categories, 79, 84, 112, 129, 141, 155, 310-312, 314

of the definitions of (-been functors, 48

examples of algebras A(fT~), 94, 95 of composition with a biset, 168 of Frobenius morphisms, 227 of functors (,4, 3a2 of functors s and •u(M), 215 of Green functors s 264

finitely generated module over a. Green functor, 114

Frobenius nlorl)hisn~s, 223 functor

from G-sets to CA, 68 of evaluation, 81

INDEX 341

functorial ideal, 291 functoriality

of M@BN, 147 of U ~ U oa X, 177 of "H(M, N) and MQN, 24 of 7fA(M, N), 146

generators and relations for A(f~2), 85, 88 for the Mackey algebra, 7

Green functors and solvable re-subgroups, 314 centre, 305 composition with a biset, 170 definition in terms of G-sets, 46 definition in terms of subgroups, 41 direct sum, 305 tensor product, 134

identification of ~A, 320, 323, 327 of "]~A(M, _N), 145 of internal homomorphisms, 11 of tensor product (first), 15 of tensor product (second), 23

injective on the orbits, 177, 186, 192

left adjoint and Morita contexts, 273 and tensor product, 272 t o M ~ M o U , 197 to Z ~-+ U OH Z, 183

Lindner construction, 94

Mackey algebra, 7

anti-automorphism, 10 axiom, 5 functors

composition with a biset, 168 definition as modules, 7 definition in terms of G-sets, 6 definition in terms of subgroups,

5 internal homomorphisms, 9 tensor product of, 9

module dual, 130

of finite type, 114 over a Green functor, 41, 47

examples, 61 over the Burnside functor, ,59 right, 129

Morita context, 100, 154 equivalence, 99

of algebras A(X2), 99, 100 theory, 1,55, 160

morphism n-linear, 29

universal property, 29 bilinear, 37, 127 Probenius, 223 of bimodules, 141 of Green functor, 41 of Mackey functors, 5, 6 of modules over a Green functor, 42

multiple of in CA, 82

natural transformation, 316 non-commutative

tensor product, 164

opposite Green fimctor, 127

product OH, 167 8 , 125 • for A~B, 134 • for CA, 317 x for ~ (M, M), 123 • for s 231

progenerator, 115, 117, 118, 155, 161, 321

projective G-algebra, 292

projective relative to, 102 solvable ~r-subgroups, 314

relative projectivity, 100 representation

of a category, 71, 81 residual rings, 274 restriction, 5

for M~N, 15, 23 for ~ (M, N), 10

342 INDEX

right adjoint to M ~ - + M o U , 210

right modules, 129

same stabilizers, 100 simple Green functors, 291

classification, 295 simple modules, 275

classification, 276 structure, 282

source algebra, 114 surjective

Morita context, 154

tensor product 7~-fold, 25

universal property, 29 and composition with a biset, 168 and left adjoints, 227 and residues, 298 and right adjoints, 242 associativity, 38 commutativity, 38 of "algebras" over b(ft2), 164 of Green functors, 134

universal property, 137 of modules over Green functors, 140 of simple modules, 298

Th~venaz's theorem, 295 trace, see transfer transfer, 5

for M Q N , 15, 23 for 7-{(M,N), 10

unitary Green functor, 46 morphism of Green functors, 41, 43

Yoshida algebra, 95 centre, 335

green functors and g-sets

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