## 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 ?

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.