LAMPLIGHTER GROUP
PETER A. LINNELL, WOLFGANG L ¨UCK, AND THOMAS SCHICK
Abstract. LetG = Z/2ZoZ be the so called lamplighter group and k a commutative ring. We show thatkGdoes not have a classical ring of quotients (i.e. does not satisfy the Ore condition). This answers a Kourovka notebook problem. Assume thatkGis contained in a ringRin which the element 1−x is invertible, withxa generator ofZ ⊂G. Then Ris not flat overkG. If k=C, this applies in particular to the algebraUGof unbounded operators affiliated to the group von Neumann algebra ofG.
We present two proofs of these results. The second one is due to Warren Dicks, who, having seen our argument, found a much simpler and more el- ementary proof, which at the same time yielded a more general result than we had originally proved. Nevertheless, we present both proofs here, in the hope that the original arguments might be of use in some other context not yet known to us.
1. Notation and Terminology
Let A be a group. Then AoZ indicates the Wreath product with base group B = L∞
i=−∞Ai, where Ai = A for all i. Thus AoZ is isomorphic to the split extensionBo Zwhere ifxis a generator forZ, thenxiA0x−i=Ai. Also we identify AwithA0. In the caseA=Z/2Zabove,AoZis often called the lamplighter group.
LetkGdenote the group algebra of the groupGover the fieldk, and letα∈kG.
Writeα=P
g∈Gagg where ag∈k. Then the support of αis{g ∈G|ag 6= 0}, a finite subset ofG.
The augmentation ideal of a group algebra will denoted by the small German letter corresponding to the capital Latin letter used to name the group. Thus ifkis a field andGis a group, thengis the ideal ofkGwhich hask-basis{g−1|g∈G\1}. For the purposes of this paper, it will always be clear over which field we are working when considering augmentation ideals.
1. Definition. A ring R satisfies the Ore condition if for any s, t ∈ R with s a non-zerodivisor, there are x, y ∈ R with x a non-zerodivisor such that sy = tx. Formally, this means thats−1t =yx−1, and the condition makes sure that a classical ring of quotients, inverting all non-zerodivisors ofR, can be constructed.
Then R is an Ore ring means thatR satisfies the Ore condition. Equivalently this means that ifS is the set of non-zerodivisors ofR, then givenr∈Rands∈S, we can findr1∈Rand s1∈S such thatrs1 =sr1. In this situation we can form the Ore localizationRS−1, which consists of elementsrs−1wherer∈Rands∈S.
Date: Jun 30 2003.
2000Mathematics Subject Classification. Primary: 16U20; Secondary: 46L99.
Key words and phrases. Ore ring, affiliated operators, flat, lamplighter group, Fox calculus.
The first author was supported in part by the Sonderforschungsbereich in M¨unster.
1
The above definition is really the right Ore condition, though for group rings the right and left Ore conditions are equivalent.
In this note, we study, which group rings satisfy the Ore condition. It is well known that this fails for a non-abelian free group.
On the other hand, abelian groups evidently satisfy the Ore condition. In this note we show that the lamplighter groups (and relatives) do not satisfy it. Note, however, that these groups are 2-step solvable, i.e. close relatives of abelian groups.
Let Gbe a group, let N(G) denote the group von Neumann algebra ofG, let U(G) denote the algebra of unbounded linear operators affiliated to N(G), and let D(G) denote the division closure ofCG in U(G). For more information on these notions, see [5,§8 and§9] and [6,§8 and§10]. In particular we have the inclusion ofC-algebras
CG⊆ D(G)⊆ U(G)
and it is natural to ask whetherD(G) andU(G) are flat overCG.
We use the following well-known and easily verified statement without further comment in this paper. Ifkis a field and g∈Ghas infinite order, then 1−gis a non-zerodivisor inkG, and in the casek=C we also have that 1−g is invertible inD(G).
IfH is the nonabelian free group of rank 2, then we have an exact sequence of CH-modules 0 →CH2 →CH → C→ 0. It was shown in [4, Theorem 1.3] (see also [5, Theorem 10.2] and [6, Theorem 10.19 and Lemma 10.39]) thatD(H) is a division ring, so when we apply⊗CHD(H) to this sequence, it becomes D(H)2→ D(H)→0→0, since (1−x) is invertible inD(H) for every element 16=x∈H, but (1−x) acts as the zero operator on C. By counting dimension, we get from this (adding the kernel) a short exact sequence 0→ D(H)→ D(H)2→ D(H)→0.
SupposeQis a ring containingD(H) which is flat overCH. Then applying⊗D(H)Q to the previous sequence, we obtain the exact sequence 0 → Q→ Q2 → Q → 0 which contradicts the hypothesis thatQis flat overCH(in the latter case we would have obtained 0→Q2→Q→0→0). In particular ifGis a group containingH, then neither D(G) norU(G) is flat overCH. SinceCGis a free CH-module, we conclude that neitherD(G) norU(G) is flat overCG.
To sum up the previous paragraph, neitherD(G) norU(G) is flat overCGwhen Gcontains a nonabelian free group. On the other hand it was proven in [8, Theorem 9.1] (see also [6, Theorem 10.84]) that ifGis an elementary amenable group which has a bound on the orders of its finite subgroups, thenD(G) andU(G) are flat over CG. Furthermore it follows from [6, Theorems 6.37 and 8.29] that ifGis amenable (in particular ifGis the lamplighter group), then at leastU(G) is “dimension flat”
overCG.
2. Original results and proof We shall prove
2. Theorem. LetH 6= 1 be a finite group and let Gbe a group containingHoZ.
Then neitherD(G) nor U(G) is flat overCG.
Closely related to this question is the problem of when the group algebrakGof the groupGover the fieldkis an Ore ring (in other words doeskGhave a classical ring of quotients; see Definition 1). Our next result answers a Kourovka Notebook
problem [7, 12.47], which was proposed by the first author. The problem there asks ifkGhas a classical quotient ring in the caseG=ZpoZwherepis a prime.
3. Theorem. LetH 6= 1 be a finite group, let k be a field and letG be a group containingHoZ. ThenkGis not an Ore ring.
4. Lemma. LetRbe a subring of the ringS and let P be a projectiveR-module.
IfP⊗RS is finitely generated as anS-module, thenP is finitely generated.
Proof. SinceP is projective, there areR-modulesQ, F withF free such thatP⊕ Q=F. LetE be a basis forF. NowP⊗RS⊕Q⊗RS=F⊗RSand sinceP⊗RS is finitely generated, there existe1, . . . , en ∈ E such thatP⊗RS⊆e1S+· · ·+enS.
We now see that every elementpofP is
(i) AnR-linear combination of elements inE ( =⇒ p⊗1 =P
e∈Ee⊗re).
(ii) AnS-linear combination ofe1, . . . , en (p⊗1 =P
ei⊗si).
SetE=e1R+· · ·+enR. Comparing coefficients, the above shows thatP ⊆Eand it follows that we have the equationP ⊕(Q∩E) =E. Therefore P is a finitely
generatedR-module as required.
5. Lemma. LetH be a nontrivial finite group, letk be a field with characteristic which does not divide|H|, and let G=H oZ. IfQ is a ring containing kGsuch thatk⊗kGQ= 0, then TorkG(k, Q)6= 0.
Proof. Letddenote the minimum number of elements required to generateG. Then we have exact sequences
0−→g−→kG−→k−→0 0−→P −→kGd−→g−→0.
Suppose to the contrary TorkG(k, Q) = 0. Then the following sequence is exact:
0−→g⊗kGQ−→kG⊗kGQ−→k⊗kGQ= 0.
Since kG has homological dimension one [1, p. 70 and Proposition 4.12], we also have 0 = TorkG2 (k, Q) = TorkG1 (g, Q). Hence we have another exact sequence (6) 0−→P⊗kGQ−→kGd⊗kGQ−→g⊗kGQ−→0.
We rewrite this to get the exact sequence
0−→P⊗kGQ−→Qd−→Q−→0
and we conclude thatP⊗kGQis a finitely generatedQ-module, which is projective since the sequence (6) splits. NowkG has cohomological dimension ≤2 [1, p. 70 and Theorem 4.6 and Proposition 4.12], hence P is a projective kG-module. (To see this, letP0 → P be a map from a projectivekG-module P0 ontoP. Because Ext2kG(k, Q) = 0, this extends to a map P0 → kGd → P. Since the image of the first arrow is P, this gives a split of the injection P →kGd, i.e. P is projective.) Therefore P is finitely generated by Lemma 4. But it is well known that P ∼= R/[R, R]⊗k as kG-modules, where 1 →R → F → G→ 1 is an exact sequence of groups andF is a free group withd generators (compare the proof of [3, (5.3) Theorem]). Consequently,Gis almost finitely presented overkas defined in [2] ifP is a finitely generatedkG-module. But this is a contradiction to [2, Theorem A or Theorem C] (here we useH 6= 1), where the structure of almost finitely presented
groups such asGis determined.
7. Corollary. Let H be a nontrivial finite group, letkbe a field with characteristic which does not divide |H|, and let G be a group containing H oZ. Let x be a generator for Z in G. IfQ is a ring containing kG and1−x is invertible in Q, thenQis not flat over kG.
Proof. Set L = H oZ. Since 1−xis invertible in Q, we have C⊗CLQ = 0, so from Lemma 5 we deduce that TorkL(k, Q)6= 0. Furthermore kGis flat overkL, consequently TorkG(k⊗kLkG, Q)6= 0 by [1, p. 2], in particularQ is not flat over
kGas required.
Proof of Theorem 2. SetL=HoZand letxbe a generator forZinL. Then 1−x is a non-zerodivisor in N(G) and therefore is invertible in U(G), and hence also invertible inD(G). The result now follows from Corollary 7 8. Lemma. Letpbe a prime, letkbe a field of characteristicp, letAbe a group of order p, letG =AoZwith base group B, let a be a generator for A, and let x∈ Gbe a generator for Z. Then there does not existα, σ ∈kG with σ /∈bkG such that (1−a)σ= (1−x)α.
Proof. Suppose there does exist α andσ as above. Observe that α∈bkG, since bkGis the kernel of the map kG→kZinduced from the obvious projection G= AoZ→Zmapping xtox, and since 1−xis not a zerodivisor inkZ.
Thus we may writeσ =τ+P
isixi where si ∈ k, τ ∈bkG and not all the si are zero, andα=P
ixiαi where αi ∈b. Then the equation (1−a)σ= (1−x)α taken modb2kGyields
X
i
(1−a)sixi =X
i
(1−x)xiαi.
Setbi= 1−x−iaxi. By equating the coefficients ofxi, we obtainsibi =αi−αi−1 modb2kGfor alli. Sinceαi6= 0 for only finitely manyi, we deduce thatPsibi= 0 modb2. Alsob/b2∼=B⊗kask-vector spaces via the map induced byb−17→b⊗1 and the elements x−iaxi⊗1 are linearly independent inB⊗k, consequently the bi are linearly independent over k modb2. We now have a contradiction and the
result follows.
9. Lemma. Letkbe a field, letH be a locally finite subgroup of the groupG, and letα1, . . . , αn∈hkG. Then there existsβ ∈kH\0 such thatβαi= 0 for alli.
Proof. LetT be a right transversal forHinG, soGis the disjoint union of{Ht|t∈ T}. For eachi, we may writeαi =P
t∈Tβittwhereβit∈h. LetBbe the subgroup generated by the supports of the βit. Then B is a finitely generated subgroup of H and thus B is a finitep-group. Alsoβit∈bfor all i, t. Setβ =P
b∈Bb. Then
βb= 0 and the result follows.
10. Lemma. Letpbe a prime, letkbe a field of characteristicp, letAbe a group of order p, and let Gbe a group containing AoZ. Then kG does not satisfy the Ore condition.
Proof. LetH =AoZ, which we may regard as a subgroup ofG, withx∈H⊂Ga generator ofZ. LetB be the base group ofH and letT be a right transversal forH inG, soGis the disjoint union of{Ht|t∈T}. Note that 1−xis a non-zerodivisor inkG. Letabe a generator forA. Suppose (1−a)σ= (1−x)αwhereα, σ∈kGand
σis a non-zerodivisor inkG. Then we may write α=P
t∈Tαtt andσ=P
t∈Tσtt withαt, σt∈kH, and then we have
(1−a)σt= (1−x)αt
for allt∈T. If σt∈bkH for allt∈T, then by Lemma 9 we see that there exists β ∈kB\0 such thatβσt= 0 for all t. This yields βσ = 0 which contradicts the hypothesis thatσis a non-zerodivisor. Therefore we may assume that there exists s∈T such thatσs∈/bkH. But now the equation
(1−a)σs= (1−x)αs
contradicts Lemma 8, and the result follows.
Proof of Theorem 3. If the characteristic ofk ispand divides|H|, then the result follows from Lemma 10. On the other hand if pdoes not divide |H|, we suppose that kG satisfies the Ore condition. ThenkG has a classical ring of quotients Q.
It is a well known fact that such a classical ring of quotients is always flat over its base (compare [9, p. 57]). In particular,Qis flat overkG. Letxbe a generator for ZinG. Since 1−xis a non-zerodivisor inkG, we see that 1−xis invertible inQ.
We now have a contradiction by Lemma 5 and the result follows.
3. Warren Dicks’ proof
In this section, we give our account of Warren Dicks’ proof and generalization of the results presented in Section 2. All credit has to go to him, all mistakes are ours. More precisely, we prove the following theorem (for elementsx, yin a group, we use the commutator convention [x, y] =xyx−1y−1):
11. Theorem. Let 2 ≤d∈Z, letG=ha, x|ad= 1,[a, xlax−l] = 1; l= 1,2, . . .i be the wreath productZ/dZoZ, and let kbe a nonzero commutative ring with unit.
If u, v ∈kG are such that u(a−1) =v(x−1), then uis a left zerodivisor inkG.
In particular,kG does not satisfy the Ore condition.
Proof. For the last statement, note that (x−1) is a non-zerodivisor inkGbecause xhas infinite order.
Recall that any presentation H = hS | Ri of a groupH gives rise to an exact sequence of leftkH-modules
(12) M
r∈R
kH−→F M
s∈S
kH−→α kH−→ k→0.
Here, is the augmentation map defined by (h) = 1 for all h ∈ H, α maps us∈L
s∈SkH (with u∈kH ands the canonical basis element corresponding to the generators∈S) tou(s−1)∈kH, and the mapF is given by the Fox calculus, i.e. ur (where u∈ kH and r is the canonical basis element corresponding to the relatorr∈R) is mapped to
X
s∈S
u∂r
∂ss.
Ifr=si1
1. . . sin
n withsi∈S andi∈ {−1,1}, then the Fox derivative is defined by
∂r
∂s :=
n
X
k=1
si1
1. . . sik−1
k−1
∂sik
k
∂s .
Here∂s/∂s= 1, ∂s−1/∂s=−s−1 and∂t/∂s= 0 ifs6=t∈S and=±1.
The above sequence can be considered as the cellular chain complex (with coeffi- cientsk) of the universal covering of the standard presentation CW-complex given by hS | Ri. Since this space is 2-connected, its first homology vanishes and its zeroth homology is isomorphic tok(by the augmentation), which implies that the sequence is indeed exact. An outline of the proof can be found in [3, II.5 Exercise 3] or in [3, IV.2, Exercises]
Now we specialize to the groupG. Let us writer0=ad andrl= [a, xlax−l] for l≥1. Supposeu, v∈kGwithu(a−1) =v(x−1). Thenα(ua−vx) = 0. Exactness implies that there exists a positive integer N and zl ∈kG (0≤l ≤N) such that F(P
lzlrl) =ua−vx. We want to prove thatuis a zerodivisor. Therefore we are concerned only with theacomponent ofF(P
0≤l≤Nzlrl). This means we first must compute∂r/∂afor all the relators in our presentation ofG. This is easily done:
∂ad
∂a = 1 +a+· · ·+ad−1 (13)
∂[a, xlax−l]
∂a = ∂(axlax−la−1xla−1x−l) (14) ∂a
= 1 +axl−axlax−la−1−axlax−la−1xla−1, (15)
the latter forl >0. Using the fact thatxlax−lcommutes withafor eachl, we can simplify:
∂[a, xlax−1]
∂a = 1 +axl−xlax−l−xl=xl(x−laxl−1)−(xlax−l−1).
Sinceuis the coefficient ofainF(P
0≤l≤Nzlrl) we see that
u=z0(1 +a+· · ·+ad−1) +
N
X
n=1
znxn(x−naxn−1)−zn(xnax−n−1).
Now let C = hxnax−n | 1 ≤ |n| ≤ Ni, a finite subgroup of the base group L∞
i=−∞Z/dZ. Then Chai =C× hai. Set γ = (1−a)P
c∈Cc. Then γ 6= 0 and βγ= 0. We conclude thatuis a left zerodivisor inkGand the result follows.
16. Corollary. Let 2≤d∈Z and let Gbe a group containing Z/dZoZ, and let x∈Gbe a generator forZ. Letk be a nonzero commutative ring with unit and let Qbe a ring containing kGsuch that1−xbecomes invertible inQ. ThenQis not flat over kG.
Proof. SincekGis free, hence flat, as leftk[Z/dZoZ]-module, by [1, p. 2] we may assume thatG=Z/dZoZ. Now tensor the exact sequence (12) overkGwith Q.
Then the resulting sequence will also be exact and a−(a−1)(x−1)−1x will be in the kernel of idQ⊗α. Therefore a−(a−1)(x−1)−1xwill be in the image of idQ⊗F. However the proof of Theorem 11 shows that if ua−vx is in the image of idQ⊗F, thenuis a zerodivisor; the only change is that we want zl∈Qrather thanzl∈kG. Sinceu= 1 in this situation, which is not a zerodivisor, the tensored
sequence is not exact.
In the casekis a subfield ofCandQ=D(G) orU(G), Corollary 16 tells us that ifGcontainsZ/dZoZ, thenD(G) andU(G) are not flat overkG.
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Department of Mathematics, Virginia Tech, Blacksburg, VA 24061-0123, USA E-mail address:linnell@math.vt.edu
URL:http://www.math.vt.edu/people/linnell/
FB Mathematik, Universit¨at M¨unster, Einsteinstr. 62, D-48149 M¨unster, Germany E-mail address:lueck@math.uni-muenster.de
URL:http://wwwmath.uni-muenster.de/math/u/lueck/
Mathematisches Institiut, Universit¨at G¨ottingen, Bunsensrt. 3-5, D-37073 G¨ottingen, Germany
E-mail address:schick@uni-math.gwdg.de URL:http://www.uni-math.gwdg.de/schick/
