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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)
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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
?
.
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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
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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
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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
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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) −→ ··· .
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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,
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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.
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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
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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)
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