Theorem 8: Let \Gamma be a \delta-hyperbolic group with respect to S. If u,v \in \Gamma are conjugate then there exists \gamma\in\Gamma such that

\gamma u\gamma^{-1}=v\\l(\gamma)\leq M(l(u),l(v))

where M depends only on \Gamma.

Proof: We work in Cay_S(\Gamma). Let \gamma\in \Gamma be such that \gamma u\gamma^{-1}=v.  Let \gamma_t \in [1,\gamma] be such that d(1,\gamma_t)=t. We want to find a bound on d(\gamma_t, v\gamma_t).

Let c=[1,\gamma u]. By Lemma 9,

d(\gamma_t,c(t))\leq 2(\delta+l(v))\\ d(v\gamma_t,c(l(\gamma u)-(l(\gamma)-t)))\leq 2(\delta+l(v))


d(c(t),c(l(\gamma u)-(l(\gamma)-t)))=l(\gamma u)-l(\gamma)+t-t\\=l(\gamma u)-l(\gamma)\leq l(u)

So d(\gamma_t,v\gamma_t)\leq 4(\delta +l(u)+l(v))= R(l(u),l(v)). Thus l(\gamma_t^{-1}v\gamma_t)\leq R. Suppose that l(\gamma)> \#B(1,R). By the Pigeonhole Principle there exist integers s>t such that \gamma_t^{-1}v\gamma_t=\gamma_s^{-1}v\gamma_s. It follows that one can find a shorter conjugating element by cutting out the section of \gamma between \gamma_t and \gamma_s.

Recall, for \gamma \in \Gamma, C(\gamma)=\{g\in\Gamma: g\gamma=\gamma g\} is the centralizer of \gamma.

Theorem 9: If \Gamma is \delta-hyperbolic with respect to S and \gamma\in\Gamma, then C(\gamma) is quasi-convex in \Gamma.

Proof: Again we work in Cay_S(\Gamma). Let g\in C(\gamma), h\in [1,g]. We need to prove that H is in a bounded neighborhood C(\gamma).

Just as in the proof of Theorem 8,

l(h^{-1}\gamma h)=d(h, \gamma h)\leq 4(\delta+2l(\gamma))

Well, g and h^{-1}\gamma h are conjugate. By Theorem 8 there exists k\in \Gamma such that

k^{-1}\gamma k=h^{-1}\gamma h \\ l(k)\leq M(l(\gamma), l(h^{-1}\gamma h))\\ \leq M(l(\gamma),4(\delta+2l(\gamma)))

But h^{-1}k\gamma=\gamma hk^{-1} so that hk^{-1}\in C(\gamma) and d(h,hk^{-1})=d(1,k^{-1})=l(k^{-1})=l(k)\leq M.

Exercise 15: Prove that


is not hyperbolic for any Anosov A.