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Groups of Piecewise Projective Homeomorphisms Groups of piecewise projective homeomorphisms Nicolas Monod ∗ ∗EPFL, 1015 Lausanne, Switzerland Submitted to Proceedings of the National Academy of Sciences of the United States of America The group of piecewise projective homeomorphisms of the line pro- PA. We write H(A)= G(A) H, which is the stabilizer of vides straightforward torsion-free counter-examples to the so-called in G(A). ∩ ∞ von Neumann conjecture. The examples are so simple that many additional properties can be established. The main result of this article is the following, which relies on a new method for proving non-amenability. Theorem 1. The group H(A) is non-amenable if A = Z. Introduction 6 The next result is a sequacious generalization of the corre- In 1924, Banach and Tarski accomplished a rather paradoxical sponding theorem of Brin–Squier about piecewise affine trans- feat. They proved that a solid ball can be decomposed into formations [16] and we claim no originality. five pieces which are then moved around and reassembled in Theorem 2. The group H does not contain any non-abelian free such a way as to obtain two balls identical with the original subgroup. Thus, H(A) inherits this property for any subring one [1]. This wellnigh miraculous duplication was based on A< R. Hausdorff’s 1914 work [2]. Thus already H = H(R) itself is a counter-example to In his 1929 study of the Hausdorff–Banach–Tarski para- the von Neumann conjecture. Writing H(A) as the directed dox, von Neumann introduced the concept of amenable union of its finitely generated subgroups, we deduce: groups [3]. Tarski readily proved that amenability is the Corollary 3. For A = Z, the groups H(A) contain finitely gen- only obstruction to paradoxical decompositions [4, 5]. How- 6 ever, the known paradoxes relied more prosaically on the ex- erated subgroups that are simultaneously non-amenable and istence of non-abelian free subgroups. Therefore, the main without non-abelian free subgroups. open problem in the subject remained for half a century to find non-amenable groups without free subgroups. Von Further properties The groups H(A) seem to enjoy a number Neumann’s name was apparently attached to it by Day in of additional interesting properties, some of which are weaker the 1950s. The problem was finally solved around 1980: forms of amenability. In the last section, we shall prove the Ol′shanski˘ı proved the non-amenability of the Tarski mon- following five propositions (and recall the terminology). Here sters that he had constructed [6, 7, 8]; Adyan showed that A< R is an arbitrary subring. his work on Burnside groups yields non-amenability [9, 10]. Proposition 4. All L2-Betti numbers of H(A) and of G(A) van- Finitely presented examples were constructed another twenty ish. ′ years later by Ol shanski˘ı–Sapir [11]. There are several more Proposition 5. The group H(A) is inner amenable. recent counter-examples [12, 13, 14]. Proposition 6. The group H is bi-orderable and hence so are all its subgroups. It follows that there is no non-trivial homo- Given any subring A < R, we shall define a group G(A) morphism from any Kazhdan group to H. and a subgroup H(A) < G(A) of piecewise projective trans- Proposition 7. Let E P1 be any subset. Then the subgroup of formations. Those will provide concrete, uncomplicated new H(A) which fixes E ⊆pointwise is co-amenable in H(A) unless examples with many additional properties. Perhaps ironically, E is dense (in which case the subgroup is trivial). our short proof of non-amenability ultimately relies on basic free groups of matrices, as in Hausdorff’s 1914 paradox, even Proposition 8. If H(A) acts by isometries on any proper though the Tits alternative [15] shows that the examples can- CAT(0) space, then either it fixes a point at infinity or it not be linear themselves. preserves a Euclidean subspace. arXiv:1209.5229v2 [math.GR] 7 Sep 2013 One can also check that H(A) satisfies no group law and Construction has vanishing properties in bounded cohomology (see below). I saw the pale student of unhallowed arts kneeling beside the thing he had put together. Mary Shelley, Frankenstein Non-amenability (introduction to the 1831 edition) An obvious difference between the actions of PSL2(A) and of 1 Consider the natural action of the group PSL2(R) on H(A) on P is that the latter group fixes whilst the former the projective line P1 = P1(R). We endow P1 with its R- does not. The next proposition shows that∞ this is the only topology making it a topological circle. We denote by G the difference as far as the orbit structure is concerned. group of all homeomorphisms of P1 which are piecewise in 1 PSL2(R), each piece being an interval of P , with finitely many pieces. We let H < G be the subgroup fixing the point Reserved for Publication Footnotes P1 corresponding to the first basis vector of R2. Thus H∞is ∈ left-orderable since it acts faithfully on the topological line P1 , preserving orientations. It follows in particular that H \{∞}is torsion-free. 1 Given a subring A< R, we denote by PA P the collec- ⊆ tion of all fixed points of all hyperbolic elements of PSL2(A). This set is PSL2(A)-invariant and is countable if A is so. We define G(A) to be the subgroup of G given by all elements that are piecewise in PSL2(A) with all interval endpoints in www.pnas.org/cgi/doi/10.1073/pnas.0709640104 PNAS Issue Date Volume Issue Number 1–5 Proposition 9. Let A< R be any subring and let p P1 . An a.e. free action of a countable group is amenable in Zim- Then ∈ \{∞} mer’s sense [20, 4.3] if and only if the associated relation is PSL2(A) p H(A) p. amenable; see [21, Thm. A]. · ⊆ {∞} ∪ · Proof of Theorem 1. Let A = Z be a subring of R. Then Thus, the equivalence relations induced by the actions of 6 ′ 1 A contains a countable subring A < A which is dense in R. PSL (A) and of H(A) on P coincide when restricted to ′ 2 Since H(A ) is a subgroup of H(A), we can assume that A P1 . \ {∞} itself is countable dense. Now H(A) is a countable group and Γ := PSL2(A) is a countable dense subgroup of PSL2(R). Proof. We need to show that given g PSL2(A) with It is proved in Th´eor`eme 3 of [22] that the equivalence gp = , there is an element h H(A) such∈ that hp = gp. 6 ∞ ∈ relation on PSL2(R) induced by the multiplication action of We assume g = since otherwise h = g will do. Equiva- Γ is non-amenable; see also Remarks 10 and 11 below. Equiv- lently, we need∞ an 6 ∞ element q G(A) fixing gp and such that −1 ∈ alently, the Γ-action on PSL2(R) is non-amenable. Viewing q = g , writing h = q g. It suffices to find a hyperbolic 1 ∞ ∞ P as a homogeneous space of PSL2(R), it follows that the element q0 PSL2(A) with q0 = g and whose fixed points 1 1 ∈ ∞ ∞ Γ-action on P is non-amenable. Indeed, amenability is pre- ξ± P separate gp from both and g , see Figure 1. In- served under extensions, see [23, 2.4] or [21, Cor. C]. This deed,∈ we can then define q to be the∞ identity∞ on the component 1 action is a.e. free since any non-trivial element has at most of P ξ± containing gp, and define q to coincide with q0 1 \ { } two fixed points. Thus the relation induced by Γ on P on the other component. 1 is non-amenable. Restricting to P , we deduce from Proposition 9 that the relation induced\ {∞} by the H(A)-action is rg also non-amenable. (Amenability is preserved under restric- r ∞ tion [18, 9.3], but here is a null-set anyway.) Thus H(A) ξ+ is a non-amenable group.{∞} r r∞ ξ− Remark 10. We recall from [22] that the non-amenability of the Γ-relation on PSL2(R) is a general consequence of the exis- gpr tence of a non-discrete non-abelian free subgroup of Γ. Thus Fig. 1. The desired configuration of ξ± the main point of our appeal to [22] is the existence of this non-discrete free subgroup, but this is much easier to prove directly in the present case of Γ = PSL2(A) than for general a b non-discrete non-soluble Γ. Let be a matrix representative of g; thus, c d Remark 11. Here is a direct argument avoiding all the above a,b,c,d A and ad bc = 1. The assumption g = references in the examples of A = Z[√2] or A = Z[1/ℓ], where ∈ − ∞ 6 ∞ 1 implies c = 0 and thus we can assume c > 0. Let q0 be ℓ is prime. We show directly that the Γ-action on P is not 6 a b + ra amenable. We consider Γ as a lattice in L := PSL2(R) given by with r A to be determined later; × c d + rc ∈ PSL2(R) in the first case and in L := PSL2(R) PSL2(Qℓ) × thus q0 = g . This matrix is hyperbolic as soon as r is in the second case, both times in such a way that the Γ-action large enough∞ to∞ ensure that the trace τ = a + d + rc is larger| | on P1 extends to the L-action factoring through the first fac- than 2 in absolute value.
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