Page 1

DOI: 10.1007/s00208-002-0397-2

Math.Ann. 325, 711–726 (2003)

MathematischeAnnalen

Higher K-theory of group-rings of virtually infinite

cyclic groups

Aderemi O. Kuku · Guoping Tang

Received: 18April 2002 / Published online: 10 February 2003 – © Springer-Verlag 2003

Abstract. F.T. Farrell and L.E. Jones conjectured in [7] that Algebraic K-theory of virtually

cyclic subgroups V should constitute ‘building blocks’ for the Algebraic K-theory of an arbi-

trary group G. In [6], they obtained some results on lower K-theory of V. In this paper, we

obtain results on higher K-theory of virtually infinite cyclic groups V in the two cases: (i) when

V admits an epimorphism (with finite kernel) to the infinite cyclic group (see 2.1 and 2.2(a),(b))

and (ii) when V admits an epimorphism (with finite kernel) to the infinite dihedral group (see

3.1, 3.2, 3.3).

Mathematics Subject Classification (2000): 19D35, 16S35, 16H05.

0. Introduction

In [7], Farrell-Jones conjectured thatAlgebraic K-theory of an arbitrary group G

can be “computed” in terms of virtually cyclic subgroups of G. So it becomes

essential to understand the K-theory of virtually cyclic subgroups as possible

building blocks for the understanding the K-theory of arbitrary group. Recall

that a group is virtually cyclic if it is either finite or virtually infinite cyclic, i.e.,

contains a finite index subgroup which is infinite cyclic. More precisely, virtually

infinite cyclic groups V are of two types, namely,

1) The group V that admits an epimorphism (with finite kernel G) to the

infinite cyclic group T =< t >, i.e., V is the semi-direct product G ?αT

where α : G −→ G is an automorphism and the action of T is given by

tgt−1= α(g) for all g ∈ G.

2) The group V which admits an epimorphism (with finite kernel) to the infinite

dihedral group D∞, i.e., V = G0∗HG1where the groups Gi,i = 0,1, and

H are finite and [Gi: H] = 2.

A.O. Kuku

Mathematics Section, The Abdus Salam International Centre for Theoretical Physics, Trieste,

Italy (e-mail: kuku@ictp.trieste.it)

G. Tang

Department of Applied Mathematics, Northwestern Polytechnical University, Xi’an Shaanxi

710072, People’s Republic of China. (e-mail: tanggp@nwpu.edu.cn)

Page 2

712 A.O. Kuku, G. Tang

In [6], Farrell-Jones studied lower K-theory of virtually infinite cyclic groups

withcopiousreferencesonworkalreadydoneforlowerK-groupsoffinitegroups.

In this paper, we focus attention on higher K-theory of virtually cyclic groups.

Let R be the ring of integers in a number field F, ? any R-order in a semi-

simple F-algebra ?. In [10], [11],A. Kuku proved that for all n ≥ 1, Kn(?) and

Gn(?) are finitely generated Abelian groups and hence that for any finite group

G, Kn(RG) and Gn(RG) are finitely generated. One consequence of this result

is that for all n ≥ 1, if C is a finitely generated free Abelian group or monoid,

then Gn(?[C]) are also finitely generated (using the fundamental Theorem for

G-theory). However we can not draw the same conclusion for Kn(?[C]) since

for a ring A, it is known that all the NKn(A) are not finitely generated unless they

are zero ([16], Proposition 4.1).

We now briefly review the results in this paper.

In§1wesetthestagebyprovingtheorems1.1and1.6whichconstitutegener-

alizationsoftheorems1.2and1.5of[6].Here,weprovetheresultsforanarbitrary

R-order ? in a semi-simple F-algebra ? (where R is the ring of integers in a

number field F) rather than for the special case ? = ZG (G finite group) treated

in [6].

In §2, we prove that if R is the ring of integers in a number field F and ?

an R-order in a semi-simple F-algebra ?, α an automorphism of ?, then for all

n ≥ 0, NKn(?,α) is s-torsion for some positive integer s and that the torsion

free rank of Kn(?α[t]) is equal to the torsion free rank of Kn(?) which is finite

by ([10], [12]). When V = G ?αT is a virtually infinite cyclic group of the first

type, we show that for all n ≥ 0, Gn(RV) is a finitely generated Abelian group

andthatforalln < −1,Kn(RV) = 0.Wealsoshowthatforalln ≥ 0,NKn(RV)

is |G|-torsion.

For a virtually infinite cyclic group V = G0∗HG1of the second type, a triple

(ZH;Z[G0−H],Z[G1−H]) arises as a special case of a triple R = (R;B,C)

where R is a ring with identity, and B, C are R-bimodules (see §3). In the case

(ZH;Z[G0−H],Z[G1−H]),theZH-bimoduleZ[Gi−H]isisomorphictoZH

as a left ZH-module but the right action is twisted by an automorphism of H.We

arethusinspiredtoconsiderthegeneralcaseR = (R;Rα,Rβ)ofR beingaunital

ring, α,β automorphisms R −→ R, Rα(resp. Rβ) the R − R-bimodule which is

R as a left R-module but with right multiplication given by a · r = aα(r)(resp.

b · r = bβ(r)).

If T is the category of triples R = (R;B,C), then, there exists a functor

ρ : T −→ Rings,

(see §3 and [3]) and an augmentation map

ρ(R) = Rρ

? : Rρ−→

?R 0

0 R

?

.

Page 3

Higher K-theory of group-rings of virtually infinite cyclic groups713

Then the Nil-groups associated to R are defined for all n ∈ Z as NKn(R) =

kernel of the maps induced by ? on the Kn-groups (see §3). We prove the im-

portant result that Rρis a twisted polynomial ring over

?R 0

0 R

?

. We also prove

that when R is a regular ring, NKn(R;Rα,Rβ) = 0 for all n ∈ Z and that if

R is quasi-regular, then NKn(R;Rα,Rβ) = 0 for all n ≤ 0. When n ≤ 1, the

statement above was proved in [4] using isomorphism between NKn(R;Rα,Rβ)

andWaldhausen’s groups?

the vanishing of lowerWaldhausen’s groups?

that if V = G0∗HG1, where Gi,i = 0,1, and H are finite and [Gi: H] = 2,

then the Nil-groups NKn(ZH;Z[G0− H],Z[G1− H]) are |H|-torsion.

NilW

n−1(R;Rα,Rβ) and the fact that?

NilW

NilW

NilW

n−1(R;Rα,Rβ)

vanishes for regular rings R [4]. In effect, we have given here another proof of

n−1(R;Rα,Rβ) based on the iso-

n−1(R;Rα,Rβ) for n ≤ 1. We then prove

morphism of NKn(R;Rα,Rβ) and?

Notes on Notations

For an exact category C we write Kn(C) for the Quillen higher K-theory πn+1

(BQC) for n ≥ 0 see [13].

If A is any ring with identity, we write, for n ≥ 0, Kn(A) = Kn(P(A)) where

P(A) is the exact category of finitely generated projective modules over A and

when A is Noetherian, we write Gn(A) for Kn(M(A)) where M(A) is the exact

category of finitely generated A-modules.

We write T =< t > for the infinite cyclic group, and Trfor the free Abelian

group of rank r. If α is an automorphism of a ring A, we shall write Aα[T] for

the α-twisted Laurent series ring, i.e., Aα[T] = A[T] additively and multipli-

cation given by (rti)(stj) = rα−i(s)ti+j. Aα[t]:=subgroup of Aα[T] generated

by A and t, that is, Aα[t] is the twisted polynomial ring. We define, for any

n ≥ 0, NKn(A,α) =ker(Kn(Aα[t])

augmentation ? : Aα[t] −→ A,t ?→ 0. If α is the identity automorphism, then

NKn(A) =ker((KnA[t]) −→ Kn(A)).

?∗

−→ Kn(A)) where ?∗is induced by the

1. Some preliminary results

Let R be the ring of integers in a number field F, ? an R-order in a semi-simple

F-algebra? andα : ? −→ ?isanR-automorphism.Thenα extendstoF-auto-

morphism on ?. Suppose that ? is a maximal element in the set of all α-invariant

R-orders in ? containing ?. Let max(?) denote the set of all two-sided maximal

ideals in ? and maxα(?) the set of all two-sided maximal α-invariant ideals in

?. Recall that a ?-lattice in ? is a ?-? submodule of ? which generates ? as

a F-vector space. The aim of this section is to prove Theorem 1.1 and 1.6 below

Page 4

714 A.O. Kuku, G. Tang

which constitute generalizations of 1.2 and 1.5 of [6]. The proof follows that of

[6] rather closely and some details are omitted.

Theorem 1.1. The set of all two-sided, α-invariant, ?-lattices in ? is a free

Abelian group under multiplication and has maxα(?) as a basis.

Proof. Let a be a two-sided, α-invariant, ?-lattice in ?. Then {x ∈ ?|xa ⊆ a}

is an α-invariant R-order containing ?. Hence, it must be equal to ? by the

maximality of ?. Similarly {x ∈ ?|ax ⊆ a} = ?.

Now let a ⊆ ? be a two-sided, α-invariant, ?-lattice in ?. Then B = ?/a is

a finite ring and, hence, Artinian. So radB is a nilpotent, α-invariant, two-sided

idealinB andB/radB issemi-simplering.HenceB/radB decomposesasadirect

sum of simple rings Bi, i.e.,

B/radB = B1⊕ B2⊕ ··· ⊕ Bn,(I)

and α : B/radB −→ B/radB induces a permutation of the factors Bi, i.e.,

α(Bi) = Bˆ α(i)

where ˆ α is a permutation of {1,2,... ,n}. So, a ∈ maxα(?) if and only if both

rad(?/a)= 0 and ˆ α is a cyclic permutation. Hence, a / ∈ maxα(?) if and only

if there exist a pair of two-sided, α-invariant, ?-lattices b and c satisfying the

following three properties:

(i) both b and c properly contain a;

(ii) b and c are both contained in ?;

(iii) bc ⊆ a.

Hence, just as in [6], we deduce the following fact:

If a is a two-sided, α-invariant, ?-lattice, then a contains

a (finite) product of elements from maxα(?).

If a be a two-sided, α-invariant, ?-lattice, then we write

(II)

(III)

(IV)

¯ a = {x ∈ ?|xa ⊆ ?}

(V)

which is also a two-sided, α-invariant, ?-lattice. We now prove the following

Lemma 1.2. If p ∈ maxα(?), then

¯ p ?= ?.(VI)

Proof. Choose a positive integer s ∈ Z such that s? ⊂ p.Applying (i)–(iii) with

a = s? we can find elements pi∈ maxα(?) such that

p1p2···pn⊂ s?.

Let us assume that n is the smallest possible integer with this property. Using

the characterization of maxα(?) given above (III and IV), we see that some pi

Page 5

Higher K-theory of group-rings of virtually infinite cyclic groups715

must be contained in p since p ∈ maxα(?).And since pi∈ maxα(?), p = pi. We

can therefore write apb ⊂ s? where a = p1···pi−1and b = pi+1···pn. Thus

1

sapb ⊂ ? and further (1

by the definition of ¯ p. Since ba is a product of n − 1 elements in maxα(?), the

minimality of n implies that ba ⊂ /s?, so1

Lemma 1.3.

If p ∈ maxα(?), then ¯ pp = ? = p¯ p.

Proof. Similar to that in step 4, page 21 of [6].

sbap)b ⊂ b. So we have (1

sbap) ⊂ ?. Hence1

sba ⊂ ¯ p

sba ⊂ /?. Thus ¯ p ?= ?.

Lemma 1.4.

If p1,p2∈ maxα(?), then p1p2= p2p1.

Proof. Similar to that of step 5 in [6].

Note that the proof in ([1], p.158) is easily adapted to yield the following

conclusion that

A two-sided, α-invariant, ?-lattice a ⊆ ? is uniquely, up to order, a product

of elements of maxα(?) and we can finish the proof as in [1], p.158.

? ?

Corollary 1.5. If every element in maxα(?) is a right projective ?-module, then

every element in max(?) is also a right projective ?-module and consequently ?

is a hereditary ring.

Proof. It is the same as the proof of Corollary 1.6 in [6].

Theorem 1.6. Let R be the ring of integers in a number field F, ? any R-order

in a semi-simple F-algebra ?. If α : ? → ? is an R-automorphism, then there

exists an R-order ? ⊂ ? such that

1) ? ⊂ ?,

2) ? is α-invariant, and

3) ? is a (right) regular ring. In fact, ? is a (right) hereditary ring.

Proof. Let S be the set consisting of all α-invariant R-orders M of ? which con-

tains ?. Then S is not empty since ? ∈ S. Choose ? to be any maximal member

of S. Such a member exists by Zorn’s Lemma. Note that this R-order ? satisfies

properties 1) and 2) by definition, and is clearly a right Noetherian ring. Hence,

it suffices to show that ? is a right hereditary ring; i.e., that every right ?-module

is either projective or has a length 2 resolution by projective right ?-modules. To

do this, it suffices to show that every maximal two-sided ideal in ? is a projective

right ?-module. Let max(?) denote the set of all two-sided maximal ideals in ?,

and maxα(?) the set of all maximal members among the two-sided α-invariant

proper ideals in ?. Note that if a ∈ maxα(?), then ?/a is a finite ring.To see this,

first observe that ?/a is finitely generated as anAbelian group under addition. If

it were not finite, then there would exist a prime p ∈ Z such that the multiplies

Page 6

716A.O. Kuku, G. Tang

of p in ?/a would form a proper two-sided α-invariant proper ideals in ?/a. But

this would contradict the maximality of a.Also ?/a is a (right)Artinan ring since

it is a finite ring. But rad(?/a) is an α-invariant two-sided ideal in ?/a. So the

maximality ofa again shows that rad(?/a)= 0. Hence, ?/a is a semi-simple ring.

We finally remark that a is a two-sided ?-lattice in ? since a has finite index in

the lattice ?.

By Corollary 1.5, it suffices to show that every element in p ∈ maxα(?) is

a right projective ?-module. Let q be the inverse of p given by Theorem 1.1;

i.e., q is a two-sided ?-lattice in ? which is α-invariant and satisfies the equa-

tions pq = qp = ?. Consequently, there exist elements a1,a2,... ,an∈ p and

b1,b2,... ,bn∈ q such that

a1b1+ a2b2··· + anbn= 1.

Now define (right) ?-module homomorphisms f : p −→ ?nand g : ?n−→ p

by

f(x) = (b1x,b2x,... ,bnx)

where x ∈ p and y = (y1,y2,... ,yn) ∈ ?n. Note that the composite g ◦f = idp .

Consequently, p is a direct summand of ?nwhich shows that p is a projective

right ?-module.

g(y1,y2,... ,yn) = a1y1+ a2y2+ ··· + anyn

? ?

2. K-theory for the first type of virtually infinite cyclic groups

The aim of this section is to prove Theorem 2.2 below. However we start by

proving the following result which we shall need to prove 2.2.

Theorem 2.1. Let A be a Noetherian ring and α an automorphism of A, Aα[t]

the twisted polynomial ring. Then

(1) Gn(Aα[t])∼= Gn(A) for all n ≥ 0.

(2) There exists a long exact sequence

··· −→ Gn(A)

Proof. 1) follows directly from Theorem 2.18 of (cf. [8], p.194). To prove the

long exact sequence in 2), we denote by A = M(Aα[t]) the category consist-

ing of finitely generated Aα[t]-modules. Consider the Serre subcategory B of

A = M(Aα[t])whichconsistsofmodulesM ∈ objAonwhicht isnilpotent,i.e.,

objB = {M ∈ objA| there exists an m ≥ 0 such that Mtm= 0}.

Applying the localization theorem to the pair (A,B) we obtain a long exact

sequence

1−α∗

−→ Gn(A) −→ Gn(Aα[t,t−1]) −→ Gn−1(A) −→ ··· .

··· −→ Kn+1(A/B)

1−α∗

−→ Kn(B)−→Kn(A) −→ Kn(A/B) −→ ··· .

Page 7

Higher K-theory of group-rings of virtually infinite cyclic groups717

By definition and 1) Kn(A) = Gn(Aα[t])∼= Gn(A). We will prove that

Kn(B)∼= Gn(A)

and

Kn(A/B)∼= Gn(Aα[T]) := Gn(Aα[t,t−1])

foralln ≥ 0.AtfirstM(A) ⊆ M(Aα[t]/tAα[t]) ⊆ B (Notethatalthought / ∈cen-

ter(Aα[t]), we have tAα[t] = Aα[t]t ? Aα[t]). Using the Devissage theorem one

gets

Kn(B)∼= Kn(M(Aα[t]/tAα[t])).

But

0 −→ tAα[t] −→ Aα[t]

is an exact sequence of homomorphisms of rings. So, we have

t=0

−→ A −→ 0

Aα[t]/tAα[t]∼= A

as rings, and so Kn(M(Aα[t]/tAα[t]))∼= Gn(A). Thus

Kn(B)∼= Kn(M(Aα[t]/tAα[t]))∼= Gn(A).

Next we prove that

Kn(A/B)∼= Gn(Aα[T]).

SinceAα[T]isadirectlimitoffreeAα[t]-modulesAα[t]t−n,itisaflatAα[t]-mod-

ule and this implies that −⊗Aα[t]Aα[T] is an exact functor from A to M(Aα[T])

and further induces an exact functor

F :

A/B −→ M(Aα[T]).

We now prove that F is an equivalence. For any M ∈objM(Aα[T]), pick a

generating set {x1,x2,··· ,xl} of the finitely generated Aα[T]-module M. Let

l?

Then M1∈ A and M1⊗Aα[t]Aα[T]∼= M. The exact sequence of Aα[t]-modules

0 −→ M1−→ M −→ M/M1−→ 0,

induces an exact sequence

M1=

i=1

xiAα[t].

0 −→ M1⊗Aα[t]Aα[T] −→ M ⊗Aα[t]Aα[T] −→ M/M1⊗Aα[t]Aα[T] −→ 0.

Since {x1,x2,··· ,xl} is a generating set for the Aα[T]-module M, then for any

x ∈ M, there exist fi ∈ Aα[T](i = 1,··· ,l) such that x =?l

i=1xifi. Thus,

Page 8

718A.O. Kuku, G. Tang

there exists n ≥ 0 such that mtn=?l

0 −→ M1⊗Aα[t]Aα[T] −→ M ⊗Aα[t]Aα[T] −→ 0

that is:

M1⊗Aα[t]Aα[T]∼= M ⊗Aα[t]Aα[T].

For any M,N ∈ A, we have, by definition

HomA/B(M,N) = lim

where M/M?and N?are t-torsion. One gets easily:

i=1mi(fitn) ∈ M1, i.e., M/M1is t-torsion.

Thus M/M1⊗Aα[t]Aα[T] = 0. It follows that

is exact.

→HomAα[t](M?,N/N?)

HomA/B(M,N) = lim

where Nt= {x ∈ N| there exists an m ≥ 0 such that xtm= 0}. Define a map

φ : HomA/B(M,N) −→ HomB(M ⊗Aα[t]Aα[T],N ⊗Aα[t]Aα[T]).

For any f ∈ HomB(M ⊗Aα[t]Aα[T],N ⊗Aα[t]Aα[T]), since M is a finitely

generated Aα[T]-module, there exists an m ≥ 0 such that

f(Mtm⊗ 1) ⊆ N ⊗ 1.

We can define Mtm

−→ N/Nt,xtm?→ n if f(xtm⊗ 1) = n ⊗ 1. This is well

defined and σ maps to f under φ.

If σ ∈ HomAα[t](M?,N/Nt) is such that its image in HomB(M ⊗Aα[t]Aα[T],

N⊗Aα[t]Aα[T])iszero,thenσ⊗1 = 0impliesthatσm⊗1 = (σ⊗1)(m⊗1) = 0

in N/Nt⊗Aα[t]Aα[T]. Hence σ(m) = 0, and so,

HomA/B(M,N)∼= HomB(M ⊗Aα[t]Aα[T],N ⊗Aα[t]Aα[T])

that is

A/B∼= M(Aα[T]).

Hence Kn(A/B)∼= Kn(M(Aα[T])) completing the proof of 2).

Theorem 2.2. Let R be the ring of integers in a number field F, ? any R-order

in a semi-simple F-algebra ?, α an automorphism of ?. Then

→HomAα[t](M?,N/N?) = lim

→HomAα[t](M?,N/Nt),

σ

(a) For all n ≥ 0

(i) NKn(?,α) is s-torsion for some positive integer s. Hence the torsion free

rank of Kn(?α[t]) is the torsion free rank of Kn(?) and is finite. If n ≥ 2,

then the torsion free rank of Kn(?α[t]) is equal to the torsion free rank of

Kn(?).

(ii) If G is a finite group of order r, then NKn(RG,α) is r-torsion, where α

is the automorphism of RG induced by that of G.

Page 9

Higher K-theory of group-rings of virtually infinite cyclic groups719

(b) Let V = G ?αT be the semi-direct product of a finite group G of order r

with an infinite cyclic group T =< t > with respect to the automorphism

α : G −→ G :

(i) Kn(RV) = 0 for all n < −1.

(ii) The inclusion RG → RV induces an epimorphism K−1(RG) → K−1

(RV). Hence K−1(RV) is finitely generated Abelian group.

(iii) For all n ≥ 0, Gn(RV) is a finitely generated Abelian group.

(iv) NKn(RV) is r-torsion for all n ≥ 0.

Proof. (a)(i)ByTheorem1.6,wecanchooseanα-invariantR-order? in? which

contains?andisregular.FirstnotethatsinceeveryR-orderisaZ-order,thereisa

non-zero integer s such that ? ⊆ ? ⊆ ?(1/s), where A(1/s) denote A⊗Z(1/s)

for anAbelian group A. Put q = s?. Then we have a Cartesian square

? −→ ?

↓

?/q −→ ?/q

Since α induces automorphisms of all the four rings in the square (I) (cf. [6]), we

have another Cartesian square

g ?→ tgt−1. Then

↓

.

(I)

?α[t]

↓

(?/q)α[t] −→ (?/q)α[t]

−→

?α[t]

↓

.

(II)

Notethatboth(?/q)α[t]and(?/q)α[t]areZ/sZ-algebra,andso,itfollowsfrom

(II) that we have a long exact Mayer-Vietoris sequence (cf. [16] or [2])

··· → Kn+1((?/q)α[t])(1/s) → Kn(?α[t])(1/s)

→ Kn((?/q)α[t])(1/s) ⊕ Kn(?α[t])(1/s)

→ Kn((?/q)α[t])(1/s) → Kn−1(?α[t])(1/s) → ··· .

Since we also have a long exact Mayer-Vietoris sequence

(III)

··· → Kn+1(?/q)(1/s) → Kn(?)(1/s) → Kn(?/q)(1/s) ⊕ Kn(?)(1/s)

→ Kn(?/q)(1/s) → Kn−1(?)(1/s) → ··· ,

then, by mapping sequence (III) to sequence (IV) and taking kernels, we obtain

another long exact Mayer-Vietoris sequence

(IV)

··· → NKn+1(?/q,α)(1/s) → NKn(?,α)(1/s)

→ NKn(?/q,α)(1/s) ⊕ NKn(?,α)(1/s) →

→ NKn(?/q,α)(1/s) → NKn−1(?,α)(1/s) → ··· .

However, by [6] ?α[t] is regular since ? is. So NKn(?,α) = 0 by Theorem 2.1

(i) since Kn(?α[t]) = Gn(?α[t]) = Gn(?) = Kn(?). Both ?/q and ?/q are

Page 10

720 A.O. Kuku, G. Tang

finite, hence quasi-regular. They are also Z/sZ-algebra, and so it follows that

(?/q)α[t] and (?/q)α[t] are also quasi-regular and Z/sZ-algebra.We now prove

that for a finite Z/sZ-algebra A, NKn(A,α) is s-torsion. Since A is finite, its

Jacobson radical J(A) is nilpotent, and by Corollary 5.4 of [15] the relative

K-groups Kn(A,J(A)) are s-torsion for any n ≥ 0. This implies that

Kn(A)(1/s)∼= Kn(A/J(A))(1/s)

from the relative K-theory long exact sequence tensored with Z?1

Kn(Aα[x])(1/s)∼= Kn((A/J(A))α[x])(1/s).

However, A/J(A) is regular and so,

s

?. Similarly,

one gets

Kn((A/J(A))α[x])(1/s)∼= Kn(A/J(A))(1/s)

Hence, we have

Kn(Aα[x])(1/s)∼= Kn(A)(1/s).

FromthefinitenessofAonegetsKn(A)isfinite(cf.[10]).HencebothKn(Aα[x])

(1/s) and Kn(A)(1/s) have the same cardinality. From the exact sequence

by 2.1(1).

0 −→ NKn(A,α) −→ Kn(Aα[x]) −→ Kn(A) −→ 0 tensored with Z

?1

s

?

,

we obtain the exact sequence

0 −→ NKn(A,α)(1/s) −→ Kn(Aα[x])(1/s) −→ Kn(A)(1/s) −→ 0.

Hence NKn(A,α)(1/s) is zero since Kn(Aα[x])(1/s) and Kn(A)(1/s) are

isomorphic. Hence, both NKn+1(?/q,α)(1/s) and NKn(?/q,α)(1/s) are zero,

and so, NKn(?,α) is s-torsion.

Since Kn(?α[t]) = Kn(?) ⊕ NKn(?,α) and NKn(?,α) is torsion, the tor-

sion free rank of Kn(?α[t]) is the torsion free rank of Kn(?). By [9] the torsion

free rank of Kn(?) is finite and if n ≥ 2 the torsion free rank of Kn(?) is the

torsion free rank of Kn(?) (see [12]).

(a)(ii) is a direct consequence of (a)(i) since if |G| = r, and we take ? = RG

then r? ⊆ ?.

(b)(i) ByTheorem 1.6, there exists an α-invariant regular ring ? in FG which

contains RG. Then for the integer s = |G|, RG ⊆ ? ⊆ RG(1/s). Put q = s?.

Then we have a Cartesian square

RG −→ ?

↓

RG/q −→ ?/q

↓

.

(VI)

Page 11

Higher K-theory of group-rings of virtually infinite cyclic groups 721

Since α induces automorphisms of all the four rings in the square (VI), we have

another Cartesian square

RV

↓

−→

?α[T]

↓

(RG/q)α[T] −→ (?/q)α[T]

.

(VII)

see [6].

SincelowerK-theoryhasexcisionproperty,itfollowsfrom[6]thatwehavelower

K-theory exact sequence

K0(RV) → K0((RG/q)α[T]) ⊕ K0(?α[T]) → K0((?/q)α[T]) →

−→ K−1(RV) → K−1((RG/q)α[T]) ⊕ K−1(?α[T]) → ··· .

However, (?/q)α[T] and (RG/q)α[T] are quasi-regular since, ?/q and RG/q

are quasi-regular (see [6]).Also, ?α[T] is regular, since ? is regular (see [6]).

Hence

(VIII)

Ki(?α[T]) = Ki((RG/q)α[T]) = Ki((?/q)α[T]) = 0

for all i ≤ −1.Thus Ki(RV) = 0 for all i < −1, from the exact sequence (VIII).

(b)(ii) The proof of b(ii) is similar to the proof of a similar statement for ZV

in [6] Corollary 1.3 and is omitted.

(b)(iii) Is a direct consequence of Theorem 2.1(2) since Gn(RG) is finitely

generated for all n ≥ 1 (see [11]).

(b)(iv) By [6], 1.3.2, we have a Cartesian square

RV

↓

−→

?α[T]

↓

(RG/q)α[T] −→ (?/q)α[T]

where ?α[T] is regular and (RG/q)α[T],(?/q)α[T] are quasi-regular (see [6]

1.1 and 1.41).

Moreover, since r annihilates RG/q and ?/q it also annihilates (RG/q)α[T],

(?/q)α[T] since for A = RG/q, or ?/q, Aα[T] is a direct limit of free Aα[t]-

module Aα[t]t−n. Hence by [16], Corollary 3.3(d), NKn(Aα[T]) is r-torsion.

(Note that [16], Corollary 3.3(d) is valid when p is any integer r – we confirmed

this from the author C. Weibel.)

We also have by [2] and [16] a long exact Mayer-Vietoris sequence

... −→ NKn+1(?/q)α[T]?1

⊕NKn(?α[T]?1

But NKn(?α[T]) = 0 since ?α[T] is regular. Hence we have NKn(RV)?1

r

?−→ NKn(RV)?1

r

r

?−→ NKn(RG/q)α[T]?1

r

r

?

?−→ NKn(?/q)α[T]?1

?−→ ...

.

(IX)

?= 0

r

from IX, and so, NKn(RV) is r-torsion.

Page 12

722 A.O. Kuku, G. Tang

3. Nil-groups for the second type of virtually infinite cyclic groups

The algebraic structure of the groups in the second class is more complicated.

We recall that a group V in the second class has the form V = G0∗HG1where

the groups Gi,i = 0,1, and H are finite and [Gi: H] = 2. We will show that

the Nil-groups in this case are torsion, too. At first we recall the definition of

Nil-groups in this case.

Let T be the category of triples R = (R;B,C), where B and C are R-bimod-

ules.A morphism in T is a triple

(φ;f,g) : (R;B,C) −→ (S;D,E)

where φ : R −→ S is a ring homomorphism and both f : B −→ D and

g : C −→ E are R−S-bimodule homomorphisms.There is a functor ρ from the

category T to the category Rings defined by

?

where TR(B ⊗RC)(resp. TR(C ⊗RB)) is the tensor algebra of B ⊗RC(resp.

C ⊗RB) and ρ(R) is the ring with multiplication given as matrix multiplication

and each entry by concatenation. There is a natural augmentation map(cf. [3])

ρ(R) = Rρ=

TR(C ⊗RB)

B ⊗RTR(C ⊗RB)

C ⊗RTR(B ⊗RC)

TR(B ⊗RC)

?

? : Rρ−→

?R 0

0 R

?

.

The Nil-group NKn(R) is defined to be the kernel of the map induced by ? on

Kn-groups.

We now formulate the Nil-groups of interest. Let V be a group in the second

class of the form V = G0∗HG1where the groups Gi,i = 0,1, and H are finite

and [Gi: H] = 2. Considering Gi− H as the right coset of H in Giwhich is

different from H, the free Z-module Z[Gi− H] with basis Gi− H is a ZH-bi-

module which is isomorphic to ZH as a left ZH-module, but the right action is

twisted by an automorphism of ZH induced by an automorphism of H. Then the

Waldhausen’s Nil-groups are defined to be NKn(ZH;Z[G0− H],Z[G1− H])

using the triple (ZH;Z[G0− H],Z[G1− H]). This inspires us to consider the

followinggeneralcase.LetR bearingwithidentityandα : R −→ R aringauto-

morphism. We denote by Rαthe R − R-bimodule which is R as a left R-module

but with right multiplication given by a · r = aα(r). For any automorphisms α

and β of R, we consider the triple R = (R;Rα,Rβ). We will prove that ρ(R) is

in fact a twisted polynomial ring and this is important for later use.

Theorem 3.1. Suppose that α and β are automorphisms of R. For the triple R =

(R;Rα,Rβ),letRρbetheringρ(R),andletγ bearingautomorphismof

defined by

?R 0

0 R

?

Page 13

Higher K-theory of group-rings of virtually infinite cyclic groups 723

γ :

?a 0

0 b

?

?−→

?β(b)

0

0

α(a)

?

.

Denote by 1α(resp. 1β) the generator of Rα(resp. Rβ) corresponding to 1. Then,

there is a ring isomorphism

?R 0

defined by mapping an element

?ai 0

?

to an element

?

Proof. By definition, each element of Rρcan be written uniquely as

?ai 0

?

It is easy to see that µ is an isomorphism from the additive group of Rρ to

?R 0

r

product of two elements such as:

?ai 0

and

?a?

We check these case by case. Note that for any a,b ∈ R, 1α· a = α(a)1αand

1β· b = β(b)1β, thus

?(1β⊗ 1α)i

µ : Rρ−→

0 R

?

γ

[x],

?

i≥0

0 bi

??(1β⊗ 1α)i

?a?

0

0

(1α⊗ 1β)i

?

+

i≥0

i0

0 b?

i

??

01β⊗ (1α⊗ 1β)i

01α⊗ (1β⊗ 1α)i

?

i≥0

?ai 0

0 bi

?

x2i+

?

i≥0

?a?

i0

0 b?

i

?

x2i+1.

?

i≥0

0 bi

??(1β⊗ 1α)i

?a?

0

0

(1α⊗ 1β)i

?

+

i≥0

i0

0 b?

i

??

01β⊗ (1α⊗ 1β)i

01α⊗ (1β⊗ 1α)i

?

.

0 R

?

[x]. To complete the proof, we only need to check that µ preserves any

ui=

0 bi

??(1β⊗ 1α)i

0

0

(1α⊗ 1β)i

?

,i ≥ 0,

vj=

j0

0 b?

j

??

01β⊗ (1α⊗ 1β)j

01α⊗ (1β⊗ 1α)j

?

,j ≥ 0.

0

0

(1α⊗ 1β)i

??aj 0

0 bj

?

Page 14

724 A.O. Kuku, G. Tang

=

?(βα)i(aj)

0

0

(αβ)i(bj)

??(1β⊗ 1α)i

0

0

(1α⊗ 1β)i

?

.

Note that the following equations hold in Rρ:

(1β⊗ 1α)i(1β⊗ 1α)j= (1β⊗ 1α)i+j

(1α⊗ 1β)i(1α⊗ 1β)j= (1α⊗ 1β)i+j

(1β⊗ 1α)i(1β⊗ (1α⊗ 1β)j) = 1β⊗ (1α⊗ 1β)i+j

(1α⊗ 1β)i(1α⊗ (1β⊗ 1α)j) = 1α⊗ (1β⊗ 1α)i+j.

This implies that µ(uiuj) = µ(ui)µ(uj) and µ(uivj) = µ(ui)µ(vj). Similarly,

we have

?

?(βα)iβ(b?

We have also equations in Rρ:

01β⊗ (1α⊗ 1β)i

0

??

1α⊗ (1β⊗ 1α)i

j)

(αβ)iα(a?

??a?

j0

0 b?

j

?

=

0

0

j)

01β⊗ (1α⊗ 1β)i

01α⊗ (1β⊗ 1α)i

?

.

(1β⊗ (1α⊗ 1β)i)(1α⊗ 1β)j= 1β⊗ (1α⊗ 1β)i+j

(1α⊗ (1β⊗ 1α)i)(1β⊗ 1α)j= 1α⊗ (1β⊗ 1α)i+j

(1β⊗ (1α⊗ 1β)i)(1α⊗ (1β⊗ 1α)j) = (1β⊗ 1α)i+j+1

(1α⊗ (1β⊗ 1α)i)(1β⊗ (1α⊗ 1β)j) = (1α⊗ 1β)i+j+1.

It follows that µ(vjui) = µ(vj)µ(ui) and µ(vivj) = µ(vi)µ(vj). Hence µ is an

isomorphism.

? ?

From Theorem 3.1 above, we obtain the following important result.

Theorem 3.2. If R is regular, then NKn(R;Rα,Rβ) = 0 for all n ∈ Z. If R is

quasi-regular then NKn(R;Rα,Rβ) = 0 for all n ≤ 0.

?R 0

a twisted polynomial ring over

0 R

mental Theorem of Algebraic K-Theory it follows that NKn(R;Rα,Rβ) = 0

for all n ∈ Z (see [13]). If R is quasi-regular, then Rρ is quasi-regular also

[6]. By the fundamental Theorem of Algebraic K-Theory for lower K-theory

NKn(R;Rα,Rβ) = 0 for any n ≤ 0.

Remark. When n ≤ 1, the result above is proved in [4] using isomorphism be-

tweenNKn(R;Rα,Rβ)andWaldhausen’sgroups?

Proof. Since R is regular then

0 R

?

?

is regular too. By Theorem 3.1, Rρis

?R 0

, and so it is regular [6]. By the funda-

NilW

n−1(R;Rα,Rβ)andthefact

that?

NilW

n−1(R;Rα,Rβ) vanishes for regular rings R [4]. In effect, we have given

Page 15

Higher K-theory of group-rings of virtually infinite cyclic groups725

here another proof of the vanishing of lowerWaldhausen’s groups?

n ≤ 1.

Now, we specialize to the case that R = ZH, the group ring of a finite group

H of order h. Let α and β be automorphisms of R induced by automorphisms of

H. Choose a hereditary order ? as in Theorem 1.6. Then we can define triples in

T ,

R = (R;Rα,Rβ),

? = (?;?α,?β),

R/h? = (R/h?;(R/h?)α,(R/h?)β),

?/h? = (?/h?;(?/h?)α,(?/h?)β).

The triples determine twisted polynomial rings

NilW

n−1(R;Rα,

Rβ) based on the isomorphism of NKn(R;Rα,Rβ) and?

NilW

n−1(R;Rα,Rβ) for

? ?

Rρcorresponding to R,

?ρcorresponding to ?,

(R/h?)ρcorresponding to R/h?,

(?/h?)ρcorresponding to ?/h?.

(1)

Hence there is a Cartesian square

RG −→ ?

↓

RG/q −→ ?/q

↓

.

(2)

which implies that the square

?R 0

0 R

↓

?

−→

?? 0

0 ?

↓

?

?R/h?

0

0

R/h?

?

−→

??/h?

0

0

?/h?

?

(3)

is a Cartesian square. By Theorem 3.1, we have the following Cartesian square

Rρ

↓

−→

?ρ

↓

(R/h?)ρ−→ (?/h?)ρ

.

(4)

Download full-text