Notes of Michah Sageev’s Cambridge lecture 10-01-2017

Uniform exponential growth for groups acting on {CAT(0)} cube complexes

Joint with A.Kar.

1. Uniform exponential growth

Let {G} be a finitely generated group. For a finite generating set {S}, let {\omega_S} denote the exponential rate of volume growth in {Cay(G,S)}.

Gromov asked wether exponential growth ({\exists S, \omega_S>1}) implies uniform exponential growth ({\omega(G):=\inf\{\omega_S\,;\,S} finite, generating set {\}>1}).

Wilson produced a counterexample.

Question. What about the subclass of groups acting on {CAT(0)} cube complexes ?

The usual trick is to produce a free subsemigroup generated by short words. Ping pong is good for that. For instance, it shows that two loxodromic isometries of a tree generate a free semigroup (up to changing them into their inverses) unless they stabilize a common line.

Theorem 1 If {a_1,\ldots,a_n} are loxodromic isometries of a {CAT(0)} square complex. Then either there exist words of length {\leq 10} that generate a free semigroup, or {a_1,\ldots,a_n} stabilize a common flat.

Corollary 2 If a finitely generated group {G} acts freely on a {CAT(0)} square complex, then either {G} is virtually abelian or {\omega(G)\geq 2^{1/10}}.

Question. What for higher dimensions ?

Question. What about torsion ?

Assume {G} is generated by two elliptics with disjoint fixed point sets. Do they generate loxodromics ? Of what length ? Unclear.

2. Loxodromic elements and hyperplanes

Let {g\in G} act loxodromically on CCC {Y}. Say a hyperplane {\hat h} is skewered by {g} if it is transverse to he axis of {g}. Then some power of {g} translates {\hat h} in the sense that some halfspace {h} is mapped into itself.

Let {sk(g)} be the set of hyperplanes skewered by {g}. A disjoint skewer set is a subset of {sk(g)} which is disjoint and invariant under {g^d}.

Lemma 3 Let {a,b} be loxodromics. Let {D} be a disjoint skewer set for {a}. Assume that no words of length {\leq \ell} generate a free semigroup. Then

  • either {D\subset sk(b)},
  • or {D\cap sk(b)=\emptyset}.

Lemma 4 Let {a,b} be loxodromics. Let {D} be a disjoint skewer set for {a}. Assume that no words of length {\leq 10} generate a free semigroup. Assume that {D\cap sk(b)=\emptyset}. Then

  1. The axis of {b} is parallel to every hyperplane in {D}.
  2. {bD\subset sk(a)}.
  3. {b^2 D=D}.

3. The parallel complex of a loxodromic isometry

Since every hyperplane skewered by {g} and every hyperplan parallel to the axis of {g} intersect, thet subset {\bigcap\{h\,;\,h } is peripheral to {g} and contains the axis of {g\}} has a product structure {E_g\times Y_g} where {E_g} is Euclidean.

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About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See http://www.math.ens.fr/metric2011/
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