The Geometric Burnsides Problem Brandon Seward University of - - PowerPoint PPT Presentation

▶
the geometric burnside s problem
SMART_READER_LITE
LIVE PREVIEW

The Geometric Burnsides Problem Brandon Seward University of - - PowerPoint PPT Presentation

Nov 18, 2022 662 likes •1.3k views

The Geometric Burnsides Problem Brandon Seward University of Michigan Geometric and Asymptotic Group Theory with Applications May 28, 2013 Brandon Seward () The Geometric Burnsides Problem GAGTA May 2013 1 / 13 Classical Problems Two

slide-1
SLIDE 1

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

slide-2
SLIDE 2

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

slide-3
SLIDE 3

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

slide-4
SLIDE 4

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

slide-5
SLIDE 5

Obtaining Geometric Reformulations

We put a metric on finitely generated groups G.

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

slide-6
SLIDE 6

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

slide-7
SLIDE 7

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

slide-8
SLIDE 8

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

slide-9
SLIDE 9

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

slide-10
SLIDE 10

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

  • n 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

slide-11
SLIDE 11

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

  • n 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

slide-12
SLIDE 12

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

  • n 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

slide-13
SLIDE 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

  • n 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

slide-14
SLIDE 14

Kevin Whyte’s Geometric Reformulations

Classical (Algebraic) Reformulated (Geometric)

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

slide-15
SLIDE 15

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

slide-16
SLIDE 16

Kevin Whyte’s Geometric Reformulations

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

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

slide-17
SLIDE 17

Kevin Whyte’s Geometric Reformulations

Classical (Algebraic) Reformulated (Geometric) Burnside’s Problem: Does every finitely generated infinite group contain Z? Geometric Burnside’s Problem: Does every finitely generated infinite group admit a TL action by Z? von Neumann Conjecture: A group is non-amenable if and

  • nly if it contains a non-abelian

free subgroup.

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

slide-18
SLIDE 18

Kevin Whyte’s Geometric Reformulations

Classical (Algebraic) Reformulated (Geometric) Burnside’s Problem: Does every finitely generated infinite group contain Z? Geometric Burnside’s Problem: Does every finitely generated infinite group admit a TL action by Z? von Neumann Conjecture: A group is non-amenable if and

  • nly if it contains a non-abelian

free subgroup. 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.

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

slide-19
SLIDE 19

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

slide-20
SLIDE 20

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

slide-21
SLIDE 21

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

slide-22
SLIDE 22

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

slide-23
SLIDE 23

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

slide-24
SLIDE 24

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

slide-25
SLIDE 25

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

slide-26
SLIDE 26

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

slide-27
SLIDE 27

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

slide-28
SLIDE 28

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

slide-29
SLIDE 29

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

slide-30
SLIDE 30

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

slide-31
SLIDE 31

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

slide-32
SLIDE 32

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

slide-33
SLIDE 33

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

slide-34
SLIDE 34

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

slide-35
SLIDE 35

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

slide-36
SLIDE 36

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 ⇐ ⇒ There exists locally finite Cay(G) with Γ ≤ Cay(G)

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 8 / 13

slide-37
SLIDE 37

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 ⇐ ⇒ There exists locally finite Cay(G) with Γ ≤ Cay(G) Orbits of ∗

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 8 / 13

slide-38
SLIDE 38

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 ⇐ ⇒ There exists locally finite Cay(G) with Γ ≤ Cay(G) Orbits of ∗ ← → Connected components of Γ

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 8 / 13

slide-39
SLIDE 39

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 ⇐ ⇒ There exists locally finite 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

slide-40
SLIDE 40

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

slide-41
SLIDE 41

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

slide-42
SLIDE 42

Sketch of Proof of Clause (i)

1 Assume G is non-amenable Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13

slide-43
SLIDE 43

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

slide-44
SLIDE 44

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

slide-45
SLIDE 45

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

slide-46
SLIDE 46

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 Γ′ Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13

slide-47
SLIDE 47

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 Γ′ 6 If d is the degree of Cay(G), then each vertex of Γ′ has degree at

least 4 and at most d

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13

slide-48
SLIDE 48

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 Γ′ 6 If d is the degree of Cay(G), then each vertex of Γ′ has degree at

least 4 and at most d

7 Claim: Γ′ is bilipschitz equivalent to every locally finite regular tree of

degree ≥ 3

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13

slide-49
SLIDE 49

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 Γ′ 6 If d is the degree of Cay(G), then each vertex of Γ′ has degree at

least 4 and at most d

7 Claim: Γ′ 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 Γ′ and Ψ

Brandon Seward () The Geometric Burnside’s Problem GAGTA May 2013 10 / 13

slide-50
SLIDE 50

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 Γ′ 6 If d is the degree of Cay(G), then each vertex of Γ′ has degree at

least 4 and at most d

7 Claim: Γ′ 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 Γ′ 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

slide-51
SLIDE 51

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 Γ′ 6 If d is the degree of Cay(G), then each vertex of Γ′ has degree at

least 4 and at most d

7 Claim: Γ′ 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 Γ′ 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

slide-52
SLIDE 52

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

slide-53
SLIDE 53

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

slide-54
SLIDE 54

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

slide-55
SLIDE 55

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

slide-56
SLIDE 56

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˝

  • s–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

slide-57
SLIDE 57

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˝

  • s–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

slide-58
SLIDE 58

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˝

  • s–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

slide-59
SLIDE 59

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˝

  • s–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

slide-60
SLIDE 60

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˝

  • s–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

slide-61
SLIDE 61

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˝

  • s–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

slide-62
SLIDE 62

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˝

  • s–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

slide-63
SLIDE 63

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