L 2 -Betti numbers and non-unitarizable groups without free subgroups
aa r X i v : . [ m a t h . G R ] F e b L -Betti numbers and non-unitarizable groups without freesubgroups D. V. Osin ∗ Abstract
We show that there exist non-unitarizable groups without non-abelian free sub-groups. Both torsion and torsion free examples are constructed. As a by-product, weshow that there exist finitely generated torsion groups with non-vanishing first L -Bettinumbers. We also relate the well-known problem of whether every hyperbolic group isresidually finite to an open question about approximation of L -Betti numbers. Let G be a group, H a Hilbert space. Recall that a representation π : G → B ( H ) is unitarizable if there exists an invertible operator S : H → H such that g → S − π ( g ) S is aunitary representation of G . A (locally compact) group G is unitarizable if every uniformlybounded representation π : G → B ( H ) is unitarizable.In 1950, Day [3] and Dixmier [4] proved that every amenable group is unitarizable. Thequestion of whether the converse holds has been open since then. A good survey of thecurrent research in this direction is given in [19].The simplest examples of a non-unitarizable groups are non-abelian free groups. (Anexplicit construction of a uniformly bounded non-unitarizable representation can be found[12]). Since unitarizability passes to subgroups, every group containing a non-abelian freesubgroup is not unitarizable as well. However, the answer to the following question wasunknown until now. Problem 1.1.
Does there exist a non-unitarizable group without non-abelian free subgroup?
Note that if there is such a group, it is a non-amenable group without non-abelian freesubgroups. The existence of such groups remained a fundamental open problem for manyyears until the first examples were constructed by Olshanskii in [13]. The aim of this noteis to answer the question affirmatively. Namely we prove the following.
Theorem 1.2.
There exists a finitely generated torsion non-unitarizable group. ∗ This work has been supported by the NSF grant DMS-0605093 L -Betti numbers of groups defined by periodic relations, and some older techniques related tohyperbolic groups [6, 15]. Though the property of being torsion is crucial for this approach,we show that the torsion group from Theorem 1.2 can be used to construct examples ofcompletely different nature. Theorem 1.3.
There exists a finitely generated torsion free non-unitarizable group withoutfree subgroups.
As a by-product, we also obtain some new results about L -Betti numbers of groups.Recall that if G is a torsion free group satisfying the Atiyah Conjecture (or even a weakerproperty ( ∗ ) introduced in [20]), then β (2)1 ( G ) > G [20, Theorem 4.1]. For finitely presented residually p -finite groups, where p is a prime, even a stronger result holds. Namely β (2)1 ( G ) > G impliesthat G is large [8]. These examples lead to a natural question of whether non-vanishing ofthe first L -Betti number always implies the existence of non-abelian free subgroups. Thefollowing theorem shows that the answer is negative. Theorem 1.4.
There exists a finitely generated torsion group with non-vanishing first L -Betti number. Moreover, Theorem 1.4 allows us to relate a question about approximation of L -Bettinumbers to one of the most intriguing open problems about hyperbolic groups. Recall thata group G is residually finite if for every element g = 1 of G there is a homomorphism ε : G → Q , where Q is finite, such that ε ( g ) = 1. By the Approximation Theorem of L¨uck[9], for every residually finite finitely presented group G and every nested sequence of finiteindex normal subgroups { N i } of G with trivial intersection, one has β (2)1 ( G ) = lim i →∞ b ( N i )[ G : N i ] , (1)where b ( N i ) is the ordinary first Betti number of N i . The following question is still open. Problem 1.5.
Does the approximation hold for any finitely generated residually finitegroup?
The other problem is well-known. For a survey of the theory of hyperbolic groups werefer to [6, 2].
Problem 1.6.
Is every hyperbolic group residually finite?
We show that if every hyperbolic group is residually finite, then the group from Theorem1.4 can be made residually finite as well. However this contradicts (1) since β ( N i ) = 0 forany subgroup of a torsion group. Thus we obtain the following. Corollary 1.7.
At least one of the two problems has a negative solution.
Acknowledgment
I am grateful to Nicolas Monod for drawing my attention to the paper[5] and stimulating discussions. I am also grateful to Jesse Peterson for explaining resultsof [20]. 2
Torsion groups with positive first L -betty numbers Recall that a group is elementary if it contains a cyclic subgroup of finite index. For everyhyperbolic group G and every element g ∈ G of infinite order, there exists a (unique)maximal elementary subgroup E ( g ) ≤ G containing g (see, e.g., [15, Lemma 1.16].Given an element g of a group G , we denote by hh g ii the normal closure of g in G , i.e., thesmallest normal subgroup of G containing g . Our main tool in this section is the followingresult about adding higher powered relations to hyperbolic groups. Up to little changesit was conjectured by Gromov [6]. It can easily be extracted from the proof of (a morecomplicated) Theorem 3 in [15]. Since the result we need does not formally follow from [15,Theorem 3], we briefly explain how to derive it from other results of [15] for convenience ofthe reader. Lemma 2.1 (Olshanskii, [15]) . Let G be a hyperbolic group, S a finite subset of G , E amaximal elementary subgroup of G , C a finite index normal cyclic subgroup of E . Supposethat C = h x i . Then for every sufficiently large integer n the following conditions hold.(1) The quotient group G/ hh x n ii is hyperbolic.(2) The image of the element x in G/ hh x n ii has order n .(3) The natural homomorphism G → G/ hh x n ii is injective on S .Proof. Let X be a finite generating set of G , W a shortest word in X ∪ X − representing x in G . By [15, Lemma 4.1] the set of all cyclic shifts of the words W ± m satisfies a smallcancellation condition, which implies properties (1)-(3) for G = h G | W n = 1 i ∼ = G/ hh x n ii by [15, Lemma 6.7] if m = m ( G, x, S ) is sufficiently large.For a background on L -Betti numberst we refer the reader to [10]. In what follows weassume that 1 / | G | = 0 if a group G has infinite order. Recall that for G = G ∗ · · · ∗ G n , wehave β (2)1 ( G ) = n − n X i =1 (cid:18) β (2)1 ( G i ) − | G i | (cid:19) (2)(see[11]). The following theorem of Peterson and Thom [20] will allow us to control first L -Betti number after adding higher-powered relations to G . Theorem 2.2. [20, Theorem 3.2] Let G be an infinite countable discrete group. Assumethat there exist subgroups G , . . . , G n of G , such that G = h G , . . . , G n | r w , . . . , r w k k i , for some elements r , . . . , r k ∈ G ∗ · · · ∗ G n and positive integers w , . . . , w k . Suppose inaddition that the order of r i in G is w i . Then, the following inequality holds: β (2)1 ( G ) ≥ n − n X i =1 (cid:18) β (2)1 ( G i ) − | G i | (cid:19) − k X j =1 w j . heorem 2.3. For every positive integer n and every ε > , there exists an n -generatedtorsion group T with β (2)1 ( T ) ≥ n − − ε . Moreover, if every hyperbolic group is residuallyfinite, then the group T can be additionally made residually finite.Proof. Roughly speaking, the main idea of the proof is to start with the free product G = Z m ∗ · · · ∗ Z m and then add periodic relations r w = 1 (one by one) for all r ∈ G , where m and all w = w ( r ) are large enough. We are going to use Theorem 2.2 to prove that thefirst L -betti number of the groups obtained at each step is close to the number of freefactors in G . A continuity argument will then help us carry over this estimate to the limitgroup. The only difficulty is to verify the hypothesis of Theorem 2.2 concerning orders of r i ’s and to ensure that the images of Z m ’s remain isomorphic to Z m on each step. This isdone by using hyperbolic groups and Lemma 2.1.Let m be an integer such that n/m < ε . Let G = G ∗ · · · ∗ G n , where G i ∼ = Z m foreach i = 1 , . . . , n . We enumerate all elements of G = { g , g , g , . . . } , and construct thegroup T by induction. Let T = G . Suppose that a group T i = h G , . . . , G n | r w , . . . , r w ki k i i ,k i ≤ i , has already been constructed for some i ≥
0. In what follows, we use the samenotation for elements of T and their images in T i . We assume that:(a) T i is hyperbolic.(b) The natural maps G l → T i are injective for l = 1 , . . . , n . In particular, we may thinkof G l ’s as subgroups of T i .(c) | r j | = w j in T i for j = 1 , . . . , k i .(d) P k i j =1 1 w j + nm < ε. (e) Elements g , . . . , g i have finite orders in T i .Observe that the inductive assumption trivially holds for T . By Theorem 2.2, conditions(b), (c), and (d) imply that β (2)1 ( T i ) > n − − ε. The group T i +1 is obtained from T i in the following way. If the image of g i +1 has finiteorder in T i , we set k i +1 = k i and T i +1 = T i . Otherwise let C be an (infinite) finite indexcyclic normal subgroup of E ( g i +1 ). Since | E ( g i +1 ) /C | is finite, there exists m > g mi +1 ∈ C . Consequently, h g mi +1 i is normal in E ( g i +1 ). Passing to h g mi +1 i , we may assumethat C = h g mi +1 i without loss of generality. Let k i +1 = k i + 1, r k i +1 = g i +1 and S = n [ l =1 G l ! ∪ k i [ j =1 h r j i . G = T i , the element x = g mi +1 , and the subset S , weobtain that for every large enough integer s , the quotient group T i +1 = h T i | r smk i +1 i = h G , . . . , G n | r w , . . . , r w ki k i , r smk i +1 i (3)is hyperbolic, | g mi +1 | = s (hence | r k i +1 | = | g i +1 | = ms ) in T i +1 , and the natural homomor-phism T i → T i +1 is injective on S . The later condition ensures that | r j | = w j in T i +1 for j = 1 , . . . , k i and the natural maps G l → T i +1 remain injective for l = 1 , . . . , n . By (d), wemay choose s such that k i X j =1 w j + 1 ms + nm < ε. Letting k i +1 = k i + 1 and w k i +1 = ms completes the inductive step.Let T = h G , . . . , G n | r w , r w , . . . i be the inductive limit of the groups T i and the natural homomorphisms T i → T i +1 . By (e)the image of every element g i has finite order in T , i.e., T is a torsion group.Note that the sequence { T i } converges to T in the topology of marked group presenta-tions. (For details about this topology we refer to [18].) Indeed this is always true wheneverwe have a sequence of normal subgroups N ≤ N ≤ . . . of a group T , T = T / ∞ S i =1 N i , and T i = T /N i . (In our case N i is the normal closure of r , . . . , r k i in T ). By semi-continuityof the first L -Betti number (see [18]), we obtain β (2)1 ( T ) ≥ lim i →∞ β (2)1 ( T i ) ≥ n − − ε. Suppose now that every hyperbolic group is residually finite. Then we adjust our con-struction as follows. Let X be a finite generating set of T and let d i denote the wordmetric on T i corresponding to the natural image of X in T i . For every i ∈ N , we choose ahomomorphism τ i : T i → Q i , where Q i is finite, such that τ i ( t ) = 1 whenever d i ( t, ≤ i and t = 1. Such a homomorphism always exists as T i is hyperbolic and hence it is residuallyfinite by our assumption.Note that passing from T i to T i +1 according to (3) we may always choose s to be amultiple of any given non-zero integer. Thus we may assume that w j is divisible by | Q i | for any i, j ∈ N , j > i . This implies that for every i ∈ N , the kernel of the naturalhomomorphism T i → T is contained in Ker( τ i ) and hence τ i factors through T i → T . Let σ i be the corresponding homomorphism T → Q i .Denote by d the word metric on T with respect to the natural image of the set X . If s is a nontrivial element of T such that d ( s,
1) = i , then there is an element t ∈ T i such that d i ( t, ≤ i and s is the natural image of t in T . According to our construction, τ i ( t ) = 1and hence σ i ( t ) = 1. Thus the group T is residually finite.5 Non-unitarizable groups without free subgroups
Given a finitely generated group H , we denote by rk ( H ) its rank. Theorem 3.1 (Epstein-Monod [5]) . Let G be a unitarizable group. Then the ratio β (2)1 ( H ) / p rk ( H ) is uniformly bounded on the set of all finitely generated subgroups of G . We are now ready to prove Theorem 1.2.
Proof of Theorem 1.2.
By Theorem 2.3 for every integer n ≥
2, there exists a group G n generated by n elements such that β (2)1 ( G n ) ≥ n −
2. Let G be the direct product of thefamily { G n | n ≥ } . Clearly G is a torsion group and is not unitarizable by Theorem 3.1.To complete the proof it remains to recall that every countable torsion group embeds into atorsion group generated by 2 elements [14], and every group containing a non-unitarizablesubgroup is non-unitarizable itself.To construct torsion free examples, we need another result about hyperbolic groups.Similarly to Lemma 2.1, it can be easily extracted from the proof of Theorem 2 in [15].We make this extraction for convenience of the reader and refer to [15] for details andterminology. Lemma 3.2 (Olshanskii, [15]) . Let G be a torsion free hyperbolic group, H a non-elementary subgroup of G , t , . . . , t m elements of G . Then there exist elements r , . . . , r m ∈ H such that the following conditions hold for the quotient group G = G/ hh r t , . . . , r m t m ii .(1) G is torsion free hyperbolic.(2) The natural image of H is a non-elementary subgroup of G .Observe that the images of the elements t , . . . , t m belong to the image of H in G .Proof. Since G is torsion free, all elementary subgroups of G are cyclic. Let g be anynon-trivial element of H such that E ( g ) = h g i . Let l be a positive integer, x , . . . , x l ∈ H elements provided by [15, Lemma 3.7]. Let W, X , X , . . . , X l be shortest words in a finiteset of generators of G representing g, r , x , . . . , x l respectively.Using [15, Lemma 4.2] and triviality of finite subgroups in G , we obtain that the setof all cyclic shifts of the words ( X W m X W m · · · X l W m ) ± satisfies a small cancellationcondition,which implies (1) and (2) for the group G = h G | X W m X W m · · · X l W m = 1 i by [15, Lemma 6.7] if m and l are large enough. Note that G ∼ = G/ hh r t ii , where t is the element of H represented by W m X W m · · · X l W m . Doing the same procedure for r , . . . , r m we prove the lemma by induction.6 roof of Theorem 1.3. We are going to construct the desired group G as a (torsion free)extention 1 → H → G → T →
1, where T is a non-unitarizable torsion group provided byTheorem 1.2 and H has no non-abelian free subgroups. It is easy to show that every sucha group G is not unitarizable and does not contain non-abelian free subgroups.More precisely, let T = h x, y | r , r , . . . i , (4)be a presentation of a non-unitarizable torsion group. Without loss of generality we mayassume T to be generated by 2 elements (see the proof of Theorem 1.2). Again we proceedby induction. Let G = h x, y, a, b i be the free group with basis x, y, a, b . In what followswe construct a series of quotients of G . As in the proof of Theorem 2.3, we keep the samenotation for elements of G and their images in these quotient groups.Clearly H = h a, b i is a non-elementary subgroup of G . Hence by Lemma 3.2, thereexist elements u , . . . , u , v ∈ H such that the quotient group G = h G | a x u , a x − u , b x u , b x − u , a y u , a y − u , b y u , b y − u , r v i is torsion free hyperbolic, and the image H of H in G is non-elementary. Without loss ofgenerality we may assume that u , . . . , u and v are words in { a ± , b ± } .We enumerate all finitely generated subgroups H = R , R , . . . of H . Suppose that forsome i ≥
1, we have already constructed a group G i = * a, b, x, y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a x u , a x − u , b x u , b x − u , a y u , a y − u , b y u , b y − u r v , . . . , r i v i w , . . . , w k i + such that the following conditions hold. By H i we denote the subgroup of G i generated by a and b (i.e., the image of H in G i ).(a) The group G i is torsion free hyperbolic.(b) The subgroup H i = h a, b i of G i is non-elementary.(c) v , . . . , v i and w , . . . , w k i are words in { a ± , b ± } .(d) For every j = 1 , . . . , i , the image of R j in G i is either cyclic or coincides with theimage of H i .Clearly these conditions hold for i = 1. Relations w , w , . . . , w k i are absent in this case.The group G i +1 is obtained from G i in two steps. First, by parts (a), (b) of the inductiveassumption and Lemma 3.2, we may choose a word v i +1 in { a ± , b ± } such that the quotientgroup K i = G i / hh r i +1 v i +1 ii is torsion free hyperbolic, and the natural image of H i in K i isnon-elementary.Further if the natural image of R i +1 in K i is cyclic, we set G i = K i and k i +1 = k i .Otherwise the image of R i +1 in K i is non-elementary. Indeed it is well-known and easy toprove that every torsion free elementary group is cyclic. Thus we can apply Lemma 3.2 to7he image of R i +1 in K i and elements a, b . Let z , z be elements of the image of R i +1 in K i such that the quotient group G i +1 = K i / hh az , bz ii is torsion free hyperbolic and theimage of R i +1 in G i +1 is non-elementary. Recall that R i +1 ≤ H = h a, b i . Hence we canassume that z , z are words in { a ± , b ± } . Since a = z − and b = z − in G i +1 , the imageof R i +1 in G i +1 coincides with the subgroup H i +1 = h a, b i of G i +1 . In particular, H i +1 is non-elementary. Note that az , bz are words in { a ± , b ± } as well. Let w k i +1 = az , w k i +2 = bz . The inductive step is completed.Let now G be the inductive limit of the sequence G , G , . . . . That is, G i = * a, b, x, y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a x u , a x − u , b x u , b x − u , a y u , a y − u , b y u , b y − u r v , r v , . . .w , w , . . . + (5)Let also H = h a, b i be the natural image of H in G . Note that the elements a x , a x − , b x , b x − , a y , a y − , b y , b y − belong to H in G , as u , . . . , u are words in { a ± , b ± } .Hence H is a normal subgroup of G . Obviously G/H ∼ = T . Indeed after imposing additionalrelations a = 1 and b = 1, the relations corresponding to the first and the third rows of (5)disappear and the second row of (5) becomes r , r , . . . (see (c)). After removing a, b fromthe set of generators, we obtain exactly the presentation (4) of T .Thus the group G splits as 1 → H → G → T →
1. Note that every finitely generatedproper subgroup of H is cyclic. Indeed, let Q be a finitely generated subgroup of H , P some finitely generated preimage of Q in H . Then P = R i for some i . The naturalhomomorphism P → Q obviously factors through the image of P = R i in G i . However theimage of R i in G i is either cyclic or coincides with H i by (d). Therefore, we obtain that Q is either cyclic or coincides with H .Let F be a nontrivial finitely generated free subgroup of G . Then F ∩ H is cyclic. Notealso that F ∩ H is normal in F and F ∩ H = 1 since F/ ( F ∩ H ) ∼ = F H/H ≤ T is a torsiongroup. Thus F has a nontrivial normal cyclic subgroup, and hence F is cyclic itself. Thisshows that G contains no non-abelian free subgroups. Further suppose that some element g = 1 has finite order in G . This means that for some n >
0, the relation g n = 1 followsfrom relations of the presentation (5). Hence it follows from some finite set of relationsof (5), i.e., g n = 1 holds in G i for some i contrary to (a). Finally we note that G is notunitarizable since it surjects onto a non-unitarizable group T . Remark . Let G be finitely generated group, X a finite generating set of G , κ ( G, l ( G ) , X )the Kazhdan constant of the left regular representation λ G with respect to X . More pre-cisely, κ ( G, l ( G ) , X ) is defined as the supremum of all ε >
0, such that for every vector u ∈ l ( G ) of norm 1, there exists x ∈ X such that k λ G ( x ) u − u k ≥ ε . Recall that a finitelygenerated group G is amenable if and only is κ ( G, l ( G ) , X ) = 0 for every finite generatingset X of G [7].Let α ( G ) = inf h X i = G κ ( G, l ( G ) , X ) = 0 , (6)where the infimum is taken over all finite generating sets X of G . A finitely generatedgroup G is called weakly amenable if α ( G ) = 0. (A similar notion of weak amenability was8onsidered in [1].) Clearly every amenable group is weakly amenable. The converse wasshown to be wrong in [16].Combining methods of [16] and [17], one can construct a (finitely generated) non-unitarizable torsion group T without free subgroup such that every non-elementary hy-perbolic group surjects onto T . In particular, every such a group is weakly amenable [16].This shows that the absence of free subgroups does not imply unitarizability even beingcombined with weak amenability. References [1] G. Arzhantseva, J. Burillo, M. Lustig, L. Reeves, H. Short, E. Ventura, Uniform non-amenability,
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