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The Geometric Burnside's Problem 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.
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