mackey functors and functorial extensions

a

n,

.

-nfixednot

Journal of Algebra 289 (2005) 1–19

www.elsevier.com/locate/jalgebr

Mackey functors and functorial extensions

Robert Hartmann1

Department of Mathematics and Computer Science, University of Leicester,University Road, Leicester LE1 7RH, UK

Received 11 August 2003

Communicated by Michel Broué

Abstract

Given a finite groupX and a normal subgroupG of X, we show that any Mackey functorMfor X induces another Mackey functorM for X associated toG. We then consider the questiowhether there exists a mapM → M extending elements fromM(G) to M(X) and compatible withthe restriction maps. In the case that the order ofG and the index ofG in X are relatively prime, wegive a sufficient condition for the existence of such a map, using canonical induction formulae 2005 Elsevier Inc. All rights reserved.

1. Introduction

Let X be a finite group, and letG be a normal subgroup ofX. A classical result in therepresentation theory of finite groups says that, if the order ofG and the index ofG in X

are relatively prime, then any irreducible complex character ofG that is stable under conjugation with elements ofX, can be extended to a character ofX. Moreover, this extensiois unique under the additional condition that the determinant of its restriction to acomplement ofG in X is trivial. However, as Example 3.7 will show, this extension isfunctorial, in the following sense.

E-mail address:[email protected].
1 Research partly supported by the European Community through Marie Curie fellowship MCFI 2002-01325.
0021-8693/$ – see front matter 2005 Elsevier Inc. All rights reserved.doi:10.1016/j.jalgebra.2005.03.025

2 R. Hartmann / Journal of Algebra 289 (2005) 1–19

n,ms. In

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withlcters.

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uctions

s ang.

e

ons are

The rings of complex characters ofX and its subgroups form a Mackey functorR for X.That is, everyR(Y ), for Y being a subgroup ofX, is aZ-module, and one has conjugatiorestriction and induction maps between these modules, satisfying a number of axioSection 2 we show, that the normal subgroupG of X induces another Mackey functorR

for X, where theZ-module for a subgroupY of X is given byR(Y ) = R(Y ∩G)Y , i.e., thevirtual characters ofY ∩G that are stable underY . Now one can consider mapse : R → R,i.e., families of mapseY : R(Y ) → R(Y ), one for each subgroupY of X. If we requireadditionally thateY followed by resYY∩G is the identity onR(Y ) then such a map can bobtained by extending characters, as described above. But, in this functorial framewogiven answer is not satisfying, since these mapseY : R(Y ) → R(Y ) obtained by extendingcharacters do not commute with the restriction maps of the Mackey functorsR andR.In Section 3 we show how to construct a different family of maps, that do commuteconjugations and restrictions. We call such a family anr-extension morphism. The essentiaingredients in its construction are the Adams operators on the ring of complex chara

Now, if M is any Mackey functor forX, we still can construct another Mackey funcM for X associated toG, where for a subgroupY of X we setM(Y ) = M(Y ∩ G)Y .Also, we still can hope to find anr-extension morphism. However, this question is muharder, and we only give a partial answer to it. First of all, we require the existenccanonical induction formula forM . Such a formula for the ring of complex characters (asome other representation rings) has been given by Boltje in [1]. We briefly recall theimportant facts about this theory in Section 4. Then we show that a canonical indformula forM induces also one forM . Finally, in Section 5 we give sufficient conditionfor the existence ofr-extension morphisms, using the canonical induction formula. Aapplication, we prove the existence of anr-extension morphism for the trivial source rin

Notation. For a finite groupG, we write H � G to denote thatH is a subgroup ofG.If H � G and x ∈ G we write xH for the conjugate subgroupxHx−1. For subgroupsH,K � G we write [G/H ] respectively[H\G/K] for a set of representatives for thcosets respectively double cosets inG.

2. Mackey functors

Let X be a finite group, and letk be any commutative ring. We recall the following twequivalent definitions of Mackey functors. The first one is due to Green, our notatiofrom [1].

2.1. Definition. (a) A k-conjugation functorM = (M,c) for X is a family of k-modulesM(Y), for Y � X, together withk-module homomorphisms

cx,Y :M(Y) → M(xY

), for x ∈ X, Y � X,

such that

cy,Y = idM(Y) for y ∈ Y, Y � X, (C1)

R. Hartmann / Journal of Algebra 289 (2005) 1–19 3

ely

ctively

ubmod-

cx1x2,Y = cx1,x2Y ◦ cx2,Y for x1, x2 ∈ X, Y � X. (C2)

(b) A k-restriction functorM = (M,c, r) for X is a k-conjugation functor(M,c) to-gether withk-module homomorphisms

rYZ :M(Y) → M(Z) for Z � Y � X,

such that

rYY = idM(Y) for Y � X, (R1)

rZW ◦ rY

Z = rYW for W � Z � Y � X, (R2)

cx,Z ◦ rYZ = r

xYxZ ◦ cx,Y for x ∈ X, Z � Y � X. (R3)

(c) A k-Mackey functorM = (M,c, r, t) for X is a k-restriction functor(M,c, r) to-gether withk-module homomorphisms

tYZ :M(Z) → M(Y) for Z � Y � X,

such that

tYY = idM(Y) for Y � X, (M1)

tYZ ◦ tZW = tYW for W � Z � Y � X, (M2)

cx,Y ◦ tYZ = txYxZ ◦ cx,Z for x ∈ X, Z � Y � X, (M3)

rYU ◦ tYZ =

∑y∈[U\Y/Z]

tUU∩yZ ◦ ryZU∩yZ ◦ cy,Z for Z � Y � X, U � Y. (M4)

We call the mapscx,Y conjugations, the mapsrYZ restrictionsand the mapstYZ transfer

or induction maps.(d) A morphismf :M → N of k-conjugation (respectively restriction, respectiv

Mackey) functorsM andN is a family ofk-module homomorphisms

fY :M(Y) → N(Y) for Y � X,

that commute with conjugations (respectively conjugations and restrictions, respeconjugations, restrictions and transfers).

Instead ofcx,Y (m), often we write justxm, and sometimes we writem|Z insteadof rY

Z (m).A is called asubfunctorof M if A(Y) ⊆ M(Y) is a submodule for allY � X, and if the

structure maps restricted to these submodules have images in the corresponding s
ules. In this case we writeA ⊆ M .


or

g

btain

e,

For subgroupsY,Z � X with Z normal inY , denote byM(Z)Y the set ofY -fixed pointsin M(Z), i.e.,

M(Z)Y = {m ∈ M(Z) | cy,Z(m) = m for all y ∈ Y

}.

The second definition of Mackey functors in terms ofX-sets is due to Dress, see fexample [4].

2.2. Definition. A k-Mackey functor forX is a bivariant functorM from the categoryX-set of finite X-sets to the categoryk-Mod of k-modules, meaning thatM consists of acovariant functorM∗ and a contravariant functorM∗ which coincide on objects, satisfyinthe following two axioms.

(A1) For any twoX-setsS andT , let iS and iT denote the respective injections fromSandT into the disjoint unionS ∪ T , then the mapsM∗(iS) ⊕ M∗(iT ) andM∗(iS) ⊕M∗(iT ) are mutually inverse isomorphisms betweenM(S ∪ T ) andM(S) ⊕ M(T ).

(A2) For any pullback square ofX-sets

Sγ

δ

T

α

Uβ

V

we have thatM∗(β) ◦ M∗(α) = M∗(δ) ◦ M∗(γ ).

2.3. Remark. If M is a Mackey functor in the sense of the second definition, then we oa Mackey functor(M,c, r, t) in the sense of the first definition by settingM(Y) = M(X/Y)

for a subgroupY of X and defining the structure maps as

cx,Y = M∗(γx,Y ), rYZ = M∗(πY

Z

)and tYZ = M∗

(πY

Z

),

whereπYZ is the natural surjectionX/Z → X/Y , xZ → xY , andγx,Y is the mapX/Y →

X/xY , gY → gx−1(xY ). Here and in the following we use the symbolM for both defin-itions; if we writeM(Y) for a subgroupY of X it is understood that we use the first onwhile M(S) for anX-setS refers to the second one.

2.4. Theorem. Let (M,c, r, t) be ak-Mackey functor forX, and letG be a normal sub-group ofX. Then

M(Y ) := M(Y ∩ G)Y , for Y � X,

cx,Y := cx,Y∩G for x ∈ X, Y � X,

rYZ := rY∩G

Z∩G for Z � Y � X,


2.1proof

n.

ect

rifythe

t YZ :=∑

y∈[Y/Z(Y∩G)]cy,Y∩G ◦ tY∩G

Z∩G for Z � Y � X,

defines ak-Mackey functor(M, c, r, t ) for X.

Proof. It is possible to prove the theorem by directly verifying the axioms of Definitionthrough elementary (though lengthy) computations. The following, more structuralhad been suggested by the referee.

We will first define a Mackey functorN in terms ofX-sets, which will turn out to bejust M as given in the statement. To defineN , let us first recall the Dress constructioFor anyX-setS and any Mackey functorM , there is a new Mackey functorMS definedby composing the bifunctorM with the endo-functorX-set → X-set, U → U × S, sothat MS(U) = M(U × S). In our situation, applying the Dress construction with respto theX-setS = X/G with G normal inX, we obtain a Mackey functorMX/G such thatMX/G(U) = M(U ×X/G) for anX-setU . Moreover, thek-moduleM(U ×X/G) is evena rightkX-module, with the action ofx ∈ X onm ∈ M(U × X/G) given by

m · x = M∗(πx)(m),

whereπx is the map ofX-setsU × X/G → U × X/G, (u, zG) → (u, zxG); in fact, ifx1, x2 ∈ X thenπx2 ◦ πx1 = πx1x2 and hence

(m · x1) · x2 = M∗(πx2)(M∗(πx1)(m)

) = M∗(πx1x2)(m) = m · (x1x2).

Now, for any additive functorF : Mod-kX → k-Mod, we claim that the compositionF ◦MX/G :X-set → k-Mod defines again a Mackey functor. In fact, we only need to vethe axioms (A1) and (A2) from Definition 2.2. But these follow immediately fromadditivity of F and the corresponding properties ofM .

Now, letN be the Mackey functor obtained from this construction forF being the fixedpoint functor−X : Mod-kX → k-Mod, so that for anX-setU we have

N(U) = M(U × X/G)X,

or in the case thatU = X/Y for some subgroupY of X,

N(X/Y) = M(X/Y × X/G)X.

SinceG is normal inX, there is an isomorphism ofX-sets

X/Y × X/G ∼=⋃

t∈X/YG

X/(Y ∩ G) ∼= X/(Y ∩ G) × X/YG.

This implies that

M(X/Y × X/G) ∼=⊕

M(Y ∩ G)

t∈X/YG


td

as ak-module, and the rightX-action on this module is such that

M(X/Y × X/G) ∼= indXYG M(Y ∩ G)

as kX-modules, where thek(YG)-module structure onM(Y ∩ G) is obtained from theconjugation action ofY via the group isomorphism

Ψ :YG/G∼→ Y/(Y ∩ G).

Now an element

∑z∈[X/YG]

z ⊗kYG mz ∈ indXYG M(Y ∩ G)

is fixed underX if and only if

mz = m ∈ M(Y ∩ G)YG = M(Y ∩ G)Y

for all z ∈ [X/YG]. It follows that

N(X/Y) = M(X/Y × X/G)X ∼= M(Y ∩ G)Y ,

so that the evaluation of the Mackey functorN is indeed the same as forM in the statemenof the theorem. It remains to show that the structure maps obtained forN are the ones statein the theorem.

The conjugation mapcx,Y onN(Y) is obtained by restricting the map

(MX/G)∗(γx,Y ) :M(X/Y × X/G) → M(X/

(xY

) × X/G)

to theX-fixed points. With the above isomorphisms,

M(Y ∩ G)Y∼→ (

indXYG M(Y ∩ G)

)X → (indX

(xY )G M(xY ∩ G

))X ∼→ M(xY ∩ G

)X,

m →∑

z∈[X/YG]z ⊗ m →

∑z∈[X/(xY )G]

z ⊗ cx,Y∩G(m) → cx,Y∩G(m).

Hence the conjugation map onN(Y) is indeedcx,Y = cY∩G.A very similar argument shows that the restrictions maps are of the formrY

Z = rY∩GZ∩G

for Z � Y . Finally, if againZ � Y andm ∈ M(Z ∩ G)Z , then the element

∑z∈[X/ZG]

z ⊗kZG m ∈ (indX

ZG M(Z ∩ G))X

is mapped under the transfer mapt YZ to


y

low-of the

∑u∈[X/YG]

∑z∈[X/ZG], z∈uYG

z ⊗kYG tY∩GZ∩G(m)

=∑

u∈[X/YG]

∑z∈[X/ZG], u−1z∈YG

u ⊗kYG cψ(u−1z),Y∩G ◦ tY∩GZ∩G(m), (1)

whereψ(u−1z) is any coset representative ofΨ (u−1zG). It is easily checked that for anu ∈ X we have a bijection of sets

{zZG ∈ X/ZG | u−1z ∈ YG

} → Y/Z(Y ∩ G),

zZG → Ψ(u−1zG

) · Z(Y ∩ G)

with inverse given by

vZ(Y ∩ G) → uvZG,

and hence the element (1) equals

∑u∈[X/YG]

∑y∈[Y/Z(Y∩G)]

u ⊗kYG cy,Y∩G ◦ tY∩GZ∩G(m).

It follows that the transfer map is given by

t YZ (m) =∑

y∈[Y/Z(Y∩G)]cy,Y∩G ◦ tY∩G

Z∩G(m)

as required. �

3. Extension morphisms

Let X be a finite group,G a normal subgroup ofX, and let (M,c, r, t) denote ak-Mackey functor forX. Define(M, c, r, t ) as in Theorem 2.4.

3.1. Lemma. The maps

d :M → M, dY = rYY∩G for Y � X

define a morphism of Mackey functors.

Proof. This can be verified by a few straightforward calculations, but we give the foling alternative proof, suggested by the referee, which also gives the interpretationlemma in terms ofX-sets.

As in Section 9 of [6], we define the morphism of Mackey functors

D :M → MX/G


the

ctors

ativeuction

at someX-setS by

DS :M(S) → MX/G(S) = M(S × X/G), m → M∗(pS)(m),

wherepS :S × X/G → S is the projection map onto the first component. In view ofdefinition of the rightkX-action onMX/G(S), it is clear that the image ofDS is fixedunderX, so thatD is in fact a morphism of Mackey functorsM → M . We will see thatD = d as stated in the lemma. In fact, the map

DY :M(Y) → MX/G(Y ) ∼=⊕

x∈X/YG

M(Y ∩ G)

is just the restriction maprYY∩G in each component, and by the isomorphism

( ⊕x∈X/YG

M(Y ∩ G)

)X

→ M(Y ∩ G)Y

given in the proof of Theorem 2.4, we see thatDY (m) = resYY∩G(m) = dY (m) for m ∈M(Y). �3.2. Definition. ForY � X andm ∈ M(Y ) = M(Y ∩ G)Y we callm ∈ M(Y) anextensionof m to M(Y) if dY (m) = m.

A morphisme : M → M of conjugation functors is called anextension morphism, if forall Y � X

dY ◦ eY = idM(Y )

.

If e is even a morphism of restriction functors, we calle anr-extension morphism.

3.3. Proposition. Assume that[X : G] ∈ k×. Then

eY : M(Y ) → M(Y), m → 1

[Y : Y ∩ G] tYY∩G(m)

defines anr-extension morphism. Even more, it defines a morphism of Mackey fune : M → M .

Proof. Again this can be shown through direct computations, but we give an alternconstruction as in Theorem 2.4, suggested by the referee. Similarly to the constrof M , we considerMX/G as a rightkX-module, and define a Mackey functorM˜ by com-posing now with the coinvariant functor−X : Mod-kX → k-Mod. For anykX-moduleV ,the moduleVX of X-coinvariants is defined to be the largest quotient ofV on whichX

acts trivially. The compositionV X ↪→ V → VX of inclusion and natural epimorphism then


tain a

e

ctors

,

ut

gives a natural map from invariants to coinvariants. Returning to our situation, we obmorphism of Mackey functorsg : M → M˜ from the natural map

M(S) = M(S × X/G)X → M˜ (S) = M(S × X/G)X

for anyX-setS. Also, similarly to the construction ofD in the proof of Lemma 3.1, thprojectionpS :S × X/G → S induces a morphism of Mackey functorsf :MX/G → M ,defined for anyX-setS by

fS :MX/G(S) → M(S), m → M∗(pS)(m).

ObviouslyfS factors through the coinvariants, and we get a morphism of Mackey funf : M˜ → M . Composing the morphismsg andf we obtain

E : M → M˜ → M.

For a subgroupY of X, the map

fY :⊕

x∈X/YG

M(Y ∩ G) → M(Y)

is just the sum of the transfer mapstYY∩G from each component, so form ∈ M(Y ) we have

EY (m) =∑

x∈X/YG

tYY∩G(m) = [Y : Y ∩ G] · [X : YG] · eY (m) = [X : G] · eY (m),

henceE = [X : G]e. In particular, if [X : G] is invertible ink then e is a morphism ofMackey functors. Moreover, applying the Mackey formula, we see that

dY

(EY (m)

) =∑

x∈X/YG

rYY∩G ◦ tYY∩G =

∑x∈X/YG

∑y∈Y/Y∩G

cy,Y∩G(m) = [X : G] · m,

henced ◦ e = dM(Y )

. �3.4. Remark. If k = Z then we can extend scalars toQ, i.e., consider theQ-Mackey func-tors QM (by that we meanQ ⊗Z M), and similarlyQM . We can apply Proposition 3.3but in general the image ofM undere will not be inM but only inQM .

Now we consider the Mackey functorR = (R, c, res, ind) that assigns to eachY � X thering R(Y ) = RC(Y ) of complex characters ofY . Moreover, assume that(|G|, [X : G]) = 1,so thatG has a complementS in X. In this situation, there is a well known result aboextensions of characters:

3.5. Theorem [5, 13.3]. For any irreducible characterχ of G that is stable underX, thereis a unique irreducible characterχ of X such thatχ |G = χ anddet(χ |S) = 1. Any other
extension ofχ is then of the formχϕ for someϕ ∈ Hom(S,C×).


t

3.6. Corollary. If χ is an irreducible character ofG andStabX(χ) = Y < X thenχ has anextensionχ to an irreducible character ofY , andindX

Y (χ) is an extension of∑

x∈[X/Y ] xχ

to a character ofX. This induces an extension morphism

e1 : R → R.

However,e1 does not commute with restrictions, as the following example shows:

3.7. Example. Let G be the quaternion group of order 8, and letX be a semidirect producof G and a cyclic groupS of order 3, such thatS acts non-trivially onG. Let χ be the irre-ducible character ofG of degree 2, thenχ ∈ R(X). Denote by 1S , ϕ andϕ2 the irreduciblecharacters ofS, thenχ = e1(χ), χϕ andχϕ2 are the different extensions ofχ to X. Onehas

resXS((e1)X(χ)

) = 2 · 1S,

but

(e1)S( ˜resXS (χ)

) = resG1 (χ) = ϕ + ϕ2.

Hence in generale1 ◦ ˜res�= res◦e1.

Let Ψ nX :R(X) → R(X) denote thenth Adams operator, for any integern, i.e.,

Ψ nX(χ)(x) = χ

(xn

)

for anyχ ∈ R(X) andx ∈ X.

3.8. Proposition. There is a uniquer-extension morphism

ext :R → R

with the property that for all cyclic subgroupsC of X and all ϕ ∈ R(C) one hasext(ϕ) =ϕ · 1. Moreover, for anyY � X, the mapextY : R(Y ) → R(Y ) is given by

extY (χ) = Ψ nY (χ1), (2)

whereχ1 ∈ R(Y ) is any extension ofχ ∈ R(Y ), andn is an integer such that

n ≡ 1 mod|G| and n ≡ 0 mod[X : G].

Proof. Let π denote the set of prime divisors of|G|, then everyx ∈ X can be written asx = xπ · xπ ′ = xπ ′ · xπ with xπ ∈ G andxπ ′ being an element ofπ ′-order.

Now letY � X. If χ1 andχ2 are two extensions ofχ ∈ R(Y ), and ify ∈ Y then

( )
Ψ n
Y (χ1)(y) = χ1 yn = χ1(yπ ) = χ(yπ ) = χ2(yπ ) = Ψ nY (χ2)(y).


ore-

ed

fol-

p

o-

Hence, ext is well defined by Eq. (2), and clearly ext commutes with conjugations. Mover, forZ � Y � X, let χ1 ∈ R(Y ) be any extension ofχ ∈ R(Y ), and letψ1 ∈ R(Z) beany extension ofψ = χ |Z∩G, then one has for anyz ∈ Z that

resYZ(extY (χ)

)(z) = extY (χ)(z) = Ψ n(χ1)(z) = χ1

(zn

) = χ1(zπ ) = χ(zπ ) = χ |Z∩G(zπ )

= ψ(zπ ) = ψ1(zn

) = Ψ n(ψ1)(z) = extZ( ˜resYZ(χ)

)(z),

hence ext is indeed anr-extension morphism. IfC is a cyclic subgroup ofX, ϕ ∈ R(C)

andy ∈ C then

extC(ϕ)(y) = ϕ(yπ) = (ϕ · 1)(yπ · yπ ′) = (ϕ · 1)(y).

Finally, to prove uniqueness, assume thate0 is another morphism satisfying the statconditions. Then, for anyY � X, anyχ ∈ R(Y ) and anyy ∈ Y one has

eY0 (χ)(y) = rY〈y〉

(eY

0 (χ))(y) = e

〈y〉0

( ˜resY〈y〉(χ))(y)

= ( ˜resY〈y〉(χ) · 1)(y) = ext〈y〉( ˜resY〈y〉(χ)

)(y) = extY (χ)(y),

hencee0 = ext. �

4. Canonical induction formulae

We will use the theory of canonical induction formulae, as described in [1]. In thelowing we will briefly recall the construction, but refer the reader to [1] for details.

Throughout, we will fix a finite groupX, andG will always denote a normal subgrouof X, as in the previous section.

4.1. Let (M,c, r) be ak-restriction functor for the finite groupX. For everyY � X wedefine

SM(Y ) :=⊕Z�X

M(Z),

and

M+(Y ) := SM(Y )/I (kY ) · SM(Y ),

whereI (kY ) denotes the augmentation ideal ofkY , i.e., the kernel of the algebra hommorphismε : kY → k mapping

∑αyy to

∑αy , wherey ∈ Y andαy ∈ k. We write

[Z,m]Y := m + I (kY ) · SM(Y )

for the elements inM+(Y ), whereZ � Y andm ∈ M(Z). ThenM+ is a Mackey functor
with the following structure maps:


toromon 5.2

etll

c+x,Y :M+(Y ) → M+(xY

), [U,m]Y → [

xU, xm]xY

,

r+YZ :M+(Y ) → M+(Z), [U,m]Y →

∑y∈[Z\Y/U ]

[Z ∩ yU, r

yUZ∩yU (ym)

]Z,

t+YZ :M+(Z) → M+(Y ), [V,m]Z → [V,m]Y .

In fact, the assignmentM → M+ is a functor from the category of restriction functorsthe category of Mackey functors. It is left adjoint to the obvious forgetful functor fthe category of Mackey functors to the category of restriction functors, see Propositiin [1]. We can also describeM+ in terms ofX-sets by

M+(S) =( ⊕

Y�X

M(Y) ⊗ kSY

)X

for anyX-setS.

From now on, we will always consider the casek = Z, unless otherwise mentioned. L(M,c, r, t) denote aZ-Mackey functor forX, andA ⊆ M a restriction subfunctor. We wimake the following

4.2. Assumptions. (i) A(Y) is a free abelian group, for everyY � X.(ii) A has a stable basis(B(Y ))Y�X , such that for allZ � Y � X andb ∈ B(Y ) the

elementrYZ (b) is a linear combination of the basis elements inB(Z) with non-negative

coefficients.

4.3. DefineM as in Theorem 2.4 with respect to the normal subgroupG. It is easy to seethat setting

A(Y ) := A(Y ∩ G)Y

for every subgroupY of X defines a restriction subfunctorA of M . If M andA satisfy 4.2,then so doM andA, sinceA(Y ),Y � X, then has the stable basis

B(Y ) ={ ∑

y∈[Y/StabY (b)]yb

∣∣ b ∈ B(Y ∩ G)

}.

4.4. ForY � X define

MY := {(Z,b) | Z � Y, b ∈ B(Z)

},

thenY acts onMY by y(Z,b) = (yZ, yb). Moreover,

(Z1, b1) =Y (Z2, b2) :⇔ (Z1, b1) = y(Z2, b2) for somey ∈ Y .


ss

h

le-

it of

h

h

defines an equivalence relation onMY . By [Z,b]Y we denote the equivalence claof (Z,b), and byMY /Y the set of equivalence classes. Note thatA+(Y ) is then thefree abelian group with basisMY /Y . Similarly, A+(Y ) is the free abelian group witbasisMY /Y , which is the set ofY -conjugacy classes of pairs(Z,b′) where Z � Y

andb′ ∈ B(Z). We denote the basis elements inA+ again by[Z,b′]Y .For b ∈ B(Z) andW � Z � Y , write rZ

W (b) as a linear combination of the basis ements inB(W), and call the (non-negative) coefficientm

(Z,b)(W,c) at c ∈ B(W) themultiplicity

of (W, c) in (Z,b). Then

(W, c) � (Z,b) :⇔ W � Z and m(Z,b)(W,c) > 0

defines a partial ordering onMY .

4.5. We define the induction morphism

b = bM :M+ → M

by

bY :M+(Y ) → M(Y), [U,m]Y → tYU (m).

Thenb is a morphism of Mackey functors (see [1, 3.1]). In fact, this is just the counthe adjunction mentioned at the end of 4.1.

A canonical induction formulaa :M → A+ is a morphism of restriction functors sucthatb ◦ a = idM .

4.6. Let p :M → A be a morphism ofZ-conjugation functors. Define

aY = a(M,A,p)Y :QM(Y) → QA+(Y ),

aY (m) = 1

|Y |∑

V <U<Y

|V |µ(V,U)[L, rU

V

(pU

(rYU (m)

))]Y,

whereµ denotes the Möbius function on the poset of subgroups ofX. Thena is a mor-phism ofQ-restriction functors.

Conversely, given a morphisma :M → A+ of Z-restriction functors, composition witthe morphism of conjugation functors

π :A+ → A,

πY

([U,m]Y) :=

{m, if Y = U,

0, otherwise,

gives a morphismp = π ◦ a :M → A of Z-conjugation functors, called theresidueof a.In fact, this defines mutually inverse isomorphisms between the set of morphismsa :M →A+ of Q-restriction functors and the set of morphismsp :M → A of Q-conjugation func-
tors (see [1, 6.1]).


-

for-

cht

4.7. Proposition (Boltje [1, 6.4]). The morphisma defined in4.6 is a canonical in-duction formula if and only if for allY � X and m ∈ M(Y) one haspY (m) − m ∈∑

Z<Y tYZ (M(Z)).

However, even if this condition is satisfied, the definition of the mapa contains denominators, so the image ofM(Y) underaY will in general not be contained inA+(Y ). Weneed an extra integrality condition:

4.8. Theorem (Boltje [1, 9.3]). Let Y � X, and assume that for a setπ of primes thefollowing condition holds:

(∗)π For all m ∈ M(Y) and all T � U � Y such thatU/T is a cyclicπ -group, and allb ∈ B(T ) which are fixed underU , the coefficients of the two elementspT (rY

T (m))

andrUT (pU(rY

U (m))) in A(T ) with respect to the basisB(T ) coincide.

Then

im(|Y |π ′ · a(M,A,p)

Y

) ⊆ A+(Y ).

In particular, if condition(∗)π is satisfied for allY � X and the setπ of all primes, thenadefines an integral canonical induction formula.

Now let us assume we are given a canonical induction formulaa :M → A+ with residuep :M → A satisfying the integrality condition(∗)π of Theorem 4.8 for the setπ of allprimes. Under suitable conditions, we will construct an integral canonical inductionmula a : M → A+ with residuep given by

p : M → A, pY = pY∩G for Y � X.

As in 4.6, we definea : M → A+ by

aY (m) = 1

|Y |∑

V <U<Y

|V |µ(V,U)[L, rU

V

(pU

(rYU (m)

))]Y, (3)

for Y � X andm ∈ M(Y ).

4.9. Theorem. Let(M,c, r, t) be aZ-Mackey functor forX, A ⊆ M a subfunctor satisfyingthe assumptions in4.2, and letp :M → A be a morphism of conjugation functors, suthat condition(∗)π from Theorem4.8 is satisfied for the setπ of all primes. Moreover, leG be a normal subgroup ofX such that the order ofG and the index inX are relativelyprime. DefineM as in Theorem2.4. Then(∗)π holds also for the Mackey functorM , itssubfunctorA and the morphismp, and the morphisma, defined by Eq.(3), is an integral
canonical induction formula.


,
Proof. Clearlyp is a morphism of conjugation functors, hence defininga by 4.6 certainlyyields a morphism of restriction functorsQM → QA+. Now, according to Proposition 4.7for a to be a canonical induction formula, we need to show that for anyY � X andm ∈QM(Y ), we have
m − pY (m) ∈∑Z<Y

tYZ(QM(Z)

).

Sincea is a canonical induction formula, again by 4.7, we know that

m − pY (m) = m − pY∩G(m) =∑

H<Y∩G

tY∩GH (mH )

for some elementsmH ∈ QM(H), thus

∑y∈[Y/Y∩G]

y(m − pY (m)

) =∑

y∈[Y/Y∩G]

∑H<Y∩G

ytY∩GH (mH )

=∑

H<Y∩G

∑y∈[Y/(Y∩G)H ]

ytY∩GH∩G(mH ) =

∑H<Y∩G

tYH (mH ).

Sincem is stable underY , and sincep is a morphism of conjugation functors, alsom −pY (m) is stable underY , so we obtain

[Y : Y ∩ G](m − pY (m)) ∈

∑H<Y∩G

tYH(QM(H)

)

and dividing by[Y : Y ∩ G] shows thata is indeed a canonical induction formula.To show integrality, we have to verify condition(∗)π from Theorem 4.8 for the setπ

of all primes. So letT � U � Y such thatU/T is cyclic, letb ∈ B(T )U , and letm ∈ M(Y )

be arbitrary. We need to show that the coefficients of the two elementspT (rYT (m)) and

rUT (pU (rY

U (m))) in A(T ) at the basis elementb coincide. We can write

b =∑

t∈T/S

tb1 (4)

for someb1 ∈ B(T ∩ G), whereS = StabT (b1).First we claim thatb1 is stable underU ∩ G. SetV = StabU∩G(b1), then clearly the

element

b′ =∑

u∈[U∩G/V ]ub1

is stable underU ∩ G, and so isb by assumption. Comparing the definition ofb′ and thedescription ofb in (4), both are linear combinations of basis elements inB(T ∩ G), so the
summands inb′ must all appear inb, andb must be a sum of conjugates ofb′. However, the


rs

o

own a

the

ula

cient

t

sum forb has[T : S] terms, and the sum forb′ has[U ∩ G : V ] terms, and these numbeare relatively prime, by hypothesis. ThusV = U ∩ G.

Now,

pT

(rYT (m)

) = pT ∩G

(rY∩GT ∩G(m)

)

and

rUT

(pU

(rYU (m)

)) = rU∩GT ∩G

(pU∩G

(rY∩GU∩G(m)

)).

Moreover,T ∩ G � U ∩ G andU ∩ G/T ∩ G is cyclic, hence the coefficients of the twelements atb1 ∈ B(T ∩G)U∩G coincide, since condition(∗)π holds forp. However, theseare the same coefficients as atb, so(∗)π holds also forp. �4.10. Remark. We just mention that there is an explicit formula fora without denomina-tors, but refer the interested reader to [1, formula (9.5a)]. Then it is easy to write dsimilar formula fora.

5. Extensions for Mackey functors

Throughout this section, we assume thatX is a finite group andG a normal subgroupof X such that(|G|, [X : G]) = 1. Let(M,c, r, t) be aZ-Mackey functor forX, let A ⊆ M

be a subfunctor, and letp :M → A be a morphism of conjugation functors satisfyinghypothesis of Proposition 4.7, and also satisfying condition(∗)π from Theorem 4.8 forthe setπ of all primes, so thatp corresponds to an integral canonical induction forma :M → A+. Let M and its subfunctorA be defined as in Theorem 2.4.

Our goal is to generalize the result of Proposition 3.8, so we want to give sufficonditions for the existence of anr-extension morphism

ext :M → M.

Our first result allows us to reduce the question to the subfunctorA:

5.1. Proposition. If there exists a morphismext∗ : A → M of restriction functors such tha

dY

(extY∗ (m)

) = m

for all m ∈ A(Y ) and allY � X, then

ext= b ◦ ext+ ◦ a

defines anr-extension morphismM → M , where

extY+([Z,m]Y

) := [Z,extZ∗ (m)

]Y

for [Z,m]Y ∈ A+(Y ) andY � X.


e

)

an

Proof. Clearly everything commutes with conjugations, and forZ,U � Y � X, andm ∈A(Z) one has

extU+(r+Y

U

([Z,m]Y)) = extU+

( ∑x∈[U\Y/Z]

[U ∩ xZ, r

xZU∩xZ

(xm

)]U

)

=∑

x∈[U\Y/Z]

[U ∩ xZ,extU∩xZ∗

(r

xZU∩xZ

(xm

))]U

=∑

x∈[U\Y/Z]

[U ∩ xZ, r

xZU∩xZ

(xextZ∗ (m)

)]U

= r+YU

([Z,extZ∗ (m)

]Y

) = r+YU

(extX+

([Z,m]Y))

.

Thus ext is a morphism of restriction functors, and

dY ◦ extY = rYY∩G ◦ bY ◦ extY+ ◦ aY = bY∩G ◦ extY∩G+ ◦ aY∩G ◦ rY

Y∩G

= bY∩G ◦ aY∩G = bY∩G ◦ aY∩G = idM(Y∩G). �We will have to assume that elements of the subfunctorA of M have extensions, mor

precisely, the need the following condition

5.2. For everyY � X and every basis elementb ∈ B(Y ∩ G) there exists a (distinguishedextensionb1 ∈ A(StabY (b)), such that for anyx ∈ X, the elementxb1 coincides with thedistinguished extension of the basis elementxb in B(xY ∩ G).

Additionally, we need an analog to the Adams operators, that is:

5.3. There is a morphismΨ :M → M of restriction functors, such thatdY (Ψ (m)) = m forall Y � X andm ∈ A(Y ), and such that we haveΨ (m1) = Ψ (m2) for any two extensionsm1,m2 ∈ M(Y) of m.

5.4. Theorem. Suppose that conditions5.2 and 5.3 are satisfied. Then there existsr-extension morphismext= extM,A : M → M .

Proof. Let Z � Y � X. Every basis element inB(Z) is of the formb′ = ∑z∈Z/S

zb, whereb ∈ B(Z ∩ G) andS = StabZ(b). Let b1 ∈ A(S) be the distinguished extension ofb to S,thenb = tZS (b1) is an extension ofb′ to Y . Now define

extY+ : A+(Y ) → M+(Y ),

extY+([

Z,b′]Y

) = [Z,Ψ

(b)]

Y.

Then clearly ext+ commutes with conjugations, and


sualropo-

lllag

nical

ene

r+YU

(extX+

([Z,b′]

Y

)) = r+YU

([Z,Ψ

(b)]

Y

)

=∑

x∈[U\Y/Z]

[xZ ∩ U, r

xZxZ∩U

xΨ(b)]

U

=∑

x∈[U\Y/Z]

[xZ ∩ U,Ψ

(r

xZxZ∩U

(x b

))]U

.

But rxZxZ∩U(xb) is an extension ofr

xZxZ∩U∩G(xb), hence the above equals

∑x∈[U\Y/Z]

extU+([

xZ ∩ U, rxZxZ∩U∩G

(x b

)]U

)

= extU+( ∑

x∈[U\Y/Z]

[xZ ∩ U, r

xZxZ∩U

(dxZ

(x b

))]U

)

= extU+( ∑

x∈[U\Y/Z]

[xZ ∩ U, r

xZxZ∩U

(xb′)]

U

)

= extU+(r+Y

U

([Z,b′]

Y

)).

Therefore ext:= b ◦ ext+ ◦ a is anr-extension morphism by Proposition 5.1.�5.5. Remark. If M = R is the character ring over the complex numbers with the uconjugation, restriction and induction maps, one can easily give another proof of Psition 3.8, using the canonical induction formula from [1]. Here the subfunctorA = Rab,where forY � X, we denote byRab(Y ) the free abelian group with basisY = Hom(Y,C×)

the group of linear characters ofY .

As an application, we will show that there is anr-extension morphism for the triviasource ring. LetF be a field of prime characteristicp, and letX be a finite group. Recathat anFX-moduleV is called atrivial source moduleif V is a direct summand ofpermutationFX-module. The images of the trivial sourceFX-modules in the Green rinof FX form a subringTF (X), called the trivial source ring (ofX overF ). ClearlyTF isa Mackey functor with the usual conjugation, restriction and induction maps. A canoinduction formula forTF has been given by Boltje in [3]. Here the subfunctorA is Rab

F ,whereRab

F (Y ) is the free abelian group with basisY(F ) = Hom(Y,F×). Of course weidentify ψ ∈ Y (F ) with the isomorphism class of the rank oneFY -moduleFψ , on whichy ∈ Y acts by multiplication withψ(y).

Now condition 5.2 just requires that everyψ ∈ Y ∩ G(F) can be extended to somψ ∈ Y (F ) in a distinguished way. In fact, we can always extendψ such that the restrictioof the extension to a complement ofY ∩ G in Y is trivial. This obviously satisfies th
conditions in 5.2.


ing

t thethankrticle.iding a

http://

30.

For the morphismΨ in 5.3 we will use the Adams operators for the trivial source ras defined by Boltje in [2]. We again choosen to be an integer such that

n ≡ 1 mod|G| and n ≡ 0 mod[X : G].By [2, Proposition 3.3], we have

Ψ nY

([Fψ ]) = [Fψn ],hence condition 5.3 is clearly satisfied. We obtain

5.6. Proposition. There exists anr-extension morphismext :TF → TF .

Acknowledgments

Most of the contents of this article are part of the author’s Diplom thesis written aUniversity of Augsburg under the supervision of Robert Boltje. The author wants toRobert Boltje for his advice and many ideas involved in the development of this aThanks go also to the referee for several valuable suggestions, especially for provshorter and more structural proof of Theorem 2.4.

References

[1] R. Boltje, A general theory of canonical induction formulae, J. Algebra 206 (1) (1998) 293–343.[2] R. Boltje, A characterization of Adams operations on representation rings, preprint, available at

math.ucsc.edu/~boltje, 2002.[3] R. Boltje, Linear source modules and trivial source modules, Proc. Sympos. Pure Math. 63 (1998) 7–[4] S. Bouc, Green Functors andG-Sets, Springer Lecture Notes in Math., vol. 1671, 1997.[5] I. Isaacs, Character Theory of Finite Groups, Academic Press, New York, 1976.
[6] J. Thevenaz, P. Webb, The structure of Mackey functors, Trans. Amer. Math. Soc. 347 (6) (1995) 1865–1961.

mackey functors and functorial extensions

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