yeshengkui.files.wordpress.comcontents 0.1 introduction . . . . . . . . . . . . . . . . . . . ....

An introduction to algebraic K-theory

Shengkui Ye

March 08, 2013 (draft only)

Contents

0.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1 K0 and applications 71.1 Motivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Idempotent matrices and de�nition of K0 . . . . . . . . . . . . . 71.3 Relations to projective modules . . . . . . . . . . . . . . . . . . . 91.4 Properties of K0 . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.5 K0 and representation of �nite groups . . . . . . . . . . . . . . . 131.6 K0 in topology: Wall�s �niteness obstruction . . . . . . . . . . . 141.7 Idempotent conjecture and Bass conjecture . . . . . . . . . . . . 171.8 Other kinds of K0 . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.8.1 Exact categories . . . . . . . . . . . . . . . . . . . . . . . 201.8.2 Isomorphism between algebraic K0 and topological K0 . . 22

2 K1 and applications 252.1 Motivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.2 De�nition and basic properties . . . . . . . . . . . . . . . . . . . 262.3 K1 in topology: Whitehead torsion . . . . . . . . . . . . . . . . . 28

2.3.1 Whitehead torsion . . . . . . . . . . . . . . . . . . . . . . 282.3.2 s-cobordism theorem . . . . . . . . . . . . . . . . . . . . . 282.3.3 Simple homotopy equivalence . . . . . . . . . . . . . . . . 30

2.4 Bass-Heller-Swan theorem . . . . . . . . . . . . . . . . . . . . . . 31

3 K2 353.1 Steinberg group and de�nition of K2 . . . . . . . . . . . . . . . . 353.2 Universal central extension . . . . . . . . . . . . . . . . . . . . . 373.3 Relative K-theory and exact sequences . . . . . . . . . . . . . . . 37

3.3.1 Relative K0 . . . . . . . . . . . . . . . . . . . . . . . . . . 373.3.2 Relative K1 . . . . . . . . . . . . . . . . . . . . . . . . . . 393.3.3 Relative K2 . . . . . . . . . . . . . . . . . . . . . . . . . . 403.3.4 Long exact sequence of K-groups . . . . . . . . . . . . . . 40

4 High algebraic K-theory 434.1 Classifying spaces with respect to families of subgroups . . . . . 43

4.1.1 G-CW complexes . . . . . . . . . . . . . . . . . . . . . . . 44

3

4 CONTENTS

4.1.2 Classifying spaces . . . . . . . . . . . . . . . . . . . . . . 444.2 Acyclic maps and homology equivalences . . . . . . . . . . . . . . 464.3 Quillen�s plus construction . . . . . . . . . . . . . . . . . . . . . . 474.4 De�nition of high algebraic K-theory . . . . . . . . . . . . . . . . 494.5 Spectrum of algebraic K-theory . . . . . . . . . . . . . . . . . . . 51

4.5.1 Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.5.2 Algebraic K-theory spectrum . . . . . . . . . . . . . . . . 52

4.6 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.6.1 Any abelian group is the center of a perfect group . . . . 544.6.2 Higher homotopy groups and homology . . . . . . . . . . 54

5 Isomorphism conjectures 575.1 K-theory spectrum of an additive category . . . . . . . . . . . . 575.2 Equivariant homology . . . . . . . . . . . . . . . . . . . . . . . . 585.3 Formulation of conjectures . . . . . . . . . . . . . . . . . . . . . . 605.4 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

6 Appendix: Background from algebraic topology 656.0.1 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . 656.0.2 Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . 68

0.1 Introduction

Roughly speaking:

� There are three kinds of K-theory: algebraic K-theory (this course�s maintopic), topological K-theory and K-theory of C�-algebras.

� Algebraic K-theory consists of low algebraic K-theory K0;K1 and K2 andhigh algebraic K-theory Ki for i � 3:

� The low algebraic K-theory is linear algebras over general rings;

� The high algebraic K-theory is generalization of low algebraic K-theory;

� For completions, there are also negative K-theory.

The course wants to introduce three-level things:

� Basic notions and facts (with detailed proofs);

� Main progress and famous theorems (with rough proofs);

� Open problems (hopefully formulated in elementary ways).

Conventions.

� A rings always means an associative ring with identity.

0.1. INTRODUCTION 5

� A map always means a function preserving the relevant structures, e.g. amap of modules (resp. rings, groups, CW complexes, etc.) is a module(resp. ring, group, cellular) homomorphism (preserving resp. identities,base-points, ect.).

6 CONTENTS

Chapter 1

K0 and applications

1.1 Motivations

Let F be a �eld andM a �nitely generated F -module. SinceM is a vector space,we could write M = Fn for some integer n = dimF (M): This means that themodule is uniquely (up to isomorphisms) determined by its "dimension". Whatis this dimension dimF ? If we denote by C the category of �nitely generatedmodules, the dimension dimF could be viewed as a function

dimF : C ! N

such that dimF (MLN) = dimF (M)

LdimF (N) and dimF (M) = dimF (N)

if M is isomorphic to N: Moreover, if we take isomorphisms classes C= � andput an additive structure on C= � by

L; the function is actually a semigroup

isomorphic.

Problem 1.1.1 Let R be an associative ring with identity. How to classify themodules over R?

If we consider only free R-modules, the same function dimR will not nec-essary work ( Rm �= Rn implies m = n?). But for general �nitely generatedmodules, what is such a function? In order to study modules, we study theirendomorphisms. The simplest case is the endomorphisms of free modules: ma-trices. We will mainly study such problem through matrices, viewed as linearalgebras over general rings.

1.2 Idempotent matrices and de�nition of K0Let R be an associative ring with identity (will be just called �ring�during thiscourse). For an integer n; the matrix ring Mn(R) = f(aij)n�n : aij 2 Rg is theset of all n� n matrices. The set of all invertible matrices is denoted by

GLn(R) = fA 2Mn(R) j AB = BA = In for some B 2Mn(R)g:

7

8 CHAPTER 1. K0 AND APPLICATIONS

De�nition 1.2.1 A matrix P = (aij)n�n 2 Mn(R) is called idempotent ifP 2 = P: Two idempotent matrices P; P 0 are conjugate if there is an invertiblematrix A 2 GLn(R) such that APA�1 = P 0:

Example 1.2.2 1) For any ring R; P = diag(Im; 0); 0 � m � n; is idempotent.2) Let G be a �nite group and R a commutative ring in which the order jGj

is invertible (eg. R = Q, the rationals), P =1

jGjP

g2G g is a 1� 1 idempotent

matrix.

We study the set of idempotent matrices of all sizes. When M is an n � nmatrix, view naturally it as the (n+ 1)� (n+ 1) matrix�

M 00 0

�:

Let Pn(R) be the set of conjugacy classes of n� n idempotent matrices. Thenthere is an injection Pn(R)! Pn+1(R): Let PR = [n�1Pn be the set of isomor-phism classes of idempotent matrices of all sizes. For an idempotent matrix P;denote by [P ] is the set of conjugacy class of P in PR:

Proposition 1.2.3 PR is an abelian semi-group (monoid) under the operation+ de�ned by

[P ] + [P 0] = [diag(P; P 0)]:

Proof. Let [P ] = [P1] and [P 0] = [P 01]; i.e. APA�1 = P1; BP

0B�1 = P 01 forsome invertible matrices A;B: Without loss of generality, assume P; P1; P 0; P 01are of the same sizes.In order to show the operation+ is well-de�ned, it su¢ ces to check [diag(P; P 0)] =

[diag(P1; P01)]; as follows. diag(A;B)diag(P; P

0)diag(A�1; B�1) = diag(P1; P01):

In order to show + is abelian, it su¢ ces to check [diag(P; P 0)] = [diag(P 0; P )]as follows �

0 InIn 0

�diag(P; P 0)

�0 InIn 0

�= diag(P 0; P ):

Proposition 1.2.4 (Grothendieck completion) Let A be a commutative semi-group (with or without unit). There is a group GA and a monoid-homomorphismf : A! GA such that for any monoid-homomorphism g : A! G to an abeliangroup G; there is a unique group homomorphism ~g : GA ! G such that g = ~g�f:

Proof. Construct AG as the classes of A� A by (x; y) � (u; v) i¤ x+ v + t =y + u+ t for some t 2 A: We need to check this is an equivalence relation. There�exivity and symmetry are obvious. For transitivity, let (u; v) � (x0; y0) byu+ y0 + t0 = v + x0 + t0 for some t0 2 A: Then

x+ y0 + v + t+ t0 = y + u+ t+ y0 + t0 = y + x0 + t+ v + t0;

1.3. RELATIONS TO PROJECTIVE MODULES 9

which (x; y) � (x0; y0): Under the operation [(x; y)] + [(u; v)] = [(x + u; y + v)];AG is an abelian group. The unit is [(a; a)] for any element a 2 A and theinverse of [(x; y)] is [(y; x)]:Fixing an element t 2 A; the map f is given by f(a) = [(a+ t; t)]: Since

f(a+ a0) = [(a+ a0 + t; t)] = [(a+ t; t)] + [(a0 + t; t)];

the map f is a group homomorphism.The map ~g is de�ned by ~g([x; y]) = g(x) � g(y): The map ~g is well-de�ned

by the de�nition of equivalence relation. For a 2 A; ~gf(a) = ~g([(a + t; t)]) =g(a + t) � g(t) = g(a); which means g = ~g � f: If g0 is another map such thatg = g0 � f; then

g0([(a; b)]) = g0([a+ t; t] + [(t; t+ b)])

= g0([a+ t; t])� g0([b+ t; t])= g0 � f(a)� g0 � f(b)= g(a)� g(b) = ~g([(a; b)]):

Exercise 1.2.5 Does this hold for noncommutative semigroup? If it�s no, howto do it? Is f necessarily injective?

De�nition 1.2.6 For a ring R; the group K0(R) is de�ned as the Grothendieckcompletion of PR:

Let�s see some examples:

Example 1.2.7 Let R = C the complex number. By Jordan canonical form,any idempotent matrix P is isomorphic to In for some integer n: Then PR �= N,the natural numbers and K0(C) �= Z, the integers.

Remark 1.2.8 Someone takes this as the basic example of K-theory and viewK0 as the generalization of �dimension�from �elds to general rings.

1.3 Relations to projective modules

Let S be a �nite set and F a free module generated by S over a ring R. In otherwords

F = fXfinite

risijsi 2 S; ri 2 Rg:

A direct summand of an R-moduleM is a submodule N such thatM = NLN 0

for some submodule N 0 of M:

De�nition 1.3.1 A projective module K is an R-mdule, which is isomorphicto a direct summand of a free module.


Proposition 1.3.2 A module K is projective i¤ for any surjection s :M ! Nof R-modules and any map f : K ! N; there is a map g : K ! M such thats � g = f; i.e. there is a commutative diagram

Kg . f #

Ms� N:

Proof. Suppose that K is isomorphic to a direct summand K 0 of a free moduleF: Let p : F ! K 0 �= K be the projection. Fix a generating set S of F: For eachelement s; choose a preimage xs 2 s�1(fp(s)): De�ne a map ~g : F ! M by~g(s) = xs: Then the composition of maps K �= K 0 ,! F

g! M could be takenas g:Conversely, choose a surjection s : F ! K by a presentation of K: For the

identity map f = idK ; the map g gives a section of K: Then F �= Im gLker s:

This shows that K is isomorphic the direct summand g(K) of F:We consider �nitely generated projective modules, i.e. a direct summand of

Rn for some integer n. Let PjR be the set of isomorphisms classes of �nitelygenerated projective modules. Under the operation of direct sum, PjR is anabelian monoid.

Theorem 1.3.3 There is an isomorphism � : PR ! PjR of monoids. Inparticular, K0(R) is isomorphic to the Grothendieck completion of isomorphismclasses of projective modules.

Proof. Let Rn be a free R-module of rank n: For an idempotent matrixP = (aij)n�n; view it as an R-linear transformation Rn ! Rn with respectto the standard basis. Then the image ImP is a projective module, sinceRn = ImP

LIm(In � P ): De�ne �([P ]) = [ImP ]; the isomorphism class of

projective module P: It is easy to check that if P 0 = APA�1 for an invertiblematrix A; then ImP is isomorphic to ImP 0: Conversely, if M is a direct sum-mand of a free module Rn: Let P be the representation matrix of the projectionRn !M with respect to the standard basis. We see that �([P ]) = [M ]:

Proposition 1.3.4 The element in K0(R) is of the form [P ] � [Q] for P;Q 2PR:

(1) Two idempotent matrices P and Q represent the same element in K0(R)i¤ they are stably conjugate, i.e. for some positive integer n we haveAdiag(P; In)A

�1 = diag(Q; In) for some integer r and invertible matrixA:

(2) Two projective modules P and Q represents the same element in K0(R)i¤ they are stably isomorphic, i.e. for some positive integer n we haveRLRn �= Q

LRn:

1.4. PROPERTIES OF K0 11

Proof. From the construction of Grothendieck completion, the �rst part ofeach statement is obvious. For the second part, [P ] = [Q] 2 K0(R) impliesthat [P ] + [e] = [Q] + [e] for some [e] 2 PR (or PjR): For projective modulee, choose the other summand (projective as well) e0 such that e

Le0 = Rn:

For idempotent matrix, it is enough to note that any idempotent matrix e isconjugate to an identity matrix after directing sum with another idempotentmatrix 1� e: Actually,�

1� e �ee 1� e

��e1� e

��1� e e�e 1� e

�=

�01

�:

1.4 Properties of K0From the construction of K0; we see that

Proposition 1.4.1 (Functoriality) For a ring homomorphism f : R ! R0;there is a group homomorphism f� : K0(R)! K0(R

0) induced by f:

Proof. Let M = (aij)n�n be an idempotent matrix with entries in R: Then(f(aij))n�n is an idempotent matrix with entries in R0: It is not hard to checkthat (f(aij))n�n is also an idempotent matrix. De�ne f�([M ]) = (f(aij))n�n:It is obvious that this is well-de�ned.The functoriality shows that K0 is a functor from the category of rings to

the category of abelian groups.

Proposition 1.4.2 (Morita invariance) Let R be a ring. For a positive in-teger n; the set of n � n matrices is denoted by Mn(R): Then the inclusion ofR into the (1; 1)-th position induces K0(R) �= K0(Mn(R)):

Proof. For each idempotent matrix P = (pij)k�k with pij 2 R: Then ~P = (~pij)is an idempotent matrix with entries in Mn(R); where ~pij is an matrix with pijin the (1; 1)-th position and zeros elsewhere.Caution: the homomorphism R ! Mn(R) does not preserves identities.

Therefore, we have to show that the map K0(R)! K0(Mn(R)) is well-de�ned.For an invertible matrix A = (aij) 2 GLk(R); let A0 = (a0ij) 2 GLk(Mn(R))such that a0ii = diag(a; In�1) a

0ij = ~aij if i 6= j: If there is an idempotent matrix

Q satisfying APA�1 = Q; then A0 ~PA0�1 = ~Q:Conversely, for each idempotent matrix P 0 with entries in Mn(R); it is an

idempotent by forgetting the matrix. This is an isomorphism on isomorphismsclasses of idempotent matrices by conjugations by permutation matrices. Forexample, assume n = 2: For an idempotent matrix�

a bc d

�


over R; its image is 2664a 0 b 00 0 0 0d 0 c 00 0 0 0

3775 ;an idempotent matrix overM2(R): The latter conjugate to the former by matrix2664

1 0 0 00 0 1 0d 1 0 00 0 0 1

3775 ;which is invertible over both R and M2(R):Since any �nitely generated module over a �eld is a �nitely dimensional

vector space, K0 of a �eld is actually the integer Z. Moreover, Let A be a(commutative) principal ideal domain (PID). The structure of �nitely generatedA-module says that any �nitely generated module M is of the form F

LT;

where F �= An is a free A-module of �nite rank and T is a torsion module(whose elements are of �nite orders). Since a submodule of a free module istorsion-free, we see that any �nitely generated module over a principal idealdomain is free. This proves the following.

Proposition 1.4.3 Let A be a principal ideal domain. Then K0(A) �= Z.

The following question will be natural:

Problem 1.4.4 Is there a ring R whose K0 is not integer?

The answer is: yes. Before we give such an example, let�s introduce somenotions.Let I be a partial order set such that for any �; � 2 I there is a 2 I such

that � < and � < . A direct system of rings is a set of rings (Ri)i2I togetherwith ring homomorphisms fij : Ri ! Rj whenever i < j such that fjkfij = fikif i < j < k:

De�nition 1.4.5 The direct limit of a direct system (Ri)i2I is a ring R togetherwith ring homomorphisms fi : Ri ! R such that fi = fj � fij and satis�es theuniversal property that for any other ring (R0; f 0i) with this property f

0i = f

0j �fij

there is a unique map f : R0 ! R such that f 0i = fi � f: The direct limit ofabelian groups could be de�ned similarly.

Proposition 1.4.6 (Direct limit) Let (fij : Ri ! Rj)i2I be a direct systemof rings. Then limK0(Ri) = K0(limRi):

Proof. By functoriality, there is a map � : limK0(Ri)! K0(limRi): We needto show this map is both injective and surjective. Since each idempotent matrixP in limRi has �nitely many entries, there exists a ring Ri such that [P ] comes

1.5. K0 AND REPRESENTATION OF FINITE GROUPS 13

from an idempotent matrix in Ri: On the other hands, if an element x represents0 2 K0(limRi): Then 0 = [P ] � [Q] for some idempotent matrices P and Q:This means P and Q are stably conjugate, i.e. there is an invertible matrix Asuch that

Adiag(P; Ir)A�1 = diag(Q; Ir)

for some identity matrix Ir: Since the entries in A and P;Q are �nitely many,such equality also holds in some ring Ri: This shows x = 0 2 K0(Ri):

Example 1.4.7 Let k be a �eld and M2n(k) ! M2n+1(k) de�ned by a !diag(a; a): The limit ring is denoted by M: Then K0(M2n(k)) = Z and fn;n+1 :K0(M2n(k))! K0(M2n+1(k)) is multiplication 2: Therefore

K0(M) = lim(Z�2!Z�2!� � �) = Z[1=2]:

1.5 K0 and representation of �nite groups

Let G be a �nite and F a �eld with the order jGj an invertible element. Recallthat a representation of G is an F -vector space with a group action of G by linearisomorphisms. The representation ring RepF (G) is the Grothendieck groupcompletion of isomorphism classes of �nite dimensional linear F -representationof G: The ring structure is given as follows: the addition is given by direct sumof representation and the multiplication is given by tensor product over F:

De�nition 1.5.1 Let M be a module over a ring R: M is called semisimple ifM is the sum of simple R-submodules. A ring R is semisimple if R itself issemisimple, viewed as a left R-module.

Lemma 1.5.2 Suppose that M is a �nitely generated left module over a semi-simple ring R. Then every submodule of M is a direct summand.

Proof. If R is semisimple, so is the �nitely many copy Rn �= RLRL� � �LR:

Since M is a quotient module of Rn for some integer n; M is semisimple. Sup-pose that M =

Li2JMi for simple submodule Mi: For a �nite generating set

S; each element s belongs to a �nite sum of Mi. Therefore, we could assumethe index set J is �nite. Let N be any submodule of M: After relabeling Mi;suppose that M = N +M1 + � � � +Mm is a decomposition with shortest m:SinceMm is simple and not contained in N +M1+ � � �+Mm�1; the intersection

(N +M1 + � � �+Mm�1) \Mm = 0:

This shows that M = NLM1

L� � �LMm is a direct sum decomposition,

which implies that N is a direct summand.

Proposition 1.5.3 Let G be a �nite and F a �eld with the order jGj an in-vertible element. Denote by RepF (G) the F -representation ring of G: ThenK0(FG) �= RepF (G):


Proof. It�s enough to prove that the monoid PjFG of isomorphism classes ofprojective modules is isomorphic to the monoid of isomorphism classes of �nitedimensional linear F -representation of G: For each �nitely generated projectiveFG-module M; G acts on M linearly. This means that M is a representationof G: Conversely, suppose that N is an F -vector space on which G acts linearly.Since N is �nitely generated, there is an exact sequence of modules

0! K ! (FG)n ! N ! 0:

Since the regular representation ofG is decomposed as a direct sum of irreduciblerepresentations, i.e. FG is a direct sum of its simple submodules, the group FGis semisimple. The previous lemma shows that every �nitely generated left FG-module A is completely reducible, i.e. every submodule is a direct summand.In particular, K is a direct summand of (FG)n and therefore N is a projectivemodule.

1.6 K0 in topology: Wall�s �niteness obstruction

Since we assume that a ring R contains an identity 1: There is always a ringa ring homomorphism f : Z ! R: By the functoriality, we have a map f� :K0(Z) = Z! K0(R):

De�nition 1.6.1 The reduced ~K0(R) is de�ned as the cokernel of f� : K0(Z) =Z! K0(R):

Corollary 1.6.2 A �nitely generated projective module P is zero in ~K0(R) i¤its is stably free, i.e. there are positive integers m an n such P

LRm �= Rn:

Proof. This is a easy corollary of Proposition 1.3.4.In this section, we will give an application of K0 in topology. Recall that

a CW complex X is �nite if there are only �nitely many cells in X: A path-connected, locally 1-connected topological space Y is called �nitely dominatedif it is a retract of a �nite complex. More precisely:

De�nition 1.6.3 A path-connected, locally 1-connected topological space Y is�nitely dominated if there exists a �nite CW complex X together with mapsi : Y ! X and r : X ! Y such that r � i is homotopic to idY :

In such case, we say that Y is dominated by X: Obviously, a �nite CWcomplex is �nitely dominated by itself. A natural question is that

Problem 1.6.4 When is a �nitely dominated CW complex homotopic to a �niteCW complex?

We want to give an obstruction, called Wall obstruction, for a �nitely dom-inated CW complex to be �nite. In order to de�ne such obstruction, we needseveral lemmas. In this section, we assume all spaces are CW complex. There-fore, a weak homotopy equivalence is actually a homotopy equivalence.

1.6. K0 IN TOPOLOGY: WALL�S FINITENESS OBSTRUCTION 15

Lemma 1.6.5 For a �nitely dominated CW complex Y; the fundamental group�1(Y ) is �nitely presented.

Proof. Without confusion, assume that r : �1(X) ! �1(Y ) and i : �1(Y ) !�1(X) satisfying r�i = Id for some �nitely presented group �1(X): Suppose thatfor a �nitely generated free group F; there is a surjection p : F ! �1(X) withker p normally generated by a �nite set of elements, i.e. G has a presentation hgi :rji. Suppose that p(wi) = i � r � p(gi): We claim that �1(Y ) has a presentationhgi : rj ; g�1i wii: Indeed, Let L be a group with such presentation. Then r factorsthrough

r : �1(X)u! L

v! �1(Y ):

Since r is onto, v is onto. Since u�i(r�p(gi)) = u�p(wi) = u�p(gi); u�i is onto.This shows that v and u � j are inverse isomorphisms of L and H; consideringv � u � i = Id�1(Y ):Let f : X ! Y be a map between CW complexes. The map f is n-connected

if relative homotopy groups �i(Y;X) = 0 for 0 � i � n+ 1:

Lemma 1.6.6 Let Y be a space dominated by a CW complex X with �niteskeletons, i.e. there are maps r : X ! Y; i : Y ! X such that r � i ' idY .Suppose that r is n-connected, i.e. �i(r) : �i(X) ! �i(Y ) is isomorphic for0 � i � n and surjective for i = n+ 1: Then ker�n+1(r) is a �nitely generatedZ�1(X)-module.

Proof. By the long exact sequence of relative homotopy groups

� � � ! �n+2(X)! �n+2(Y )! �n+2(Y;X)! �n+1(X)! �n+1(Y )! � � �

and the fact that �n+1(X) ! �n+1(Y ) is surjective, it su¢ ces to show that�n+2(Y;X) is �nitely generated over Z�1(X): Note that �i(Y;X) = 0 fori � n+1: By considering the universal covering spaces ~X and ~Y , �n+2( ~Y ; ~X) =�n+2(Y;X): The relative Hurewitz theorem tells that �n+2( ~Y ; ~X) = Hn+2( ~Y ; ~X):By long exact sequences

� � � ! Hn+2( ~X)rn+2! Hn+2( ~Y )! Hn+2( ~Y ; ~X)! Hn+1( ~X)

rn�1! Hn+1( ~Y )! � � �

and

� � � ! Hn+1( ~Y )in+1! Hn+1( ~X)! Hn+1( ~X; ~Y )! Hn( ~Y )

in! Hn( ~X)! � � �

splitted by r� and i�, we see that Hn+2( ~Y ; ~X) �= Hn+1( ~X; ~Y ) is a direct sum-mand of Hn+1( ~X): Since r is n-connected, we may assume X and Y have thesame n-skeleton for considering their homology groups. The relative homologygroup Hn+2( ~Y ; ~X) �= Hn+1( ~X; ~Y ) is a quotient of the Z�1(X)-free modulesCn+1( ~X; ~Y ) with a basis the set of all (n+ 1)-cells in X � Y; which is �nite byassumption. Therefore, �n+2(Y;X) is �nitely generated.

Lemma 1.6.7 A �nitely dominated CW-complex Y is of �nite type, i.e. it ishomotopy equivalent to a CW-complex with �nite skeleta.


Proof. We may assume that Y is connected. By Lemma 1.6.5, the fundamentalgroup of Y is �nitely presented, as a retract of a �nitely presented group. Usethe notations in the proof of Lemma 1.6.5. For each element g�1i wi in �1(X);attach a 2-cell to X to kill this element. The resulting space is denoted byX1; :which is �nite as well. In other words, X1 is the pushout of the followingdiagram

_S1 _fi! X# #_D2 ! X1;

where fi : S1 ! X is a representative of the element g�1i wi 2 �1(X): Sinceg�1i wi has trivial image in �1(Y ); the retraction r extends toX1: By the previouslemma applied to the map r : X1 ! Y; we get that ker�2(r) is �nitely generated.For each generator, attach a 2-cell to X1 to get a new �nite CW complex X2:Repeating such process, we get a CW complex of �nite type homotopy equivalentto Y:Suppose that Y is dominated by a �nite CW-complexX: The previous lemma

implies that Y is of �nite type. Let M be a Z�1(Y )-module. We have theisomorphism H�(Y ;M) �= i� � r�(H�(Y ;M)): Note that r�(H�(Y ;M)) is asubmodule of H�(X; r�M) and that X is �nite. Therefore, for some su¢ cientlarge number N; the cohomology group Hq(Y ;M) = 0 for any q � N: Supposethat (C�(Y ); d) is the cellular chain complex of Y: TakeM = Im dN+2: Accordingto the vanishing HN+2(Y ;M) = 0 and the exact sequence

HomZ�1(Y )(CN+1;M)! HomZ�1(Y )(CN+2;M)! HomZ�1(Y )(CN+3;M);

we see that there is a projection f : CN+1 ! Im dN+2. This shows that Im dN+2is a �nitely generated projective Z�1(Y )-module.Let YN be theN -skeleton of Y: Then the relative homology groupHN+1(Y; YN )

is the quotient group of CN+1(Y ) by Im dN+2: By the previous argument, we seethat Im dN+2 is a direct summand of CN+1(Y ): This implies that �N+1(Y; YN ) �=HN+1(Y; YN ) is a projective Z�1(Y )-module.The Wall obstruction w(Y ) of Y is de�ned as

(�1)N+1[�N+1(Y; YN )] 2 ~K0(Z�1(Y )):

Proposition 1.6.8 w(Y ) is well-de�ned.

Proof. Suppose that N 0 > N is another integer. Since the higher homologygroups are zeros, the sequence

0! Im dN 0+2 ! CN 0+1 ! � � � ! CN+2 ! Im dN+2 ! 0

is exact. This implies that each Im di+1 = ker di (i = N + 2; :::; N 0 + 2) isprojective. ThereforeX

(�1)iCi + (�1)N+1 Im dN+2 + (�1)N0+2 Im dN 0+2 = 0 2 K0(Z�1(Y ))

1.7. IDEMPOTENT CONJECTURE AND BASS CONJECTURE 17

and (�1)N+1[�N+1(Y; YN )] = (�1)N0+1[�N+1(Y; YN 0 )] 2 ~K0(Z�1(Y )): It�s not

hard to see that relative homotopy group �N+1(Y; YN ) is independent of thehomotopy types of YN : This shows that w(Y ) is well-de�ned.Wall [9; 10] proves the following.

Theorem 1.6.9 A �nitely dominated CW-complex Y is homotopy equivalentto a �nite CW-complex if and if w(Y ) = 0: Moreover, for any �nitely presentedgroup � and element x 2 ~K0(Z�); there exists a �nitely dominated CW-complexY with �1(Y ) = � and w(Y ) = x:

Proof. It is clear that if Y is homotopic to a �nite CW complex we have w(Y ) =0; since we could choose Im dN+2 = 0 for su¢ ciently largeN: Conversely, assumethat w(Y ) = 0: This means that �N+1(Y; YN ) is stably free, i.e. for some integersm and n there is an isomorphism

�N+1(Y; YN )M

(Z�)m �= (Z�)n:

Form the bouquet Y 0 = Y _ni=1 SNi of m N -spheres and YN and extend theinclusion YN ! Y to Y 0 by mapping the spheres to the based point. Thenthe relative homotopy group �N+1(Y; Y 0) �= (Z�)n; by considering the relativehomology group HN+1( ~Y ; ~Y 0) = HN+1( ~Y ; ~YN )

L(Z�)n: For each generator of

such free module, attach an (N + 1)-cell to Y 0; getting Y 00 = Y 0 [SN (_DN+1):Since for each integer i the relative homology group Hi( ~Y ; ~Y 00) = 0 ; the relativehomotopy group �i(Y; Y 00) is zero. Therefore, Y is homotopy equivalent to theresulted CW complex Y 00, which is �nite: The other part the theorem could beconstructed similarly.

Example 1.6.10 ~K0(Z�) 6= 0 for the quaternion group � = f�1;�i;�j;�kg.Therefore, there is a �nitely dominated CW complex with fundamental group �,which is not �nite.

A well-known problem is the following.

Conjecture 1.6.11 Let � be a torsion-free �nitely presented group �: Then~K0(Z�) = 0: In other words, any �nitely dominated CW-complex with funda-mental group � is homotopy equivalent to a �nite CW-complex.

1.7 Idempotent conjecture and Bass conjecture

Kaplansky asks the following.

Conjecture 1.7.1 (Idempotent conjecture) Let G be a torsion-free groupand ZG the group ring over integers. Then there are no idempotents, i.e. x =x2; in Z[G] except 0 and 1:

In this section, we will discuss Bass�s conjecture on K0 of group rings, whichis closed related to the idempotent conjecture. Let G be a group and ZG thegroup ring. Denote by [ZG;ZG] the abelian subgroup of ZG generated by ab�bafor all a; b 2 ZG:


Lemma 1.7.2 As abelian groups, there is an isomorphism � : ZG=[ZG;ZG]!L[s]2conj(G) Z[s]; the free abelian group generated by the set conj(G) of conjugacy

classes of elements in G:

Proof. Let f : ZG !L

[s]2conj(G) Z[s] be a abelian group homomorphismde�ned by mapping each g to its conjugacy class [g]: For any two elementsa =

Pagg; b =

Pbhh in ZG; the di¤erence ab � ba is a linear combination of

gh� hg (when expanding ab; if there is a term gh; there must be a term hg inthe expansion of ba). Since hg = h(gh)h�1; the map f factors through a map�: The inverse of � is de�ned by mapping [s] to the image of s in ZG=[ZG;ZG]:The inverse is well-de�ned, since ghg�1 � h = ghg�1 � g�1(gh) 2 [ZG;ZG]:For an idempotent matrix P = (aij)n�n with entries in ZG; its Hattori-

Stallings rank rP is de�ned by

rP = �ni=1aii + [ZG;ZG] =

M[s]2conj(G)

rP (s)[s] 2 ZG=[ZG;ZG]:

Since the trace is invariant under conjugations, this gives a well-de�ned map

HS : K0(ZG)! ZG=[ZG;ZG]:

H. Bass [?] made the following conjecture.

Conjecture 1.7.3 (Bass conjecture) Let G be a group. Then the image ofHS lies in the identity component of ZG=[ZG;ZG] =

L[s]2conj(G) Z[s]: In other

words, for any idempotent matrix P = (aij)n�n; the Hattori-Stallings rank hasrP (s) = 0 for any s 6= 1:

Similar conjectures can be formulated for any commutative ring R insteadof Z, with further requirement that any �nite-order element in G has an ordernon-invertible in R: For example, R = C and G is torsion-free. H. Bass proveConjecture 1.7.3 is true for linear groups. For more updated information, seeEmmanouil�s textbook [6].We will show that Bass conjecture imply idempotent conjecture. Surpris-

ingly, the proof depends on C�-algebras. Let�s recall some basic things fromfunctional analysis.Let G be a group. Then

l2(G) = ff : G! C jXg2G

jf(g)j2 < +1g

is a Hilbert space with inner product

hf1; f2i =Xx2G

f1(x)f2(x);

where f2(x) is the complex conjugation of f2(x). Let

B(l2(G)) = fT : l2(G)! l2(G) j T is linear and the norm k T k= suphx;xi�1

hT (x); T (x)i < +1g

be the set of all bounded linear operators over the Hilbert space l2(G):

1.7. IDEMPOTENT CONJECTURE AND BASS CONJECTURE 19

De�nition 1.7.4 The reduced group C�-algebra C�r (G) is the completion ofC[G] in B(l2(G)) with respect to the operator norm, i.e. T 2 C�r (G) i¤ thereexists ai 2 CG such that k T �ai k! 0 ; the group von Neumann algebra NG isthe completion of C[G] in B(l2(G)) with respect to the weak operator topology,i.e. T 2 NG i¤ there exists ai 2 CG such that h(T � ai)x; yi ! 0 for anyx; y 2 l2(B).

Let S be a subset of B(l2(G)): The commutant B0 is de�ned as fa 2B(l2(G)) j as = sa for any s 2 Sg: If B is selfadjoit, B0 is also selfadjoit,since for any x; y 2 B(l2(G)) we have

ha�sx; yi = hx; s�ayi = hx; as�yi = hsa�x; yi:

A famous result is that NG = (NG)00; i.e. the von Neumann algebra equals itsbi-commutant. The following lemma is essential for our later argument.

Lemma 1.7.5 For an idempotent e 2 NG, there exists projection f 2 NG(means f2 = f = f�; the adjoint of f) such that ef = f and fe = e:

Proof. Note that the image of e : l2(G)! l2(G) is closed, since it is the kernelof 1 � e. De�ne the element f as the projection onto the image of e: Then wehave:

(1) Im f = Im e and the complement Im f? are both NG0-invariant. Notethat e 2 CG � NG and for any A 2 NG0 we have eA = Ae: Thereforefor any v 2 Im e; v0 2 Im e?;

Av = Ae(v) = eA(v) 2 Im e

andhv;Av0i = hA�v; v0i = 0;

which implies Av0 2 Im e?:

(2) f 2 NG00 = NG: For any A 2 NG0 and any x 2 v 2 Im e; v0 2 Im e?; wehave

Af(v + v0) = A(v) = f(A(v) +A(v0)) = fA(v + v0):

It�s not hard to see that f satis�es the prescribed property.

Proposition 1.7.6 De�ne a function tr : NG ! R (or C) by a 7�! ha(1); 1i;called von Neumann trace. Then tr(ab) = tr(ba):

Proof. Assume that for g 2 G � l2(G); a(g) =Pahgh 2 l2(G) and b(g) =P

bhgh: There exist ai 2 CG such that ai ! a in the weak operator topology.Therefore

ahg = ha(g); hi = limhai(g); hi = limhai(1); hg�1i = ha(1); hg�1i = a(hg�1)1:


Then

hab(1); 1i = ha(Xh

bh1h); 1i = hXh

bh1Xh0

ah0hh0; 1i

=Xh

bh1a1h =Xh

ah�11bh1 =Xk

ak1bk�11 =Xk

ak1b1k = hba(1); 1i:

Proof of Bass conjecture implies Idempotent conjecture. Let E =Pegg be an idempotent element in the group ring ZG: Suppose that the Bass

conjecture is true. Let " : ZG! Z be the augmentation map. Then

"(E) =X

eg = e1 +Xg 6=1

eg = e1 +Xg 6=1

rg = e1:

Since "(E) is also an idempotent element in Z, e1 = 0 or 1: After replacing E by1�E; we may assume e1 = 0: A profound result of Kaplansky, called KaplanskyPositivity, says that when e1 = 0; E = 0: The idea is as follows. Embed ZG intoCG; which is a subring of the group von Neumann algebra NG: In the largerring NG; there is a trace t : NG ! C de�ned by g 7�! hg(1); 1il2(G): It is nothard to see that e1 = t(E): Therefore,

e1 = t(E) = t(fE) = t(Ef) = t(f) = t(f�f)

= hf(1); f(1)i = 0:

This implies that f(1) = 0: Suppose that Ai ! f with Ai 2 CG: Then forany g 2 G and y 2 l2(G); we have hAig; yi ! hf(g); yi; and hAi; yg�1i !hf(1); yg�1i = 0: Since hAig; yi = hAi; yg�1i; we have f(g) = 0 for any g 2 G:This means that f = 0 and thus E = 0:There is a weak version of Bass conjecture:

Conjecture 1.7.7 (weak Bass conjecture) Let G be a group. For any idem-potent matrix P = (aij)n�n; the sum of all nontrivial Hattori-Stallings ranksvanishes, i.e. �s 6=1rP (s) = 0:

Berrick, Chatterji and Mislin give an interesting characterization of weakBass conjecture in terms of self-maps of manifolds, as follows.

Theorem 1.7.8 The weak Bass conjecture holds for a �nitely presentable groupG if and only if every pointed homotopy idempotent self-map of a closed, smoothand oriented manifold inducing the identity on the fundamental group G, hasits Lefschetz number equal to the L2-Lefschetz number of the induced map on itsuniversal cover.

1.8 Other kinds of K01.8.1 Exact categories

In this section, we give some other kinds of K0: Since it was introduced byGrothendieck, the concept of K0 has been used in several situations. Whenever

1.8. OTHER KINDS OF K0 21

we talk about a category, we always assume the category is small, i.e. the classof objects is a set. Note that in any abelian category there is a notion of exactsequences for which the Five-Lemma and Snake Lemma are valid.

De�nition 1.8.1 A category with exact sequences is a full subcategory P ofan abelian category A such that P is closed under extension, i.e. if 0 ! P1 !P ! P2 ! 0 are exact sequence in A with P1; P2 2 Ob(P); then P 2 Ob(P):

Example 1.8.2 The following are categories with exact sequences.1) Abelian categories (themselves);2)Let R be a ring and ModR the category of left modules. The subcategory

Pr(R) of �nitely generated projective modules;Proof. Proof. If 0 ! P1 ! P2 ! P3 ! 0 is an exact sequence, then P2 �=P1LP2 is a projective module.

Assume F = Rn or Cn and X is a compact space. A vector bundle over X isa map p : E ! X such that for any x 2 X there is an open set U containing xsuch that p�1(U) is homeomorphic to U �Fn by a map fU : p�1(U)! U �Fnwith the property that pr1�fU = p: Furthermore, on the intersection of two opensets U and V we need fU � f�1V : U \ V � F ! U \ V � F is a homeomorphismin which the �rst coordinate is �xed and the second one a linear isomorphismof vector spaces.For two vector bundles (E; p) and (E0; p0); a map between them is a function

f : E ! E0 with the property p = f �p0 such that f : p�1(x)! p0�1(x) is linearhomomorphic. The category of vector bundles is denoted by Vect(X); which isan additive category.

Example 1.8.3 3)Let X be a compact Hausdor¤ space and Vect(X) the cate-gory of vector bundles. Then Vect(X) is an exact category with exact sequencesde�ned by the split sequences.

De�nition 1.8.4 Let P be a category with exact sequences. Then K0(P) isde�ned as the free abelian group generated by objects of P modulo the followingrelations:1) [P ] = [P 0] if there is an isomorphism P �= P 0 in P:2) [P ] = [P1] + [P2] if there is an exact sequence 0! P1 ! P ! P2 ! 0 in

P:

De�nition 1.8.5 Let X be a compact Hausdor¤ space. The topological K-theory of X is de�ned as Ki(X) = K0(Vect(�

iX)); where �iX is the i-th sus-pension of X: For general locally compact Hausdor¤ space Y; de�ne the K-theoryof Y as Ki(Y ) = lim

XKi(X); where the inverse limit is taken over compact sub-

space of Y:

Proposition 1.8.6 fKigi�0 is a generalized cohomology theory from the cate-gory of pairs of CW complex to the category of abelian groups. In other words,it satis�es the Eilenberg-Steenrod axioms for cohomology theory: homotopy in-variance, long exact sequence of homology groups and excision.


De�nition 1.8.7 Recall that a C�-algebra A is a subalgebra of the set B(H) ofall bounded linear operators over a Hilbert space H such that

1. A is closed in the norm topology of operators;

2. A is self-adjoint, i.e. a 2 A implies the adjoint operator a� 2 A.

Exercise 1.8.8 Let A be a C�-algebra. The matrix ring Mn(A) is also a C�-algebra for each integer n � 1; where the adjoint is de�ned as (aij)� = (a�ji): Sup-pose that Pn(A) = fp 2Mn(A) j p2 = p = p�g, the set of projections in Mn(A):Denote by P (A) = [Pn(A): Two projections p 2Mn(A); q 2Mm(A) are equiv-alent � (Murray-von Neumann equivalence) if there exists u 2 Mm;n(A) suchthat p = u�u and q = uu�:

1. Show that " � " is an equivalence relation and the equivalent classesP (A)= � is a commutative semigroup under the operation [P ] + [q] =[diag(p; q)]: The operator K-group Kop

0 (A) is de�ned as the Grothendieckgroup completion of P (A)= �. (hint: show that u = qu = up = qup byconsidering z�z = 0 for z = (1 � q)u: Use this to show the relation istransitive.)

2. For two equivalent projections p = u�u and q = uu�; show that�q 1� q

1� q q

�;

�u 1� q

1� p u�

�are invertible (actually unitary) and that p; q are stably conjugate.

3. Similar to Lemma 1.7.5, show that for an idempotent matrix e 2 Mn(A)there is a projection f such that ef = f and fe = e: Furthermore, showthat Ane is isomorphic to Anf:

4. Show that the operator K-group Kop0 (A) is isomorphic to the algebraic

K-group K0(A):

1.8.2 Isomorphism between algebraic K0 and topologicalK0

Let X be a compact Hausdor¤ space and C(X) the ring of real-valued continousfunctions. Swan prove

Theorem 1.8.9 Ktop0 (X) �= K0(C(X)):

Outline of proof. Let p : E ! X be a vector bundle. De�ne

�(E;X) = fs : X ! E j p � s = idXg

be the set of continuous sections. It is clear that �(E;X) is a C(X)-module.

1.8. OTHER KINDS OF K0 23

(1) �(E;X) is a �nitely generated projective C(X)-module.

Since X is compact, X is a union of �nitely many open sets fUigki=1: Foreach such Ui; fU : p�1(U) ! Ui � F is a trivial vector bundle over Ui:Suppose eij : Ui ! F (j = 1; : : : ; n) are the constant functions determinedby a standard basis of F: By the partition of unity, there are functionsfi : X ! R such that sup p(fi) � Ui and

Pfi = 1: Then f

Pi eijfignj=1

generates �(E;X): Suppose that fsjgni=1 is a set of generators of �(E;X):De�ne a map

� : X � F ! E

of vector bundles by (x; v1; � � � ; vn) 7�! (x;Pn

i=1 visi): Since the sj(x)spans p�1(x); � is surjective. De�ne a vector E0 = ker�; i.e. E0x = ker�x:It is not hard to check that this is a vector bundle. Then

�(X;E0)M

�(X;E) = �(X;X � F ) �= F;

which shows that �(X;E) is a projective module.

(2) For each �nitely generated projective module P over C(X); there is avector bundle E such that �(X; p) �= P:Suppose that for some projective module Q; we have P

LQ �= C(X;F ):

View P as functions over X: Let

E = f(x; v1; � � � ; vn) 2 X � F : 9s 2 P; s(x) = (v1; � � � ; vn)g

and p : E ! X, projection to the �rst component.

We want to mention two famous results in K0 to conclude this chapter. Formore details, see the �rst chapter of Rosenberg�s book [18] for �rst theorem andLam�s book [13] for the second one.

Theorem 1.8.10 (K0 in number theory) Let F be a number �eld and R thering of algebraic integers in F: In other words, F is �nite algebraic extension ofthe rationals Q and R is the integral closure of Z in F: Then ~K0(R) is isomorphicto the ideal class group of R:

Theorem 1.8.11 (Quillen-Suslin�s theorem) Let F [x1; : : : ; xn] be a poly-nomial ring over a �eld F: Then every �nitely generated projective module overa polynomial ring is a free module. In particular, K0(F [x1; : : : ; xn]) = Z.

Chapter 2

K1 and applications

2.1 Motivations

Let F be a �eld and GLn(F ) the general linear group of size n (n � 2). Inlinear algebra, there are three kinds of matrix operations:

(1) Adding a multiplication of one row (resp. column) to another. The matrixis of the form �

1 r0 1

�;

(2) Interchanging two rows (resp. columns). The matrix is of the form�0 11 0

�;

(3) Multiplying a number r 2 Fnf0g to a row (resp. column). The matrix isof the form �

r 00 1

�:

If we slightly modify the matrix in (2) to be the form�0 �11 0

�; (2�)

we see that the determinant of matrices in (1) and (2�) are both 1: It is nothard to see that any invertible matrix A 2 GLn(F ) could be reduced to be ofthe form (3) by matrices in (1) and (2�).

Exercise 2.1.1 Check this.

This shows that the kernel of the determinant der : GLn(F ) ! Fnf0g isactually the subgroup E generated by matrices of forms (1) and (3). In otherwords, the determinant is a process of taking matrix operations of kinds (1) and(2�).

25


Problem 2.1.2 Could this process be done for any general ring?

2.2 De�nition and basic properties

Let R be a ring. In this section, we will de�ne algebraic K-group K1(R): Let nbe a positive integer and Mn(R) the set of all n�n matrices. Denote by In theidentity matrix. The general linear group GLn(R) = fA 2Mn(R) j there existsB 2 Mn(R) such that AB = BA = Ing is the set of all invertible matrices.There is an injection GLn(R)! GLn+1(R) de�ned by

A 7!�A 00 1

�:

Then the stable general linear group GL(R) is de�ned as [n�1GLn(R); thecolimit of all GLn(R) with these injective maps. For an element r 2 R and anyintegers i; j such that 1 � i 6= j � n; denote by eij(r) the elementary n � nmatrix with 1 in the diagonal positions and r in the (i; j)-th position and zeroselsewhere. The group En(R) is generated by all such eij(r); i.e.

En(R) = heij(r) j 1 � i 6= j � n; r 2 Ri:

Denote by In the identity matrix and by [a; b] the commutator aba�1b�1:The following lemma displays the commutator formulas forEn(R) (cf. Lemma

9.4 in [?]).

Lemma 2.2.1 Let R be a ring and r; s 2 R: Then for integers i; j; k; l with1 � i; j; k; l � n; the following holds:

(1) eij(r + s) = eij(r)eij(s);

(2) [eij(r); ejk(s)] = eik(rs) for distinct integers i; j; k;

(3) [eij(r); ekl(s)] = In for j 6= k and i 6= l:

Similarly, let E(R) = [n�2En(R):

Lemma 2.2.2 (Whitehead lemma) For any ring R; E(R) is the subgroupof GL(R) generated by commutators, i.e. E(R) = [GL(R);GL(R)]:

Proof. According to the commutator formula in Lemma 2.2.1, we see thatevery elementary matrix eij(r) = [eik(1); ekj(r)] is a commutator for su¢ cientlarge k: This shows that the left hand is contained in the right hand. On thecontrary, let A;B 2 GL(R): Without loss of generality, assume that for someinteger n; A;B 2 GLn(R): Then we have in GL(R) that

ABA�1B�1 =

�ABAB�1 0

0 In

�=

�A 00 A�1

��B 00 B�1

��A�1B�1 0

0 BA

�:

2.2. DEFINITION AND BASIC PROPERTIES 27

For any matrix C 2 GLn(R);�C 00 C�1

�=

�0 In�In 0

��0 �C�1C 0

�:

However, for any invertible matrix D 2 GLn(R); we have�0 �D�1

D 0

�=

�In �D�1

0 In

��In 0D In

��In �D�1

0 In

�2 E(R):

This shows that diag(C;C�1) 2 E2n(R) and ABA�1B�1 2 E(R):

De�nition 2.2.3 Let R be a ring. The algebraic K-group K1(R) is de�ned asGL(R)=E(R):

It should be noted that in K1(R); the product of two matrices is the sameas the direct sum of two matrices, i.e. by Whitehead lemma we have [AB] =[diag(AB; I)] = [diag(A;B)][diag(B;B�1)] = [diag(A;B)] for anyA;B 2 GLn(R):

Example 2.2.4 We make some computations for special rings.

1. Let R be a �eld. Then GLn(R) = SLn(R)R� and when n � 3; SLn(R) =En(R) by row and column operations. The process is rough as follows.Suppose A = (aij) 2 SLn(R): By multiplying permutation matrices, wecould assume a11 6= 0: Then

�ni=2ei1(ai1a�111 ) �A ��ni=2ei1(a�111 a1i)

will have nondiagonal entries vanishing in �rst row and column. Repeat-ing such an argument, A could be reduced to be a diagonal matrix bymultiplying matrices in En(R): Then multiply A by matrices of the formdiag(a; a�1) to reduce A to be the identity matrix. Therefore, K1(R) = R

�;the invertible elements in R: More generally, for a commutative local ringR, K1(R) = R

�:

2. Let R = Z, the integers. Suppose that A = (aij) 2 SLn(Z): Consideringthe �rst column, we see that great common divisor (gcd) is 1: By Euclideanalgorithm, we could get a11 = �1 after timing elementary matrices inEn(Z): Then

�ni=2ei1(ai1a�111 ) �A ��ni=2ei1(a�111 a1i)

will have nondiagonal entries vanishing in �rst row and column. Repeatingsuch argument, A could be reduced to be the identity matrix I as previousexample. Therefore for n � 2 we have SLn(Z) = En(Z) and K1(Z) = Z� =Z=2; the two element group. More generally, for an Euclidean domain R;K1(R) = R

�; the units of R:

Similar to K0; K1 also has functorial property and Morita invariance.


Proposition 2.2.5 Let f : R1 ! R2 be a ring homomorphism. Then there isa group homomorphism K1(R1)! K1(R2) induced by f:

Proof. The map f induces maps GLn(R1)! GLn(R2) and En(R1)! En(R2)and therefore indues a map K1(R1)! K1(R2):

Proposition 2.2.6 Let n be a positive integer and f : R ! Mn(R) the inclu-sion of R into the (1; 1)-th position. Then f indues an isomorphism K1(R) !K1(Mn(R)):

Proof. This is similar to the Morita invariance of K0 (using the map A ! A0

de�ned in the proof of Lemma 1.4.2).

Corollary 2.2.7 Let G be a �nite group: Then K1(CG) is (C�)jconj(G)j; wherejconj(G)j is the number of conjugacy classes.

Proof. According to Wendburn-Artin theorem, CG �=LMni(C): The corol-

lary follows the previous proposition.

2.3 K1 in topology: Whitehead torsion

In this section, we will give two applications of K1 to topology.

2.3.1 Whitehead torsion

De�nition 2.3.1 Let G be a group. The Whitehead group Wh(G) is the quo-tient group of K1(ZG) by the image of f�g j g 2 Gg:

Of course, when G is trivial the Whitehead groupWh(G) = 0; sinceK1(Z) =f�1g:

Example 2.3.2 When G = C5 the cyclic group of 5 elements, Wh(G) 6= 0:

Proof. Let t be a generator of G: De�ne a ring homomorphism ZG ! C byt 7! e

2�i5 : This induces a group homomorphism K1(ZG) ! K1(C) = C�: Since

each element in G has norm 1 in C�, it�s enough to �nd an element in K1(ZG)whose norm is not 1: Choose a = 1� t� t�1 2 (ZG)�: It is directly checked that� has image 1� 2 cos 2�

5; which shows that � has an in�nite order in K1(ZG):

2.3.2 s-cobordism theorem

In this subsection, we consider h-cobordism of manifolds, as follows.

De�nition 2.3.3 An h-cobordism over a closed manifoldsM is a compact man-ifold W whose boundary is the disjoint union M [M 0 such that the inclusionsM ,!W;W 0 ,!W are both homotopy equivalent.

2.3. K1 IN TOPOLOGY: WHITEHEAD TORSION 29

Example 2.3.4 The trivial h-cobordism is W =M � [0; 1] and M 0 =M:

The inclusion i : M ,! W has a lift i : ~M ! ~W to the universal covering,which induces a Z[�1(M)]-chain homotopy equivalence i� : C�( ~M) ! C�( ~W ).Here �1(M) acts on the free Z-module Ci( ~M) by deck transformation on basisfor each i. Let

(C�; c�) = C��1( ~M)LC�( ~W );

the contractible mapping cone of i�. Denote by Cod =L

i odd Ci and Cev =Li even Ci: Let r� : C� ! C�+1 be any chain contraction, i.e. ci+1ri + ciri+1 =

idCi for all i: We have an isomorphism (see the following lemma)

(c� + r�) : Cod ! Cev

of Z[�1(M)]-modules. For a �xed bases of C�( ~M) and C�( ~W ); there is a matrixrepresents the isomorphism (c� + r�): This gives an element in K1(Z[�1(M)]);which is mapped to be an element in Wh(�1(M)):

De�nition 2.3.5 (Whitehead torsion) For an h-cobordism (W ;M;M 0); de-�ne the Whitehead torsion as

�(M ,!W ) = [(c� + r�)] 2Wh(�1(M)):

Lemma 2.3.6 The element �(i) is well-de�ned, i.e. it is independent of choicesof the basis and of the contraction r�:

Proof. Suppose that �� : C� ! C�+1 is another chain contraction and c�+ �� :Cev ! Cod. De�ne

�n = (rn+1 � �n+1)�n and �n = (�n+1 � rn+1)rn:

It could be directly checked that (id+��)od and (id+��)ev and the compositions(c�+r�)(id+��)od(c�+��) and (c�+��)(id+��)ev(c�+r�) are upper triangularmatrices (this also shows (c�+r�) is an isomorphism, since each such a triangularmatrix is invertible).The following theorem gives a criterion when an h-cobordism is trivial.

Theorem 2.3.7 (s-cobordism theorem) Suppose thatMn (n � 5) is a closed(resp. di¤erentiable) manifold and (W ;M;M 0) is an h-cobordism over M: ThenW is homeomorphic (resp. di¤eomorphic) to M� [0; 1] relative to M if and onlyif �(M ,!W ) = 0 2Wh(�1(M)):

Outline of proof. We consider the smooth case. For the compact manifoldW there is a handlebody decomposition

W =M � [0; 1] [n�5i=0 [ijj=1 [�ij D

i+1j �Dn�i;

where �ij : Si �Dn�i ! M is an embedding, similar to cellular decomposition

of a CW complex. For 1 � i � n � 4; there is a decomposition involving only


�ij and �i+1j : Such decomposition, called normal form, is unique up to di¤eo-

morphisms relative to M: If �(M ,!W ) = 0; this means a matrix representing�(M ,!W ) could be reduced to the identity matrix by elementary matrices upto multiplication by an element in �1(M): For such elementary matrices andmultiplication by an element in �1(M); we could modify the decomposition innormal form of W to get a new manifold W 0 with decomposition in normalform which is di¤eomorphic to W relative to M such that the representationmatrix of W 0 is the identity. According to the uniqueness of the decompositionin normal, W 0 is di¤eomorphic to M � [0; 1]: For more details, see the proof ofTheorem 1.1 in [8].When �1(M) is trivial, i.e. M is simply connected, the Whitehead torsion

is trivial. Therefore, any cobordism over M is di¤eomorphic to M � [0; 1].

Corollary 2.3.8 (Poincaré conjecture for n � 6) Let M be a closed topo-logical manifold of the homotopy type of the sphere Sn; n � 6: Then M is home-omorphic to Sn:

Proof. Let W be obtained by removing two small disjoint n-ball D1 and D2:Then (W ;Sn�1; Sn�1) is an h-cobordism. Since the fundamental group of Sn�1

is trivial,W is homeomorphic to Sn�1�[0; 1] �xing Sn�1�0 by the s-cobordismtheorem. By Alexander trick, the homeomorphism @D1 = S

n�1 � 0 f! Sn�1 �1 = @D2 could be extended to a homeomorphism D1 ! D2 by sending tx !tf(x) for x 2 @D1; t 2 [0; 1]: This implies that M =W [D1 [D2 is Sn:

Conjecture 2.3.9 Let G be a torsion-free �nitely presented group. Then anyh-cobordism (W ;M;M 0) over a closed manifold Mn (n � 5) with �1(M) = Gis homeomorphic to M � [0; 1]: This is equivalent to that Wh(G) = 0:

2.3.3 Simple homotopy equivalence

Let A be a subcomplex of a CW complex X: The inclusion A ,! X is calledelementary expansion, written A %e X; if X = (A [f Bn) [g Bn+1 for someattaching map f : Sn�1 ! A and the attaching map g : Sn ! A [f Bn; whichmaps one hemi-sphere of Sn identically onto the n-cell and the other hemi-sphereinto A: In other words, X is the pushout

Sn+ ! A# #

Bn+1 ! X:

More generally, we say that X collapses to A or A expands to X if

A%e X1 %e X2 %e � � � %e X:

The inclusion A ,! X is simple homotopy equivalence if X can be obtainedfrom A by a series of expansions and collapses �xing A: Suppose that A ,! Xis simple homotopy equivalence. Using the same way as previous subsection toconsider the mapping cone, we can de�ne a torsion �(A ,! X) 2Wh(�1(A)):

2.4. BASS-HELLER-SWAN THEOREM 31

Theorem 2.3.10 Assume that (X;A) is a pair of CW complexes and the in-clusion A ,! X is a homotopy equivalence. Then the inclusion is a simplehomotopy equivalence if and only if �(A ,! X) = 0 2Wh(�1(A)):

Proof. Suppose that A ,! X is simple homotopy equivalence. By de�nition,torsions are additive for the composition of two homotopy equivalences (notethat for inclusions A ,! X1 ,! X; C�(X;A) = C�(X;X1)

LC�(X1; A)). There-

fore, it is enough to prove the case when A ,! X a one-step expansion. Butin such case, the chain complex of the mapping cone is concentrated in twoconsecutive dimensions and the nontrivial di¤erential is id : Z�1(A)! Z�1(A):This means that �(A ,! X) = 0: The other direction is similar to the proof ofs-cobordism theorem, see [5] or Theorem 2.21 in [8] for more details.Conjecture 2.3.9 in the previous subsection implies that for a homotopy pair

(X;A) with a torsion-free fundamental group is always a simple homotopy pair.

2.4 Bass-Heller-Swan theorem

In this section, we consider the relations betweenK1(R);K1(R[t]) andK1(R[t; t�1]):

The maps R ,! R[t] and R[t] ! R; t 7�! 1; induce injection K1(R) !K1(R[t]):Denote byNK1(R) the cokernel. Therefore,K1(R[t]) �= K1(R)

LNK1(R):

Lemma 2.4.1 (Higman trick) Let R be a ring. Then

1. Any B 2 GL(R[t]) can be reduced, modulo GL(R) and E(R[t]); to a matrixof the form 1+At; where A is a nilpotent matrix over R; i.e. Am = 0 forsome integer m:

2. Any B 2 GL(R[t; t�1]) can be reduced, modulo GL(R) and E(R[t; t�1]);

to a matrix of the form�tn

1

�(1 + A(t� 1)); where A is a matrix over

R and A(1�A) is nilpotent. Moreover, A = P +N;PN = NP for someidempotent matrix P , nilpotent matrix N:

Proof. For the �rst part, assume B = B0 + tB1 + � � �+ tdBd with Bi a matrixover R for each i: Since

B ��B 00 1

��B td�1Bd0 1

��

�B � tdBd td�1Bd�t 1

�mod GL(R) and E(R[t]);

we could assume that d � 1: If d = 0; there is nothing to prove. Since B isinvertible, B0 is also invertible. After replacing B by B�10 B; we could assumeB0 = I: Assume that B�1 = C0 + C1t+ � � � + trCr for some integer r: By theequality

I = (I +B1t)(C0 + C1t+ � � �+ trCr);


we get that C0 = I; C1 = �B1 and inductively Ci = (�1)iBi1: Since Cr+1 = 0;Br+11 = 0:For the second, we could still assume B = B0+ tB1 after timing some power

of t: The map R[t; t�1]! R; t 7! 1; induces B0�B1 is an invertible matrix. Wecould assume that

B = I +A(t� 1) = (I �A) +At:

Similar trick as in (1) shows that (I �A)A is nilpotent.Suppose that (I �A)rAr = 0: Since (1� x)r and xr are prime to each other

in Z[x], there exists polynomial p(x) and q(x) in Z[x] such that p(x)xr+q(x)(1�x)r = 1: Let P = p(A)Ar and N = A� P: Then

P (1� P ) = p(A)Ar(I � p(A)Ar) = p(A)q(A)Ar(1�A)r = 0

and N = (A� I)+ (I�P ) = (A� I)(�I+ q(A)(I�A)r�1) = �(A� I)Arp(A);which is divisible by both A and I �A: Therefore N is nilpotent.

Theorem 2.4.2 Let R be a ring. Denote by Nil(R) the category whose objectsare pairs (P;A); where P is a �nitely generated free R-module and A a nilpo-tent endomorphism of P ; and morphisms are R-homomorphism T : P ! P 0

such that A0T = TA: Nil(R) is an exact category with exact sequences the splitsequences. Then

1. K1(R[t]) �= K1(R)L ~K0(Nil(R)); where ~K0(Nil(R)) �= NK1(R) is the

quotient of K0(Nil(R)) by the image of pairs with A = 0:

2. K1(R[t; t�1]) �= K1(R)

LK0(R)

LNK1(R)

LNK1(R):

Outline of proof. (1) We will show that there is an isomorphism f :NK1(R) �= ~K0(Nil(R)): By (1) of the previous lemma, each element in NK1(R)is presented by 1+At for some nilpotent matrix An�n over R: De�ne f [1+At] =[(Rn; A)]; the element in ~K0(Nil(R)) represented by the pair (Rn; A): Check thisis well-de�ned as follows. If I +A0t is conjugate to I +At by an invertible ma-trix in GL(R); then their images are isomorphic. The map f is also a grouphomomorphism, since

f([I +At] + [I +A0t]) = f([diag(I +At; I +A0t)])

= f([I + diag(A;A0)t])

= [(Rn; A)] + [Rn; A0]:

In order to show f is isomorphic, it su¢ ces to de�ne its inverse by assign a pair(P = Rn; A) the class [I + At]): If 0 ! (P1; A1)

�1! (P2; A2)�2! (P3; A3) ! 0 is

exact, (I +A2t)�1 = �1(I +A1t) and (I +A3t)�2 = �2(I +A2t): Without lossof generality, we assume that P1 has a basis fvigmi=1 and P3 has a basis fuigni=1such that fui; �1(vj)g is a basis of P3 (by viewing P1 and P3 as submodulesof P2). Assume furthermore that the automorphisms A1 and A3 have matrixrepresentations A1 = (aij) and A3 = (bij): Now A3(�1(vi)) = �1(A1(vi)) =

2.4. BASS-HELLER-SWAN THEOREM 33

Pj aija1(vj) and A3(ui) =

Pj cij�1(vj) +

Pj bijuj : Therefore, A2 has the

matrix form �A 0C B

�=

�A 00 B

� �I 0

B�1C I

�:

This implies that [I +A2t] = [I +A1t][I +A2t] in K1(R[t]):(2) For a �nitely generated projective R-module P; suppose that for some Q

we have PLQ �= Rn: Let t

NP denote the automorphism t

NidP

LidR[t;t�1]

NRQ

onR[t; t�1]N

R(PLQ) �= R[t; t�1]n:De�ne a map � : K1R

LK0R! K1(R[t; t

�1])by

[A] + [P ] 7! [A] + tNP:

We will de�ne a converse map = (1;2) : K1(R[t; t�1]) ! K1R

LK0R as

follows. For an invertible matrix A over R[t; t�1]; let 1(A) be the invertiblematrix over R induced by ring homomorphism R[t; t�1] ! R; t 7�! 1: Forinvertible matrix A : (R[t; t�1])n ! (R[t; t�1])n; there exists some integer msuch that tmA is a matrix over R[t]; which could be viewed as a map R[t]n !R[t]n:We claim that [R[t]n= Im tmA]�m[R[t]n=tR[t]n] is a well-de�ned elementin K0(R). It could be checked that � � = id; which means K0(R)

LK1(R) is

a direct summand. By (2) of the previous lemma, any [B] 2coker(�) could bereduced to the form [I +(P +N)t] with P idempotent and N nilpotent. De�nea map coker(�)! ~K0(Nil(R))

L ~K0(Nil(R));

[B] 7�! ((Rn; PN); (Rn; (I � P )N)):

Similar to (1), this is an isomorphism.

De�nition 2.4.3 Let R be a left Noetherian ring. R is called (left) regular ifevery �nitely generated (left) R-module M has a �nite projective resolution, i.e.there exists a long exact sequence

0! Pn ! Pn�1 ! � � �P0 !M ! 0

with Pi �nitely generated projective.

For example, R = Z or F a �eld.In order to prove when R is regular the nil-group is trivial, we need the

following lemma.

Lemma 2.4.4 (Resolution theorem) Suppose that P ,! M are two exactsubcategories of an abelian category A: Suppose that

1. each object M in M has a �nite resolution by objects in P; i.e. there isan exact sequence

0! Pn ! Pn�1 ! � � � ! P0 !M ! 0

with Pi in P:


2. if 0 ! M1 ! M2 ! M3 ! 0 is an exact sequence in A with M2;M3 inM (resp. P ), then M1 lies in M (resp. P ).

Then K0(P) = K0(M):

Proof. Since P is an exact subcategory ofM, there is a map K0(P)! K0(M):Conversely, for an object M inM having a resolution, de�ne its inverse imageasP(�1)i[Pi]:

Proposition 2.4.5 Let R be a regular ring. Then NK1(R) = 0: In particular,K1(R[t]) �= K0(R) and K1(R[t; t

�1]) �= K1(R)LK0(R):

Outline of proof. According to the previous theorem, it�s enough to provethat ~K0(Nil(R)) = 0: Extend the category Nil(R) by adding objects (P; 0) withP a �nitely generated projective R-module and 0 the trivial map. The newcategory is denoted by Nil(R)P ; which is also an exact category. It is not hardto see that ~K0(Nil(R)) = ~K0(Nil(R)P ); where ~K0 is de�ned as K0 modulo outthe objects with trivial endomorphism. Each object (P; �) in Nil(R)P has a�ltration

0 = Im�n � Im�n�1 � � � � Im� � P:Extend Nil(R)P to be a new categoryM containing all such Im�i and quotientsIm�i= Im�i+1: By the assumption that R is regular, each object in M has aprojective resolution in Nil(R)P : Therefore, the object (Im ai= Im�i+1; 0) has aprojective resolution

0! (Pn; 0)! (Pn�1; 0)! � � � ! (P0; 0)! (Im ai= Im�i+1; 0)! 0:

This implies that inK0(M) we have [(Im ai= Im�i+1; 0)] =P(�1)i[(Pi; 0)]: The

Resolution theorem says that K0(M) = K0(Nil(R)P ). But

[(P; a)] =X(�1)i[(Im ai= Im�i+1; 0)] 2 K0(M) = K0(Nil(R)P ):

This implies that [(P; 0)] = 0 2 ~K0(Nil(R)) and ~K0(Nil(R)) = 0:For more details on the proof, see Weibel [22], Chapter III, and Rosenberg

[18], Chapter 3. For K0; we have the following result:

Theorem 2.4.6 (Grothendieck) Let R be a regular ring. Then K0(R[t]) =K0(R) = K0(R[t; t

�1]):

Corollary 2.4.7 Let � be any group. We have

Wh(� � Z) �= Wh(�)M

~K0(Z�)M

(NK1(Z�))2:

In particular, Wh(Zn) = 0 for any positive integer n:

Proof. This is clear from the isomorphism in Bass-Heller-Swan theorem. Whenring of integers Z is regular, the group ring Z[t; t�1] is also regular. Inductively,we prove that Wh(Zn) = 0:

Chapter 3

K2

Let R = Z be the ring of integers and n � 3 an integers. By the commutatorformula (cf. Lemma 2.2.1), we know that the special linear group SLn(Z) isgenerated by feij(1) j 1 � i 6= j � ng (actually all i; j with j i � j j= 1 areenough).

Problem 3.0.8 Is the group SLn(Z) (n � 3) �nitely presented? In other words,does there exist �nitely many generators fgigki=1 and relators frjgmj=1 such that

SLn(Z) �= hg1; g2; � � � ; gk j r1; r2; � � � ; rmi?

In this chapter, we will propose a possible way to solve this problem. Therough idea is as follows. First, de�ne a group Stn(Z) with generators xij(1)corresponding to eij(1) for each pair (i; j) and relations in Lemma 2.2.1. Thenstudy the kernel of Stn(Z)! SLn(Z): It is clear that when this kernel is �nite,the special linear group SLn(Z) is �nitely presented.

3.1 Steinberg group and de�nition of K2Let R be a ring and n an integer, En(R) the elementary subgroup of generallinear group generated by elementary matrices. As shown in Lemma 2.2.1,elements in En(R) satisfy several relations. A natural question will be: Arethere other relations in En(R)?

De�nition 3.1.1 The Steinberg group Stn(R) is a group generated by fxij(r) jr 2 R; 1 � i 6= j � ng subject to the following relations (called Steinbergrelations)1) xij(r + r0) = xij(r)xij(r0) for any r; r0 2 R;2) [xij(r); xjk(r0)] = xij(r)xjk(r

0)xij(r)�1xjk(r

0)�1 = xik(rr0) for distinct

i; j; k:

3) [xij(r); xst(r0)] = I for i 6= t and j 6= s:

35

36 CHAPTER 3. K2

Note that the relation 1) implies that xij(r)�1 = xij(�r) and then therelation 2) implies that

[xji(r); xkj(r0)] = xki(�rr0):

From the de�nition, there is a group homomorphism 'n : Stn(R)! En(R)de�ned by xij(r) 7! eij(r): There is a natural map Stn(R) ! Stn+1(R) byinclusion of xij(r) (Caution: this is not always injective as GLn!). Denoteby St(R) = limStn(R); the stable Steinberg group. Then we have a map' : St(R)! E(R):

De�nition 3.1.2 The kernel of ' : St(R)! E(R) is K2(R):

We will show that K2(R) is abelian and actually the center of St(R): Thefollowing lemma is useful for later argument.

Lemma 3.1.3 Let Nn (n � 3) be the subgroup of Stn(R) generated by xij(r);1 � i < j � n; r 2 R: Then 'n maps Nn injectively onto the upper triangularmatrix in En(R):

Proof. Let K1 be the subgroup of Stn(R) generated by x1i(r) for i = 2; : : : ; nand r 2 R: By Steinberg relation 3), K1

�= Rn�1. Therefore, 'n(K1) is isomor-phic to the subgroup generated by e1i(r) for i = 2; : : : ; n and r 2 R: Similarly,the subgroup Kj (j = 2; : : : ; n � 1) generated by xji(r) for i = j + 1; : : : ; nand r 2 R is mapped isomorphic to the subgroup generated by eji(r) fori = j + 1; : : : ; n and r 2 R:Denote by K12 the subgroup generated by K1 and K2: By the commutative

diagram1! K1 ! K12 ! K2 ! 1

#�= # #�=1! 'n(K1) ! 'n(K12)! 'n(K2) ! 1;

the subgroup K12 is isomorphic to 'n(K12): Inductively, we know that 'n is anisomorphism on Nn:

Theorem 3.1.4 K2(R) is the center of St(R): Therefore, there is a long exactsequence

1! K2(R)! St(R)! GL(R)! K1(R)! 1:

Proof. Let x be an element in the center of St(R): Since ' is surjective, '(x) isin the the center of En(R); which is trivial when n � 3. Therefore, x 2 K2(R):Conversely, let y 2 K2(R): Write y as a product of xij(r): Suppose that N isan integer larger than the subscripts i and j in the product. Then y normalizesthe subgroup KN generated by all xiN (r) for i = 1; : : : ; N � 1 and r 2 R:Choose z 2 KN : We have '(yzy�1z�1) = I; which implies yz = zy by Lemma3.1.3, since yzy�1 lies in KN as well. Similarly, y commutes with the subgroupK 0N generated by xNi(r) for i = 1; : : : ; N � 1 and r 2 R: However, Ki and

K 0i (i = N;N + 1; : : :) generates St(R) by Steinberg relations (eg. xij(r) =

[xiN (1); xNj(r)]). Therefore y is in the center of St(R): The exact sequence isobvious by mapping St(R) into GL(R) through E(R):

3.2. UNIVERSAL CENTRAL EXTENSION 37

Example 3.1.5 Let R be any ring. Consider x = x12(1)x21(�1)x12(1): Since'(x4) = I 2 E(R); x4 2 K2(R): When R = Z, x4 is order 2 element in K2(Z):

Remark 3.1.6 We could de�ne K2;n(R) as the kernel of 'n : Stn(R) !En(R): There is a natural map fn : K2;n(R) ! K2(R): The stability of K2

says that when R = Z and n � 4; the map fn is isomorphic (cf. [20]). There-fore, the special linear group SLn(Z) is �nitely presented and the presentationcould be given by Steinberg relations and x4 in the previous example.

3.2 Universal central extension

Let�s see some group theoretic de�nitions. A central extension of a group G is anexact sequence of groups 1! A! K ! G! 1 such that A is in the center ofK.A universal central extension is a central extension 1! A! U ! G! 1 suchthat for any central extension 1 ! A ! K ! G ! 1 there is a commutativediagram

1! A! K ! G! 1# # #

1! A! K ! G! 1

where the vertical arrow on G is the identity.

Proposition 3.2.1 For a group G, there is a universal central extension if andonly G is perfect, i.e. G = [G;G].

Proposition 3.2.2 When n � 5; the group homomorphism 'n : Stn(R) !En(R) is a universal central extension. Moreover, ' : St(R) ! E(R) is auniversal central extension.

3.3 Relative K-theory and exact sequences

In this section, I will always be an ideal of a ring R: We will de�ne the relativeK-groups that will �t into a long exact sequence.

3.3.1 Relative K0

De�nition 3.3.1 Let I be a ring with or without identity. Denote by I+ =ILZ the ring with identity de�ned by

(x;m)(y; n) = (xy +my + nx;mn):

The identity is I+ is (0; 1)

Remark 3.3.2 If I has already an identity e, then I+ ! I �Z; (x;m) 7! (x+me;m) is an isomorphism. Further more, the construction of I+ is functorial.

38 CHAPTER 3. K2

De�nition 3.3.3 Let I be a ring with or without an identity. There is a splitexact sequence

0! I ! I+ ! Z! 0;

given by Z!I+; n 7! (0; n): De�ne K0(I) = ker(K0(I+)! K0(Z) = Z): If I isan ideal of a ring R; the relative K0-group K0(R; I) is de�ned as K0(I):

It is clear that when I has an identity, the previous de�nition of K0(I) is thesame as that de�ned earlier. We give another characterization of relative K0:

De�nition 3.3.4 For an ideal I in a ring R; let D = R�IR be a ring f(x; y) 2R � R : x� y 2 Ig; where the addition and multiplication are de�ned pairwise.Denote by q : R�I R! R the projection onto the �rst component.

Proposition 3.3.5 K0(R; I) �= Ker(K0(q) : K0(D)! K0(R)):

Proposition 3.3.6 De�ne f : I+ ! D by (x; n) 7�! (n � 1R; n � 1R + x): Thecommutative diagram

I+f! D

g # q #Z ! R

induces a map K0(R; I)! Ker(K0(q)):

(1) Surjectivity For [x]� [y] 2 Ker(K0(q)) (x; y are idempotent matrices overD), i.e. [q(x)] = [q(y)] 2 K0(R); there exists an invertible matrices Aover R and integer r such that

Adiag(q(x); Ir)A�1 = diag(q(y); Ir)

(cf. Proposition 1.3.4). Suppose that the size of diag(x; Ir) is n, we mayassume that

diag(A;A�1)diag(q(x); Ir; 0n)diag(A;A�1) = diag(q(y); Ir; 0n):

Since q is surjective and diag(A;A�1) 2 E2n(R), there exists an elemen-tary matrix B 2 E2n(D) such that q(B) = diag(A;A�1): Therefore

[x]� [y] = [BxB�1]� [y] = [diag(BxB�1; 1�BxB�1)]� [diag(y; 1�BxB�1)]= [diag(I; 0)]� [diag(y; 1�BxB�1)]= [diag(I; 0)]� [diag(y �BxB�1; 1)];

which is an element in K0(R; I):

(2) Injectivity Let [x] � [y] 2 K0(R; I) for idempotent matrices x; y over I+:As above, after direct sum both elements with 1�y we may assume y = Irfor some integer r: By de�nition of K0(R; I); we may assume g(x) = Ir:

3.3. RELATIVE K-THEORY AND EXACT SEQUENCES 39

Now x = (x0; Ir) and y = (Ir; Ir) over I+: Suppose that [f(x)] = [f(y)] 2Ker(K0(q)): This means

[(Ir; x0)] = [(Ir; Ir)] 2 K0(D);

where x0 is the I-part of x: Then there exists (A1; A2) 2 GL(D) suchthat A2x0A

�12 = Ir: Therefore, A

�11 A2x0A

�12 A1 = Ir: Note that A

�11 A2 �

Irmod I and A�11 A2 2 GL(I): This implies [x] = [Ir] 2 K0(R; I):

For the functoriality of the construction of I+; we see that there is a sequenceK0(I) ! K0(R) ! K0(R=I): Moreover, we will prove later that this is exact.In the language of D; the map K0(R; I)! K0(R) is induced by the projectionto the second component R�I R! R.

3.3.2 Relative K1

For any ideal I in a ring R; the quotient map R ! R=I induces group homo-morphism qn : GLn(R) ! GL(R=I): De�ne GLn(R; I) = ker qn as the relativegeneral linear group. The relative elementary group is de�ned as the normalsubgroup generated by eij(r) with r 2 I and 1 � i 6= j � n: Taking direct limit,we de�ne GL(R; I) = [n�1GLn(R; I) and E(R) = [n�1En(R; I).

Proposition 3.3.7 E(R; I) is a normal subgroup of GL(R; I). Furthermore,[GL(R; I);GL(R; I)] � E(R; I):

Proof. Let A 2 GLn(R; I) and B 2 En(R; I): Then�ABA�1 00 In

�=

�A 00 A�1

� �B 00 In

� �A�1 00 A

�:

Since in the proof of Whitehead Lemma,

diag(A;A�1) =

�In A� In0 In

� �In 0In In

� �In �A�1(A� In)0 In

� �In 0�In In

� �In 0

�(A� In) In

�:

Since A � In has entries in the ideal I; the second product lies in E2n(R; I):Therefore the �rst product lies in E2n(R; I): Let B 2 GLn(R; I); the secondequality also implies that

diag(ABA�1B�1; In; In) = [diag(A;A�1; In);diag(B; In; B

�1)] 2 E2n(R; I):

De�nition 3.3.8 For an ideal I in a ring R; the relative K1 is de�ned as

K1(R; I) = GL(R; I)=E(R; I):

40 CHAPTER 3. K2

3.3.3 Relative K2

Let D = R �I R be the ring de�ned in De�nition 3.3.4. Denote by p1; p2 :D ! R the two projections to the �rst and second components. These are ringhomomorphisms and induce maps of K2:

De�nition 3.3.9 For an ideal I in a ring R; the relative K2-group K2(R; I) isde�ned as

kerK2(p2)=[ker St(p1); ker St(p2)]:

Proposition 3.3.10 [ker St(p1); ker St(p2)] is a normal subgroup of St(D) con-taining in kerK2(p2):

3.3.4 Long exact sequence of K-groups

Theorem 3.3.11 There is a long exact sequence

K2(R)q2! K2(R=I)

@2! K1(R; I)i1! K1(R)

q1! K1(R=I)@1! K0(R; I)

i0! K0(R)q0! K0(R=I):

Proof. Suppose that q : R! R=I is the quotient map. First, we need to de�neall the maps involved. For an element [A] 2 K1(R=I); let A 2 GLn(R=I) be aninvertible matrix and A0 its preimage in Mn(R): De�ne a D-module Rn�A Rnas f(x; y) 2 Rn �Rn : q(y) = q(xA0)g with D-action by (a; b) � (x; y) = (ax; by)for (a; b) 2 D: It is not hard to see that Rn �A Rn is independent of choices ofA0 and that

Rn �A RnM

Rn �A�1 Rn �= R2n �diag(A;A�1) R2n

= f(x; y) 2 R2n �R2n : x� Cy 2 Ing�= (R�I R)2n;

where C is an invertible matrix whose image is diag(A;A�1): Therefore, Rn�ARn is a projective D-module. De�ne @1([A]) = [Rn �A Rn]� [Dn] 2 K0(R; I):This is a well-de�ned map (check this!). It is clear that any composition of twomaps is zero. We prove the theorem in several cases.

(1) exactness at K0(R): Suppose that for some idempotent matrices x; y; theelement [x] � [y] 2 K0(R) satis�es q0([x]) = q0([y]) 2 K0(R=I): There-fore for some invertible matrix A 2 GL(R=I) and integer r; we haveAdiag(q(x); Ir)A

�1 = diag(q(y); Ir) (cf. Proposition 1.3.4). Suppose thatthe size of diag(x; Ir) is n, we may assume that

diag(A;A�1)diag(q(x); Ir; 0n)diag(A;A�1) = diag(q(y); Ir; 0n):

However, diag(A;A�1) 2 E2n(R=I); which could be lifted to an invertiblematrix of En(R): Denote by B 2 En(R) a lifting of diag(A;A�1): ThenBxB�1 � y is a matrix over I: The element

[(BxB�1; BxB�1)]� [(BxB�1; y)] 2 K0(R; I)

has image [x]� [y] in K0(R):

3.3. RELATIVE K-THEORY AND EXACT SEQUENCES 41

(2) exactness at K0(R; I): Let [P ] � [Dn] be an element in ker(K0(R; I) !K0(R)): View K0(R; I) � K0(D) with P a projective module over D suchthat i0([P ]) = i0(D

n): Then each projection p1; p2 has the image of Pa free module in K0(R): After directing sum with free modules, we mayassume that P has the same rank in p1(K0(D)) and p2(K0(D)): This isto say that P is of the form Rn �A Rn: Therefore, [P ] � [Dn] lies in theimage of K1(R=I):

(3) exactness at K1(R=I): Suppose that [A] 2 K1(R=I) has trivial image inK0(R; I): In other words, [Rn �A Rn] is stably free, i.e for some integerm we have [Rn�ARn]

LDm �= Dm+n: After replacing A by diag(A; Im);

we may assume that there is an isomorphism

f : Dn = (R�I R)n = Rn �In Rn ! Rn �A Rn:

Let feigni=1 be a standard basis ofRn:De�ne matricesB;C by (eiB; eiC) =f(ei; ei): In other words, the i-th rows of B and C are de�ned as the �rstand second component of f(ei; ei): Since f is invertible, B;C are invertible(actually f�1(x; y) = (xB�1; yC�1)). By the de�nition of Rn �A Rn; wehave q(B�1C) = A: Therefore, [A] lies in the image of q1:

(4) exactness at K1(R; I) and K1(R): Let St(R; I) be the kernel of St(R)!St(R=I): It could be directly checked that St(R; I) is the normal subgroupof St(R) generated by xij(r) with r 2 I (check this!). Therefore thenatural map St(R; I)! E(R; I) is surjective. Consider the commutativediagram

St(R; I) ! St(R) ! St(R=I)# # #

GL(R; I) ! GL(R) ! GL(R=I):

The snake lemma for groups (cf. the following exercise) implies that thereis an exact sequence

K2(R)! K2(R=I)! K1(R; I)! K1(R)! K1(R=I):

Exercise 3.3.12 Suppose that

A1 ! A2 ! A3 ! 1f1 # f2 # f3 #

1! B1 ! B2 ! B3

is commutative diagram of groups with exact rows and assume that the imageIm fi is normal in Bi for each i: Then it induces an exact sequence of groups

ker f1 ! ker f2 ! ker f3 ! cokerf1 ! cokerf2 ! cokerf3:

42 CHAPTER 3. K2

Chapter 4

High algebraic K-theory

Roughly, there are three kinds of K-theory: algebraic K-theory, topologicalK-theory and K-theory of operator algebras. We have already de�ned hightopological K-theory in De�nition 1.8.5, i.e. for a compact CW complex Xthe Ktop

i (X) is de�ned as Ktop0 (�iX); the Grothendieck group of isomorphism

classes of vector bundles over the i-th suspension ofX:Moreover, these fKtopi gi�0

form a cohomology theory (cf. Proposition 1.8.6). Similarly, for a C�-algebraA; we de�ne the operator K-theory Kop

0 (A) and show that it is isomorphic toalgebraic K-group K0(A) in Exercise 1.8.8. The high operator K-theory couldbe de�ned as follows. The stable general linear group GL(A) is a topologicalgroup with topology coming from lim

n!1Mn(A): For each positive integer i; the

i-th operator K-group is de�ned as

Kopi (A) = �i�1(GL(A)):

In particular, Kop1 (A) = GL(A)=GL(A)0 is the coset of connected component

containing the identity: Such K-groups will �t into a six-term exact sequence(cf. [3]). A natural question is the following:

Problem 4.0.13 Is there a de�nition of high algebraic K-theory to extend thelong exact sequence

K2(R)q2! K2(R=I)

@2! K1(R; I)i1! K1(R)

q1! K1(R=I)@1! K0(R; I)

i0! K0(R)q0! K0(R=I)?

This question will be answered by Quillen�s plus construction (to de�ne highalgebraic K-theory).

4.1 Classifying spaces with respect to families ofsubgroups

Let G be a group.

43

44 CHAPTER 4. HIGH ALGEBRAIC K-THEORY

4.1.1 G-CW complexes

A G-CW complex is a G-space X together with a �ltration

; = X�1 � X0 � X1 � � � � � X

such that X = colimn!1 [Xn . The (n+ 1)-skeleton is obtained by attached(n + 1)-dimensional G-cells G=H � Dn+1 to its n-skeleton, i.e. Xn+1 is thepushout the following diagram

qi2IG=Hi � Sn ! Xn# #

qi2IG=Hi �Dn+1 ! Xn+1;

where I is a index set.

Example 4.1.1 The real line X = R with group action of Z = hti by tx = x+1is a two-dimensional G-CW complex with X0 = Z and the 1-G-cell Z� [0; 1].

4.1.2 Classifying spaces

De�nition 4.1.2 Suppose that F a family of subgroups which is closed undertaking subgroup groups and conjugations, i.e. H < K;K 2 F implies H 2 Fand gKg�1 2 F for any g 2 G:

Example 4.1.3 F could be

1. the family of �nite subgroups Fin,

2. the family of cyclic subgroups Cyc;

3. the family of virtually cyclic subgroups Vcyc(Recall that a subgroup H < Gis called virtually cyclic if H contains a cyclic subgroup of �nite index.);

4. the family of trivial subgroup T r; or

5. the family of all subgroups All:

De�nition 4.1.4 A classifying space with respect to a family F is a G-CWcomplex EF (G) such that the �xed point set EF (G)H is contractible if H 2 F orempty if H =2 F . (In particular, EF (G) is contractible when H is the trivialsubgroup.)

Theorem 4.1.5 There exists a classifying space EF (G) and EF (G) is uniqueup to G-homotopy equivalences.

Proof. The existence is by a similar argument of bar resolution, as follows. Weconstruct a simplicial EF (G): The n-simplex is

� = g0S0(g1S1; :::; gnSn);

4.1. CLASSIFYING SPACESWITH RESPECT TO FAMILIES OF SUBGROUPS45

where each Si 2 F and giSi is a left coset. We further require that

Si�1gi � giSi (*)

for each i: The boundary map is de�ned as

d0� = g0g1S1(g2S2; :::; gnSn);

dn� = g0S0(g1S1; :::; gn�1Sn�1);

di� = g0S0(g1S1; :::; gi�1Si�1; gigi+1Si+1; :::; gnSn)

for 1 � i � n� 1: d0 and di is well-de�ned since (*). The group G acts on � byg� = gg0S0(g1S1; :::; gnSn): Note that all the boundary maps are G-equivariant.We will check the �xed point set of a subgroup inG: If g� = �; then g 2 g0S0g�10 :Then shows that if H =2 F ; EF (G)H is empty. When S 2 F , the �xed point setEF (G)

S consists of simplices � such that Sg0 � g0S0: Since for any such �; thesimplex

S(g0S0; g1S1:::gnSn) 2 EF (G):

This gives a map s : EF (G)S ! EF (G) such that d0s = id: Intuitively, eachn-cell � could be contracted to the point S along the above (n + 1)-simplex.Therefore, EF (G)S is contractible.The uniqueness part follows the following lemma.

Lemma 4.1.6 Let X and Y be two G-CW complexes and F a family of sub-groups. Suppose that all the isotropy groups of X are in F and for eachH 2 F , the �xed point Y H is contractible. Then the is a unique equivariantmap f : X ! Y up to G-homotopy equivalence.

Proof. Let X = [Xi be a �ltration by its skeletions. We will construct amap inductively on its skeletons. For X0 = tG=Hi � S0 with Hi 2 F , wede�ne a G-equivariant map f0 : X0 ! Y by assign the coset eHi � S0 a pointin Y Hi ; which is contractible, for each i. This completes the �rst step in theinduction. Suppose that fn : Xn ! Y is constructed. For an (n + 1)-cellG=H � Dn+1; suppose that � : eH � Sn ! Y the restriction of fn to theboundary of eH �Dn+1: Since H acts on eH trivially, the image Im� � Y H ;which is contractible. We could have an extension of � to a map eH �Dn ! Y:This gives an equivariant map G=H �Dn+1 ! Y; which is an extension of fn:From the construction, we see that for each cell G=H � Dn+1; its image in Ylie in the �xed point set Y H ; which is contractible. If there are two such map,we could construct a G-homotopy between them using an inductive approachas above.

Example 4.1.7 When F consists of trivial subgroup, EF (G) is usually denotedby E(G); a CW complex on which G acts freely. The quotient space BG isusually called a classifying space for the group G: Since the quotient map EG!BG is a covering map, �1(BG) = G and �i(BG) = 0 for i � 2: The space BGis also called Eilenberg-Mac Lane�s K(G; 1) space.


Example 4.1.8 If F1 � F2 are two families of subgroups of a group G, thereis a unique G-equivariant map f : EF1(G) ! EF2(G) by Lemma 4.1.6. Sucha map f is called assembly map, which will play an important role in the for-mulations of Farrell-Jones conjecture and Baum-Connes conjecture discussed innext chapter.

For each CW complex X; a universal cover space ~X is a free G-CW complex(i.e. stabilizers are trivial). The previous lemma shows there is a map G-map~X ! EG: When passing to the quotient spaces, we call the map X ! BG aclassifying map.

De�nition 4.1.9 For a group G and a ZG-module M; the homology of G withcoe¢ cients M is de�ned as the homology of BG: In other words, H�(G;M) =H�(BG;M):

4.2 Acyclic maps and homology equivalences

In this section, we introduce some basic facts on homology equivalences. A CWcomplex X is called acyclic if H�(X;Z) = H�(pt;Z); the reduced homologygroup ~H�(X;Z) = 0 for any �: Recall that for any map f : X ! Y betweenCW complexes, there is a �bration p : Ef ! Y with i : X ,! Ef a homotopyequivalence and f = p � i: The �ber of p is called the homotopy �ber of f;denoted by F (f) or Ff : We call a map f : X ! Y between CW complexes isacyclic if the homotopy �ber F (f) is acyclic.

Theorem 4.2.1 Let f : X ! Y be a cellular map between CW complexes. Thefollowing are equivalent:

1. f is acyclic.

2. Hi(f) : Hi(X;M)! Hi(Y ;M) is isomorphic for any Z[�1(Y )]-module Mand i � 0.

3. Hi(f) : Hi(X;Z[�1(Y )])! Hi(Y ;Z[�1(Y )]) is isomorphic for any i � 0.

Proof. 1 =) 2. Since the �ber F (f) is acyclic, �1(Y ) acts trivially onH�(F (f);M): By Serre spectral sequence, we have that

Hp(Y ;Hq(F (f);M)) =) Hp+q(X;M):

Since the homotopy �ber F (f) acts trivially onM; we know thatHq(F (f);M) =0 when q 6= 0 and M when q = 0: This shows that Hi(f) is isomorphic.2 =) 3 is obvious.3 =) 1. Let ~Y be the universal covering space of Y and �X the pull back of

f : X ! Y: Then F (f) is homotopy equivalent to the homotopy �ber of �X ! ~Y :The condition 3 shows that �X ! ~Y is an integral homology equivalence. Since

4.3. QUILLEN�S PLUS CONSTRUCTION 47

~Y is simply connected, F (f) is connected, which implies that H0(F (f)) = Z.Consider the Serre spectral sequenceHp( ~Y ;Hq(F (f);Z)) =) Hp+q( �X;Z): SinceH1( �X;Z)! H1( ~Y ;Z) is isomorphic, we have thatH0( ~Y ;H1(F (f);Z)) = 0: Thisimplies H1(F (f);Z) = 0: Inductively, we prove that for any integer i � 1; thehomology group Hi(F (f);Z) = 0: This �nishes the proof.Given a CW complex X; how can we classify all acyclic maps f : X ! Y ?

A necessary condition is that the kernel of �1(f) : �1(X) ! �1(Y ) is a perfectsubgroup of �1(X); i.e. ker(�1(f)) = [ker(�1(f)); ker(�1(f))]: The reason is asfollowing.

Proposition 4.2.2 For a CW complex X and a acyclic cellular map f : X !Y; the kernel ker�1(f) is a perfect normal subgroup in �1(X).

Proof. Since the homotopy �ber F (f) is acyclic, the map f induces an epimor-phism on fundamental groups by the long exact sequence of homotopy groups

� � � ! �2(Y )! �1(F (f))! �1(X)! �1(Y )! �0(Ff )! �0(X):

SinceH1(�1(F (f));Z) = 0; �1(F (f)) = [�1(F (f)); �1(F (f))]: Therefore, ker�1(f)is perfect as well, as the image of �1(F (f)):Conversely, for any perfect normal subgroup P C �1(X), Quillen�s plus

construction gives a acyclic map f : X ! Y with ker�1(f) = P: This will bediscussed in more details in next section.

4.3 Quillen�s plus construction

Quillen (1969) proves the following theorem.

Theorem 4.3.1 (Quillen�s plus construction) Let X be a connected CWcomplex and P a perfect normal subgroup of �1(X); i.e. P = [P; P ]: Then thereexists a CW complex Y and an inclusion f : X ! Y such that �1(Y ) = �1(X)=Pand for any Z[�1(X)=P ]-module M and integer i � 0; the map f induces anisomorphism

Hi(f) : Hi(X;M)! Hi(Y ;M):

The map f is unique up to homotopy equivalences:

In general, we call Y in the above theorem a plus construction of X withrespect to P; denoted by X+

P :Proof. Existence. We �rst consider the case �1(X) = P: Let S be a set ofgenerators of P: For each generator s 2 S; assume that fs : S1 ! X is aloop representing s: Attach a 2-cell along each fs to get a CW complex W =X[s2Se1s: Then �1(W ) = �1(X)=P = 1: By the long exact sequence of homologygroups

� � � ! H2(X;Z)! H2(W ;Z)! H2(W;X;Z)! H1(X;Z)! H1(W ;Z)! � � �


and the fact that H1(X;Z) = 0; the map H2(W ;Z) ! H2(W;X;Z) is surjec-tive. Since W is simple connected, the Hurewicz map �2(W ) ! H2(W ;Z) isisomorphic. By the construction ofW; the relative homology group H2(W;X;Z)is a free abelian group with a basis indexed by S. Therefore, we could �nd mapsgs : S

2s !W such that the composition

H2(_sS2s )! H2(W )! H2(W;X;Z)

is isomorphic. For each gs; attach a 3-cell to W: Let Y be the pushout

_sS2s ! W# #

_sB3s ! Y:

Then �1(Y ) is trivial. For any abelian group M , the following commutativediagram with exact rows

� � � ! Hq(_sS2s ;pt;M) ! H2(_sB3s ;pt;M) ! Hq(_sB3s ;_sS2s ;M)! � � �# # #

� � � ! Hq(W;X;M) ! Hq(Y;X;M) ! Hq(Y;W ;M)! � � �

shows that H�(Y;X;M) = 0: This implies that the inclusion map X ! Y isacyclic.For the general case, let X1 be a covering space of X with �1(X1) = P and

YX1the CW space Y constructed above. Take the new space Y as the pushout

X1 ! YX1

# #X ! Y:

The fundamental group �1(Y ) = �1(X)=P by van Kampen theorem. The ex-cision property shows that H�(Y;X;M) = H�(YX1

; X1;M) for any coe¢ cientsover Y: This shows that the inclusion X ! Y is acyclic.Uniqueness. Suppose that there is another acyclic map f 0 : X ! Y 0 with

ker g = P: Take Z as the pushout

Xf! Y

f 0 # # g0Y 0

g! Z:

Using similar arguments as that of the previous paragraph, the maps g; g0 areacyclic. By van Kampen theorem, ker�1(g) is trivial. Therefore, the long exactsequence of homotopy groups

� � � ! �2(Z)! �1(Fg)! �1(Y0)! �1(Z)! � � �

shows that homotopy �2(Z) ! �1(Fg) is surjective. Since H1(�1(Fg)) = 0;�1(Fg) = 0: This shows that Fg is contractible and g is a homotopy equiva-lence. Similarly, g0 is a homotopy equivalence, which shows that Y and Y 0 arehomotopy equivalent.

4.4. DEFINITION OF HIGH ALGEBRAIC K-THEORY 49

4.4 De�nition of high algebraic K-theory

We use Quillen�s plus construction to de�ne higher algebraic K-theory. Let Rbe an associated ring with identity. Denote by GL(R) the stable general lineargroup and by E(R) the stable elementary subgroup of GL(R): According toWhitehead�s lemma, we know that E(R) is a perfect normal subgroup. SupposeBG is the classifying space of a group G, i.e. BG is a CW complex such that�1(BG) = G and �i(BG) = 0 for i � 1: For i � 1; we de�ne the algebraicK-group of R as

Ki(R) = �i(BGL(R)+):

The following proposition shows that the lower algebraic K-groups de�ned herecoincide with that de�ned before using algebraic approach.

Proposition 4.4.1 K1(R) = GL(R)=E(R); K2(R) = H2(E(R)); K3(R) =H3(St(R)):

Proof. According to the construction, we have that K1(R) = �1(BGL(R)+) =

GL(R)=E(R): Let X ! BGL(R)+ be the universal covering space. It is nothard to see that the pullback of the following diagram

BE(R) ! X# #

BGL(R) ! BGL(R)+

is BE(R) (view BGL(R) as a subcomplex of BGL(R)+ and consider the decktransformation group or use the fact that the homotopy �ber is invariant underpullback). Since the bottom arrow induces homology isomorphism with coe¢ -cients Z�1(BGL(R)+), the upper horizontal arrow preserves integral homologygroups. Since X is simply connected, we see that the map BE(R) ! X isacyclic and a Quillen�s plus construction by Theorem 4.2.1. Therefore,

K2(R) = �2(BGL(R)+) �= �2(X) �= H2(X) �= H2(E(R)):

Actually for any i � 2; we have Ki(R) = �i(X):Let 1 ! K2(R) ! St(R) ! E(R) ! 1 be the universal central extension.

Considering plus constructions there is a commutative diagram

BK2(R) ! BSt(R) ! BE(R)# #

F ! BSt(R)+ ! BE(R)+;

where F is the homotopy �ber of the lower right arrow. Since the composi-tion BK2(R) ! BSt(R) ! BSt(R)+ is homotopic to identity, there is a map� : BK2(R) ! F: By long exact sequences of homotopy groups of both rows,we see that � induces isomorphism of fundamental groups (note: H2(St(R) =H1(St(R))) = 0). The comparison theorem for spectral sequence shows that� induces homology isomorphism. Since fundamental groups of BK2(R) and


F are both abelian, � is a homotopy equivalence (by a result of E. Dror [2]).Therefore

BK2(R)! BSt(R)+ ! BE(R)+

is a �bration and

K3(R) = �3(BGL(R)+) = �3(BE(R)

+)

= �3(BSt(R)+) = H3(BSt(R)

+)

= H3(St(R)):

The previous proposition says that for lower dimensions, algebraic K-groupsare isomorphic to homology groups of some group-valued functors on rings. Itis natural to ask that whether such phenomenon holds for any dimension.

Problem 4.4.2 (Open) Let n � 4 be a integer, Rng the category of rings andGp the category of groups. Is there a functor

Gn : Rng! Gp

such that for a ring R there is an isomorphism Hn(Gn(R)) = Kn(R):

Let I be an ideal of R and q : R! R=I the natural quotient map. We havea natural map q0 : BGL(R)+ ! BGL(R=I)+: Denote by F (R; I) be homotopy�ber of q0 and de�ne the relative K-group by

Ki(R; I) = �i(F (R; I))

for each integer i � 1:

Proposition 4.4.3 Let I be an ideal of a ring R: Then there is a long exactsequence

� � � ! K2(R=I)! K1(R; I)! K1(R)! K1(R=I)! K0(R; I)! K0(R)! K0(R=I):

Proof. This follows the long exact sequence of homotopy groups for a �bration.

Another interesting example of Quillen�s plus construction is the following.Let �n be the symmetric group of n letters and �1 = lim

n!+1�n the stable

symmetric group. It is not hard to see that the stable alternating group A1 isa perfect normal subgroup of �1: A famous theorem of Barratt-Kahn-Priddyand Quillen says that

Proposition 4.4.4 �i((B�1)+) �= �si (S0) the stable homotopy groups of spheres.

4.5. SPECTRUM OF ALGEBRAIC K-THEORY 51

4.5 Spectrum of algebraic K-theory

4.5.1 Spectrum

De�nition 4.5.1 A spectrum is a sequence of CW complexes E = fEig togetherwith structure maps �i : S ^ Ei ! Ei+1 for each i: A spectrum E is called an-spectrum if the adjoint Ei ! Ei+1 of �i is a homotopy equivalence. A mapof spectra f : E! E0 is a sequence of maps fn : En ! E0n which are compatiblewith structure maps, i.e. f(n+ 1) � �n = �0n+1 � (f(n) ^ idS1):

For an integer n; the map �i induces a map �n(Ei)! �n+1(Ei+1) by

Sn+1 = S1 ^ Sn ! S1 ^ Ei�i! Ei+1:

The (stable) homotopy group of a spectrum E is de�ned as

�si (E) = limn!1

�n+i(En):

Exercise 4.5.2 For an -spectrum E, the homotopy group ��i(E) = �k(Ei+k)for any k:

A map f : E ! E0 is weak homotopy equivalent if f induces isomorphismson homotopy groups. A homotopy between maps f; g : E ! E0 is a maph : [0; 1]+ ^ E ! E0 such that h(0;�) = f and h(1;�) = g: A homotopyequivalence between two spectra could be de�ned in a similar way as that ofspaces. A weak homotopy equivalence is a homotopy equivalence between twospectra.

Example 4.5.3 Let S = fSig with �i : S ^ Si ! Si+1 the identity map. Thisis called sphere spectrum. Generally, for a CW complex X; fSi ^ Xgi�1 is aspectrum. Homotopy groups of such a spectrum are the stable homotopy groupsof X:

Example 4.5.4 Let G be an abelian group, K(G;n) a space with �n(K(G;n)) =G and �i(K(G;n)) = 0; i 6= n: It�s not hard to see that K(G;n) ' K(G;n+1)for n � 1 (by comparing homotopy groups). The map �n : S ^ K(G;n) !K(G;n + 1) is the adjoint of this homotopy equivalence. The spectrum KG =fK(G;n)gn�1 is called Eilenberg-Mac Lane spectrum.

Example 4.5.5 Let X be a CW complex and E = fEig a spectrum. Themapping space spectrum is de�ned as fMap(X;Ei)g with structure map �i :S1 ^Map(X;Ei)! Map(X;S1 ^ Ei); z ^ � 7! (x 7! z ^ �(x)):

De�nition 4.5.6 Let CWh be the homotopy category of pairs of pointed CWcomplexes, i.e. objects are pair of CW complexes with based points and mor-phisms are homotopy classes of maps. Let S : CWh ! CWh be a functor de�nedby (X;x0)! (SX; x0) (suspension). A generalized reduced homology theory h�is a series of functors hn : CW

h ! Ab to the category of abelian groups andnatural equivalences @n : hn ! hn+1 � S; n 2 Z, satisfying


1. Exactness: for every pair (X;A) 2 T the following is exact

hn(A; x0)! hn(X;x0)! hn(X [ CA; �);

where CA is the mapping cone of A and (A; x0)! (X;x0) and (X;x0)!(X [ CA; �) are induced by inclusions.

2. Wedge axiom: for every collection f(X�; xa)g�2I of pointed spaces theinclusions i� : X� ! _�2IX� induce an isomorphismL

�2I hn(X�)! hn(_�2IX�);

for any integer n 2 Z.

Generalized cohomology theory could be de�ned similarly by replacing con-variance by contra-variance and replacing the direct sum in wedge axiom byproduct.

Theorem 4.5.7 (Brown�s Representability) h� be a reduced homology (resp.h� is a reduced cohomology) theory if and only if there exists a spectrum E suchthat hn(X) = �si (X ^E) (resp. hn(X) = �s�n(Map(X;E))).

Example 4.5.8 Let G be an abelian group. The ordinary reduced (co-)homologytheory with coe¢ cient G (trivial action of fundamental groups) is representedby Eilenberg-Mac Lane spectrum. In other words, for any CW complex X;~Hi(X;G) = �si (X ^ KG); ~Hi(X;G) = �s�i(Map(X;KG)) �= [X;K(G; i)]; theabelian group of homotopy classes of maps.

Example 4.5.9 The (complex) topological K-theory spectrum isKU = fBU;U;BU;U; � � � g;where U is the stable unitary group. Therefore, for any CW complex X; the i-threduced topological K-theory of X is �s�i(Map(X;KU)) = [X;KUi]:

For each (co-)homology theory, the Atiyah-Hizebruch spectral sequence isuseful for computation.

Lemma 4.5.10 (Atiyah-Hizebruch spectral sequence) Let h� (resp. h�)be a (co-)homology theory. For for any CW complex X; there is a spectralsequence

Hp(X;hq(pt)) =) hp+q(X);

and Hp(X;hq(pt)) =) hp+q(X):

4.5.2 Algebraic K-theory spectrum

In this section, for a ring R we will construct a spectrum E such that the stablehomotopy group �i(E) = Ki(R):

4.5. SPECTRUM OF ALGEBRAIC K-THEORY 53

De�nition 4.5.11 Let R be a ring. The cone ring CR = fA = (rij)1;1 jeach row and column has only �nitely many nonzero entries in A g; the ring ofall in�nite matrices (rij) over R such that there are only �nitely many nonzeroentries in each row or column. The mR is the ring of �nite matrices limMn(R),which is an ideal of CR: The suspension �R is the quotient ring CA=mA:

Inductively, we could de�ne the i-th suspension ring SiR:

Lemma 4.5.12 The stable general linear group GL(CA) is acyclic. Therefore,GL(CA)+ is contractible.

Theorem 4.5.13 There is a functor from the category of rings to the cate-gory of (0-connected) spectra which associates to a ring R the spectrum KR =fBGL(SrR)+; r � 0g such that

�si (KR) = Ki(R); i � 1:

Moreover, K0(R)�BGL(R)+ ' BGL(SR)+: In particular, Ki(R) = Ki+1(SR)for i � 0:

Proof. Let GL(mR) denote the relative stable general linear group for the idealmA: It is not hard to see that GL(mR) is isomorphic to GL(R): Then we havea exact sequence

1! GL(mR)! GL(CR)! GL(SR):

By Lemma 4.5.12, GL(CR) = [GL(CR);GL(CR)] = E(CR): Therefore, thefollowing sequence is exact

1! GL(mR)! E(CR)! E(SR)! 1:

Applying plus constructions, we have a commutative diagram

BGL(mR) ! BE(CR) ! BE(SR)# � # #F ! BE(CR)+ ! BE(SR)+;

where F is the homotopy �ber the bottom arrow. The comparison theorem forspectral sequence shows that � induces homology isomorphism and thereforethe induced map �� : BGL(mR)+ ! F is (integral) homology equivalent. Sincefundamental groups of BGL(mR)+ and F are both abelian, �� is a homotopyequivalence (by a result of E. Dror [2]). From �ber sequence

� � � ! BE(CR)+ ! BE(SR)+ ! F ! BE(CR)+ ! BE(SR)+

and the fact that BE(CR)+ is contractible (cf. Lemma 4.5.12), we have that

BGL(R)+ ' BE(SR)+:


Since BE(SR)+ is the universal covering space of BGL(SR)+; using a similar�ber sequence as above (i.e. BE(SR)+ ! BGL(SR)+ ! K1(SR)) we havethat

BE(SR)+ ' BGL(SR)+0 ;

the component of loops contractible to a point. The structure map �0 : S1 ^BGL(R)+ ! BGL(SR)+ is de�ned as the adjoint ofBGL(R)+ ! BE(SR)+ !BGL(SR)+: Other structure maps �i (i > 0) could be de�ned similarly. Notethat for any i � 1 and r � 0;

�i+r(BGL(SrR)+) = Ki+r(S

rR) �= �i+r+1(BGL(Sr+1R)+) = Ki+r+1(Sr+1R):

Therefore �i(KR) = Ki(R); i � 1: In order to prove the second statement, it isenough to note that K1(SR) = K0(R) (cf. Wagoner [21], Prop. 5.1).

Remark 4.5.14 For a CW complex X; let pn : X ! X [n] be the n-stage ofa Postnikov tower, i.e. �i(X

[n]) = �i(X) for i � n and �i(X [n]) = 0 fori > n: Denote by X(n) the homotopy �ber of pn: Then X(n) is n-connected, i.e.�i(X(n)) = 0 for i � n: The spectrum KR in the previous theorem could be re-placed by the 0-connected -spectrum f�r(BGL(R)+); r < 0; BGL(SrR)+(r); r �0g or the (�1)-connected -spectrum f�r(K0(R)�BGL(R)+); r < 1; BGL(SrR)+(r�1); r �1g:

4.6 Applications

4.6.1 Any abelian group is the center of a perfect group

Proof. For an abelian group A; let

0! B ! F =L

� Z!A! 0

be a presentation of A: For a ring R; denote by �R the suspension of R: Thenthe center of St(

PP(��Z)) is K2(

PP(��Z)) �= K0(��Z) = F: The group

St(PP

(��Z))=B is a perfect group, which has center A:This result is from [A J Berrick: Torsion generators for all abelian groups,

J Algebra 139 (1991), 190-194].

4.6.2 Higher homotopy groups and homology

This is a main result of Hausmann [On the homotopy of Non-nilpotent spaces,Math. Z. 1981]

Theorem 4.6.1 Let Pi(i � 2) and Qi(i � 1) be two given series of abeliangroups. Then exists a CW complex X such that �i(X) = Pi (i � 2) andHi(X;Z) = Qi (i � 1):

4.6. APPLICATIONS 55

The proof is based on the following lemma. Take Y = �i�2K(Pi; i+ 1) andY 0 = Y: Then Y 0 is a central covering of some acyclic space Z: Take U =_i�1M(Qi; i) be the wedge of Moore spaces. Note that here a Moore M(G; 1)means all homology groups vanishes except H1 = G: This is di¤erent from theusual de�nitions. According to Kan-Thurston�s theorem we have that for somegroup G and �; BG! Y and B� ! U are Quillen�s plus construction. Denoteby F the homotopy �ber of BG ! Y; which is an acyclic space. Consideringthe Purple sequence for BG! Y 0 ! F ! BG! Y; we know that Y 0 ! F isa central covering space. This implies that �i(F ) = �i(Y 0) = Pi for any i � 2:Now take X = F �B�, which has the designed property.Actually, with more techniques, we have the following.

Lemma 4.6.2 A space X is a loop space, i.e. X = Y for some CW complexY; i¤ X is a central covering of an acyclic space Z such that �1(Z) acts triviallyon �i(Z) i � 2.

Chapter 5

Isomorphism conjectures

From the long exact sequence of relative K-theoryThe following Farrell-Jones conjecture, which was originally asked by Farrell

and Jones in [7], says that the algebraic K-theory of a group ring is isomorphicto an equivariant homology theory. Here we follow the formulations of Davisand Lück [1]. For more details, see the survey article of Lück and Reich [14].

5.1 K-theory spectrum of an additive category

We assume thatAssumption. There is a functor Ka : Ad ! Spectra from the categoryof additive categories to the category of spectra such that for a ring R andthe category ProjR of �nitely generated projective R-modules, Ka(ProjR) is analgebraic K-theory spectrum of R: We also require that if F : A1 ! A2 isan equivalence of categories (i.e. f is a fully faithful and essentially surjectivefunctor), then Ka(f) : K(A1)! K(A2) is a homotopy equivalence.More details could be �nd in Pertersen and Weibel [17].Let G be a group and X a G-space. Form a category (actually a groupoid) �X

with objects x; x 2 X and morphisms mor �X(x; x0) = fg j gx = x0g: For a ring R;

R �X denote the additive category with the same objects as �X; but morphismsmorR �X(x; x

0) = Rmor �X(x; x0); the free module spanned by mor �X(x; x

0): Thecategory R �XL is the formal sum of R �X as follows. The objects are n-tuples

(x1; x2; : : : ; xn) with xi 2 ob(R �X); n = 0; 1; 2; : : : ; and morphisms are given bymatrix form. The category R �XL is a symmetric monoidal category.

Given a category C, de�ne its idempotent completion P(C) to be the followingcategory. An object in P(C) is an endomorphism p : x ! x in C which is anidempotent, i.e. p � p = p. A morphism in P(C) from p : x ! x to q : y ! yis a morphism f : x ! y in C satisfying q � f � p = f . The identity on theobject p : x! x in P(C) is given by the morphism p : x! x in C. If C has thestructure of a R-category or of a symmetric monoidal R-category, then so doesP(C) in the obvious way.

57

58 CHAPTER 5. ISOMORPHISM CONJECTURES

Let C be a symmetric monoidal R-category, all of whose morphisms are iso-morphisms (eg. Iso(P(R �XL)) subcategory of P(R �XL) with the same objects

and morphisms are isomorphisms). Then its group completion is the followingsymmetric monoidal R-category. An object in C^ is a pair (x; y) of objects inC. A morphism in C^ from (x; y) to (x0; y0) is given by equivalence classes oftriples (z; f; g) consisting of an object z in C and isomorphisms f : x

Lz ! x0

and g : yLz ! y0. We call two such triples (z; f; g) and (z0; f 0; g0) equiva-

lent if there is an isomorphism h : z ! z0 such that f 0 � (idxLh) = f and

g0 � (idxLh) = g: The sum in C^ is given by

(x; y)L(x0; y0) := (x

Lx0; y

Ly0):

The group completion Iso(P(R �XL))^ will be our main interest. It is not

hard to see that the construction of Iso(P(R �XL))^ is functorial with respect

to X: When X = G=H; the left coset of H,

Lemma 5.1.1 When X = G=H; the left coset of H, Ka(Iso(P(R �XL))^) is an

algebraic K-theory spectrum of the group ring RH:

Proof. Let Y = eH the left coset of the trivial element. Then �Y is a categorywith only one object eH and morphisms indexed by elements in H: Moreover,R �YL is the category of free RH-modules and P(R �XL) the category of �-

nitely generated projective RH-modules. Note that the inclusion �Y ,! �X isan equivalent of categories. Therefore, P(R �YL) ! P(R �XL)) are equivalent.

This implies that Ka(Iso(P(R �XL))^) is an algebraic K-theory spectrum of the

group ring RH:

De�nition 5.1.2 (Orbit category) Let G be a group. The orbit categoryOr(G) is a category whose objects are G=H for subgroups H < G and mor-phisms the G-maps.

For each object G=H = X, we assign the additive category Iso(P(R �XL))^:

Compositing with Ka, we get a covariant functor

K : Or(G)! �spectra;

such that �si (K(G=H)) = Ki(RH) for each i 2 Z.

5.2 Equivariant homology

Let Or(G) be the orbit category of a group G and E : Or(G) ! �spectra acovariant functor (eg. K is as de�ned in the end of last section).

De�nition 5.2.1 For a G-CW complex X; the functor

X? : Or(G)! CW;G=H 7�! XH ;

5.2. EQUIVARIANT HOMOLOGY 59

is a contravariant. In other words, X? assign each coset G=H the �xed pointset of H:

Let A be a G-CW complex. The cone Cone(A) = A � [0; 1]=(A � 1) isnaturally a G-CW complex with G action trivially on [0; 1]: Note that when Ais the empty set, Cone(A) consists of one point.

De�nition 5.2.2 For a contravariant functor Y : Or(G)! CW and a covari-ant Z : Or(G) ! CW; the tensor product over Or(G) is de�ned as the CWcomplex

YN

Or(G) Z = [Y (G=H)� Z(G=H)= �;

where the equivalence relation � is de�ned as (x; f�y) � (f�x; y) for any f :G=H ! G=K and x 2 Y (G=K) and y 2 Z(G=H):

Denote by En : Or(G) ! CW the n-th component of E. According to thede�nition,

S1 ^ (YN

Or(G) En) ! YN

Or(G)(S1 ^ En)

s ^ (yNe) 7�! y

N(s ^ y)

is a homeomorphism. Therefore, we could de�ne YN

Or(G) E as a spectrumcomponent-wise. For a subgroup H < G and an H-CW complex, the induc-tion G �H X is a G-CW complex de�ned as the quotient G � X=f(g; x) �(gh; h�1x); h 2 Hg:

De�nition 5.2.3 Let (X;A) be a pair of G-CW complexes and E : Or(G) !�spectra a covariant functor. The equivariant homology group HG

n (X;A;E)is de�ned as the stable homotopy group

�sn((X [A Cone(A))?N

Or(G) E):

As usual, HGn (X;E) means HG

n (X; ;;E):

Theorem 5.2.4 The functors HGn (�;�;E) forms a G-homology theory for pairs

of G-CW complexes. In other words,

1. Two G-homotopic maps induce the same homomorphisms;

2. For pairs of G-CW complexes, there are natural long exact sequences.

3. Excision property holds for G-CW complexes, i.e. for a pushout diagram

X0 ! X1# #X2 ! X

with all arrows are inclusions of G-CW complexes, the inclusions induce

HGn (X1; X0;E) �= HG

n (X;X2;E):


4. Disjoint union axiom holds for any collection fXigi2I of G-CW complexes,i.e. L

i2I HGn (Xi;E) �= HG

n (ti2IXi;E):

Moreover, H?�(�;E) is an equivariant homology theory, i.e. for each group

G; HG� (�;E) is a G-homology theory and for an injection H ,! G and an

H-CW complexes X there is a natural isomorphism

HGn (G�H X;E) �= HH

n (X;EjH)

for each integer n: In particular, HGn (G=H;E) �= �n(E(G=H)):

Proof. By straightforward checking of de�nitions in terms of spectra, we couldeasily see that HG

n (X;A;E) is a G-equivariant homology theory. The last iso-morphism follows from the de�nition as well.

5.3 Formulation of conjectures

Let G be a group and Or(G) the category whose objects are G=H for subgroupsH < G and morphisms the G-maps. Let R be a ring and

KR : Or(G)! �spectra

be a functor from Or(G) to the category of -spectra such that for each integeri the stable homotopy group �si (KR(G=H)) = Ki(RH); the algebraic K-groupsthe group ring RH:Denote by Evcyc(G) aG-CW complex such that the �xed point setEvcyc(G)H

is contractible for any virtually cyclic subgroup H < G and empty if H is notvirtually cyclic. Such space is unique up to homotopy equivalence. Since one-point space could be a classifying space for the family of all subgroups. Thetrivial map Evcyc(G)! pt (cf. Lemma 4.1.6) induced a map of homology groups

HG� (Evcyc(G);KR)! HG

� (pt;KR);

called the Farrell-Jones assembly map.

Conjecture 5.3.1 (Farrell-Jones conjecture) Let G be a group and R be aring. Then for each integer i; the assembly map

HGi (Evcyc(G);KR)! HG

i (pt;K) = Ki(RG)

is an isomorphism.

The following lemma, called transitivity principal, will be useful for compar-ing equivariant homology of two di¤erent families.

5.3. FORMULATION OF CONJECTURES 61

Lemma 5.3.2 (Transitivity Principal) Let F � G be two families of sub-groups of a group G and H?

i () an equivariant homology theory as de�ned inTheorem 5.2.4. Suppose that for each group H 2 G and integer i, there is anisomorphism

HHi (EH\F (H))! HH

i (pt)

induced by the assembly map EH\F (H)! pt: Then the assembly map EF (G)!EG(G) induces isomorphism

HGi (EF (G))! HG

i (EG(G))

for each integer i:

Proof. Note that EG(G) � EF (G) is a new model for the classifying spaceEF (G): We view the assembly map as the projection pr1 : EG(G) � EF (G) !EG(G): We prove the lemma by induction on the skeleton of EG(G): For eachgroup H 2 G, the G-map

G�H EF (G) ! G=H � EF (G);(g; x) 7! (gH; gx)

is a G-homeomorphism. Since EF (G) could be a model for EH\F (H); we havethat

HGi (G=H � EF (G)) �= HG

i (G�H EF (G)) �= HHi (EH\F (H)) (1)

! HHi (pt)

�= HGi (G=H)

is an isomorphism for each integer i: After taking disjoint union, we get isomor-phism

HGi (EG(G)

(0) � EF (G))! HGi (EG(G)

(0)); (2)

where EG(G)(0) is the 0-th skeleton of EG(G): This �nishes the induction step.Suppose that the previous isomorphism holds for the n-th skeleton EG(G)(n):By the de�nition of G-CW complex, we have a pushout

qi2IG=Hi � Sn ! EG(G)(n)

# #qi2IG=Hi �Dn+1 ! EG(G)

(n+1):

The isomorphism (1) and a �ve lemma argument show that the isomorphism(2) holds for EG(G)(n+1): The proof is �nished by taking direct limit EG(G) =limnEG(G)

(n):

Lemma 5.3.3 When R is regular and G is torsion-free, the assembly mapE(G)! Evcyc(G) will induce isomorphism HG

i (E(G);KR)! HGi (Evcyc(G);KR):

Outline of proof. By the transitivity principal, it is enough to prove thatassembly maps induce isomorphisms

HHi (E(H);KR)! HH

i (pt;KR) = Ki(RH)


for each virtually cyclic subgroup H: Since G is torsion-free, every H is iso-morphic to Z (cf. the classi�cation of virtually cyclic group in [14], Lemma2.15). Therefore, the left-hand side is �i(S1+ ^ KR); which is isomorphic toKi(R)

LKi�1(R) by Atiyah-Hizebruch spectral sequence. The high-dimensional

analog of Bass-Heller-Swan theorem implies that the above maps are isomor-phisms.Therefore, we have the following special case of Farrell-Jones conjecture:

Conjecture 5.3.4 Let G be a torsion-free group and R be a regular ring, theassembly map

Hi(BG;K(R)) = �si (BG+ ^K(R))! Ki(RG)

is an isomorphism for each i; where K(R) is the algebraic spectrum of R:

Remark 5.3.5 The Bass-Heller-Swan theorem also implies that the regular-ness of R is necessary in general. Moreover, when R is a regular such that theorder of each �nite-order element in G is invertible in R; the assembly mapHGi (EFin(G);KR) ! HG

i (Evcyc(G);KR) is isomorphic for each i: In such acase, the left-hand side of Farrell-Jones conjecture could be replaced by HG

i (EFin(G);KR)(cf. [14], Prop. 2.4).

Similarly, we can consider the Baum-Connes conjecture. Let l2(G) be theHilbert space on G: The group algebra CG acts on l2(G) by left translation. Thisgives an injection CG! B(l2(G)) to the set of all bounded linear operators ofl2(G): The completion of CG in B(l2(G)) with respect to the operator norm isthe reduced group C�-algebra C�r (G): Let

Ktop : Or(G)! �spectra

be a functor from Or(G) to the category of -spectra such that for each integeri the stable homotopy group �i(Ktop(G=H)) = Ktop

i (C�rH); the topological K-groups the C�-algebra C�r(H): Denote by EFin(G) a G-CW complex such thatthe �xed point set EFin(G)H is contractible for any �nite subgroup H < Gand empty if H is not �nite. The trivial map EFin(G)! pt induced a map ofhomology groups

HG� (EFin(G);Ktop)! HG

� (pt;Ktop);

called the Baum-Connes assembly map.

Conjecture 5.3.6 (Baum-Connes conjecture) Let G be a group. Then foreach integer i; the assembly map

HGi (EFin(G);Ktop)! HG

i (pt;Ktop) = Ktopi (C�r(G))

is an isomorphism.

5.4. APPLICATION 63

5.4 Application

The Farrell-Jones conjecture has many applications. For example:

Lemma 5.4.1 If the Farrell-Jones conjecture is true for a torsion-free group G;then the Wall obstructions ~K0(ZG) = 0 and the Whitehead group Wh(G) = 0:

Proof. For simplicity, we assume that the spectrum K in the formation ofFarrell-Jones conjecture is (�1)-connected, i.e. �i(K(G=H)) = 0 for i < 0: TheAtiyah-Hizebruch spectral sequence implies that

K0(ZG) �= H0(BG;KZ) �= H0(BG;�0(KZ)) = K0(Z):

Similarly,

K1(ZG) �= H0(BG;�1(KZ))M

H1(BG;�0(KZ))

= K1(Z)M

Gab:

Therefore, ~K0(ZG) = 0 and Wh(G) = K1(ZG)=h�Gabi = 0:This implies that Bass conjecture holds for any torsion-free group G and that

any �nitely dominated CW complex with fundamental group G will be �niteand that a homotopy equivalence between CW complexes with fundamentalgroups G is simple homotopy equivalence. Moreover, the Farrell-Jones andBaum-Connes conjectures can imply many other conjecture conjectures, suchas Borel conjecture, Novikov conjecture, zero-in-the spectrum conjecture and soon. For more details, see Lück and Reich [14].

Chapter 6

Appendix: Backgroundfrom algebraic topology

In this appendix, we outline the main results in algebraic topology that will beused frequently in this course. Most of them could be found in textbooks likeHatcher [12] or Swizer [19].

6.0.1 Main results

Until otherwise stated, spaces considered in this course are (pointed) CW com-plexes and the maps are (pointed) cellular maps in the following sense.

De�nition 6.0.2 (CW complexes) A CW complex X is a Hausdor¤ spacetogether with �ltration

X0 � X1 � � � � � Xn � � � �

such that X = [Xn; X0 is a disjoint union of points and Xn is constructedfrom Xn�1 by attaching n-cells. In other words Xn is the pushout out

_i2ISn�1_fi! Xn�1

# #_i2IDn ! Xn;

where fi : Sn�1 ! Xn�1 is called attaching map.The Xn is called n-skeleton. A map f : X ! Y of CW complexes is cellular

if f(Xn) � Yn for each n:

De�nition 6.0.3 (Homotopy equivalence) Let f; g : X ! Y be two maps.They are homotopic, denoted by f � g; if there exists a map H : X � [0; 1]! Ysuch that H(�; 0) = f and H(�; 1) = g: Two spaces X;Y are said to be homotopyequivalence if there exists maps f : X ! Y and f 0 : Y ! X such that g�f � idXand f � f 0 � idY :

65

66CHAPTER 6. APPENDIX: BACKGROUND FROMALGEBRAIC TOPOLOGY

De�nition 6.0.4 (Homotopy groups) Let X be a CW complex, Y a sub-complex and n � 0 be an integer. The relative homotopy group �n(X;Y ) is theset of homotopy equivalence classes Map�((D

n; @Dn); (X;Y ))= � :When n � 1;�n(X; pt) is a group. Without confusions, we will write �n(X) for short.

Note that there is an action of �1(X) on �i(X) for i � 2:

Proposition 6.0.5 (long exact sequence of homotopy groups) Let f : Y !X be an inclusion of CW complexes. The following sequence is exact

� � � ! �n(Y )! �n(X)! �n(X;Y )! �n�1(Y )! �n�1(X)! � � �

For a map f : X ! Y; the mapping cylinder Mf = Y [f (X � [0; 1]); wheref is viewed as a map on X � 0. Note that the relative homotopy groups andrelative homology groups could be de�ned for any map f : X ! Y by replacingY by the mapping cylinder Mf ; which is homotopy equivalent to Y: Moreover,for any map f : X ! Y; we could de�ne the mapping cone Cone(f) asMf=Y �1:The co�ber sequence of f is the following

X ! Y ! Cone(f)! �X ! �Y ! � � � :

Proposition 6.0.6 (Whitehead theorem) Suppose that a map f : X ! Ybetween two CW complexes induces isomorphism �i(f) : �i(X) ! �i(Y ) foreach integer i; i.e. f is a weak homotopy equivalence. Then f is a homotopyequivalence.

Proposition 6.0.7 (van Kampen theorem) Suppose that a CW complex Xis a union of two subcomplexes X1 and X2 with the intersection X1 \X2 pathconnected. Denote by i1 : �1(X1\X2)! �1(X1) and i2 : �1(X1\X2)! �1(X2)the group homomorphisms induced by inclusions of CW complexes. Then

�1(X) = �1(X1) � �1(X2)=hi1(a)i2(a)�1 j a 2 �1(X1 \X2)i:

De�nition 6.0.8 (Covering space) A map p : (X;x) ! (Y; y) between CWcomplexes is a covering map if there exists open cover fUig of Y such that foreach Ui the preimage p�1(Ui) is a disjoint union of open sets in X, each ofwhich is mapped homeomorphically onto Ui by p. Y is called a covering spaceof Y:

Associating the subgroup p�(�1(X;x)) in �1(Y; y); there is a one-to-one cor-respondence between all the di¤erent connected covering spaces of Y and theconjugacy classes of subgroups of �1(Y; y): When p�(�1(X;x)) is a normal sub-group in �1(Y; y); �1(Y; y)=p�(�1(X;x)) is the deck transformation group of thecovering map p: If a group G acts properly discontinuously on a space X; thequotient map X ! X=G is a covering map.

De�nition 6.0.9 (Homology groups) Let X be a CW complex and ~X itsuniversal covering space. The free Z�1(X)-module Cn( ~X) is spanned by the

67

set of n-dimensional cells in X: For a Z�1(X)-module M; the homology groupHi(X;M) (resp. cohomology group Hi(X;M)) with coe¢ cients M are de-�ned as the i-th homology group of the chain complex C�( ~X)

NZ�1(X)M (resp.

HomZ�1(X)(C�(~X);M)).

Let Y be a subcomplex a CW complex X: The relative homology groupHi(X;Y ;Z) (resp. cohomology group Hi(X;Y ;Z)) is de�ned as the reducedhomology (resp. cohomology) groups of X=Y:

Proposition 6.0.10 (long exact sequence of homology groups) Let Y bea subcomplex a CW complex X: There exists a long exact sequence of homologygroups

� � � ! Hn(Y ;Z)! Hn(X;Z)! Hn(X;Y ;Z)! Hn�1(Y ;Z)! Hn�1(X;Z)! � � � :

Similar sequence exists for cohomology groups.

Proposition 6.0.11 (Hurewicz theorem) (1) Suppose that n � 1: If aCW complex X is n-connected, i.e. �i(X) = 0 for i � n; then the Hurewiczmap �i(X)! Hi(X) is isomorphic for i � n+ 1: Moreover,

H1(X) = �1(X)=[�1(X); �1(X)];

i.e. H1(X) is the abelianization of �1(X): A general version is:

(2) Let X be a CW complex, A a subcomplex of X and n � 1: Suppose that(X;A) is n-connected, i.e. �i(X;A) = 0 for i � n: Then Hi(X) = 0 fori � n and Hn+1(X;A) is isomorphic to �i(X;A) modulo out the action of�1(A):

De�nition 6.0.12 (Serre�s �bration) Let n � 0 and Dn a disk. A Serre�bration is a map p : E ! B such that for any homotopy gt : Dn ! B there isa lifting ~gt : Dn ! E; i.e. ~gt = p�gt: If B is path connected, all the �bers p�1(b)(b 2 B) are homotopy equivalent to each other. We denote by F = p�1(b) the�ber if b is the based point.

Covering maps and bundle maps are special kinds of Serre �bration. Gen-erally, for any map f : X ! Y between CW complexes, we could associate a�bration f 0 : Ef ! Y (with �ber Ff ) such that Ef is homotopy equivalent toX: The �ber Ff is called the homotopy �ber of f: Moreover, we have a �brationsequence

� � � ! Ff ! X ! Y ! Ff ! X ! Y;

where any three successive terms are homotopy equivalent to a �bration.

Proposition 6.0.13 (Homotopy and homology of �brations) Let p : E !B be a Serre �bration with �ber F .(i) There is a long exact sequence of homotopy groups

� � � ! �n(F; x)! �n(E; x)! �n(B; y)! �n�1(F )! � � � ! �0(E; x)! 0:

68CHAPTER 6. APPENDIX: BACKGROUND FROMALGEBRAIC TOPOLOGY

(ii) For local coe¢ cients G on X; when the fundamental group �1(B) actstrivially on H�(F ;G); there is a spectral sequence

Hp(B;Hq(F ;G)) =) Hp+q(X;G):

6.0.2 Techniques

1. Kill elements in homotopy groups

Let X be a CW complex. Suppose that f : Sn ! X is a representative ofan element [f ] 2 �n(X): By attaching an (n+ 1)-cell to X to kill [f ]; wemean the process of taking the pushout out

Snf! X

# #Dn+1 ! X 0:

2. Extend maps

If f : X ! Y is a map of CW complexes and X 0 = X [Sn Dn+1: Supposethat the sphere Sn has image f(Sn) homotopic to a point. Then f couldextend to be a map f 0 : X 0 ! Y such that f 0jX = f:

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