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CHAPTER II

THE GROTHENDIECK GROUP K0

There are several ways to construct the Grothendieck group of a mathematicalobject. We begin with the group completion version, because it has been the mosthistorically important. After giving the applications to rings and topological spaces,we discuss -operations in4. In sections 6 and 7 we describe the Grothendieckgroup of an exact category, and apply it to the K-theory of schemes in

8. This

construction is generalized to the Grothendieck group of a Waldhausen categoryin9.

1. The Group Completion of a monoidBothK0(R) andK

0(X) are formed by taking the group completion of an abelianmonoidthe monoidP(R) of f.g. projectiveR-modules and the monoid VB(X) ofvector bundles over X, respectively. We begin with a description of this construc-tion.

Recall that an abelian monoidis a set Mtogether with an associative, commu-tative operation + and an additive identity element 0. A monoid map f: M Nis a set map such that f(0) = 0 and f(m + m

) = f(m) + f(m

). The most famousexample of an abelian monoid is N = {0, 1, 2,...}, the natural numbers with additiveidentity zero. IfA is an abelian group then not only is A an abelian monoid, butso is any additively closed subset ofAcontaining 0.

The group completion of an abelian monoid M is an abelian group M1M,together with a monoid map [ ]: M M1Mwhich is universal in the sense that,for every abelian group A and every monoid map : M A, there is a uniqueabelian group homomorphism : M1M A such that ([m]) = (m) for allm M.

For example, the group completion of N is Z. If A is an abelian group thenclearly A1A= A; ifM is a submonoid ofA (additively closed subset containing

0), then M1M is the subgroup ofA generated by M.Every abelian monoid Mhas a group completion. One way to contruct it is to

form the free abelian group F(M) on symbols [m],m M, and then factor out bythe subgroupR(M) generated by the relations [m + n] [m] [n]. By universality,if M N is a monoid map, the map M N N1N extends uniquely to ahomomorphism from M1M to N1N. Thus group completion is a functor fromabelian monoids to abelian groups. A little decoding shows that in fact it is leftadjoint to the forgetful functor, because of the natural isomorphism

Hom abelianmonoids

(M, A) =Homabeliangroups

(M1M, A).

Typeset by AMS-TEX

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II. THE GROTHENDIECK GROUP K0 3

Example 1.4. LetXbe a topological space. The set [X, N] of continuous mapsX N is a semiring under pointwise + and. The group completion of [X, N] isthe ring [X, Z] of all continuous maps X Z.

IfXis (quasi-)compact, [X, Z] is a free abelian group. Indeed, [X, Z] is a sub-group of the group Sof all bounded set functions from X to Z, and S is a freeabelian group (S is a Specker group; see [Fuchs]).

Example 1.5 (Burnside Ring). Let G be a finite group. The setM of (iso-morphism classes of) finite G-sets is an abelian monoid under disjoint union, 0being the empty set. Suppose there are c distinct G-orbits. Since every G-setis a disjoint union of orbits, M is the free abelian monoid Nc, a basis ofM beingthe classes of the c distinct orbits ofG. Each orbit is isomorphic to a coset G/H,where H is the stabilizer of an element, and G/H= G/H iffH and H are con-

jugate subgroups ofG, so c is the number of conjugacy classes of subgroups ofG.Therefore the group completion A(G) ofM is the free abelian group Zc, a basis

being the set of allc coset spaces [G/H].The direct product of two G-sets is again aG-set, soMis a semiring with 1 the

1-element G-set. ThereforeA(G) is a commutative ring; it is called the BurnsideringofG. The forgetful functor from G-sets to sets induces a map M N andhence an augmentation map : A(G) Z. For example, if G is cyclic of primeorderp, then A(G) is the ring Z[x]/(x2 =px) and x= [G] has(x) =p.

Example 1.6. (Representation ring). LetGbe a finite group. The set RepC(G)of finite-dimensional representations : G GLnC (up to isomorphism) is anabelian monoid under . By Maschkes Theorem, CGis semisimple andRepC(G) =Nr, where ris the number of conjugacy classes of elements ofG. Therefore the group

completionR(G) ofRepC(G) is isomorphic to Zr as an abelian group.The tensor product VCWof two representations is also a representation, so

RepC(G) is a semiring (the element 1 is the 1-dimensional trivial representation).Therefore R(G) is a commutative ring; it is called the Representation ringofG.For example, ifG is cyclic of prime order p then R(G) is isomorphic to the groupringZ[G], a subring ofQ[G] = Q Q(),p = 1.

Every representation is determined by its character : G C, and irreduciblerepresentations have linearly independent characters. ThereforeR(G) is isomorphicto the ring of all complex characters : G C, a subring ofM ap(G, C).

Definition. A (connected) partially ordered abelian group (A, P) is an abelian

groupA, together with a submonoid P ofA which generatesA (so A = P1

P) andP (P) = {0}. This structure induces a translation-invariant partial orderingonA: a bifa b P. Conversely, given a translation-invariant partial order onA, let P be{a A:a 0}. Ifa, b 0 then a + b a 0, so P is a submonoid ofA. IfP generatesAthen (A, P) is a partially ordered abelian group.

IfMis an abelian monoid, M1Mneed not be partially ordered (by the imageof M), because we may have [a] + [b] = 0 for a, b M. However, interestingexamples are often partially ordered. For example, the Burnside ring A(G) andRepresentation ring R(G) are partially ordered (by G-sets and representations).

When it exists, the ordering on M1M is an extra piece of structure. Forexample,Zr is the group completion of both Nr andM=

{0} {

(n1

,...,nr

)

Nr :n1,...,nr > 0}. However, the two partially ordered structures on Zr are different.

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4 II. THE GROTHENDIECK GROUP K0

EXERCISES

1.1The group completion of a non-abelian monoid Mis a group

M, together with

a monoid map M

Mwhich is universal for maps from M to groups. Show thatevery monoid has a group completion in this sense, and that ifM is abelian thenM=M1M. IfMis the free monoid on a set X, show that the group completionofM is the free group on the set X.Note: The results in this section fail in for non-abelian monoids. Proposition 1.1fails for the free monoid on X. Corollary 1.2 can also fail: an example of a cancel-

lation monoid Mwhich does not inject intoMwas given by Malcev in 1937.1.2 IfM=M1M2, show thatM1Mis the product group (M11 M1)(M12 M2).1.3 IfM is the filtered colimit of abelian monoids M, show that M1M is thefiltered colimit of the abelian groups M1 M.

1.4 Mayer-Vietoris for group completions. Suppose that a sequence L

M1M2 Nof abelian monoids is exact in the sense that whenever m1 M1 and

m2 M2 agree in N then m1 and m2 are the image of a common L. IfL is cofinal in M1, M2 and N, show that there is an exact sequence L

1L(M11 M1) (M12 M2) N1N, where the first map is the diagonal inclusion andthe second map is the difference map (m1, m2) m1 m2.1.5 Classify all abelian monoids which are quotients ofN ={0, 1, . . .} and showthat they are all finite. How many quotient monoids M = N/ of N have melements and group completionM= Z/nZ?1.6Here is another description of the Burnside ring A(G) of a finite groupG. Foreach subgroup H, and finite G-set X, let H(X) denote the cardinality ofX

H.

(a) Show that Hdefines a ring homomorphism A(G) Z, and = 1 .(b) Deduce that the productof the H(over the c conjugacy classes of sub-

groups) induces an injection ofA(G) into the product ringc

1 Z.(c) Conclude thatA(G) Q = c1 Q.

1.7 (T-Y Lam) Let : G H be a homomorphism of finite groups. Showthat the restriction functor from H-sets to G-sets (gx = (g)x) induces a ringhomomorphism : A(H) A(G). If X is a G-set, we can form the H-setHGX = H X/{(h,gx)(h(g), x))}. Show that HG induces a group ho-momorphism : A(G) A(H). If is an injection, show that the FrobeniusReciprocityformula holds: (

(x)

y) =x

(y) for all x

A(G),y

A(H).

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2. K0 of a ringLet R be a ring. The set P(R) of isomorphism classes of f.g. projective R-

modules, together with direct sum and identity 0, forms an abelian monoid. TheGrothendieck group ofR,K0(R), is the group completion P

1

P ofP(R).When R is commutative, K0(R) is a commutative ring with 1 = [R], because

the monoid P(R) is a commutative semiring with productR. This follows fromthe following facts: distributes over; PRQ=Q RP and PRR=P; ifP, Qare f.g. projective modules then so is PRQ (by Ex. I.2.7).

For example, let k be a field or division ring. Then the abelian monoid P(k) isisomorphic to N = {0, 1, 2, . . .}, soK0(k) = Z. The same argument applies to showthat K0(R) = Z for every local ring R by (I.2.2), and also for every PID (by theStructure Theorem for modules over a PID). In particular, K0(Z) = Z.

The Eilenberg SwindleI.2.8 shows why we restrict to finitely generated projec-tives. If we includedR, then the formulaP

R

=R would imply that [P] = 0

for every f.g. projective R-module, and therefore that K0(R) = 0.K0 is a functor from rings to abelian groups, and from commutative rings to

commutative rings. To see this, suppose that R Sis a ring homomorphism. ThefunctorRS: P(R) P(S) (sending P toPRS) yields a monoid map P(R) P(S), hence a group homomorphism K0(R) K0(S). If R, S are commutativerings thenRS: K0(R) K0(S) is a ring homomorphism, becauseS: P(R)P(S) is a semiring map:

(PRQ) RS=(PRS) S(Q RS).The free modules play a special role in understanding K0(R) because they are

cofinal in P(R). By Corollary 1.3 every element ofK0(R) can be written as [P]

[Rn] for somePandn. Moreover, [P] = [Q] inK0(R) iffP, Qare stably isomorphic:P Rm=Q Rm for some m. In particular, [P] = [Rn] iffP is stably free. Themonoid L of isomorphism classes of free modules is N iffR satisfies the InvariantBasis Property of Chapter I, 1. This yields the following information aboutK0(R).

Lemma 2.1. The monoid map N P(R) sending n to Rn induces a grouphomomorphismZ K0(R). We have:

(1) Z K0(R) is injective iffR satisfies the Invariant Basis Property (IBP);(2) Suppose thatR satisfies the IBP (e.g., R is commutative). Then

K0(R) = Z iff every f.g. projectiveR-module is stably free.

Example 2.1.1. Suppose that R is commutative, or more generally that thereis a ring map R F to a field F. In this caseZ is a direct summand ofK0(R),because the map K0(R)K0(F)=Z takes [R] to 1. A ring with K0(R) = Q isgiven in Exercise 2.12 below.

Example 2.1.2 (Simple rings). Consider the matrix ring R = Mn(F) over afield F. We saw in Example I.1.1 that every R-module is projective (because it isa sum of copies of the projective module V=Fn), and that length is an invariantof finitely generated R-modules. Thus length is an abelian group isomorphism

K0(Mn(F)) = Z sending [V] to 1. SinceR has lengthn, the subgroup ofK0(R) =

Z generated by the free modules has index n. In particular, the inclusion Z

K0(R)

of Lemma 2.1 does not split.

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Example 2.1.3. (Karoubi) We say a ring R is flasqueif there is an R-bimoduleM, f.g. projective as a right module, and a bimodule isomorphism : R M=M.If R is flasque then K0(R) = 0. This is because for every P we have a naturalisomorphismP

(P

RM)

=P

R(R

M)

=(P

RM).

IfR is flasque and the underlying right R-module structure on M is R, we saythat R is an infinite sum ring. The right module isomorphism R2=R underlyingmakesR a direct sum ring (Ex. I.1.7). The Cone Rings of Ex. I.1.8, and the ringsEndR(R

) of Ex. I.1.7, are examples of infinite sum rings, and hence flasque rings;see exercise 2.15.

IfR =R1 R2 then P(R)= P(R1) P(R2). As in Example 1.2, this impliesthat K0(R) =K0(R1) K0(R2). Thus K0 may be computed componentwise.

Example 2.1.4 (Semisimple rings). Let R be a semisimple ring, with sim-ple modules V1,...,Vr (see Ex. I.1.1). Schurs Lemma states that each Di =

HomR(Vi, Vi) is a division ring; the Artin-Wedderburn Theorem states that

R =Mn1(D1) Mnr(Dr),

where dimDi(Vi) = ni. By (2.1.2),K0(R)= K0(Mni(Di)) = Zr.

Another way to see that K0(R) = Zr is to use the fact that P(R) = Nr: the Krull-Schmidt Theorem states that every f.g. (projective) module M is V11 Vrrfor well-defined integers 1,...,r.

Example 2.1.5 (Von Neumann regular rings). A ring R is said to be vonNeumann regular if for every r

R there is an x

R such that rxr = r. Since

rxrx= rx, the element e = rx is idempotent, and the idealrR = eR is a projectivemodule. In fact, every finitely generated right ideal ofR is of the formeR for someidempotent, and these form a lattice. Declaringe e ifeR = eR, the equivalenceclasses of idempotents in R form a lattice: (e1 e2) and (e1 e2) are defined tobe the idempotents generating e1R + e2R and e1R e2R, respectively. Kaplanskyproved in [Kap58] that every projective R-module is a direct sum of the moduleseR. It follows that K0(R) is determined by the lattice of idempotents (modulo)inR. We will see several examples of von Neumann regular rings in the exercises.

Many von Neumann regular rings do not satisfy the (IBP), the ring E ndF(F)

of Ex. I.1.7 being a case in point.

We call a ring R unit-regularif for every rR there is a unit xR such thatrxr = rx. Every unit-regular ring is Von Neumann regular, has stable range 1, andsatisfies the (IBP) (Ex. I.1.13). In particular, ZK0(R). It is unknown whetheror not for every simple unit-regular ring R the groupK0(R) is strictly unperforated,meaning that wheneverx K0(R) andnx = [Q] for someQ, thenx = [P] for someP. Goodearl [Gdrl1] has given examples of simple unit-regular rings R in whichthe groupK0(R) is strictly unperforated, but has torsion.

An example of a von Neumann regular ring R having the IBP and stable range2, and K0(R) = Z Z/nis given in [MM82].

Suppose thatR is the direct limit of a filtered system{Ri}of rings. Then everyf.g. projectiveR-module is of the form P

iRiR for somei and some f.g. projective

Ri-module Pi. Any isomorphism PiRi R= PiRi R may be expressed using

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finitely many elements ofR, and hence Pi Ri Rj= PiRi Rj for some j. Thatis, P(R) is the filtered colimit of the P(Ri). By Ex. 1.3 we have

K0(R) =lim K0(Ri).This observation is useful when studying K0(R) of a commutative ring R, becauseR is the direct limit of its finitely generated subrings. As finitely generated com-mutative rings are noetherian with finite normalization, properties ofK0(R) maybe deduced from properties ofK0 of these nice subrings. IfR is integrally closedwe may restrict to finitely generated normal subrings, so K0(R) is determined byK0 of noetherian integrally closed domains.

Here is another useful reduction; it follows immediately from the observation thatifI is nilpotent (or complete) then idempotent lifting (Ex. I.2.2) yields a monoidisomorphismP(R)

=P(R/I). Recall that an ideal Iis said to becompleteif every

Cauchy sequence n=1 xn with xn In converges to a unique element ofI.Lemma 2.2. IfI is a nilpotent ideal ofR, or more generally a complete ideal,

then

K0(R) =K0(R/I).In particular, ifR is commutative thenK0(R) =K0(Rred).

Example 2.2.1 (0-dimensional commutative rings). Let R be a commu-tative ring. It is elementary that Rred is Artinian iff Spec(R) is finite and discrete.More generally, it is known (see Ex. I.1.13) that the following are equivalent:

(i) Rred is a commutative von Neumann regular ring (2.1.5);

(ii) R has Krull dimension 0;

(iii) X= Spec(R) is compact, Hausdorff and totally disconnected. (For example,to see that a commutative von Neumann regular R must be reduced, observethat ifr2 = 0 then r = rxr= 0.)

WhenR is a commutative von Neumann regular ring, the moduleseR are com-ponentwise free; Kaplanskys result states that every projective module is compo-nentwise free. By I.2, the monoid P(R) is just [X, N],X= Spec(R). By (1.4) thisyieldsK0(R) = [X, Z]. By Lemma 2.2, this proves

Pierces Theorem 2.2.2. For every 0-dimensional commutative ringR:

K0(R) = [Spec(R), Z].

Example 2.2.3 (K0 does not commute with infinite products). Let Rbe an infinite product of fields

Fi. Ris von Neumann regular, soX= Spec(R) is

an uncountable totally disconnected compact Hausdorff space. By Pierces Theo-rem,K0(R) =[X, Z]. This is contained in but not equal to the product

K0(Fi) =

Z.

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Rank andH0

Definition. When R is commutative, we write H0(R) for [Spec(R), Z], thering of all continuous maps from Spec(R) to Z. Since Spec(R) is quasi-compact,

we know by (1.4) that H0(R) is always a free abelian group. IfR is a noetherianring, then Spec(R) has only finitely many (sayc) components, and H0(R) = Zc. IfR is a domain, or more generally if Spec(R) is connected, then H0(R) = Z.

H0(R) is a subring ofK0(R). To see this, consider the submonoid L of P(R)consisting of componentwise free modules Rf. Not only isL cofinal in P(R), butL P(R) is a semiring map: RfRg=Rfg; by (1.3),L1Lis a subring ofK0(R).Finally,L is isomorphic to [Spec(R), N], so as in (1.4) we have L1L =H0(R). Forexample, Pierces theorem (2.2.2) states that if dim(R) = 0 then K0(R) =H0(R).

Recall from I.2 that the rank of a projective module gives a map from P(R)to [Spec(R), N]. Since rank(P Q) = rank(P) + rank(Q) and rank(P Q) =rank(P) rank(Q) (by Ex. I.2.7), this is a semiring map. As such it induces a ringmap

rank: K0(R) H0(R).Since rank(Rf) =ffor every componentwise free module, the composition H0(R) K0(R) H0(R) is the identity. Thus H0(R) is a direct summand ofK0(R).

Definition 2.3. The idealK0(R) of the ring K0(R) is defined as the kernel ofthe rank map. By the above remarks, there is a natural decomposition

K0(R) =H0(R)

K0(R).

We will see later (in4,6) thatK0(R) is a nil ideal. Since H0(R) is visibly areduced ring,K0(R) is the nilradical ofK0(R).

Lemma 2.3.1. IfR is commutative, let Pn(R) denote the subset of P(R) con-sisting of projective modules of constant rankn. There is a map Pn(R)K0(R)sendingP to[P][Rn]. This map is compatible with the stabilization map Pn(R) Pn+1(R) sendingP to P R, and the induced map is an isomorphism:

lim Pn(R) =K0(R).Proof. This follows easily from (1.3).

Corollary 2.3.2. LetR be a commutative noetherian ring of Krull dimensiond or more generally any commutative ring of stable ranged + 1 (Ex. I.1.5). For

everyn > d the above maps are isomorphisms: Pn(R) =K0(R).Proof. If P and Q are f.g. projective modules of rank > d, then by Bass

Cancellation (I.2.3b) we may conclude that

[P] = [Q] in K0(R) iff P=Q.

Here is another interpretation of

K0(R): it is it is the intersection of the kernels

ofK0(R)

K0(F) over all maps R

F, F a field. This follows from naturality

of rank and the observation thatK0(F) = 0 for every field F.

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This motivates the following definition for a noncommutative ring R: letK0(R)denote the intersection of the kernels of K0(R) K0(S) over all maps R S,where S is a simple artinian ring. If no such mapRSexists, we set

K0(R) =

K0(R). We define H0(R) to be the quotient of K0(R) byK0(R). When R iscommutative, this agrees with the above definitions of H0 andK0, because themaximal commutative subrings of a simple ring Sare fields.

H0(R) is a torsionfree abelian group for every ring R. To see this, note thatthere is a set Xof maps R Sx through which every other R S factors. Sinceeach K0(Sx) K0(S) is an isomorphism,K0(R) is the intersection of the kernelsof the maps K0(R) K0(Sx),x X. Hence H0(R) is the image ofK0(R) in thetorsionfree group

xXK0(Sx)

= x Z =M ap(X, Z).Example 2.4 (Whitehead group W h0). If R is the group ring Z[G] of a

groupG, the (zero-th)Whitehead groupW h0(G) is the quotient ofK0(Z[G]) by the

subgroupK0(Z) = Z. The augmentation map: Z[G] Z sending Gto 1 inducesa decomposition K0(Z[G])= Z W h0(G), and clearlyK0(Z[G]) W h0(G). Itfollows from a theorem of Swan ([Bass, XI(5.2)]) that ifG is finite thenK0(ZG) =W h0(G) and H0(ZG) =Z. I do not know whether or notK0(ZG) =W h0(G) forevery group.

The group W h0(G) arose in topology via the following result of C.T.C. Wall.We say that a CW complex X is dominated by a complex K if there is a mapf: K Xhaving a right homotopy inverse; this says that X is a retract ofK inthe homotopy category.

Theorem 2.4.1 (Wall Finiteness Obstruction). Suppose thatX is dom-

inated by a finite CW complex, with fundamental group G = 1(X). This datadetermines an element w(X) of W h0(G) such that w(X) = 0 iff X is homotopyequivalent to a finite CW complex.

Hattori-Stallings trace map

For any associative ring R, let [R, R] denote the subgroup ofR generated by theelements [r, s] =rs sr,r, s R.

For each n, the trace of an n n matrix provides an additive map from Mn(R)to R/[R, R] invariant under conjugation; the inclusion ofMn(R) in Mn+1(R) via

g

g 00 0 is compatible with the trace map. It is not hard to show that the traceMn(R) R/[R, R] induces an isomorphism:

Mn(R)/[Mn(R), Mn(R)] =R/[R, R].

IfP is a f.g. projective, choosing an isomorphismP Q =Rn yields an idempo-tent e in Mn(R) such that P =e(R

n) and Aut(P) =eMn(R)e. By Ex. I.2.3, anyother choice yields an e1 which is conjugate to e in some larger Mm(R). Thereforethe trace of an automorphism ofPis a well-defined element ofR/[R, R], indepen-dent of the choice ofe. This gives thetrace mapAut(P) R/[R, R]. In particular,the trace of the identity map ofP is the trace ofe; we call it the traceofP.

IfP is represented by an idempotent matrixfthenP

P is represented by theidempotent matrix e0 0f so the trace ofPP is trace(P)+trace(P). Therefore the

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trace is an additive map on the monoid P(R). The mapK0(R) R/[R, R] inducedby universality is called the Hattori-Stallings trace map, after the two individualswho first studied it.

When R is commutative, we can provide a direct description of the ring mapH0(R) Robtained by restricting the trace map to the subring H0(R) ofK0(R).Any continuous mapf: Spec(R) Z induces a decompositionR = R1 Rc byEx. I.2.4; the coordinate idempotentse1,...,ec are elements ofR. Since trace(eiR)is ei, it follows immediately that trace(f) is

f(i)ei. The identity trace(f g) =

trace(f)trace(g) which follows immediately from this formula shows that trace is aring map.

Proposition 2.5. IfR is commutative then the Hattori-Stallings trace factorsas

K0(R) rank H0(R) R.

Proof. The product over all p in Spec(R) yields the commutative diagram:K0(R)

K0(Rp)

trace

traceR

diagonalinclusion

Rp.

The kernel of the top arrow isK0(R), so the left arrow factors as claimed. Example 2.5.1 (Group rings). Let k be a commutative ring, and suppose

that R is the group ring kG of a group G. If g and h are congugate elements of

G then h g [R, R] because xgx1

g = [xg,x1

]. From this it is not hardto see that R/[R, R] is isomorphic to the free k-modulek[g] on the set G/ ofconjugacy classes of elements ofG. We write

trace(P) =

rP(g)[g].

The coefficients rP(g) of trace(P) are therefore functions on the set G/ for eachP.

IfGis finite, then any f.g. projective kG-modulePis also a projective k-module,and we may also form the trace map Autk(P) k and hence the characterP: G k by the formula P(g) = trace(g). Hattori proved that ifZG(g) denotesthe centralizer ofg

Gthen Hattoris formula holds:

(2.5.2) P(g) = |ZG(g)| rP(g1).Corollary 2.5.3. IfG is a finite group, the ringZGhas no idempotents except

0 and 1.

Proof. Let e be an idempotent element of ZG. P(1) is the rank of the Z-module P = eZG, which must be less than the rank|G| ofZG. Since rP(1)Z,this contradicts Hattoris formula P(1) = |G| rP(1).

Bass has conjectured that for every group Gand every f.g. projective ZG-moduleP we have rP(g) = 0 for g= 1 and rP(1) = rankZ(PZGZ). ForG finite, thisfollows from Hattoris formula and Swans theorem (cited in 2.3.2) thatK0 = W h0.See [Bass76].

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Example 2.5.4. Suppose that k is a field of characteristic 0 and that G is afinite group with c conjugacy classes, so that kG/[kG,kG]= kc. By Maschkestheorem, kG is a product of simple k-algebras: S1 Sc so kG/[kG,kG] iskc. By (2.1.4) K

0(kG)

=Zc. Hattoris formula (and some classical representation

theory) shows that the trace map from K0(kG) to kG/[kG,kG] is isomorphic tothe natural inclusion ofZc in kc.

Determinant

Suppose now thatR is a commutative ring. Recall from I.3 that the determinantof a fin. gen. projective module Pis an element of the Picard group Pic(R).

Proposition 2.6. The determinant induces a surjective group homomorphism

det: K0(R) Pic(R)

Proof. By the universal property ofK0, it suffices to show that det(P Q) =det(P) R det(Q). We may assume that P and Q have constant rank m and n,respectively. Thenm+n(P Q) is the sum over all i, j such that i+j =m +nof (iQ) (jP). If i > m or j > n we haveiP= 0 orjQ = 0, respectively.Hencem+n(P Q) = (mP) (nQ), as asserted.

Definition 2.6.1. Let SK0(R) denote the subset of K0(R) consisting of theclasses x= [P] [Rm], where Phas constant rank m andmP=R. This is thekernel of det:K0(R) Pic(R), by Lemma 2.3.1 and Proposition 2.6.

SK0(R) is an ideal ofK0(R). To see this, we use Ex. I.3.4: ifx = [P] [Rn]and Qhas ranknthen det(x

Q) = (det P)n(det Q)m(det Q)m =R.

Corollary 2.6.2. For every commutative ringR there is a surjective ring ho-momorphism with kernelSK0(R):

rank det: K0(R) H0(R) Pic(R)

Corollary 2.6.3. IfR is a 1-dimensional commutative noetherian ring, thenthe classification of f.g. projectiveR-modules in I.3.4 induces an isomorphism:

K0(R) =H0(R) Pic(R).

Morita Equivalence

We say that two rings R and S are Morita equivalent if mod-R and mod-S areequivalent as abelian categories. That is, if there exist additive functors T andU

mod-RT

U

mod-S

such that U T= idR and T U= idS. SetP = T(R) and Q = U(S); P is an R-Sbimodule and Q is a S-R bimodule via the maps R= EndR(R)

TEndS(P) andS= End

S(S)

U

End

R(Q). BothU T(R)

=P

SQ

=R andT U(S)

=Q

RP=

Sare bimodule isomorphisms. The following result is taken from [Bass, II.3].

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Define n: GLn(S/I) K0(R) by Milnor patching: n(g) is [P] [Rn], where Pis the projectiveR-module obtained by patching free modules alongg as in (I.2.6).The formulas of Ex. I.2.9 imply that n(g) = n+1

g0

01

andn(g)+n(h) =n(gh).

Therefore the{n} assemble to give a group homomorphism from GL(S/I) toK0(R). The following result now follows from (I.2.6) and Ex. 1.4.Theorem 2.9 (Mayer-Vietoris). Given a Milnor square as above, the se-

quence

GL(S/I) K0(R) K0(S) K0(R/I) K0(S/I)

is exact. The image ofis the double coset space

GL(S)\GL(S/I)/GL(R/I) = GL(S/I)/

wherex gxh forx GL(S/I), g GL(S) andh GL(R/I).Example 2.9.1. IfR is the coordinate ring of the node over a field k (I.3.10.2)

then K0(R)= Z k. If R is the coordinate ring of the cusp over k (I.3.10.1)then K0(R)= Z k. Indeed, the coordinate rings of the node and the cusp are1-dimensional noetherian rings, so 2.6.3 reduces the Mayer-Vietoris sequence to theUnits-Pic sequence I.3.10.

We conclude with a useful construction, anticipating several later developments.

Definition 2.10. Let T : P(R) P(S) be an additive functor, such as thebasechange or transfer of 2.8.1. P(T) is the category whose objects are triples

(P,,Q), where P, Q P(R) and : T(P) T(Q) is an isomorphism. A mor-phism (P,,Q) (P, , Q) is a pair ofR-module maps p: P P, q:Q Qsuch that T(p) = T(q). Anexact sequencein P(T) is a sequence

() 0 (P, , Q) (P,,Q) (P, , Q) 0

whose underlying sequences 0P P P 0 and 0Q QQ 0are exact. We define K0(T) to be the abelian group with generators the objects ofP(T) and relations:

(a) [(P,,Q)] = [(P, , Q)] + [(P, , Q)] for every exact sequence ();(b) [(P1, , P 2)] + [(P2, , P 3)] = [(P1,,P3)].

IfT is the basechange f, we write K0(f) for K0(T).

It is easy to see that there is a map K0(T) K0(R) sending [(P,,Q)] to[P][Q]. If T is a basechange functor f associated to f : R S, or moregenerally if the T(Rn) are cofinal in P(S), then there is an exact sequence:

(2.10.1) GL(S) K0(T) K0(R) K0(S).

The construction ofand verification of exactness is not hard, but lengthy enoughto relegate to exercise 2.17. If f : R

R/I then K

0(f) is the group K

0(I) of

Ex. 2.4; see Ex. 2.4(e).

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EXERCISES

2.1 Let R be a commutative ring. If A is an R-algebra, show that the functorR: P(A) P(R) P(A) yields a mapK0(A) ZK0(R) K0(A) makingK0(A)into a K0(R)-module. IfAB is an algebra map, show that K0(A) K0(B) isa K0(R)-module homomorphism.

2.2 Projection Formula. LetR be a commutative ring, and Aan R-algebra whichas an R-module is f.g. projective of rank n. By Ex. 2.1,K0(A) is aK0(R)-module,and the basechange map f: K0(R) K0(A) is a module homomorphism. Weshall write x fy for the product in K0(A) ofxK0(A) and y K0(R); this isan abuse of notation when Ais noncommutative.

(a) Show that the transfer map f: K0(A) K0(R) of Example 2.8.1 is aK0(R)-module homomorphism, i.e., that the projection formulaholds:

f(x

fy) =f(x)

y for every x

K0(A), y

K0(R).

(b) Show that both compositionsff andff are multiplication by [A].(c) Show that the kernels offf and ff

are annihilated by a power ofn.

2.3 Excision forK0. IfI is an ideal in a ring R, form the augmented ring R Iand let K0(I) = K0(R, I) denote the kernel ofK0(R I) K0(R).

(a) IfR S is a ring map sending I isomorphically onto an ideal ofS, showthat K0(R, I)= K0(S, I). Thus K0(I) is independent of R. Hint. Showthat GL(S)/GL(S I) = 1.

(b) IfI J= 0, show that K0(I+ J) =K0(I) K0(J).(c) Ideal sequence. Show that there is an exact sequence

GL(R) GL(R/I)

K0(I) K0(R) K0(R/I).(d) If R is commutative, use Ex. I.3.6 to show that there is a commutative

diagram with exact rows, the vertical maps being determinants:

GL(R) GL(R/I) K0(I) K0(R) K0(R/I) R (R/I) Pic(I) Pic(R) Pic(R/I).

2.4 K0I. IfI is a ring without unit, we define K0(I) as follows. LetR be a ringwith unit acting upon I, form the augmented ring R

I, and let K0(I) be the

kernel ofK0(R I) K0(R). ThusK0(R I) =K0(R) K0(I) by definition.(a) IfIhas a unit, show that RI=RIas rings with unit. SinceK0(RI) =

K0(R) K0(I), this shows that the definition of Ex. 2.3 agrees with theusual definition ofK0(I).

(b) Show that a mapI Jof rings without unit induces a map K0(I) K0(J)(c) LetM(R) denote the unionMn(R) of the matrix groups, where Mn(R)

is included inMn+1(R) as the matrices g

000

. M(R) is a ring without unit.

Show that the inclusion ofR = M1(R) in M(R) induces an isomorphism

K0(R) =K0(M(R)).(d) Ifk is a field, show that R = k

M

(k) is a von Neumann regular ring.Then show that H0(R) = Z and K0(R) = Z Z.

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(e) Iff : R R/I, show that K0(I) is the group K0(f) of 2.10. Hint: Usef0 : R I R and Ex. 2.3(c).

2.5 Radical ideals. Let Ibe a radical ideal in a ring R (see Ex. I.1.12, I.2.1).

(a) Show that K0(I) = 0, and that K0(R) K0(R/I) is an injection.(b) IfIis a complete ideal,K0(R) =K0(R/I) by Lemma 2.2. IfRis a semilocal

but not local domain, show that K0(R) K0(R/I) is not an isomorphism.2.6 Semilocal rings. A ring R is called semilocal if R/J is semisimple for someradical idealJ. Show that ifR is semilocal then K0(R) = Zn for some n >0.2.7 Show that iff :R Sis a map of commutative rings, then:

ker(f) contains no idempotents(= 0) H0(R) H0(S) is an injection.Conclude that H0(R) = H0(R[t]) = H0(R[t, t

1]).

2.8 Consider the following conditions on a ring R (cf. Ex. I.1.2):

(IBP)R satisfies the Invariant Basis Property (IBP);(PO) K0(R) is a partially ordered abelian group (see1);(III) For all n, ifRn=Rn P then P = 0.

Show that (III) (P O) (IBP). This implies thatK0(R) is a partially orderedabelian group ifR is either commutative or noetherian. (See Ex. I.1.4.)

2.9 Rim squares. Let G be a cyclic group of prime order p, and = e2i/p aprimitive pth root of unity. Show that the map ZG Z[] sending a generatorof G to induces an isomorphism K0(ZG)= K0(Z[]) and hence W h0(G)=Pic(Z[]). Hint: Form a Milnor square with ZG/I=Z, Z[]/I= Fp, and consider

the cyclotomic units u= i11, 1

i < p.

2.10LetR be a commutative ring. Prove that(a) If rank(x) > 0 for some xK0(R), then there is an n >0 and a fin. gen.

projective modulePso thatnx= [P]. (This says that the partially orderedgroupK0(R) is unperforated in the sense of [Gdearl].)

(b) IfP, Q are f.g. projectives such that [P] = [Q] in K0(R), then there is ann >0 such that P P=Q Q (ncopies ofP,n copies ofQ).

Hint: First assume that R is noetherian of Krull dimension d < , and use Bass-Serre Cancellation. In the general case, write R as a direct limit.

2.11 A (normalized) dimension function for a von Neumann regular ring R is agroup homomorphism d : K0(R)

R so that d(Rn) = n and d(P) > 0 for every

nonzero f.g. projective P.(a) Show that wheneverP Qany dimension function must haved(P) d(Q)(b) IfR has a dimension function, show that the formula (r) =d(rR) defines

a rank function : R [0, 1] in the sense of Ex. I.1.13. Then show thatthis gives a 1-1 correspondence between rank functions onR and dimensionfunctions on K0(R).

2.12 Let R be the union of the matrix rings Mn!(F) constructed in Ex. I.1.13.Show that the inclusion Z K0(R) extends to an isomorphism K0(R) = Q.2.13LetR be the infinite product of the matrix rings Mi(C),i = 1, 2,...

(a) Show that every f.g. projectiveR-moduleP is componentwise trivial in thesense that P= Pi, the Pi being f.g. projective Mi(C)-modules.

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(b) Show that the map fromK0(R) to the group

K0(Mi(C)) =

Z of infinitesequences (n1, n2,...) of integers is an injection, and that K0(R) = H0(R)is isomorphic to the group of bounded sequences.

(c) Show thatK0

(R) is not a free abelian group, even though it is torsionfree.Hint: Consider the subgroup Sof sequences (n1,...) such that the power of2 dividingni approaches asi ; show thatSis uncountable but thatS/2S is countable.

2.14BivariantK0. IfR and R are rings, let Rep(R, R) denote the set of isomor-

phism classes ofRR bimodulesMsuch thatM is finitely generated projective asa rightR-module. EachMgives a functor RM fromP(R) to P(R) sendingP toPRM. This induces a monoid map P(R) P(R) and hence a homomorphismfromK0(R) toK0(R

). For example, iff: R R is a ring homomorphism and Ris considered as an element ofRep(R, R), we obtain the mapRR. Show that:

(a) Every additive functorP(R) P(R

) is induced from anM inRep(R, R

);(b) If K0(R, R) denotes the group completion ofRep(R, R), then MR N

induces a bilinear map fromK0(R, R) K0(R, R) to K0(R, R);

(c) K0(Z, R) isK0(R), and ifM Rep(R, R) then the mapRM :K0(R) K0(R

) is induced from the product of (b).(d) If R and R are Morita equivalent, and P is the R-R bimodule giving

the isomorphism mod-R=mod-R, the class ofP in K0(R, R) gives theMorita isomorphism K0(R) =K0(R).

2.15In this exercise, we connect the definition 2.1.3 of infinite sum ring with a moreelementary description due to Wagoner. IfR is a direct sum ring, the isomorphismR2

=R induces a ring homomorphism

:R

R

EndR(R

2)

=EndR(R) =R.

(a) Suppose thatR is an infinite sum ring with bimodule M, and write r rfor the ring homomorphism REndR(M)= R arising from the left action ofRon the right R-module M. Show that r r =r for allr R.

(b) Conversely, suppose that R is a direct sum ring, and that R R is a ring

map so that r r =r for allr R. Show that R is an infinite sum ring.(c) (Wagoner) Show that the Cone Rings of Ex. I.1.8, and the rings EndR(R

)of Ex. I.1.7, are infinite sum rings. Hint: R= i=1 R, so a version of theEilenberg Swindle I.2.8 applies.

2.16 For any ring R, letJbe the (nonunital) subring ofE= EndR(R) of all fsuchthat f(R) is finitely generated (Ex. I.1.7). Show that M(R) Jn induces anisomorphismK0(R)=K0(J). Hint: For the projection en :R

Rn

, Jn =enEmaps onto Mn(R) = enEen with nilpotent kernel. But J= Jn.2.17This exercise shows that there is an exact sequence (2.10.1) whenT is cofinal.

(a) Show that [(P,,Q)]+[(Q, 1, P)] = 0 and [(P, T(), Q)] = 0 inK0(T).(b) Show that every element ofK0(T) has the form [(P,,R

n)].(c) Use cofinality and the maps() = [(Rn, , Rn)] of 2.10(b), from Aut(T Rn)

toK0(T), to show that there is a homomorphism K1(S) K0(T).(d) Use (a), (b) and (c) to show that (2.10.1) is exact atK0(T).(e) Show that (2.10.1) is exact at K0(R).

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3. K(X), KO(X) and KU(X) of a topological spaceLet Xbe a paracompact topological space. The sets VBR(X) and VBC(X) of

isomorphism classes of real and complex vector bundles over Xare abelian monoids

under Whitney sum. By Construction I.4.2, they are commutative semirings under. Hence the group completions K O(X) ofVBR(X) andKU(X) ofVBC(X) arecommutative rings with identity 1 = [T1]. If the choice ofR or C is understood,we will just write K(X) for simplicity.

Similarly, the setVBH(X) is an abelian monoid under , and we writeKSp(X)for its group completion. Although it has no natural ring structure, the constructionof Ex. I.4.18 endowsK Sp(X) with the structure of a module over the ring K O(X).

For example if denotes a 1-point space then K() = Z. IfX is contractible,then KO(X) =K U(X) =Z by I.4.6.1. More generally, K(X)=K(Y) wheneverX andYare homotopy equivalent by I.4.6.

The functorK(X) is contravariant inX. Indeed, iff: Y Xis continuous, theinduced bundle constructionE fEyields a monoid mapf: VB(X) VB(Y)and hence a ring homomorphism f: K(X) K(Y). By the Homotopy InvarianceTheorem I.4.5, the map f depends only upon the homotopy class off in [Y, X].

For example, the universal map X induces a ring map from Z =K() intoK(X), sending n >0 to the class of the trivial bundle Tn over X. IfX=thenany point of X yields a map X splitting the universal map X . Thusthe subring Z is a direct summand of K(X) when X=. (But if X = thenK() = 0.) For the rest of this section, we will assume X= in order to avoidawkward hypotheses.

The trivial vector bundles Tn and the componentwise trivial vector bundles Tf

form sub-semirings ofVB(X), naturally isomorphic to N and [X, N], respectively.When X is compact, the semirings N and [X, N] are cofinal in VB(X) by theSubbundle Theorem I.4.1, so by Corollary 1.3 we have subrings

Z [X, Z] K(X).

More generally, it follows from Construction I.4.2 that dim: VB(X)[X, N] is asemiring map splitting the inclusion [X, N]VB(X). Passing to Group Comple-tions, we get a natural ring map

dim: K(X) [X, Z]

splitting the inclusion of [X, Z] inK(X).

The kernel of dim will be written asK(X), or asKO(X) orKU(X) if we wishto emphasize the choice ofR orC. ThusK(X) is an ideal inK(X), and there is anatural decomposition

K(X) =K(X) [X, Z].Warning. IfX is not connected, our group

K(X) differs slightly from the no-

tation in the literature. However, most applications will involve connected spaces,where the notation is the same. This will be clarified by Theorem 3.2 below.

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d (mod 8) 1 2 3 4 5 6 7 8

KO(Sd) Z/2 Z/2 0 Z 0 0 0 Z

KS p(Sd) 0 0 0 Z Z/2 Z/2 0 Z

Both of the idealsKO(Sd) andKU(Sd) are of square zero.Remark 3.1.2. The complexification maps Z =KO(S4k) KU(S4k) =

Z are multiplication by 2 if k is odd, and by 1 if k is even. (The forgetful

mapsKU(S4k)KO(S4k) have the opposite parity.) Similarly, the maps Z=KU(S4k) KS p(S4k)= Z are multiplication by 2 if k is odd, and by 1 if k iseven. (The forgetful maps KSp(S4k) KU(S4k) have the opposite parity.) Thesecalculations are taken from [MTo, IV.5.12 and IV.6.1].

Let B O (resp. BU, B Sp) denote the direct limit of the Grassmannians Grassn.

As noted after (3.1.0) and in I.4.10.1, the notation reflects the fact that GrassnisOn (resp.Un,S pn), and the maps in the direct limit correspond to the standardinclusions, so that we have BO O = On, BU U = Un and BS pSp =

Spn.

Theorem 3.2. For every compact spaceX:

KO(X) =[X, Z BO] andKO(X) =[X,BO];KU(X) =[X, Z BU] andKU(X) =[X,BU];

KSp(X) =[X, Z BS p] and KSp(X) =[X,BSp].

In particular, the homotopy groups n(BO) =KO(Sn), n(BU) =KU(Sn) andn(BS p) = KS p(S

n) are periodic and given in Example 3.1.1.

Proof. IfXis compact then we have [X,BO] = lim [X,BOn] and similarly for[X,BU] and [X,BSp]. The result now follows from Theorem 3.1 for connected X.For non-connected compact spaces, we only need to show that the maps [X,BO] KO(X), [X,BU]KU(X) and [X,BSp] KS p(X) of Theorem 3.1 are stillisomorphisms.

SinceXis compact, every continuous mapX Z is bounded. Hence the rank ofevery vector bundleEis bounded, say rank E nfor somen N. Iff=nrank Ethen F =E

Tf has constant rank n, and [E]

rank E= [F]

n. Hence every

element ofK(X) comes from some VBn(X).To see that these maps are injective, suppose thatE, F VBn(X) are such that

[E] n = [F] n. By (1.3) we have E Tf = F Tf in VBn+f(X) for somef [X, N]. Iff p, pN, then adding Tpf yieldsE Tp =F Tp. Hence Eand Fagree in VBn+p(X).

Definition 3.2.1 (K0). For every paracompactXwe write KO0(X) for [X, ZBO],KU0(X) for [X, Z BU] and KSp0(X) for [X, Z BS p]. By Theorem 3.2,we have KO0(X)= KO(X), KU0(X)= KU(X) and KSp0(X)= KSp(X) forevery compact X. Similarly, we shall write

KO0(X),

KU0(X) and KS p0(X) for

[X,BO], [X,BU] and [X,BSp]. When the choice ofR,C or H is clear, we will just

writeK0(X) andK0(X).

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IfY is a subcomplex ofX, we define relative groups K0(X, Y) = K0(X/Y)/Z

and

K0(X, Y) =

K0(X/Y).

WhenXis paracompact but not compact, K0(X) and K(X) are connected by

stabilization and the map (3.1.0):KO(X) lim VBn(X) =lim[X,BOn] [X,BO] =KO0(X)and similarly forKU(X) and KSp(X). We will see in Example 3.7.2 and Ex. 3.2that the left map need not be an isomorphism. Here is an example showing thatthe right map need not be an isomorphism either.

Example 3.2.2. (McGibbon) LetXbe the infinite bouquet of odd-dimensionalspheres S3 S5 S7 . By homotopy theory, there is a map f: X BO3whose restriction toS2p+1 is essential of orderp for each odd prime p. IfEdenotesthe 3-dimensional vector bundle fE3 on X, then the class of f in lim

[X,BOn]

corresponds to [E]3 KO(X). In fact, since X is a suspension, we havelim[X,BOn] =KO(X) by Ex. 3.8.

Each (n + 3)-dimensional vector bundleETn is nontrivial, since its restrictionto S2p+1 is nontrivial whenever 2p > n+ 3 (again by homotopy theory). Hence

[E] 3 is a nontrivial element ofKO(X). However, the corresponding elementinKO0(X) = [X,BO] is zero, because the homotopy groups ofBO have no oddtorsion.

Proposition 3.3. If Y is a subcomplex of a CW complex X, the followingsequences are exact:

K0(X/Y) K0(X) K0(Y),K0(X, Y) K0(X) K0(Y).

Proof. Since Y X is a cofibration, we have an exact sequence [X/Y,B][X, B] [Y, B] for every connected space B; see III(6.3) in [Wh]. This yieldsthe first sequence (B is BO, BU or BS p). The second follows from this and the

classical exact sequenceH0(X/Y; Z) H0(X; Z) H0(Y; Z). Change of structure field 3.4. IfX is any space, the monoid (semiring)

map VBR(X) VBC(X) sending [E] to [E C] (see Ex. I.4.5) extends by uni-versality to a ring homomorphism KO(X)KU(X). For example,KO(S8n)KU(S8n) is an isomorphism butKO(S8n+4) =Z embeds inKU(S8n+4) = Z as asubgroup of index 2.

Similarly, the forgetful map VBC(X) VBR(X) extends to a group homomor-phism KU(X) KO(X). As dimR(V) = 2 dimC(V), the summand [X, Z] ofKU(X) embeds as 2[X, Z] in the summand [X, Z] ofKO(X). SinceEC =EEas real vector bundles (by Ex. I.4.5), the composition KO(X) KU(X) KO(X)is multiplication by 2. The composition in the other direction is more complicated;

see Exercise 3.1. For example, it is the zero map onKU(S8n+4) = Z but is multi-plication by 2 on

KU(S8n) = Z.

There are analogous maps KU(X) KS p(X) and KS p(X) KU(X), whoseproperties we leave to the exercises.

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Higher TopologicalK-theory

Once we have a representable functor like K0, standard techniques in infinite loopspace theory all us to expand it into a generalized cohomology theory. Rather than

get distracted by infinite loop spaces now, we choose to adopt a rather pedestrianapproach, ignoring the groups Kn forn >0. For this we use the suspensions SnXofX, which are all connected paracompact spaces.

Definition 3.5. For each integer n > 0, we defineKOn(X) and KOn(X)by:

KOn(X) =KO0(SnX) = [SnX,BO]; KOn(X) =KOn(X) KO(Sn).Replacing O by U yields definitionsKU

n(X) =KU0(SnX) = [SnX,BU] and

KUn(X) =KUn(X)KU(Sn); replacing O by Sp yields definitions forKS pn(X) and KS pn(X). When the choice ofR, C orH is clear, we shall drop

the O, U and Sp,, writing simplyKn(X) and Kn(X).We shall also define relative groups as follows. IfY is a subcomplex ofX, and

n >0, we set Kn(X, Y) =Kn(X/Y).Based maps 3.5.1. Note that our definitions do not assume Xto have a base-

point. If X has a nondegenerate basepoint and Y is an H-space with homotopyinverse (such asBO,BUorBSp), then the group [X, Y] is isomorphic to the group0(Y) [X, Y], where the second term denotes homotopy classes ofbasedmapsfromX toY; see pp. 100 and 119 of [Wh]. For such spaces Xwe can interpret theformulas for KOn(X), KUn(X) and KS pn(X) in terms of based maps, as isdone in [Atiyah, p.68].

IfX denotes the disjoint union ofXand a basepoint, then we have the usualformula for an unreduced cohomology theory: Kn(X) =K(Sn(X)). This easilyleads (see Ex. 3.11) to the formulas for n 1:

KOn(X) =[X, nBO], KUn(X) =[X, nBU] and K Spn(X) =[X, nBS p].

Theorem 3.6. If Y is a subcomplex of a CW complex X, we have the exactsequences (infinite to the left):

K2(Y) K1(X/Y) K1(X) K1(Y) K0(X/Y) K0(X) K0(Y), K2(Y)K1(X, Y)K1(X)K1(Y)K0(X, Y)K0(X)K0(Y).

Proof. Exactness at K0(X) was proven in Proposition 3.3. The mapping conecone(i) of i: Y X is homotopy equivalent to X/Y, and j: X cone(i) inducescone(i)/X SY. This gives exactness at K0(X, Y). Similarly, cone(j) SYand cone(j)/cone(i) SX, giving exactness at K1(Y). The long exact sequencesfollows by replacing Y X bySY SX.

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Characteristic Classes 3.7. The total Stiefel-Whitney class w(E) of a realvector bundle Ewas defined in chapter I,4. By (SW3) it satisfies the productformula: w(E F) =w(E)w(F). Therefore if we interpret w(E) as an element of

the abelian group Uof all formal sums 1+a1+ in H

(X; Z/2) we get a get a grouphomomorphismw: KO(X) U. It follows that each Stiefel-Whitney class inducesa well-defined set mapwi: KO(X) Hi(X; Z/2). In fact, sincew vanishes on eachcomponentwise trivial bundleTf it follows thatw([E] [Tf]) = w(E). Hence eachStiefel-Whitney class wi factors through the projection K O(X) KO(X).

Similarly, the total Chern class c(E) = 1 +c1(E) + satisfies c(E F) =c(E)c(F), so we may think of it as a group homomorphism from KU(X) to the

abelian group U of all formal sums 1 + a2 +a4 + in H(X; Z). It followsthat the Chern classes ci(E) H2i(X; Z) of a complex vector bundle define setmaps ci: KU(X) H2i(X; Z). Again, since c vanishes on componentwise trivialbundles, each Chern class ci factors through the projection KU(X)

KU(X).Example 3.7.1. For even spheres the Chern class cn: KU(S2n)H2n(Sn; Z)is an isomorphism. We will return to this point in Ex. 3.6 and in4.

Example 3.7.2. The map lim[R P, BOn]KO(R P) of (3.1.0) cannot beonto. To see this, consider the element = 1 [E1] ofKO(R P), whereE1 is thecanonical line bundle. Sincew() =w(E1) = 1 +x we have w() = (1 + x)1 =i=0 x

i, and wi() = 0 for every i 0. Axiom (SW1) implies that cannot equal[F] dim(F) for any bundle F.

Similarly, lim[C P, BUn]

KU(C P) cannot be onto; the argument is similar,

again using the canonical line bundle: ci(1 [E1]) = 0 for every i 0.

EXERCISES

3.1 Let X be a topological space. Show that there is an involution of VBC(X)sending [E] to the complex conjugate bundle [E] of Ex. I.4.6. The correspond-ing involution c on KU(X) can be nontrivial; use I(4.9.2) to show that c is mul-

tiplication by1 onKU(S2) = Z. (By Bott periodicity, this implies that cis multiplication by (1)k onKU(S2)= Z.) Finally, show that the compositeKU(X) KO(X) KU(X) is the map 1 + c sending [E] to [E] + [E].3.2IfXiis the disjoint union of spacesXi, show thatK(Xi) =

K(Xi). Then

construct a spaceXsuch that the map lim

VBn(X)K(X) of (3.1.0) is not onto.3.3 External products. Show that there is a bilinear map K(X1)K(X2)

K(X1 X2) for everyX1 andX2, sending [E1] [E2] to [1 (E1) 2 (E2)], wherepii: X1 X2 Xi is the projection. Then show that if X1 = X2 = X thecomposition with the diagonal map : K(XX) K(X) yields the usual productin the ring K(X), sending [E1] [E2] to [E1 E2].3.4 Recall that the smash product X Y of two based spaces is the quotientX Y /X Y, where X Y is the union ofX {} and{} Y. Show that

Kn(X Y) =

Kn(X Y)

Kn(X)

Kn(Y).

3.5 Show that K U2(

)

KUn(X)

KUn2(X) induces a periodicity iso-

morphism: KUn(X) KUn2(X) for all n. Hint: S2 SnX Sn+2X.

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3.6 Let Xbe a finite CW complex with only even-dimensional cells, such as CPn.Show that KU(X) is a free abelian group on the set of cells of X, and thatKU(SX) = Z, so that KU1(X) = 0. Then use Example 3.7.1 to show that

the total Chern class injects the groupKU(X) into H2i(X; Z).3.7 Chern character forCPn. LetE1be the canonical line bundle on CP

n, and letxdenote the class [E1] 1 inK U(CPn). Use Chern classes and the previous exerciseto show that{1, [E1], [E1 E1], . . . , [En1 ]}, and hence{1, x , x2, . . . , xn}, forms abasis of the free abelian group KU(CPn). Then show thatxn+1 = 0, so that thering KU(CPn) is isomorphic to Z[x]/(xn+1). We will see in Ex. 4.11 below thatthe Chern character ch maps the ring KU(CPn) isomorphically onto the subringZ[t]/(tn+1) ofH(CPn; Q) generated by t= ec1(x) 1.3.8Consider the suspension X=SYof a paracompact space Y. Use Ex. I.4.16 to

show that lim[X,BOn] =KO(X).

3.9 IfX is a finite CW complex, show by induction on the number of cells thatbothK O(X) andKU(X) are finitely generated abelian groups.

3.10 Show that KU(RP2n) = KU(RP2n+1) = ZZ/2n. Hint: Try the totalStiefel-Whitney class, using 3.3.

3.11 Let X be a compact space with a nondegenerate basepoint. Show thatKOn(X) =[X, nBO] =[X, n1O] andK Un(X) =[X, nBU] =[X, n1U]for alln 1. In particular,K U1(X) =[X, U] and KO1(X) =[X, O].3.12 Let X be a compact space with a nondegenerate basepoint. Show thatthe homotopy groups of the topological groups GL(RX) = Hom(X,GL(R)) andGL(CX) = Hom(X,GL(C)) are (for n >0):

n1GL(RX) = K On(X) and n1GL(C

X) =KUn(X).

3.13IfE X is a complex bundle, there is a quaternionic vector bundleEH Xwith fibers ExC H, as in Ex. I.4.5; this induces the map KU(X) KS p(X)mentioned in 3.4. Show that EH X, considered as a complex vector bundle, isisomorphic to the Whitney sum E E. Deduce that the compositionKU(X)KS p(X) KU(X) is multiplication by 2.3.14 Show that H CH is isomorphic to H H as an H-bimodule, on generators1 1 j j. This induces a natural isomorphismVC H =VVof vector spacesoverH. IfE Xis a quaternionic vector bundle, with underlying complex bundleuE X, show that there is a natural isomorphism (uE)H=EE. Conclude thatthe compositionK Sp(X) KU(X) KSp(X) is multiplication by 2.3.15Let Ebe the complex conjugate bundle of a complex vector bundle EX;see Ex. I.4.6. Show that EH=EHas quaternionic vector bundles. This shows thatKU(X) KS p(X) commutes with the involution c of Ex. 3.1.

Using exercises 3.1 and 3.14, show that the composition K Sp(X) KO(X) KS p(X) is multiplication by 4.

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4. Lambda and Adams OperationsA commutative ringKis called a-ringif we are given a family of set operations

k: K K fork 0 such that for all x, y K: 0(x) = 1 and 1(x) = x for allx K; k(x + y) = ki=0i(x)ki(y) = k(x) + k1(x)1y+ + k(y).This last condition is equivalent to the assertion that there is a group homo-

morphism t from the additive group ofKto the multiplicative group W(K) =1 + tK[[t]] given by the formula t(x) =

k(x)tk.

Example 4.1.1 (Binomial Rings). . The integers Z and the rationals Q are

-rings with k(n) =nk

. IfK is any Q-algebra, we define

xk

= x(x1)(xk+1)k!

forx Kand k 1; again the formula k(x) = (xk) makes Kinto a -ring.More generally, a binomial ringis a subring K of a Q-algebraKQ such that for

all x

K and k

1,

xk K. We make a binomial ring into a -ring by settingk(x) = xk. If K is a binomial ring then formally t is given by the formula

t(x) = (1 + t)x. For example, ifXis a topological space, then the ring [X, Z] is a

-ring withk(f) =fk

, the function sending xto

f(x)k

.

The notion of -semiring is very useful in constructing -rings. Let M be asemiring (see1); we know that the group completion M1M of M is a ring.We call M a semiringif it is equipped with operations k: M M such that0(x) = 1,1(x) =x and k(x + y) =

i(x)ki(y).

IfM is a -semiring then the group completionK=M1M is a -ring. To seethis, note that sendingx Mto the power series

k(x)tk defines a monoid map

t: M

1 + tK[[t]]. By universality ofK, this extends to a group homomorphism

t fromKto 1 + tK[[t]], and the coefficients oft(x) define the operations k(x).

Example 4.1.2 (Algebraic K0). Let R be a commutative ring and set K=K0(R). If P is a f.g. projective R-module, consider the formula

k(P) = [kP].The decompositionk(P Q)= (iP) (k1Q) given in ch.I,3 shows thatP(R) is a -semiring. Hence K0(R) is a -ring.

Since rank(kP) = rankPk , it follows that rank: K0(R) [Spec(R), Z] is amorphism of-rings, and hence thatK0(R) is a -ideal ofK0(R).

Example 4.1.3 (Topological K0). Let Xbe a topological space and let Kbe eitherKO(X) orKU(X). IfE Xis a vector bundle, let k(E) be the exteriorpower bundle

kEof Ex. I.4.3. The decomposition of

k(E

F) given in Ex. I.4.3

shows that the monoid VB(X) is a -semiring. Hence KO(X) and KU(X) are-rings, and KO(X) KU(X) is a morphism of-rings.

Since dim(kE) = dimEk , it follows that KO(X) [X, Z] and KU(X)[X, Z] are -ring morphisms, and thatKO(X) andKU(X) are -ideals.

Example 4.1.4 (Representation Ring). Let G be a finite group, and con-sider the complex representation ring R(G) constructed in Example 1.6. R(G) isthe group completion of RepC(G), the semiring of finite dimensional representa-tions ofG; as an abelian group R(G)= Zc, where c is the number of conjugacyclasses of elements inG. The exterior powers i(V) of a representation V are alsoG-modules, and the decomposition of k(V

W) as complex vector spaces used in

(4.1.2) shows that RepC(G) is a -semiring. Hence R(G) is a -ring.

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Ifd = dimC(V) then dimC(kV) =

dk

, so dimC is a -ring map from R(G) to

Z. The kernel

R(G) of this map is a -ideal ofR(G).

Example 4.1.5. Let Xbe a scheme, or more generally a locally ringed space

(Ch. I,5). We will define a ringK0(X) in7 below, using the category VB(X).As an abelian group it is generated by the classes of vector bundles on X. We willsee in8 that the operations k[E] = [kE] are well-defined on K0(X) and make itinto a -ring. (The formula fork(x + y) will follow from Ex. I.5.4.)

Positive structures

Not every -ring is well-behaved. In order to avoid pathologies, we introducea further condition, satisfied by the above examples: the -ring K must have apositive structure and satisfy the Splitting Principle.

Definition 4.2.1. By a positive structure on a -ring K we mean: 1) a -

subringH0 ofKwhich is a binomial ring; 2) a -ring surjection : K H0 whichis the identity on H0 ( is called the augmentation); and 3) a subset P K (thepositiveelements), such that

(1) N = {0, 1, 2, } is contained in P.(2) P is a -sub-semiring ofK. That is,Pis closed under addition, multipli-

cation, and the operations k.(3) Every element of the kernelKof can be written aspqfor somep,q P.(4) IfpP then (p) =nN. Moreover,i(p) = 0 for i > n and n(p) is a

unit ofK.

Condition (2) states that the group completion P

1

P ofP is a -subring ofK;by (3) we have P1P = Z K. By (4), (p) > 0 for p= 0, so P (P) = 0;thereforeP1Pis a partially ordered abelian group in the sense of1. An element P with () = 1 is called a line element; by (4), 1() = and is a unit ofK.That is, the line elements form a subgroup L of the units ofK.

The -rings in examples (4.1.2)(4.1.5) all have positive structures. The-ringK0(R) has a positive structure with

H0 =H0(R) = [Spec(R), Z] and P = {[P] : rank(P) is constant};

the line elements are the classes of line bundles, so L = Pic(R). Similarly, the

-ringsKO(X) and KU(X) have a positive structure in which H0 is H0(X, Z) =[X, Z] andP is {[E] : dim(E) is constant}, as long as we restrict to compact spacesor spaces with 0(X) finite, so that (I.4.1.1) applies. Again, line elements are theclasses of line bundles; for KO(X) and KU(X) we have L = H1(X; Z/2) andL= H2(X; Z), respectively. For R(G), the classes [V] of representations V are thepositive elements; H0 is Z, and L is the set of 1-dimensional representations ofG.Finally, ifXis a scheme (or locally ringed space) then in the positive structure onK0(X) we have H

0 =H0(X; Z) andP is {[E] : rank(E) is constant}; see I.5.1. Theline bundles are again the line elements, so L = Pic(X) = H1(X, OX) by I.5.10.1.

There is a natural group homomorphism det from K toL, which vanishes onH0. Ifp

Pwe define det(p) =n(p), where (p) =n. The formula for n(p + q)

and the vanishing ofi(p) for i > (p) imply that det: P L is a monoid map,

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i.e., that det(p+q) = det(p) det(q). Thus det extends to a map from P1P to L.As det(n) = (nn) = 1 for every n 0, det(Z) = 1. By (iii), defining det(H0) = 1extends det to a map fromK toL. WhenK isK0(R) the map det was introducedin

2. ForK O(X), det is the first Stiefel-Whitney class; for K U(X), det is the first

Chern class.Having described what we mean by a positive structure on K, we can now state

the Splitting Principle.

Definition 4.2.2. TheSplitting Principlestates that for every positive elementp in K there is a extension K K (of-rings with positive structure) such thatpis a sum of line elements in K.

The Splitting Principle for KO(X) and KU(X) holds by Ex. 4.12. Using alge-braic geometry, we will show in 8.8 that the Splitting Principle holds for K0(R) aswell asK0 of a scheme. The Splitting Principle also holds for R(G); see [AT, 1.5].

The importance of the Splitting Principle lies in its relation to special -rings, anotion we shall define after citing the following motivational result from [FL, ch.I].

Theorem 4.2.3. IfK is a-ring with a positive structure, the Splitting Prin-ciple holds iffKis a special-ring.

In order to define special -ring, we need the following technical example:

Example 4.3 (Witt Vectors). For every commutative ring R, the abeliangroup W(R) = 1 +tR[[t]] has the structure of a commutative ring, natural in R;W(R) is called the ring of (big)Witt vectorsofR. The multiplicative identity of theringW(R) is (1 t), and multiplicationis completely determined by naturality,formal factorization of elements ofW(R) asf(t) = i=1(1 riti) and the formula:

(1 rt) f(t) = f(rt).It is not hard to see that there are universal polynomials Pn in 2n variables

so that:

(

aiti) (

bjt

j) =

cntn, with cn= Pn(a1, . . . , an; b1, . . . , bn).

Grothendieck observed that there are operations k on W(R) making it into a-ring; they are defined by naturality, formal factorization and the formula

k

(1 rt) = 0 for all k 2.Another way to put it is that there are universal polynomials Pn,k such that:

k(

aiti) =

bnt

n, with bn= Pn,k(a1,...,ank).

Definition 4.3.1. A special -ring is a -ring Ksuch that the group homo-morphismt fromK toW(K) is a -homomorphism. In effect, a special -ring isa -ringK such that

k(1) = 0 fork= 0, 1 k(xy) isPk(1(x),...,k(x); 1(y),...,k(y)), and

n

(k

(x)) = Pn,k(1

(x),...,nk

(x)).

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Example 4.3.2. The formula n(s1) =sn defines a special -ring structure onthe polynomial ringU= Z[s1,...,sn,...]; see [AT]. Clearly ifx is any element in anyspecial -ringKthen the map U Ksending sn ton(x) is a -homomorphism.The -ringUcannot have a positive structure by Theorem 4.6 below, since U hasno nilpotent elements except 0.

Adams operations

For every augmented-ringKwe can define theAdams operationsk: K K fork 0 by setting 0(x) =(x),1(x) = x, 2(x) = x2 22(x) and inductively

k(x) =1(x)k1(x)2(x)k2(x)+ +(1)kk1(x)1(x)+(1)k1kk(x).

From this inductive definition we immediately deduce three facts:

if is a line element then k

() =

k

; ifI is a -ideal withI2 = 0 then k(x) = (1)k1kk(x) for allx I; For every binomial ring Hwe have k = 1. Indeed, the formal identity

xk1i=0(1)i

xi

= (1)k+1kxk shows thatk(x) = x for allx H.

The operationsk are named after J.F. Adams, who first introduced them in 1962in his study of vector fields on spheres.

Here is a slicker, more formal presentation of the Adams operations. Definek(x) to be the coefficient oftk in the power series:

t(x) =

k(x)tk =(x) td

dtlog

t(x).

The proof that this agrees with the inductive definition ofk(x) is an exercise informal algebra, which we relegate to Exercise 4.6 below.

Proposition 4.4. Assume K satisfies the Splitting Principle. Each k is aring endomorphism ofK, andjk =jk for allj, k 0.

Proof. The logarithm in the definition oft implies that t(x + y) = t(x) +t(y), so each

k is additive. The Splitting Principle and the formula k() =k

for line elements yield the formulas k(pq) =k(p)k(q) and j(k(p)) =jk(p)for positive p. The extension of these formulas to Kis clear.

Example 4.4.1. Consider the -ring KU(S2n) =Z Z of 3.1.1. On H0 =Z,k = 1, but onKU(Sn) = Z,k is multiplication bykn/2. (See [Atiyah, 3.2.2].)

Example 4.4.2. Consider KU(R P2n), which by Ex. 3.10 is Z Z/2n. I claimthat for allx KU(X):

k(x) =

x ifk is odd

0 ifk is even.

To see this, note that

KU(R P2n) = Z/2n is additively generated by ( 1), where

is the nonzero element ofL = H2(R P2n; Z) = Z/2. Since 2 = 1, we see thatk( 1) = (k 1) is 0 ifk is even and ( 1) ifk is odd. The assertion follows.

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-operations

Associated to the operations k are the operations k: K K. To constructthem, note that if we set s = t/(1

t) then K[[t]] = K[[s]] and t = s/(1 +s).

Therefore we can rewrite s(x) =i(x)si as a power series t(x) = k(x)tkin t. By definition, k(x) is the coefficient oftk in t(x). Since t(x) = s(x) wehave t(x+y) =t(x)t(y). In particular

0(x) = 1, 1(x) = x and k(x+y) =i(x)ki(y). That is, the -operations satisfy the axioms for a-ring structure

on K. An elementary calculation, left to the reader, yields the useful identity:

Formula 4.5. k(x) = k(x+k 1). This implies that2(x) = 2(x) + xand

k(x) =k(x + k 1) =k(x) +

k 11

k1(x) + +

k 1k 2

2(x) + x.

Example 4.5.1. IfH is a binomial ring then for allx Hwe have

k(x) =

x + k 1

k

= (1)k

xk

.

Example 4.5.2. k(1) = 1 for all k . More generally, if is a line element thenk() = for all k 1.

Lemma 4.5.3. Ifp P is a positive element with(p) =n, thenk(p n) = 0for allk > n. In particular, if K is a line element thenk( 1) = 0 for everyk >1.

Proof. Ifk > n thenq= p + (k n1) is a positive element with (q) =k 1.Thusk(p n) =k(q) = 0.

Ifx K, the -dimensiondim(x) ofx is defined to be the largest integer nfor which n(x (x)) = 0, provided n exists. For example, dim(h) = 0 for everyh H0 and dim() = 1 for every line element (except = 1 of course). Bythe above remarks ifp P and n = (p) then dim(p) = dim(p n) n. Thesupremum of the dim(x) for x Kis called the -dimensionofK.

Examples 4.5.4. IfR is a commutative noetherian ring, the Serre Cancellation

I.2.4 states that every element of

K0(R) is represented by [P] n, where P is a

f.g. projective module of rank< dim(R). Hence K0

(R) has -dimension at mostdim(R).

Suppose thatXis a CW complex with finite dimension d. The Real CancellationTheorem I.4.3 allows us to use the same argument to deduce that KO(X) has -dimension at mostd; the Complex Cancellation Theorem I.4.4 shows that K U(X)has -dimension at most d/2.

Corollary 4.5.5. IfKhas a positive structure in whichNis cofinal inP, thenevery element ofKhas finite-dimension.

Proof. Recall that N is cofinal in P means that for every p there is a p so

thatp +p =n for somen N. Therefore everyx Kcan be written as x = pm

for some p P withm= (p). By Lemma 4.5.3, dim(x) m.

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Theorem 4.6. If every element of K has finite -dimension (e.g., K has a

positive structure in whichN is cofinal inP), then

Kis a nil ideal. That is, every

element of

K is nilpotent.

Proof. Fix xK, and set m= dim(x), n= dim(x). Then both t(x) =1+xt+2(x)t2+ +m(x)tm andt(x) = 1xt+ +n(x)tn are polynomialsin t. Since t(x)t(x) = t(0) = 1, the polynomials t(x) and t(x) are unitsin the polynomial ring K[t]. By (I.3.12), the coefficients of these polynomials arenilpotent elements ofK.

Corollary 4.6.1. The idealK0(R) is the nilradical ofK0(R) for every com-mutative ringR.

IfX is compact (or connected and paracompact) thenKO(X) andKU(X) arethe nilradicals of the ringsKO(X) andKU(X), respectively.

Example 4.6.2. The conclusion of Theorem 4.6 fails for the representation ringR(G) of a cyclic group of order 2. Ifdenotes the 1-dimensional sign representation,

then L ={1, } andR(G)= Z is generated by ( 1). Since ( 1)2 = (2 2 + 1) = (2)( 1), we see that ( 1) is not nilpotent, and in fact thatR(G)n = (2n1) R(G) for every n 1. The hypothesis of Corollary 4.5.5 fails herebecausecannot be a summand of a trivial representation. In fact dim(1) = ,because n(1 ) = (1 )n = 2n1(1 ) for all n 1.

The-Filtration

The -filtrationon Kis a descending sequence of ideals:

K=F0K F1K FnK .

It starts withF0K=KandF1K=K(the kernel of). The first quotientF0/F1

is clearlyH0 =K/ K. Forn 2,FnKis defined to be the ideal ofKgenerated bythe products k1(x1) km(xm) with xiK and ki n. In particular,FnKcontains k(x) for allx Kand k n.

It follows immediately from the definition that FiFj Fi+j . For j = 1, this

implies that the quotients FiK/Fi+1 K are H-modules. We will prove that the

quotient F1/F2 isL:

Theorem 4.7. If K satisfies the Splitting Principle, then the map 1induces a group isomorphism, split by the map det:

L = F1K/F2K.

Corollary 4.7.1. For every commutative ring R, the first two ideals in the-filtration ofK0(R) areF

1 =K0(R) andF2 =SK0(R). (See 2.6.2.) In particular,

F0/F1=H0(R) and F1/F2=Pic(R).

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Proposition 4.9. Letk, n 1 be integers. Ifx FnK then modulo Fn+1 K:

k(x) knx; and k(x) (1)kkn1x.

Proof. Ifis a line element then modulo ( 1)2 we have

k( 1) = (k1 + ... + + 1)( 1) k( 1).

Therefore if1, , m are line elements and m n we have

k((1 1) (n 1)) kn(1 1) (n 1) modulo Fn+1 K.

The Filtered Splitting Principle implies that k(x) knx modulo Fn+1 K forevery x F

nK. For

k

, we use the inductive definition ofk

to see that kn

x =(1)k1kk(x) for every x FnK. The Filtered Splitting Principle allows us toconsider the universal case W= Ws of Exercise 4.4. Since there is no torsion inFnW/F

n+1 W, we can divide bykto obtain the formulak

n1x= (1)k1k(x). Theorem 4.10 (Structure ofK Q). Suppose thatKhas a positive struc-

ture in which every element has finite-dimension e.g., ifN is cofinal inP). Then:

(1) The eigenvalues ofk onKQ=K Q are a subset of{1, k , k2, k3,...} foreachk;

(2) The subspaceK(n)Q =K

(n,k)Q of eigenvectors for

k =kn is independent ofk;

(3) K(n)Q is isomorphic to FnKQ/Fn+1 KQ=(FnK/Fn+1 K) Q;(4) K

(0)Q

=H0 Q andK(1)Q =L Q;(5) The ringKQ is isomorphic to the graded ringK(0)Q K(1)Q K(n)Q .

Proof. For every positive p, consider the universal -ring UQ = Q[s1,...] ofExample 4.3.2, and the mapUQ KQsendings1 top and sk tok(p). If(p) =nthen si maps to zero for i > n and each si

ni

maps to a nilpotent element by

Theorem 4.6. The image A of this map is a -ring which is finite-dimensional overQ, soA is an artinian ring. ClearlyFN A= 0 for some largeN. Consider the linear

operation Nn=0(k kn) on A; by Proposition 4.9 it is trivial on each Fn/Fn+1 ,so it must be zero. Therefore the characteristic polynomial of k on A divides(t kn), and has distinct integer eigenvalues. This proves (1) and that KQ is thedirect sum of the eigenspacesK

(n,k)Q for

k. Ask preserves products, Proposition4.9 now implies (3) and (4). The rest is immediate from Theorem 4.7.

Chern class homomorphisms

The formalism in3 for the Chern classes ci: KU(X) H2i(X; Z) extends tothe current setting. Suppose we are given a -ringKwith a positive structure anda commutative graded ring A = A0 A1 . Chern classeson K with valuesin A are set maps c

i: K

Ai for i

0 with c

0(x) = 1, satisfying the following

axioms:

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(CC0) The ci send H0 to zero (for i 1): ci(h) = 0 for every h H0.

(CC1) (Dimension) ci(p) = 0 whenever pis positive and i (p).(CC2) (Sum Formula) For every x, y inKand every n:

cn(x + y) =n

i=0

ci(x)cni(y).

(CC3) (Normalization) c1: L A1 is a group homomorphism. That is, for , :

c1() = c1() + c1(

).

Thetotal Chern classofxis the elementc(x) = ci(x) of the completionA= A

i

ofA. In terms of the total Chern class, (CC2) becomes the product formula

c(x + y) =c(x)c(y).

Example 4.11.1. The Stiefel-Whitney classes wi: KO(X) Ai =Hi(X; Z/2)and the Chern classesci: KU(X) Ai =H2i(X; Z) are Chern classes in this sense.

Example 4.11.2. Associated to the -filtration on K we have the associatedgraded ring GrK with Gr

iK = F

i/F

i+1 . For a positive elementp in K, define

ci(p) to be i(p (p)) modulo Fi+1 . The multiplicative formula for t implies

that ci(p+ q) = ci(p) +ci(q), so that the ci extend to classes ci: K

GrK.

The total Chern class c: K GrKis a group homomorphism with torsion kerneland cokernel, because by Theorem 4.10 and Ex. 4.10 the induced map cn: K

(n)Q

GrnKQ

=K(n)Q is multiplication by (1)n(n 1)!.The Splitting Principle implies the following Splitting Principle (see [FL, I.3.1]).

Chern Splitting Principle. Given a finite set{pi} of positive elements ofK,there is a-ring extensionK K in which eachpisplits as a sum of line elements,and a graded extension A A such that the ci extend to maps ci: K (A)isatisfying (CC1) and (CC2).

The existence of Chern roots is an important consequence of this SplittingPrinciple. Suppose thatp Kis positive, and that in an extensionK ofKwe canwritep = 1+ + n,n = (p). TheChern roots ofpare the elements ai= c1(i)in (A)1; they determine the ck(p) in A

k. Indeed, because c(p) is the product ofthe c(i) = 1 +ai, we see that ck(p) is the k

th elementary symmetric polynomialk(a1,...,an) of the ai in the larger ring A

. In particular, the first Chern class isc1(p) =

ai and the top Chern class is cn(p) =

ai.

A famous theorem of Isaac Newton states that every symmetric polynomial inn variables t1,...,tn is in fact a polynomial in the symmetric polynomials k =k(t1,...,tn), k = 1, 2, . Therefore every symmetric polynomial in the Chernroots ofp is also a polynomial in the Chern classes c

k(p), and as such belongs to

the subring A ofA. Here is an elementary application of these ideas.

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Proposition 4.11.3. Suppose that K satisfies the Splitting Principle. Thencn(

kx) = kncn(x) for allx K. That is, the following diagram commutes:K

cn

Ank kn

K cn An.

Corollary 4.11.4. IfQ A thencn vanishes onK(i)Q , i =n.

Chern character

As an application of the notion of Chern roots, suppose given Chern classesci: K Ai, where for simplicity A is an algebra over Q. If p K is a positiveelement, with Chern rootsai, define ch(p) to be the formal expansion

ch(p) =ni=0

exp(ai) =k=0

1

k!

ni=0

aki

of terms in A. Thekth term 1k!

aki is symmetric in the Chern roots, so it is a

polynomial in the Chern classesc1(p),...,ck(p) and hence belongs to Ak. Therefore

ch(p) is a formal expansion of terms in A, i.e., an element ofA =

Ak. Forexample, ifis a line element ofKthench() is just exp(c1()). From the definition,it is immediate that ch(p + q) =ch(p) + ch(q), sochextends to a map from P1P

to A. Since ch(1) = 1, this is compatible with the given map H0

A0, and so it

defines a map ch: K A, called the Chern characteron K. The first few terms inthe expansion of the Chern character are

ch(x) =(x) + c1(x) +1

2[c1(x)

2 c2(x)] +16

[c1(x)3 3c1(x)c2(x) + 3c3(x)] +

An inductive formula for the term in ch(x) is given in Exercise 4.14.

Proposition 4.12. IfQ Athen the Chern character is a ring homomorphismch: K A.

Proof. By the Splitting Principle, it suffices to verify thatch(pq) =ch(p)ch(q)when p and q are sums of line elements. Suppose that p = i and q = mjhave Chern roots ai = c1(i) and bj = c1(mj), respectively. Since pq =

imj ,

the Chern roots ofpqare the c1(imj) = c1(i) + c1(mj) =ai+ bj . Hence

ch(pq) =

ch(imj) =

exp(ai + bj) =

exp(ai) exp(bj) =ch(p)ch(q).

Corollary 4.12.1. Suppose that K has a positive structure in which everyxK has finite-dimension(e.g., N is cofinal inP). Then the Chern characterlands in A, and the induced map from KQ =K(n)Q to A is a graded ring map.

That is, thenth

termchn: KQ An

vanishes onK

(i)

Q fori =n.

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Example 4.12.2. The universal Chern character ch: KQ KQ is the identitymap. Indeed, by Ex. 4.10(b) and Ex. 4.14 we see that chn is the identity map on

each K(n)Q .

The following result was proven by M. Karoubi in [Kar63]. (See Exercise 4.11for the proof when Xis a finite CW complex.)

Theorem 4.13. IfXis a compact topological space andH denotes Cech coho-mology, then the Chern character is an isomorphism of graded rings.

ch: KU(X) Q =

H2i(X; Q)

Example 4.13.1 (Spheres). For each even sphere, we know by Example 3.7.1

that cn maps

KU(S2n) isomorphically onto H2n(S2n; Z) =Z. The inductive for-

mula for chn shows that in this case ch(x) = dim(x) + (1)ncn(x)/(n 1)! forall x KU(X). In this case it is easy to see directly that ch: KU(S

2n

) Q=H2(S2n; Q)EXERCISES

4.1 Show that in K0(R) orK0(X) we have

k([P] n) =

(1)i

n + i 1i

[kiP].

4.2For every groupG and every commutative ring A, letRA(G) denote the groupK0(AG,A) of Ex. 2.14, i.e., the group completion of the monoid Rep(AG,A) ofall AG-modules which are f.g. projective as A-modules. Show that thek makeRA(G) into a -ring with a positive structure given by Rep(AG,A). Then showthatRA(G) satisfies the Splitting Principle, so that (by 4.2.3) it is a special -ring.Hint: The line elements are the characters.

When p = 0 in A, show that p = in RA(G), where : A A is theFrobenius (a) =ap. To do this, reduce to the case in which is a character andshow that k(g) =(gp) =(g)p.

4.3 Suppose that a -ringK is generated as an H-algebra by line elements. Show

that Fn =Kn for all n, so the -filtration is the adic filtration defined by theideal

K. Then show that ifKis any -ring satisfying the Splitting Principle every

element xofFnKcan be written in an extension K ofKas a product

x= (1 1) (m 1)of line elements withm n. In particular, show that everyx F2can be writtenas a sum of terms (i 1)(j 1)in K.4.4 Universal special-ring. Let Ws denote the Laurent polynomial ringZ[u1, u

11 ,...,us, u

1s ], and : Ws Z the augmentation defined by (ui) = 1.

(a) Show that Ws is a -ring with a positive structure, the line elements beingthe monomialsu =

unii . This implies thatWs is generated by the group

L = Zs of line elements, so by Exercise 4.3 FnWs isWn.(b) Show that eachFnW/F

n+1 Wis a torsionfree abelian group.

(c) If K is a special -ring show that any family

{1,...,s

} of line elements

determines a -ring mapWs Ksending ui toi.

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4.5 A line element is called ample for K if for every xK there is an integerN = N(x) such that for every n N there is a positive element pn so thatnx = pn (pn). (The terminology comes from Algebraic Geometry; see 8.8.4

below.) IfKhas an ample line element, show that every element ofKis nilpotent.4.6 Verify that the inductive definition ofk agrees with the t definition of

k.

4.7Ifp is prime, use the Splitting Principle to verify thatp(x) xp modulo p forevery x K.4.8Adamse-invariant. Suppose given a mapf: S2m1 S2n. The mapping coneC(f) fits into a cofibration sequenceS2n

i C(f)

j S2m. Associated to this is the

exact sequence:

0

KU(S2m)

j

KU(C)

i

KU(S2n) 0.

Choose x, yKU(C) so that i(x) generatesKU(S2n) =Z and y is the image ofa generator ofKU(S2m) = Z. Since j is a ring map, y 2 = 0.

(a) Show by applying k that xy = 0, and that ifm= 2n then x2 = 0. (Whenm= 2n, x2 defines the Hopf invariant off; see the next exercise.)

(b) Show that k(x) = knx+aky for appropriate integers ak. Then show (forfixed xand y) that the rational number

e(f) = ak

km knis independent of the choice ofk .

(c) Show that a different choice ofx only changese(f) by an integer, so thate(f)is a well-defined element ofQ/Z; e(f) is called the Adamse-invariantoff.

(d) If f and f are homotopic maps, it follows from the homotopy equivalencebetween C(f) andC(f) that e(f) =e(f). By considering the mapping coneoffg, show that the well-defined set map e: 2m1(S2n) Q/Zis a grouphomomorphism. J.F. Adams used this e-invariant to detect an importantcyclic subgroup of2m1(S

2n), namely the image of J.

4.9 Hopf Invariant One. Given a continuous mapf: S4n1 S2n, define an integerH(f) as follows. Let C(f) be the mapping cone off. As in the previous exercise,we have an exact sequence:

0 KU(S4n) jKU(C(f)) iKU(S2n) 0.Choose x, yKU(C(f)) so that i(x) generatesKU(S2n) = Zand y is the imageof a generator ofKU(S4n) =Z. Since i(x2) = 0, we can write x2 =H y for someinteger H; this integerH=H(f) is called the Hopf invariantoff.

(a) Show that H(f) is well-defined, up tosign.(b) If H(f) is odd, show that n is 1, 2, or 4. Hint: Use Ex. 4.7 to show that

the integera2 of the previous exercise is odd. Considering e(f), show that 2n

dividespn 1 for every odd p.It turns out that the classical Hopf maps S3 S2, S7 S4 and S15 S8 allhave Hopf invariantH(f) = 1. In contrast, for every even integerHthere is a mapS4n1 S2n with Hopf invariant H.

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4.10Operations. A natural operation on -rings is a map : K Kdefined forevery-ringKsuch thatf = ffor every-ring mapf: K K. The operationsk,k, and k are all natural operations on -ring.

(a) IfKsatisfies the Splitting Principle, generalize Proposition 4.9 to show thatevery natural operation preserves the -filtration ofK and that there areintegers n= n(), independent ofK, such that for everyx FnK

(x) nx modulo Fn+1 K.

(b) Show that for =k and x Fn we have:

k(x)=

0 if n < k

(1)k1(k 1)! if n= k

n= 0 if n > k(c) Show thatxk k gives a ring map from the power series ring Z[[x1, x2, ]]

to the ring C of all natural operations on -rings. In fact this is a ringisomorphism; see [Atiyah, 3.1.7].

4.11 By Example 4.13.1, the Chern character ch: KU(Sn) Q H2(Sn; Q) isan isomorphism for every sphere Sn. Use this to show that ch: KU(X) QH2(X; Q) is an isomorphism for every finite CW complex X.

4.12Let Kbe a-ring. Given aK-moduleM, construct the ringKM in whichM2 = 0. Given a sequence ofK-linear endomorphisms k ofM with 1(x) = x,

show that the formulae k

(x) = k(x) extend the-ring structure onKto a-ringstructure on K M. Then show thatK Mhas a positive structure ifK does,and that K M satisfies the Splitting Principle whenever Kdoes. (The elementsin 1 + Mare to be the new line elements.)

4.13 Hirzebruch characters. Suppose that A is an H0-algebra and we fix a powerseries(t) = 1+1t+2t

2 + inA0[[t]]. Suppose given Chern classesci: K Ai.Ifp Kis a positive element, with Chern roots ai, define ch(p) to be the formalexpansion

ch(p) =ni=0

(ai)k=0

k

ni=0

aki

of terms inA. Show thatch(p) belongs to the formal completion AofA, and that

it defines a group homomorphism ch: K A. This map is called the Hirzebruchcharacterfor .

4.14 Establish the following inductive formula for the nth term chn in the Cherncharacter:

chn 1n

c1chn1+ 1i!ni

cichni+ + (1)n(n 1)! cn = 0.

To do this, set x=

ti in the identity (x + ai) =xn + c1xn1 + + cn.

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5. K0 of a Symmetric Monoidal CategoryThe idea of group completion in 1 can be applied to more categories than just thecategories P(R) in

2 and VB(X) in

3. It applies to any category with a direct

sum, or more generally any natural product making the isomorphism classesof objects into an abelian monoid. This leads us to the notion of a symmetricmonoidal category.

Definition 5.1. Asymmetric monoidal categoryis a categoryS, equipped witha functor : S S S, a distinguished object e and four basic natural isomor-phisms:

es =s, se =s, s(tu) =(st)u, and st =ts.

These basic isomorphisms must be coherent in the sense that two natural iso-

morphisms of products ofs1, . . . , sn built up from the four basic ones are the samewhenever they have the same source and target. (We refer the reader to [Mac] forthe technical details needed to make this definition of coherent precise.) Coher-ence permits us to write expressions without parentheses like s1 sn unam-biguously (up to natural isomorphism).

Example 5.1.1. Any category with a direct sum is symmetric monoidal; thisincludes additive categories like P(R) and VB(X) as we have mentioned. Moregenerally, a category with finite coproducts is symmetric monoidal withst= st.Dually, any category with finite products is symmetric monoidal with st= s t.

Definition 5.1.2 (K0S). Suppose that the isomorphism classes of objects ofS form a set, which we call Siso. If S is symmetric monoidal, this set Siso is anabelian monoid with product and identitye. The group completion of this abelianmonoid is called the Grothendieck group ofS, and is written as K0 (S), or simplyas K0(S) if is understood.

From1 we see that K0 (S) may be presented with one generator [s] for eachisomorphism class of objects, with relations that [st] = [s] + [t] for each s andt. From Proposition 1.1 we see that every element ofK0 (S) may be written as adifference [s] [t] for some objects s and t.

Examples 5.2. (1) The category P(R) of f.g. projective modules over a ringRis symmetric monoidal under direct sum. Since the above definition is identical tothat in2, we see that we have K0(R) =K0 (P(R)).

(2) Similarly, the category VB(X) of (real or complex) vector bundles over atopological space Xis symmetric monoidal, with being the Whitney sum . Fromthe definition we see that we also have K(X) = K0 (VB(X)), or more explicitly:

KO(X) = K0 (VBR(X)), KU(X) = K0 (VBC(X)).

(3) IfR is a commutative ring, let Pic(R) denote the category of algebraic linebundles L over R and their isomorphisms (I.3). This is a symmetric monoidalcategory with =R, and the isomorphism classes of objects already form agroup, so K0Pic(R) = Pic(R).

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Finite Sets 5.2.1. Let Setsfin denote

kbook.ii

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