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 2 G set d(a; b) = ja−1bj = 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 2 G set d(a; b) = ja−1bj = 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 2 G set d(a; b) = ja−1bj = 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 2 G set d(a; b) = ja−1bj = 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 2 G set d(a; b) = ja−1bj = 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) = jg −1ghj = jhj for every g 2 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) supx2X d(x; x ∗ h) < 1 for every h 2 H. Brandon Seward () The Geometric Burnside's Problem GAGTA May 2013 4 / 13 (TL1) is clear. For (TL2) we have d(g; gh) = jg −1ghj = jhj for every g 2 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) supx2X d(x; x ∗ h) < 1 for every h 2 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) = jg −1ghj = jhj for every g 2 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) supx2X d(x; x ∗ h) < 1 for every h 2 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) supx2X d(x; x ∗ h) < 1 for every h 2 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) = jg −1ghj = jhj for every g 2 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.