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Addendum Median Structures on Asymptotic Cones and Homomorphisms Into Mapping Class Groups e Proc. London Math. Soc. (3) 102 (2011) 555–562 C 2011 London Mathematical Society doi:10.1112/plms/pdr008 Addendum Median structures on asymptotic cones and homomorphisms into mapping class groups (Proc. London Math. Soc. (3) (2011) 503–554) Jason Behrstock, Cornelia Drut¸u and Mark Sapir The goal of this addendum to [1] is to show that our methods together with a result of Bestvina, Bromberg and Fujiwara [3, Proposition 5.9] yield a proof of the following theorem. Theorem 1. If a finitely presented group Γ has infinitely many pairwise non-conjugate homomorphisms into MCG(S), then Γ virtually splits (virtually acts non-trivially on a simplicial tree). This theorem is a particular case of a result announced by Groves.† From private emails received by the authors, it is clear that the methods used by Groves are significantly different. Note that the same new methods allow us to give another proof of the finiteness of the set of homomorphisms from a group with property (T) to a mapping class group [1, Theorem 1.2], which is considerably shorter than our original proof; see Corollary 6 and the discussion following it. Theorem 1.2 in [1] may equally be obtained from Theorem 1 and the fact that every group with property (T) is a quotient of a finitely presented group with property (T) (see [11, Theorem p. 5]). The property of the mapping class groups given in Theorem 1 can be viewed as another ‘rank 1’ feature of these groups. In contrast, note that a recent result of [8] shows that the rank 2 lattice SL3(Z) contains infinitely many pairwise non-conjugate copies of the triangle group Δ(3, 3, 4) = a, b | a3 = b3 =(ab)4 =1. Also, as was pointed out to us by Kassabov, although the group SL3(Z[x]) has property (T) (see [12]), it has infinitely many pairwise non-conjugate homomorphisms into SL3(Z) induced by ring homomorphisms Z[x] → Z. The following proposition contains one of the main auxiliary results in [3] and the key ingredient missed in our treatment of groups with many homomorphisms into mapping class groups in [1]. Proposition 2 (Bestvina, Bromberg and Fujiwara [3, Proposition 5.9]). There exists an explicitly defined finite index torsion-free subgroup BBF(S) of MCG(S) such that the set of all sub-surfaces of S can be partitioned into a finite number of subsets C1,C2,...,Cs, each of which is an orbit of BBF(S), and any two sub-surfaces in the same subset overlap and have the same complexity. 2000 Mathematics Subject Classification 20F65 (primary), 20F69, 20F38, 22F50 (secondary). The research of the first author was supported in part by an Alfred P. Sloan Fellowship. The research of the second author was supported in part by the ANR project ‘Groupe de recherche de G´eom´etrie et Probabilit´es dans les Groupes’. The research of the third author was supported in part by NSF grant DMS-0700811. †Groves first announced a version of this result at MSRI in 2007 (see http://www.math.cornell.edu/ ∼vogtmann/MSRI/groves daniel415.pdf). More recent announcements by Groves have included stronger versions of Theorem 1. 556 JASON BEHRSTOCK, CORNELIA DRUT¸ U AND MARK SAPIR The proof of this important result, explained to us by Bestvina, is surprisingly simple: the subgroup BBF(S) is the subgroup of mapping classes from MCG(S) acting as an identity on the factor π1(S)/B over certain characteristic subgroup B of π1(S) of finite index which is explicitly constructed. We consider the set of colors K = {1, 2,...,s}, and we color each sub-surface of S contained in the subset Ci by i. Note that the whole surface S has a color which is different from that of any proper sub-surface. Recall that for every sequence of sub-surfaces U from ΠU/ω, we defined an R-tree TU (see [1, Notation 4.4]) and that there is an equivariant bi-Lipschitz embedding ψ of AM into 1 U∈ΠU/ω TU (see [ , Corollary 4.17]). LetCk be the set of all sub-surfaces of S with the given ∈ U U color k K.Letπk be the projection of U∈ΠU/ω T onto U∈ΠCk/ω T . Remark 3. 1 By [ , Lemma 2.1], we have that U∈ΠU/ω TU can be written as TU. k∈K U∈ΠCk/ω In what follows, we use the notion of tree-graded space introduced in [6]. Theorem 4. ∈ AM Consider an arbitrary color k K and the imageTk = πkψ( ). U ∈ U × { V} For every sub-surface ΠCk/ω, consider the tree TU = T V∈ΠCk/ω\{U} a , where aV is the point in TV that is the projection of ∂U to TV. The space Tk is tree-graded with respect to TU and with transversal trees reduced to singletons. In particular, it is an R-tree. ⊂ Proof. Step 1. We prove by induction on n that, for any finite subsetF ΠCk/ω of AM AM cardinality n, the projection πF ( )of onto the finite product U∈F TU is an R-tree. The case n = 1 is obvious, the case n = 2 follows from [1, Theorem 4.21, (2)], since the sub-surfaces in F pairwise overlap. Assume that the statement is proved for n and consider F ⊂ ΠCk/ω, F of cardinality n +1. Both ψ(AM) and its projections are geodesic spaces. For ψ(AM) this follows from Proposition 4.18, while for projections it follows from the fact that the distance is 1.To prove that πF (ψ(AM)) is a real tree, it suffices therefore to prove that it is 0-hyperbolic, that is, for every geodesic triangle its three edges have a common point. By Lemma 4.30, the subset πF (ψ(AM)) is a median, thus it suffices to prove that, for an arbitrary triple of points ν, ρ and σ in πF (ψ(AM)) and every geodesic g joining ν, ρ in πF (ψ(AM)), the median point μ of the triple is on g. Assume that there exist U and V such that the projection of μ on TU × TV is not (v,u). Assume that it is (x, u), with x = v (the other case is similar). Consider the projection on the product Y∈F \{V} TY. By the inductive hypothesis, AM πF (ψ( )) projects onto a real tree, in particular, there exists μ on g such that its projection on Y∈F \{V} TY coincides with that of μ. In particular, πU(μ )=πU(μ)=x. This implies that the projection on TU × TV of both μ and μ is (x, v) (the unique point with first coordinate x). This implies that all coordinates of μ and μ are equal, thus the two points coincide. Assume now that, for every pair U and V in F , the projection of μ on TU × TV is (v,u). Fix such a pair. By the inductive hypothesis and an argument as above, there exists μ1 ∈ g such ∈ that its projection on Y∈F\{U} TY coincides with that of μ. Similarly, there exists μ2 g × such that its projection on Y∈F \{V} TY coincides with that of μ. Then on TU TV the point μ1 projects onto some (x, u)andμ2 projects onto some (v,y). This implies that there HOMOMORPHISMS INTO MAPPING CLASS GROUPS 557 exists some μ on g between μ1 and μ2 projecting on TU × TV in (v,u). Note that, for every Y ∈ F \{U, V}, the projection of μ coincides with that of μ1 and μ2, hence with that of μ. It follows that μ = μ. ⊂ We now prove by induction on n that, for any finite F ΠCk/ω of cardinality n,the AM AM projection πF ( )of onto the finite product U∈F TU is tree-graded with respect F × { } U to the trees TU = TU V∈F \{U} aV , where aV is the projection of ∂ to TV. It only AM F remains to prove that πF ( ) is complete and that it is covered by TU. Both statements are AM F proved simultaneously when proving that πF ( ) equals the union U∈F TU . Clearly, the union is contained in πF (AM). Conversely, consider a point x =(x1,...,xn+1)inπF (AM). The inductive hypothesis applied to (x1,...,xn) and (x2,...,xn+1) implies that, for each n-tuple, there exists U ∈ F such that, for every V = U, the corresponding coordinate is πV(U), that is, the point in TV which is the projection of ∂U to TV. Assume that in (x1,...,xn), the surface U corresponds to the first coordinate, and that in (x2,...,xn+1) the surface U × corresponds to the last coordinate. The projection (x1,xn+1)ofx on TU TU is either of the U U { U }× form (πU( ),xn+1)or(x1,πU ()). In the first case, x is in V∈F \{U} πV( ) TU and × { U } in the second case, x is in TU V∈F \{U} πV( ) . Step 2. We now prove the statements on Tk. First, we prove that Tk is a real tree, using an approximation argument similar to that in the proof that AM is a median space [1, Theorem 4.25]. Since Tk is a complete geodesic space, it suffices to prove that it is zero hyperbolic. Thus, it suffices to prove that, for every triple α, β and γ and μ being its median point, μ is on any geodesic g joining α and β in ψ(AM).
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