The Geometric Burnside’s Problem

Brandon Seward University of Michigan

Geometric and Asymptotic Group Theory with Applications May 28, 2013

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 1 / 13 Both have negative answers (1) Golod–Shafarevich 1964 (2) Olshanskii 1980

These famous problems have inspired various reformulations. We will consider a geometric reformulation due to Kevin Whyte.

Classical Problems

Two classic problems of group theory: (1) Burnside’s Problem: Does every finitely generated infinite group contain Z as a subgroup? (2) von Neumann Conjecture: A group is non-amenable if and only if it contains a non-abelian free subgroup.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 2 / 13 These famous problems have inspired various reformulations. We will consider a geometric reformulation due to Kevin Whyte.

Classical Problems

Two classic problems of group theory: (1) Burnside’s Problem: Does every finitely generated infinite group contain Z as a subgroup? (2) von Neumann Conjecture: A group is non-amenable if and only if it contains a non-abelian free subgroup.

Both have negative answers (1) Golod–Shafarevich 1964 (2) Olshanskii 1980

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 2 / 13 Classical Problems

Two classic problems of group theory: (1) Burnside’s Problem: Does every finitely generated infinite group contain Z as a subgroup? (2) von Neumann Conjecture: A group is non-amenable if and only if it contains a non-abelian free subgroup.

Both have negative answers (1) Golod–Shafarevich 1964 (2) Olshanskii 1980

These famous problems have inspired various reformulations. We will consider a geometric reformulation due to Kevin Whyte.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 2 / 13 Fix a finite generating set S. For a, b ∈ G set d(a, b) = |a−1b| = the S-word length of a−1b.

We obtain geometric reformulations by replacing subgroup containment with a similar (but not equivalent) geometric property: the existence of a translation-like action.

Obtaining Geometric Reformulations

We put a metric on finitely generated groups G.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 3 / 13 For a, b ∈ G set d(a, b) = |a−1b| = the S-word length of a−1b.

We obtain geometric reformulations by replacing subgroup containment with a similar (but not equivalent) geometric property: the existence of a translation-like action.

Obtaining Geometric Reformulations

We put a metric on finitely generated groups G. Fix a finite generating set S.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 3 / 13 We obtain geometric reformulations by replacing subgroup containment with a similar (but not equivalent) geometric property: the existence of a translation-like action.

Obtaining Geometric Reformulations

We put a metric on finitely generated groups G. Fix a finite generating set S. For a, b ∈ G set d(a, b) = |a−1b| = the S-word length of a−1b.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 3 / 13 the existence of a translation-like action.

Obtaining Geometric Reformulations

We put a metric on finitely generated groups G. Fix a finite generating set S. For a, b ∈ G set d(a, b) = |a−1b| = the S-word length of a−1b.

We obtain geometric reformulations by replacing subgroup containment with a similar (but not equivalent) geometric property:

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 3 / 13 Obtaining Geometric Reformulations

We put a metric on finitely generated groups G. Fix a finite generating set S. For a, b ∈ G set d(a, b) = |a−1b| = the S-word length of a−1b.

We obtain geometric reformulations by replacing subgroup containment with a similar (but not equivalent) geometric property: the existence of a translation-like action.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 3 / 13 Observation If G is finitely generated and H ≤ G, then the natural right action of H on G is a TL action. (TL1) is clear. For (TL2) we have d(g, gh) = |g −1gh| = |h| for every g ∈ G.

Translation-Like Actions

Definition [Kevin Whyte, 1999] Let H be a group and let (X , d) be a metric space. A right action, ∗, of H on X is a translation-like action (TL action) if

(TL1) the action is free (x ∗ h = x =⇒ h = 1H )

(TL2) supx∈X d(x, x ∗ h) < ∞ for every h ∈ H.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 4 / 13 (TL1) is clear. For (TL2) we have d(g, gh) = |g −1gh| = |h| for every g ∈ G.

Translation-Like Actions

Definition [Kevin Whyte, 1999] Let H be a group and let (X , d) be a metric space. A right action, ∗, of H on X is a translation-like action (TL action) if

(TL1) the action is free (x ∗ h = x =⇒ h = 1H )

(TL2) supx∈X d(x, x ∗ h) < ∞ for every h ∈ H.

Observation If G is finitely generated and H ≤ G, then the natural right action of H on G is a TL action.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 4 / 13 For (TL2) we have d(g, gh) = |g −1gh| = |h| for every g ∈ G.

Translation-Like Actions

Definition [Kevin Whyte, 1999] Let H be a group and let (X , d) be a metric space. A right action, ∗, of H on X is a translation-like action (TL action) if

(TL1) the action is free (x ∗ h = x =⇒ h = 1H )

(TL2) supx∈X d(x, x ∗ h) < ∞ for every h ∈ H.

Observation If G is finitely generated and H ≤ G, then the natural right action of H on G is a TL action. (TL1) is clear.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 4 / 13 Translation-Like Actions

Definition [Kevin Whyte, 1999] Let H be a group and let (X , d) be a metric space. A right action, ∗, of H on X is a translation-like action (TL action) if

(TL1) the action is free (x ∗ h = x =⇒ h = 1H )

(TL2) supx∈X d(x, x ∗ h) < ∞ for every h ∈ H.

Observation If G is finitely generated and H ≤ G, then the natural right action of H on G is a TL action. (TL1) is clear. For (TL2) we have d(g, gh) = |g −1gh| = |h| for every g ∈ G.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 4 / 13 Burnside’s Problem: Geometric Burnside’s Problem: Does every finitely generated Does every finitely generated infinite group contain Z? infinite group admit a TL action by Z?

von Neumann Conjecture: Geometric von Neumann Conjecture: A group is non-amenable if and A finitely generated group is only if it contains a non-abelian non-amenable if and only if free subgroup. it admits a TL action by a non-abelian free group.

Kevin Whyte’s Geometric Reformulations

Classical (Algebraic) Reformulated (Geometric)

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 5 / 13 Geometric Burnside’s Problem: Does every finitely generated infinite group admit a TL action by Z?

von Neumann Conjecture: Geometric von Neumann Conjecture: A group is non-amenable if and A finitely generated group is only if it contains a non-abelian non-amenable if and only if free subgroup. it admits a TL action by a non-abelian free group.

Kevin Whyte’s Geometric Reformulations

Classical (Algebraic) Reformulated (Geometric)

Burnside’s Problem: Does every finitely generated infinite group contain Z?

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 5 / 13 von Neumann Conjecture: Geometric von Neumann Conjecture: A group is non-amenable if and A finitely generated group is only if it contains a non-abelian non-amenable if and only if free subgroup. it admits a TL action by a non-abelian free group.

Kevin Whyte’s Geometric Reformulations

Classical (Algebraic) Reformulated (Geometric)

Burnside’s Problem: Geometric Burnside’s Problem: Does every finitely generated Does every finitely generated infinite group contain Z? infinite group admit a TL action by Z?

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 5 / 13 Geometric von Neumann Conjecture: A finitely generated group is non-amenable if and only if it admits a TL action by a non-abelian free group.

Kevin Whyte’s Geometric Reformulations

Classical (Algebraic) Reformulated (Geometric)

Burnside’s Problem: Geometric Burnside’s Problem: Does every finitely generated Does every finitely generated infinite group contain Z? infinite group admit a TL action by Z?

von Neumann Conjecture: A group is non-amenable if and only if it contains a non-abelian free subgroup.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 5 / 13 Kevin Whyte’s Geometric Reformulations

Classical (Algebraic) Reformulated (Geometric)

Burnside’s Problem: Geometric Burnside’s Problem: Does every finitely generated Does every finitely generated infinite group contain Z? infinite group admit a TL action by Z?

von Neumann Conjecture: Geometric von Neumann Conjecture: A group is non-amenable if and A finitely generated group is only if it contains a non-abelian non-amenable if and only if free subgroup. it admits a TL action by a non-abelian free group.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 5 / 13 However he was unable to resolve the Geometric Burnside’s Problem.

Theorem [BS, 2011] The answer to the Geometric Burnside’s Problem is yes. That is, every finitely generated infinite group admits a TL action by Z.

In fact, we can say something stronger than both theorems above.

Reformulations Have Positive Answers

Kevin Whyte resolved the Geometric von Neumann Conjecture

Theorem [Kevin Whyte, 1999] The Geometric von Neumann Conjecture is true. That is, a finitely generated group is non-amenable if and only if it admits a TL action by a non-abelian free group.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 6 / 13 Theorem [BS, 2011] The answer to the Geometric Burnside’s Problem is yes. That is, every finitely generated infinite group admits a TL action by Z.

In fact, we can say something stronger than both theorems above.

Reformulations Have Positive Answers

Kevin Whyte resolved the Geometric von Neumann Conjecture

Theorem [Kevin Whyte, 1999] The Geometric von Neumann Conjecture is true. That is, a finitely generated group is non-amenable if and only if it admits a TL action by a non-abelian free group.

However he was unable to resolve the Geometric Burnside’s Problem.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 6 / 13 In fact, we can say something stronger than both theorems above.

Reformulations Have Positive Answers

Kevin Whyte resolved the Geometric von Neumann Conjecture

Theorem [Kevin Whyte, 1999] The Geometric von Neumann Conjecture is true. That is, a finitely generated group is non-amenable if and only if it admits a TL action by a non-abelian free group.

However he was unable to resolve the Geometric Burnside’s Problem.

Theorem [BS, 2011] The answer to the Geometric Burnside’s Problem is yes. That is, every finitely generated infinite group admits a TL action by Z.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 6 / 13 Reformulations Have Positive Answers

Kevin Whyte resolved the Geometric von Neumann Conjecture

Theorem [Kevin Whyte, 1999] The Geometric von Neumann Conjecture is true. That is, a finitely generated group is non-amenable if and only if it admits a TL action by a non-abelian free group.

However he was unable to resolve the Geometric Burnside’s Problem.

Theorem [BS, 2011] The answer to the Geometric Burnside’s Problem is yes. That is, every finitely generated infinite group admits a TL action by Z.

In fact, we can say something stronger than both theorems above.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 6 / 13 In fact: (i) G is non-amenable if and only if it admits a transitive TL action by every finitely generated non-abelian free group. (ii) G has finitely many ends if and only if it admits a transitive TL action by Z.

Theorem [BS, 2011] (Cayley Graph Version) Every finitely generated infinite group G has a locally finite Cayley graph admitting a regular spanning tree. In fact, for k > 2 we have: (i) G is non-amenable if and only if it has a locally finite Cayley graph admitting a k-regular spanning tree; (ii) G has finitely many ends if and only if it has a locally finite Cayley graph admitting a 2-regular spanning tree.

Stronger Results

Theorem [BS, 2011] Every finitely generated infinite group G admits a transitive TL action by some free group (possibly Z).

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 7 / 13 (ii) G has finitely many ends if and only if it admits a transitive TL action by Z.

Theorem [BS, 2011] (Cayley Graph Version) Every finitely generated infinite group G has a locally finite Cayley graph admitting a regular spanning tree. In fact, for k > 2 we have: (i) G is non-amenable if and only if it has a locally finite Cayley graph admitting a k-regular spanning tree; (ii) G has finitely many ends if and only if it has a locally finite Cayley graph admitting a 2-regular spanning tree.

Stronger Results

Theorem [BS, 2011] Every finitely generated infinite group G admits a transitive TL action by some free group (possibly Z). In fact: (i) G is non-amenable if and only if it admits a transitive TL action by every finitely generated non-abelian free group.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 7 / 13 Theorem [BS, 2011] (Cayley Graph Version) Every finitely generated infinite group G has a locally finite Cayley graph admitting a regular spanning tree. In fact, for k > 2 we have: (i) G is non-amenable if and only if it has a locally finite Cayley graph admitting a k-regular spanning tree; (ii) G has finitely many ends if and only if it has a locally finite Cayley graph admitting a 2-regular spanning tree.

Stronger Results

Theorem [BS, 2011] Every finitely generated infinite group G admits a transitive TL action by some free group (possibly Z). In fact: (i) G is non-amenable if and only if it admits a transitive TL action by every finitely generated non-abelian free group. (ii) G has finitely many ends if and only if it admits a transitive TL action by Z.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 7 / 13 Every finitely generated infinite group G has a locally finite Cayley graph admitting a regular spanning tree. In fact, for k > 2 we have: (i) G is non-amenable if and only if it has a locally finite Cayley graph admitting a k-regular spanning tree; (ii) G has finitely many ends if and only if it has a locally finite Cayley graph admitting a 2-regular spanning tree.

Stronger Results

Theorem [BS, 2011] (TL Action Version) Every finitely generated infinite group G admits a transitive TL action by some free group (possibly Z). In fact: (i) G is non-amenable if and only if it admits a transitive TL action by every finitely generated non-abelian free group. (ii) G has finitely many ends if and only if it admits a transitive TL action by Z.

Theorem [BS, 2011] (Cayley Graph Version)

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 7 / 13 In fact, for k > 2 we have: (i) G is non-amenable if and only if it has a locally finite Cayley graph admitting a k-regular spanning tree; (ii) G has finitely many ends if and only if it has a locally finite Cayley graph admitting a 2-regular spanning tree.

Stronger Results

Theorem [BS, 2011] (TL Action Version) Every finitely generated infinite group G admits a transitive TL action by some free group (possibly Z). In fact: (i) G is non-amenable if and only if it admits a transitive TL action by every finitely generated non-abelian free group. (ii) G has finitely many ends if and only if it admits a transitive TL action by Z.

Theorem [BS, 2011] (Cayley Graph Version) Every finitely generated infinite group G has a locally finite Cayley graph admitting a regular spanning tree.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 7 / 13 (ii) G has finitely many ends if and only if it has a locally finite Cayley graph admitting a 2-regular spanning tree.

Stronger Results

Theorem [BS, 2011] (TL Action Version) Every finitely generated infinite group G admits a transitive TL action by some free group (possibly Z). In fact: (i) G is non-amenable if and only if it admits a transitive TL action by every finitely generated non-abelian free group. (ii) G has finitely many ends if and only if it admits a transitive TL action by Z.

Theorem [BS, 2011] (Cayley Graph Version) Every finitely generated infinite group G has a locally finite Cayley graph admitting a regular spanning tree. In fact, for k > 2 we have: (i) G is non-amenable if and only if it has a locally finite Cayley graph admitting a k-regular spanning tree;

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 7 / 13 Stronger Results

Theorem [BS, 2011] (TL Action Version) Every finitely generated infinite group G admits a transitive TL action by some free group (possibly Z). In fact: (i) G is non-amenable if and only if it admits a transitive TL action by every finitely generated non-abelian free group. (ii) G has finitely many ends if and only if it admits a transitive TL action by Z.

Theorem [BS, 2011] (Cayley Graph Version) Every finitely generated infinite group G has a locally finite Cayley graph admitting a regular spanning tree. In fact, for k > 2 we have: (i) G is non-amenable if and only if it has a locally finite Cayley graph admitting a k-regular spanning tree; (ii) G has finitely many ends if and only if it has a locally finite Cayley graph admitting a 2-regular spanning tree.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 7 / 13 Sketch of Equivalence of Two Versions

Let G be a finitely generated group. Let F be a free group freely generated by T , |T | < ∞. We have the following correspondances / equivalences:

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 8 / 13 Sketch of Equivalence of Two Versions

Let G be a finitely generated group. Let F be a free group freely generated by T , |T | < ∞. We have the following correspondances / equivalences:

Actions ∗ of F on G

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 8 / 13 Sketch of Equivalence of Two Versions

Let G be a finitely generated group. Let F be a free group freely generated by T , |T | < ∞. We have the following correspondances / equivalences:

Actions ∗ of F on G ←→ Directed T -labelled graphs Γ with vertex set G such that for each g ∈ G and t ∈ T there is a t-edge leaving g and a t-edge entering g

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 8 / 13 Sketch of Equivalence of Two Versions

Let G be a finitely generated group. Let F be a free group freely generated by T , |T | < ∞. We have the following correspondances / equivalences:

Actions ∗ of F on G ←→ Directed T -labelled graphs Γ with vertex set G such that for each g ∈ G and t ∈ T there is a t-edge leaving g and a t-edge entering g (TL1) The action ∗ is free

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 8 / 13 Sketch of Equivalence of Two Versions

Let G be a finitely generated group. Let F be a free group freely generated by T , |T | < ∞. We have the following correspondances / equivalences:

Actions ∗ of F on G ←→ Directed T -labelled graphs Γ with vertex set G such that for each g ∈ G and t ∈ T there is a t-edge leaving g and a t-edge entering g (TL1) The action ∗ is free (TL2) supg∈G d(g, g ∗ f ) < ∞ for each f ∈ F

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 8 / 13 Sketch of Equivalence of Two Versions

Let G be a finitely generated group. Let F be a free group freely generated by T , |T | < ∞. We have the following correspondances / equivalences:

Actions ∗ of F on G ←→ Directed T -labelled graphs Γ with vertex set G such that for each g ∈ G and t ∈ T there is a t-edge leaving g and a t-edge entering g (TL1) The action ∗ is free ⇐⇒ Each connected component of Γ is isomorphic to CayT (F ) (TL2) supg∈G d(g, g ∗ f ) < ∞ for each f ∈ F

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 8 / 13 Sketch of Equivalence of Two Versions

Let G be a finitely generated group. Let F be a free group freely generated by T , |T | < ∞. We have the following correspondances / equivalences:

Actions ∗ of F on G ←→ Directed T -labelled graphs Γ with vertex set G such that for each g ∈ G and t ∈ T there is a t-edge leaving g and a t-edge entering g (TL1) The action ∗ is free ⇐⇒ Each connected component of Γ is isomorphic to CayT (F ) (TL2) supg∈G d(g, g ∗ f ) < ∞ ⇐⇒ There exists locally finite for each f ∈ F Cay(G) with Γ ≤ Cay(G)

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 8 / 13 Sketch of Equivalence of Two Versions

Let G be a finitely generated group. Let F be a free group freely generated by T , |T | < ∞. We have the following correspondances / equivalences:

Actions ∗ of F on G ←→ Directed T -labelled graphs Γ with vertex set G such that for each g ∈ G and t ∈ T there is a t-edge leaving g and a t-edge entering g (TL1) The action ∗ is free ⇐⇒ Each connected component of Γ is isomorphic to CayT (F ) (TL2) supg∈G d(g, g ∗ f ) < ∞ ⇐⇒ There exists locally finite for each f ∈ F Cay(G) with Γ ≤ Cay(G) Orbits of ∗

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 8 / 13 Sketch of Equivalence of Two Versions

Let G be a finitely generated group. Let F be a free group freely generated by T , |T | < ∞. We have the following correspondances / equivalences:

Actions ∗ of F on G ←→ Directed T -labelled graphs Γ with vertex set G such that for each g ∈ G and t ∈ T there is a t-edge leaving g and a t-edge entering g (TL1) The action ∗ is free ⇐⇒ Each connected component of Γ is isomorphic to CayT (F ) (TL2) supg∈G d(g, g ∗ f ) < ∞ ⇐⇒ There exists locally finite for each f ∈ F Cay(G) with Γ ≤ Cay(G) Orbits of ∗ ←→ Connected components of Γ

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 8 / 13 Sketch of Equivalence of Two Versions

Let G be a finitely generated group. Let F be a free group freely generated by T , |T | < ∞. We have the following correspondances / equivalences:

Actions ∗ of F on G ←→ Directed T -labelled graphs Γ with vertex set G such that for each g ∈ G and t ∈ T there is a t-edge leaving g and a t-edge entering g (TL1) The action ∗ is free ⇐⇒ Each connected component of Γ is isomorphic to CayT (F ) (TL2) supg∈G d(g, g ∗ f ) < ∞ ⇐⇒ There exists locally finite for each f ∈ F Cay(G) with Γ ≤ Cay(G) Orbits of ∗ ←→ Connected components of Γ ←→ Transitive TL Action by F CayT (F ) is a spanning sub- graph of some Cay(G)

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 8 / 13 Notice that it implies: Theorem [BS, 2011] The answer to the Geometric Burnside’s Problem is yes. That is, every finitely generated infinite group admits a TL action by Z.

Recall Results

Recall the earlier stated result: Theorem [BS, 2011] (TL Action Version) Every finitely generated infinite group G admits a transitive TL action by some free group (possibly Z). In fact: (i) G is non-amenable if and only if it admits a transitive TL action by every finitely generated non-abelian free group. (ii) G has finitely many ends if and only if it admits a transitive TL action by Z. We will next prove this theorem.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 9 / 13 Recall Results

Recall the earlier stated result: Theorem [BS, 2011] (TL Action Version) Every finitely generated infinite group G admits a transitive TL action by some free group (possibly Z). In fact: (i) G is non-amenable if and only if it admits a transitive TL action by every finitely generated non-abelian free group. (ii) G has finitely many ends if and only if it admits a transitive TL action by Z. We will next prove this theorem. Notice that it implies: Theorem [BS, 2011] The answer to the Geometric Burnside’s Problem is yes. That is, every finitely generated infinite group admits a TL action by Z.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 9 / 13 2 By GvNC (proved by Whyte), G admits TL action by some non-abelian free group 3 Let Γ be the graph associated to this action 4 Γ is a regular forest of degree ≥ 4 and is a subgraph of some Cay(G) 5 Carefully add edges from Cay(G) to Γ to obtain a tree Γ0 6 If d is the degree of Cay(G), then each vertex of Γ0 has degree at least 4 and at most d 7 Claim: Γ0 is bilipschitz equivalent to every locally finite regular tree of degree ≥ 3 8 Fix k ≥ 2. By claim, ∃ 2k-regular tree Ψ with vertex set G such that the identity map G → G induces a bilipschitz bijection between Γ0 and Ψ

9 The graph Ψ tells the free group Fk how to act on G 10 This action is transitive and TL. QED

Sketch of Proof of Clause (i)

1 Assume G is non-amenable

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13 3 Let Γ be the graph associated to this action 4 Γ is a regular forest of degree ≥ 4 and is a subgraph of some Cay(G) 5 Carefully add edges from Cay(G) to Γ to obtain a tree Γ0 6 If d is the degree of Cay(G), then each vertex of Γ0 has degree at least 4 and at most d 7 Claim: Γ0 is bilipschitz equivalent to every locally finite regular tree of degree ≥ 3 8 Fix k ≥ 2. By claim, ∃ 2k-regular tree Ψ with vertex set G such that the identity map G → G induces a bilipschitz bijection between Γ0 and Ψ

9 The graph Ψ tells the free group Fk how to act on G 10 This action is transitive and TL. QED

Sketch of Proof of Clause (i)

1 Assume G is non-amenable 2 By GvNC (proved by Whyte), G admits TL action by some non-abelian free group

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13 4 Γ is a regular forest of degree ≥ 4 and is a subgraph of some Cay(G) 5 Carefully add edges from Cay(G) to Γ to obtain a tree Γ0 6 If d is the degree of Cay(G), then each vertex of Γ0 has degree at least 4 and at most d 7 Claim: Γ0 is bilipschitz equivalent to every locally finite regular tree of degree ≥ 3 8 Fix k ≥ 2. By claim, ∃ 2k-regular tree Ψ with vertex set G such that the identity map G → G induces a bilipschitz bijection between Γ0 and Ψ

9 The graph Ψ tells the free group Fk how to act on G 10 This action is transitive and TL. QED

Sketch of Proof of Clause (i)

1 Assume G is non-amenable 2 By GvNC (proved by Whyte), G admits TL action by some non-abelian free group 3 Let Γ be the graph associated to this action

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13 5 Carefully add edges from Cay(G) to Γ to obtain a tree Γ0 6 If d is the degree of Cay(G), then each vertex of Γ0 has degree at least 4 and at most d 7 Claim: Γ0 is bilipschitz equivalent to every locally finite regular tree of degree ≥ 3 8 Fix k ≥ 2. By claim, ∃ 2k-regular tree Ψ with vertex set G such that the identity map G → G induces a bilipschitz bijection between Γ0 and Ψ

9 The graph Ψ tells the free group Fk how to act on G 10 This action is transitive and TL. QED

Sketch of Proof of Clause (i)

1 Assume G is non-amenable 2 By GvNC (proved by Whyte), G admits TL action by some non-abelian free group 3 Let Γ be the graph associated to this action 4 Γ is a regular forest of degree ≥ 4 and is a subgraph of some Cay(G)

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13 6 If d is the degree of Cay(G), then each vertex of Γ0 has degree at least 4 and at most d 7 Claim: Γ0 is bilipschitz equivalent to every locally finite regular tree of degree ≥ 3 8 Fix k ≥ 2. By claim, ∃ 2k-regular tree Ψ with vertex set G such that the identity map G → G induces a bilipschitz bijection between Γ0 and Ψ

9 The graph Ψ tells the free group Fk how to act on G 10 This action is transitive and TL. QED

Sketch of Proof of Clause (i)

1 Assume G is non-amenable 2 By GvNC (proved by Whyte), G admits TL action by some non-abelian free group 3 Let Γ be the graph associated to this action 4 Γ is a regular forest of degree ≥ 4 and is a subgraph of some Cay(G) 5 Carefully add edges from Cay(G) to Γ to obtain a tree Γ0

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13 7 Claim: Γ0 is bilipschitz equivalent to every locally finite regular tree of degree ≥ 3 8 Fix k ≥ 2. By claim, ∃ 2k-regular tree Ψ with vertex set G such that the identity map G → G induces a bilipschitz bijection between Γ0 and Ψ

9 The graph Ψ tells the free group Fk how to act on G 10 This action is transitive and TL. QED

Sketch of Proof of Clause (i)

1 Assume G is non-amenable 2 By GvNC (proved by Whyte), G admits TL action by some non-abelian free group 3 Let Γ be the graph associated to this action 4 Γ is a regular forest of degree ≥ 4 and is a subgraph of some Cay(G) 5 Carefully add edges from Cay(G) to Γ to obtain a tree Γ0 6 If d is the degree of Cay(G), then each vertex of Γ0 has degree at least 4 and at most d

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13 8 Fix k ≥ 2. By claim, ∃ 2k-regular tree Ψ with vertex set G such that the identity map G → G induces a bilipschitz bijection between Γ0 and Ψ

9 The graph Ψ tells the free group Fk how to act on G 10 This action is transitive and TL. QED

Sketch of Proof of Clause (i)

1 Assume G is non-amenable 2 By GvNC (proved by Whyte), G admits TL action by some non-abelian free group 3 Let Γ be the graph associated to this action 4 Γ is a regular forest of degree ≥ 4 and is a subgraph of some Cay(G) 5 Carefully add edges from Cay(G) to Γ to obtain a tree Γ0 6 If d is the degree of Cay(G), then each vertex of Γ0 has degree at least 4 and at most d 7 Claim: Γ0 is bilipschitz equivalent to every locally finite regular tree of degree ≥ 3

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13 9 The graph Ψ tells the free group Fk how to act on G 10 This action is transitive and TL. QED

Sketch of Proof of Clause (i)

1 Assume G is non-amenable 2 By GvNC (proved by Whyte), G admits TL action by some non-abelian free group 3 Let Γ be the graph associated to this action 4 Γ is a regular forest of degree ≥ 4 and is a subgraph of some Cay(G) 5 Carefully add edges from Cay(G) to Γ to obtain a tree Γ0 6 If d is the degree of Cay(G), then each vertex of Γ0 has degree at least 4 and at most d 7 Claim: Γ0 is bilipschitz equivalent to every locally finite regular tree of degree ≥ 3 8 Fix k ≥ 2. By claim, ∃ 2k-regular tree Ψ with vertex set G such that the identity map G → G induces a bilipschitz bijection between Γ0 and Ψ

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13 10 This action is transitive and TL. QED

Sketch of Proof of Clause (i)

1 Assume G is non-amenable 2 By GvNC (proved by Whyte), G admits TL action by some non-abelian free group 3 Let Γ be the graph associated to this action 4 Γ is a regular forest of degree ≥ 4 and is a subgraph of some Cay(G) 5 Carefully add edges from Cay(G) to Γ to obtain a tree Γ0 6 If d is the degree of Cay(G), then each vertex of Γ0 has degree at least 4 and at most d 7 Claim: Γ0 is bilipschitz equivalent to every locally finite regular tree of degree ≥ 3 8 Fix k ≥ 2. By claim, ∃ 2k-regular tree Ψ with vertex set G such that the identity map G → G induces a bilipschitz bijection between Γ0 and Ψ

9 The graph Ψ tells the free group Fk how to act on G

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13 Sketch of Proof of Clause (i)

1 Assume G is non-amenable 2 By GvNC (proved by Whyte), G admits TL action by some non-abelian free group 3 Let Γ be the graph associated to this action 4 Γ is a regular forest of degree ≥ 4 and is a subgraph of some Cay(G) 5 Carefully add edges from Cay(G) to Γ to obtain a tree Γ0 6 If d is the degree of Cay(G), then each vertex of Γ0 has degree at least 4 and at most d 7 Claim: Γ0 is bilipschitz equivalent to every locally finite regular tree of degree ≥ 3 8 Fix k ≥ 2. By claim, ∃ 2k-regular tree Ψ with vertex set G such that the identity map G → G induces a bilipschitz bijection between Γ0 and Ψ

9 The graph Ψ tells the free group Fk how to act on G 10 This action is transitive and TL. QED

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13 This generalizes a theorem of Papsoglu (1995) which states that all locally finite regular trees of degree ≥ 4 are bilipschitz equivalent.

Justification of Claim

The claim follows from

Theorem [BS, 2011]

Let Γ1 and Γ2 be two trees. If every vertex of Γ1 and Γ2 has degree at least 3 and if the vertices of Γ1 and Γ2 have uniformly bounded degree, then Γ1 and Γ2 are bilipschitz equivalent.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 11 / 13 Justification of Claim

The claim follows from

Theorem [BS, 2011]

Let Γ1 and Γ2 be two trees. If every vertex of Γ1 and Γ2 has degree at least 3 and if the vertices of Γ1 and Γ2 have uniformly bounded degree, then Γ1 and Γ2 are bilipschitz equivalent.

This generalizes a theorem of Papsoglu (1995) which states that all locally finite regular trees of degree ≥ 4 are bilipschitz equivalent.

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 11 / 13 2 Fix Cay(G) 3 Theorem of Erd˝os–Gr¨unwald–Weiszfeld essentially (modulo small trick) implies there is an Eulerian path P on Cay(G) 4 View P as a function from Z to G 5 Construct bipartite graph Γ on Z ∪ G by joining each n ∈ Z to P(n) ∈ G with an edge 6 Coarsen Γ to obtain new bipartite graph and apply Hall’s Marriage (Selection) Theorem to get A ⊆ Z such that consecutive members of S are a uniformly bounded distance apart and P : S → G is a bijection 7 S inherits a total order from Z and the bijection P : S → G passes this total order to G 8 This total order tells Z how to act on G 9 This action is transitive and TL. QED

Sketch of Proof of Clause (ii)

1 Assume G is countably infinite and has finitely many ends (1 or 2 ends)

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 12 / 13 3 Theorem of Erd˝os–Gr¨unwald–Weiszfeld essentially (modulo small trick) implies there is an Eulerian path P on Cay(G) 4 View P as a function from Z to G 5 Construct bipartite graph Γ on Z ∪ G by joining each n ∈ Z to P(n) ∈ G with an edge 6 Coarsen Γ to obtain new bipartite graph and apply Hall’s Marriage (Selection) Theorem to get A ⊆ Z such that consecutive members of S are a uniformly bounded distance apart and P : S → G is a bijection 7 S inherits a total order from Z and the bijection P : S → G passes this total order to G 8 This total order tells Z how to act on G 9 This action is transitive and TL. QED

Sketch of Proof of Clause (ii)

1 Assume G is countably infinite and has finitely many ends (1 or 2 ends) 2 Fix Cay(G)

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 12 / 13 4 View P as a function from Z to G 5 Construct bipartite graph Γ on Z ∪ G by joining each n ∈ Z to P(n) ∈ G with an edge 6 Coarsen Γ to obtain new bipartite graph and apply Hall’s Marriage (Selection) Theorem to get A ⊆ Z such that consecutive members of S are a uniformly bounded distance apart and P : S → G is a bijection 7 S inherits a total order from Z and the bijection P : S → G passes this total order to G 8 This total order tells Z how to act on G 9 This action is transitive and TL. QED

Sketch of Proof of Clause (ii)

1 Assume G is countably infinite and has finitely many ends (1 or 2 ends) 2 Fix Cay(G) 3 Theorem of Erd˝os–Gr¨unwald–Weiszfeld essentially (modulo small trick) implies there is an Eulerian path P on Cay(G)

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 12 / 13 5 Construct bipartite graph Γ on Z ∪ G by joining each n ∈ Z to P(n) ∈ G with an edge 6 Coarsen Γ to obtain new bipartite graph and apply Hall’s Marriage (Selection) Theorem to get A ⊆ Z such that consecutive members of S are a uniformly bounded distance apart and P : S → G is a bijection 7 S inherits a total order from Z and the bijection P : S → G passes this total order to G 8 This total order tells Z how to act on G 9 This action is transitive and TL. QED

Sketch of Proof of Clause (ii)

1 Assume G is countably infinite and has finitely many ends (1 or 2 ends) 2 Fix Cay(G) 3 Theorem of Erd˝os–Gr¨unwald–Weiszfeld essentially (modulo small trick) implies there is an Eulerian path P on Cay(G) 4 View P as a function from Z to G

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 12 / 13 6 Coarsen Γ to obtain new bipartite graph and apply Hall’s Marriage (Selection) Theorem to get A ⊆ Z such that consecutive members of S are a uniformly bounded distance apart and P : S → G is a bijection 7 S inherits a total order from Z and the bijection P : S → G passes this total order to G 8 This total order tells Z how to act on G 9 This action is transitive and TL. QED

Sketch of Proof of Clause (ii)

1 Assume G is countably infinite and has finitely many ends (1 or 2 ends) 2 Fix Cay(G) 3 Theorem of Erd˝os–Gr¨unwald–Weiszfeld essentially (modulo small trick) implies there is an Eulerian path P on Cay(G) 4 View P as a function from Z to G 5 Construct bipartite graph Γ on Z ∪ G by joining each n ∈ Z to P(n) ∈ G with an edge

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 12 / 13 7 S inherits a total order from Z and the bijection P : S → G passes this total order to G 8 This total order tells Z how to act on G 9 This action is transitive and TL. QED

Sketch of Proof of Clause (ii)

1 Assume G is countably infinite and has finitely many ends (1 or 2 ends) 2 Fix Cay(G) 3 Theorem of Erd˝os–Gr¨unwald–Weiszfeld essentially (modulo small trick) implies there is an Eulerian path P on Cay(G) 4 View P as a function from Z to G 5 Construct bipartite graph Γ on Z ∪ G by joining each n ∈ Z to P(n) ∈ G with an edge 6 Coarsen Γ to obtain new bipartite graph and apply Hall’s Marriage (Selection) Theorem to get A ⊆ Z such that consecutive members of S are a uniformly bounded distance apart and P : S → G is a bijection

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 12 / 13 8 This total order tells Z how to act on G 9 This action is transitive and TL. QED

Sketch of Proof of Clause (ii)

1 Assume G is countably infinite and has finitely many ends (1 or 2 ends) 2 Fix Cay(G) 3 Theorem of Erd˝os–Gr¨unwald–Weiszfeld essentially (modulo small trick) implies there is an Eulerian path P on Cay(G) 4 View P as a function from Z to G 5 Construct bipartite graph Γ on Z ∪ G by joining each n ∈ Z to P(n) ∈ G with an edge 6 Coarsen Γ to obtain new bipartite graph and apply Hall’s Marriage (Selection) Theorem to get A ⊆ Z such that consecutive members of S are a uniformly bounded distance apart and P : S → G is a bijection 7 S inherits a total order from Z and the bijection P : S → G passes this total order to G

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 12 / 13 9 This action is transitive and TL. QED

Sketch of Proof of Clause (ii)

1 Assume G is countably infinite and has finitely many ends (1 or 2 ends) 2 Fix Cay(G) 3 Theorem of Erd˝os–Gr¨unwald–Weiszfeld essentially (modulo small trick) implies there is an Eulerian path P on Cay(G) 4 View P as a function from Z to G 5 Construct bipartite graph Γ on Z ∪ G by joining each n ∈ Z to P(n) ∈ G with an edge 6 Coarsen Γ to obtain new bipartite graph and apply Hall’s Marriage (Selection) Theorem to get A ⊆ Z such that consecutive members of S are a uniformly bounded distance apart and P : S → G is a bijection 7 S inherits a total order from Z and the bijection P : S → G passes this total order to G 8 This total order tells Z how to act on G

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 12 / 13 Sketch of Proof of Clause (ii)

1 Assume G is countably infinite and has finitely many ends (1 or 2 ends) 2 Fix Cay(G) 3 Theorem of Erd˝os–Gr¨unwald–Weiszfeld essentially (modulo small trick) implies there is an Eulerian path P on Cay(G) 4 View P as a function from Z to G 5 Construct bipartite graph Γ on Z ∪ G by joining each n ∈ Z to P(n) ∈ G with an edge 6 Coarsen Γ to obtain new bipartite graph and apply Hall’s Marriage (Selection) Theorem to get A ⊆ Z such that consecutive members of S are a uniformly bounded distance apart and P : S → G is a bijection 7 S inherits a total order from Z and the bijection P : S → G passes this total order to G 8 This total order tells Z how to act on G 9 This action is transitive and TL. QED

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 12 / 13 Further Questions

Question [Kevin Whyte] Geometric Gersten Conjecture: A group is not word hyperbolic if and only if it admits a TL action by some Baumslag–Solitar group.

Question Does every Cayley graph of every group with finitely many ends admit a bi-infinite Hamiltonian path?

Question Is exponential growth equivalent to the existence of a TL action by some non-abelian free semigroup?

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 13 / 13