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Home , Counterexample, Von Neumann conjecture

Groups of piecewise projective homeomorphisms

Nicolas Monod1

Section de Mathématiques, École Polytechnique Fédérale de Lausanne, 1015 Lausanne, Switzerland

Edited by Gregory A. Margulis, Yale University, New Haven, CT, and approved February 5, 2013 (received for review October 22, 2012) The of piecewise projective homeomorphisms of the line The main result of this article is the following, for which we provides straightforward torsion-free to the so- introduce a method for proving amenability. called von Neumann . The examples are so simple that many additional properties can be established. Theorem 1. The group H(A) is nonamenable if A ≠ Z. The next result is a sequacious generalization of the corre- free groups | paradoxical decomposition | von Neumann problem sponding theorem of Brin–Squier (16) about piecewise affine transformations, and we claim no originality. n 1924, Banach and Tarski (1) accomplished a rather para- Idoxical feat. They proved that a solid ball can be decomposed Theorem 2. The group H does not contain any nonabelian free sub- fi group. Thus, H(A) inherits this property for any subring A < R. into ve pieces, which are then moved around and reassembled = R in such a way as to obtain two balls identical to the original one Thus, already H H( ) itself is a to the . Writing H(A) as the directed union of its (1). This wellnigh miraculous duplication was based on Haus- fi dorff’s (2) 1914 work. nitely generated subgroups, we deduce Corollary 3. In his 1929 study of the Hausdorff–Banach–Tarski paradox, Corollary 3. For A ≠ Z, the groups H(A) contain finitely generated von Neumann (3) introduced the concept of amenable groups. subgroups that are simultaneously nonamenable and without non- Tarski (4, 5) readily proved that amenability is the only ob- abelian free subgroups. struction to paradoxical decompositions. However, the known paradoxes relied more prosaically on the existence of nonabelian Additional Properties. The groups H(A) seem to enjoy a number free subgroups. Therefore, the main open problem in the subject of additional interesting properties, some of which are weaker remained for half a century to find nonamenable groups without forms of amenability. In the last section, we shall prove the fol- free subgroups. Von Neumann’s (3) name was apparently at- lowing five propositions (and recall the terminology). Here, A < R tached to it by Day in the 1950s. The problem was finally solved is an arbitrary subring. around 1980: Ol′shanskiĭ (6–8) proved the nonamenability of the Tarski monsters that he had constructed, and Adyan (9, 10) Proposition 4. All L2-Betti numbers of H(A) and G(A) vanish. showed that his work on Burnside groups yields nonamenability. Finitely presented examples were constructed another 20 y later Proposition 5. The group H(A) is inner amenable. by Ol′shanskiĭ–Sapir (11). There are several more recent coun- terexamples (12–14). Proposition 6. The group H is biorderable, and hence, so are all of its Given any subring A < R,weshalldefine a group G(A)anda subgroups. It follows that there is no nontrivial homomorphism subgroup H(A) < G(A) of piecewise projective transformations. from any Kazhdan group to H. Those groups will provide concrete, uncomplicated examples with many additional properties. Perhaps ironically, our short Proposition 7. Let E ⊆ P1 be any subset. Then, the subgroup of proof of nonamenability ultimately relies on basic free groups H(A), which fixes E pointwise, is coamenable in H(A) unless E is of matrices, as in Hausdorff’s (2) 1914 paradox, although the dense (in which case, the subgroup is trivial). Tits (15) alternative shows that the examples cannot be linear themselves. Proposition 8. If H(A) acts by isometries on any proper CAT(0) space, then either it fixesapointatinfinity or it preserves a Euclidean subspace. Construction One can also check that H(A) satisfies no group law and has vanishing properties in bounded cohomology (see below). I saw the pale student of unhallowed arts kneeling beside the thing he Nonamenability had put together. An obvious difference between the actions of PSL2(A) and H(A) on P1 is that the latter group fixes ∞, whereas the former does Mary Shelley, introduction to the 1831 edition of Frankenstein not. The next proposition shows that this difference is the only Consider the natural action of the group PSL2(R)onthe one as far as the orbit structure is concerned. projective line P1 = P1(R). We endow P1 with its R-topology, 1 making it a topological circle. We denote by G the group of all Proposition 9. Let A < R be any subring, and let p ∈ P \{∞}. Then, 1 homeomorphisms of P that are piecewise in PSL2(R), each piece 1 · ⊆ ∞ ∪ · : being an interval of P with finitely many pieces. We let H < G PSL2ðAÞ p f g HðAÞ p be the subgroup fixing the point ∞∈P1 corresponding to the first basis vector of R2. Thus, H is left-orderable, because it acts Thus, the equivalence relations induced by the actions of PSL2(A) P1 P1 ∞ faithfully on the topological line P1\{∞}, preserving orientations. and H(A) on coincide when restricted to \{ }. It follows in particular that H is torsion-free. 1 Given a subring A < R, we denote by PA ⊆ P the collection of Author contributions: N.M. designed research, performed research, and wrote the paper. all fixed points of all hyperbolic elements of PSL2(A). This set is The author declares no conflict of interest. PSL2(A)-invariant and countable if A is so. We define G(A)tobe the subgroup of G given by all elements that are piecewise in This article is a PNAS Direct Submission. PSL2(A) with all interval endpoints in PA. We write H(A) = G(A) Freely available online through the PNAS open access option. ∩ H, which is the stabilizer of ∞ in G(A). 1E-mail: nicolas.monod@epfl.ch.

4524–4527 | PNAS | March 19, 2013 | vol. 110 | no. 12 www.pnas.org/cgi/doi/10.1073/pnas.1218426110 Downloaded by guest on October 1, 2021 Proof. We need to show that, given g ∈ PSL2(A)withgp ≠∞, nonamenable (see also Remark 10 and Remark 11). Equiva- ∈ = 1 there is an element h H(A), such that hp gp. We assume lently, the Γ-actiononPSL2(R) is nonamenable. Viewing P as ∞≠∞ = g , because otherwise, h g will do. Equivalently, we a homogeneous space of PSL2(R), it follows that the Γ-action on need an element q ∈ G(A) fixing gp andsuchthatq∞≠g∞, P1 is nonamenable. Indeed, amenability is preserved under −1 writing h ≠ q g.Itsuffices to find a hyperbolic element q0 ∈ extensions (ref. 21, Corollary C or ref. 23, 2.4). This action is a. 1 PSL2(A)withq0∞ = g∞ and fixed points ξ± ∈ P that separate e.-free, because any nontrivial element has, at most, two fixed gp from both ∞ and g∞ (Fig. 1). Indeed, we can then define q points. Thus, the relation induced by Γ on P1 is nonamenable. 1 1 to be the identity on the component of P \{ξ±} containing gp Restricting to P \{∞}, we deduce from Proposition 9 that the fi and de ne qto coincide with q0 on the other component. relation induced by the H(A) action is also nonamenable. ab Let be a matrix representative of g; thus, a, b, c, d, ∈ A, [Amenability is preserved under restriction (ref. 18, 9.3), but here, cd {∞} is a null-set anyway.] Thus, H(A) is a nonamenable group.□ − = ∞≠∞ ≠ and ad bc 1. The assumption g implies c 0, and ab+ ra Remark 10. We recall from ref.22that the nonamenability of the thus, we can assume c > 0. Let q0 be given by with cd+ rc Γ-relation on PSL2(R) is a general consequence of the existence of Γ r ∈ A to be determined later; thus, q0∞≠g∞. This matrix is a nondiscrete nonabelian free subgroup of . Thus, the main point hyperbolic as soon as jrj is large enough to ensure that the trace of our appeal to ref.22is the existence of this nondiscrete free τ = a + d + rc is larger than 2 in absolute value. We only need to subgroup, but this existence is much easier to prove directly in the show that a suitable choice of r will ensure the above condition present case of Γ = PSL2(A) than for general nondiscrete 1 on ξ±. Notice that ∞ and g∞ lie in the same component of P \{ξ±}, nonsoluble Γ. because q0 preserves these components and sends ∞ to g∞.In fi conclusion, it suf ces to prove the following two claims: (i)as Remark 11. Here is a direct argumentpffiffiffi avoiding all of the above ref- jrj → ∞,theset{ξ±}convergesto{∞, g∞}; and (ii) changing erences in the examples of A = Z½ 2 or A = Z[1/ℓ], where ℓ is the sign of r (when jrj is large) will change the component of prime. We show directly that the Γ-action on P1 is not amenable. P1 ∞ ∞ ξ \{ , g } in which ± lie (we need it to be the component of gp). We consider Γ as a lattice in L: = PSL2(R) × PSL2(R) in the first The claims can be proved by elementary dynamical considera- case and L: = PSL2(R) × PSL2(Qℓ) in the second case, both times tions; we shall instead verify them explicitly. in such a way that the Γ-action on P1 extends to the L-action x ± 1 The fixed points ξ± are represented by the eigenvectors , factoring through the first factor. If the Γ-action on P were ame- pffiffiffiffiffiffiffiffiffiffiffiffi c where x± = λ± − d − rc and λ ± = ðτ ± τ2 − 4Þ=2 are the eigen- nable, so would be the L-action (by coamenability of the lattice). values. Now, limr→+∞λ+ =+∞ implies limr→+∞λ− = 0, because However, of course, L does not act amenably, because the stabilizer λ+λ− = 1; therefore, limr→+∞x− = −∞.Similarly,limr→−∞x+ =+∞ of any point contains the (nonamenable) second factor of L. (Fig. 1 depicts the case r > 0). Thus, we already proved claim (ii) The nondiscreteness of A was essential in our proof, thus ex- and half of claim (i). Because g∞ = [a:c], it only remains to cluding A = Z. verify that both limr→+∞x+ and limr→−∞x− converge to a,whichis a direct computation. □ Problem 12. Is H(Z) amenable? We recall that a measurable equivalence relation with count- The group H(Z) is related to Thompson’sgroupF, for which the able classes is amenable if there is an a.e. defined measurable question of (non)amenability is a notorious open problem. Indeed, assignment of a mean on the orbit of each point in such a way F seems to be historically the first candidate for a counterexample that the means of two equivalent points coincide. We refer, e.g., to the so-called von Neumann conjecture. The relation is as fol- to refs. 17 and 18 for background on amenable equivalence re- lows: if we modify the definition of H(Z) by requiring that the 1 lations. It follows from this definition that any relation produced breakpoints be rational, then all its elements are automatically C , by a measurable action of a (countable) is and the resulting group is conjugated to F. The corresponding re- amenable, by push-forward of the mean [ref. 19, 1.6(1)]. An a.e. lation holds between G(Z) and Thompson’sgroupT. These facts free action of a countable group is amenable in Zimmer’s sense are attributed to a remark of Thurston around 1975, and a very (ref. 20, 4.3) if and only if the associated relation is amenable detailed exposition can be found in ref. 24. (ref. 21, Theorem A). H Is a Free Group Proof of Theorem 1. Let A ≠ Z be a subring of R. Then, A contains We shall largely follow ref. 16 (§3), the main difference being a countable subring A′ < A, which is dense in R. Because H(A′) that we replace commutators by a nontrivial word in the second is a subgroup of H(A), we can assume that A itself is countable derived subgroup of a free group on two generators. ∈ ≠ dense. Now, H(A) is a countable group, and Γ: = PSL2(A)isa The support supp(g)ofanelementg H denotes the set {p: gp fi countable dense subgroup of PSL2(R). p}, which is a nite union of open intervals. Any subgroup of H It is proved in Théorème 3 in ref. 22 that the equivalence fixing some point p ∈ P1 has two canonical homomorphisms to the R relation on PSL2(R) induced by the multiplication action of Γ is metabelian stabilizer of p in PSL2( ) given by left and right germs. Therefore, we deduce the following elementary fact, wherein 〈f, g〉 denotes the subgroup of H generated by f and g.

Lemma 13. If f, g ∈ H have a common fixed point p ∈ P1, then any element of the second derived subgroup 〈f, g〉″ acts trivially on

a neighborhood of p. □ Theorem 2 is an immediate consequence of the following more precise statement.

Theorem 14. Let f, g ∈ H. Either 〈f, g〉 is metabelian or it contains a free abelian group of rank two.

Proof. We suppose that 〈f, g〉 is not metabelian, so that there is a word w in the second derived subgroup of a free group on two Fig. 1. The desired configuration of ξ±. generators such that w(f, g) ∈ H is nontrivial. We now follow

Monod PNAS | March 19, 2013 | vol. 110 | no. 12 | 4525 Downloaded by guest on October 1, 2021 faithfully the proof of theorem 3.2 in ref. 16, replacing [f, g]by or if it is =1 but the second derivative is >0. Then, define the set H+ w(f, g). For the reader’s convenience, we sketch the argument; of positive elements of H to consist of all transformations with first the details are on p. 495 in ref. 16 or p. 232 in ref. 25. Applying nontrivial germ (starting from ∞ along the orientation) that is Lemma 13 to all endpoints p of the connected components of positive. Now, H+ is a conjugacy invariant subsemigroup, and H\{e} −1 supp(f) ∪ supp(g), we deduce that the closure of supp(w(f, g)) is is H+ ⊔ H+ ;thus,H+ defines a biinvariant total order. contained in supp(f) ∪ supp(g). This fact implies that some ele- Suppose now that we are given a homomorphism from a Kazh- ment of 〈f, g〉 will send any connected component of supp(w(f, g)) dan group to H. Its image is then a Kazhdan subgroup K < H. to a disjoint interval. The needed element might depend on the Kazhdan’s property implies that K is finitely generated. It has been connected component. However, upon replacing w〈f, g) by an- known for a long time that any nontrivial finitely generated bio- ∈ 〈 〉″ other nontrivial element w1 f, g with a minimal number of rderable group has a nontrivial homomorphism to R: this fact intersecting components with supp(f) ∪ supp(g), some element follows ultimately from Hölder’s (32) 1901 work by looking at 〈 〉 h of f, g sends the whole of supp(w1) to a set disjoint from it. maximal convex subgroups and is explained in ref. 33 (§2). How- = −1 The corresponding conjugate w2: hw1h will commute with ever, this circumstance is impossible for a Kazhdan group. □ w1, and indeed, these two elements generate freely a free □ 1 abelian group. Lemma 16. For any p ∈ P \{∞}, there is a sequence {gn} in H(Z), such 1 As pointed out to us by Cornulier, the above argument can be that gnq converges to ∞ uniformly for q in compact subsets of P \{p}. pushed so that w1 and h generate a wreath product Z ≀ Z (com- pare ref. 26, Theorem 21 for the piecewise linear case). Proof. It suffices to show that, for any open neighborhoods U and V of p and ∞, respectively, in P1, there is g ∈ H(Z), which maps P1\U Lagniappe into V. Because the collection of pairs of fixed points of hyperbolic 2 1 1 Proof of Proposition 4. We refer to ref. 27 for the L -Betti numbers elements of PSL2(Z)isdenseinP × P ,wecanfind hyperbolic βn ∈ N Γ = ∈ Z fi ð2Þ, n . Fix a large integer n, and let G(H)orH(A). Choose matrices h1, h2 PSL2( ) with repelling xed points ri in U\{p}and ⊆ + Λ < Γ asetF PA of n 1 distinct points, and let be the pointwise attracting fixed points ai in V\{∞}, such that the cyclic order is ∞, Λ fi stabilizer of F. Any intersection *ofany( nite) number of con- a1, r1, p, r2, a2.Now,wedefine g tobeasufficiently high power of h1 Λ fi jugates of is still the pointwise stabilizer of a nite set F*con- on the interval [a1, r1] (for the above cyclic order), h2 on the interval ≥ + fi Λ taining m n 1 points. The de nition of G(A)showsthat *isthe [r2, a2], and the identity elsewhere. □ product of m infinite groups. The Künneth formula (ref. 27, §2) βi Λ* = = ... − implies ð2Þð Þ 0 for all i 0, , m 1. In this situation, The- Proof of Proposition 7. Let K be the pointwise stabilizer of a non- βi Γ = ≤ − □ ⊆ P1 fi fi orem 1.3 in ref. 28 asserts ð2Þð Þ 0 for all i m 1. dense subset E ; it suf ces to nd a mean invariant under A subgroup K of a group J is called coamenable if there is a J- H(A)″. Let {gn} be the sequence provided by Lemma 16 for p,an invariant mean on J/K. Equivalent characterizations, generalizations, interior point of the complement of E. Any accumulation point of and unexpected examples can be found in refs. 29 and 30. the sequence of point-masses at gnK in H(A)/K will do. Indeed, Recall that a group J is inner amenable if there is a conjugacy- because any g ∈ H(A)″ is trivial in a neighborhood of ∞, we have −1 ∈ □ invariant mean on J\{e}. It is equivalent to exhibit such a mean that gn ggn K for n large enough. is invariant under the second derived subgroup J″, because the The existence of two (or more) commuting coamenable sub- latter is coamenable in J. Thus, Proposition 5 is a consequence of groups is also a weak form of amenability. It is the key in the the stronger fact that H(A) is {asymptotically commutative}-by- argument cited below. metabelian in a sense inspired by ref. 31 as follows. Proof of Proposition 8. Consider two disjoint nonempty open sets in Proposition 15. Let A < R be any subring. For any finite set S ⊆ P1. The pointwise stabilizers of their complement commute with ″ ∈ H(A) , there is a nontrivial element hS H(A) commuting with each other and are coamenable by Proposition 7. In this situation, each element of S. Corollary 2.2 in ref. 34 yields the desired conclusion. □ Indeed, any accumulation point of this net of point-masses at The properties used in this section show immediately that ″ hS is H(A) -invariant. H(A) fulfills the criterion of ref. 35, Theorem 1.1 and thus, sat- isfies no group law. Proof of Proposition 15. By the argument of Lemma 13, there is Combining Theorems 1 and 2 with the main result of ref. 36, we ∞ aneighborhoodof on which all elements of S are trivial. Thus, it conclude that the wreath product Z ≀ H is a torsion-free non- fi suf ces to exhibit a nontrivial element hS of H(A), which is sup- unitarisable group without free subgroups. We can replace it by Z ported in this neighborhood. Notice that PSL2( ) contains hyper- a finitely generated subgroup upon choosing a nonamenable, fi- fi ξ ∞ bolic elements with both xed points ± arbitrarily close to and nitely generated subgroup of H. This construction provides some 21 1 n ’ – on the same side. For instance, conjugate by for new examples to Dixmier s problem, unsolved since 1950 (37 39). 11 01 Finally, we mention that our argument from Proposition 6.4 in fi ∈ N ξ n suf ciently large n . We choose such an element h0 with ± in ref. 40 applies to show that the bounded cohomology H ðHðAÞ; VÞ fi b the given neighborhood and de ne hS to be trivial on the compo- vanishes for all n ∈ N and all mixing unitary representations V. P1 ξ ∞ nent of \{ ±} containing and coincide with h0 on the other More generally, it applies to any semiseparable coefficient module component. □ V, unless all finitely generated subgroups of H(A)″ have invariant A group is called biorderable if it carries a biinvariant total order. vectors in V (see ref. 40 for details and definitions). This vanishing The construction below is completely standard (compare, e.g., ref. should be contrasted with the fact that amenability is characterized 25, p. 233 for a first-order version of our second-order argument). by the vanishing of bounded cohomology with all dual coefficients. P1 ∞ fi Proof of Proposition 6. Choose an orientation of \{ } and de ne a ACKNOWLEDGMENTS. This work was supported in part by the European (right) germ at a point p to be positive if either its first derivative is >1 Research Council and the Swiss National Science Foundation.

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