Embedding of hyperbolic groups into products of binary trees

Buyalo, Sergei; Dranishnikov, Alexander; Schroeder, Viktor

doi:10.1007/s00222-007-0045-2

Cited by 27 publications

(40 citation statements)

References 16 publications

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“…In the case of hyperbolic spaces the embedding theorem of Section 8 can be improved to the following [28] Theorem 42. Every visual hyperbolic space X admits a quasi-isometric embedding into the product of n + 1 copies of the binary metric tree where n = dim ∂ ∞ X is the topological dimension of the boundary at infinity.…”

Section: Proposition 41mentioning

confidence: 97%

Asymptotic dimension

Bell

Dranishnikov

2008

Topology and its Applications

135

175

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Section: Proposition 41mentioning

confidence: 97%

Asymptotic dimension

Bell

Dranishnikov

2008

Topology and its Applications

135

175

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“…A positive answer to this question gives information on the asymptotic geometry of the group: it bounds the asymptotic dimension, the dimensions of the asymptotic cones, and the uniform Hilbert space compression. Many groups are known to be quasi-isometrically embeddable into a product of finitely-many trees, such as hyperbolic groups [BDS07], some relatively hyperbolic groups [MS13], mapping class groups [Hum] and virtually special groups. On the other hand, Thompson's group F , the discrete Heisenberg group and wreath products are known for not satisfying this property [Pau01].…”

Section: Introductionmentioning

confidence: 99%

Hyperplanes of Squier’s cube complexes

Genevois

2018

Algebr. Geom. Topol.

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To any semigroup presentation P = Σ | R and base word w ∈ Σ + may be associated a nonpositively curved cube complex S(P, w), called a Squier complex, whose underlying graph consists of the words of Σ + equal to w modulo P where two such words are linked by an edge when one can be transformed into the other by applying a relation of R. A group is a diagram group if it is the fundamental group of a Squier complex. In this paper, we describe hyperplanes in these cube complexes. As a first application, we determine exactly when S(P, w) is a special cube complex, as defined by Haglund and Wise, so that the associated diagram group embeds into a right-angled Artin group. A particular feature of Squier complexes is that the intersections of hyperplanes are "ordered" by a relation ≺. As a strong consequence on the geometry of S(P, w), we deduce, in finite dimensions, that its univeral cover isometrically embedds into a product of finitely-many trees with respect to the combinatorial metrics; in particular, we notice that (often) this allows to embed quasi-isometrically the associated diagram group into a product of finitelymany trees, giving information on its asymptotic dimension and its uniform Hilbert space compression. Finally, we exhibit a class of hyperplanes inducing a decomposition of S(P, w) as a graph of spaces, and a fortiori a decomposition of the associated diagram group as a graph of groups, giving a new method to compute presentations of diagram groups. As an application, we associate a semigroup presentation P(Γ) to any finite interval graph Γ, and we prove that the diagram group associated to P(Γ) (for a given base word) is isomorphic to the right-angled Artin group A(Γ). This result has many consequences on the study of subgroups of diagram groups. In particular, we deduce that, for all n ≥ 1, the right-angled Artin group A(Cn) embeds into a diagram group, answering a question of Guba and Sapir. greater than five.Recall that Γ ⊂ Λ is an induced subgraph of Λ if any vertices x, y ∈ Γ are linked by an edge in Λ if and only if they are linked by an edge in Γ.Proof. Let n ≥ 5 be an odd integer. Suppose by contradiction that there exist n hyperplanes J 1 , . . . , J n such that J i and J k are transverse if and only if k = i ± 1 (modulo n). Suppose that J 1 ≺ J 2 ; the case J 2 ≺ J 1 will be completely symmetric.Since J 2 and J 3 are transverse, either J 2 ≺ J 3 or J 3 ≺ J 2 . But we already know that J 1 ≺ J 2 so that J 2 ≺ J 3 would imply J 1 ≺ J 3 and a fortiori that J 1 and J 3 are transverse. Therefore, J 2 ≺ J 3 . Similarly, we deduce that J 2 ≺ J 4 , J 5 ≺ J 4 , and so on. Thus, J 2k+1 ≺ J 2k for all 0 ≤ k ≤ n−1 2 . In particular, J n ≺ J n−1 since n is odd. Then, because J n and J 1 are transverse, either J 1 ≺ J n or J n ≺ J 1 . In the first case, we deduce from J n ≺ J n−1 that J 1 and J n−1 are transverse, a contradiction. In the second case, we deduce from J 1 ≺ J 2 that J n and J 2 are transverse, a contradiction.Remark 4.6. Corollary 4.5 does not hold for induced cycles of even length. For ever...

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“…Let H 3 (R) denote the Heisenberg group over the ring R. If K is a field of characteristic zero, the automorphism group of the algebraic group H 3 (K) is easily seen (working with the Lie algebra) to be K-isomorphic to GL 2 (K) K 2 (where the K 2 corresponds to inner automorphisms). If an automorphism corresponds to (A, v) ∈ GL 2 (K) K 2 , its action on the center of H 3 (K) is given by multiplication by det(A) 2 .…”

Section: Splittings Of the Exponential Radicalmentioning

confidence: 99%

“…It was proved in [2] that every word-hyperbolic group quasi-isometrically embeds into a finite product of binary trees, and therefore the question has a positive answer for products of rank one nonpositively curved simply connected symmetric space. However, I do not know the answer for any irreducible nonpositively curved simply connected symmetric space of higher rank, for example, SL 3 (R)/SO 3 (R).…”

Section: Introductionmentioning

confidence: 99%