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Lemma 29: Suppose $\{f_i':C_i\longrightarrow\Gamma^{H}\}$ is a finite set of infinite degree elvations and $\Delta \subseteq \Gamma^{H}$ is compact. Then for all sufficiently large $d>0$, there exists an intermediate covering $\Gamma_d$ such that

(a) $\Delta$ embeds in $\Gamma_d$

(b) every $f'_i$ descends to an elevation $\hat{f_i}:\hat{C_i}\longrightarrow \Gamma_d$ of degree $d$

(c) the $\hat{f_i}$ are pairwise distinct

Proof: We claim that the images of $f_i'$ never share an infinite ray (a ray is an isometric embedding of $[0,\infty)$). Neither do two ends of the same elevation $f_i'$. Let’s claim by passing to the universal cover of $\Gamma$, a tree $T$.

For each $i$, lift $f_i'$ to a map $\tilde{f_i}:\mathbb{R}\longrightarrow T$. If $f'_i$ and $f'_j$ share an infinite ray then there exists $h\in H$ such that $\tilde{f_i}$ and $h\tilde{f_j}$ overlay in an infinite ray. The point is that $\tilde{f_i}, \tilde{f_j}$ correspond to cosets $g_if_{\ast}(\pi_1(C))$ and $g_jf_{\ast}(\pi_1(C))$. But this implies that

$g_if_{\ast}(\pi_1(C))=hg_jf_{\ast}(\pi_1(C))$

This implies that $Hg_if_{\ast}(\pi_1(C))=Hg_jf_{\ast}(\pi_1(C))$. So $f'_i=f'_j$. A similar argument implies that the two ends of $f'_i$ do not overlap in an infinite ray. This proves the claim.

Let $\Gamma'$ be the core of $\Gamma^{H}$. Enlarging $\Delta$ if necessary, we can assume that

(i) $\Gamma'\subseteq\Delta$;

(ii) $\Delta$ is a connected subgraph;

(iii) for each $i$, for some $x_i\in C'_i$, $f'_i(x_i)\in\Delta$;

(iv) for each $i$, $|im(f'_i)\cap \delta\Delta|=2$.

For each $i$ identifying $C'_i$ with $\mathbb{R}$ so that $C$ is identified with $\mathbb{R}/\mathbb{Z}$ and $x_i$ is identified with $0$. Let

$\Delta_d=\Delta\cup(\bigcup_i f'_i([-d/2,d/2]))$

For all sufficiently large $d$,

$f'_i(\pm d/2)\notin\Delta$

Now, the restriction of $\Delta_d \longrightarrow \Delta$ factors through $\Delta_d/\sim\longrightarrow\Gamma$, where $f'_i(d/2)\sim f'_i(-d/2)$. This is a finite-to-one immersion, so, by theorem 5, we can complete it to a finite-sheeted covering map as required. $\Box$

Theorem 19: $D$ is LERF.

Recall the set-up from the previous lecture. We built a graph of spaces $\mathscr{X}$ for $D$.

Proof: Let $H\subset D$ be finitely generated. Let $X_H$ be the corresponding covering space of $X_{\mathscr{X}}$ and let $\Delta\subseteq X_H$ be compact. Because $H$ is finitely generated, there exists a subgraph of spaces $X'$ such that $\pi_1(X') =H$. We can take $X'$ large enough so that $\Delta \subseteq X'$. We can enlarge $\Delta$ so that it contains every finite-degree edge space of $X'$. Also enlarge $\Delta$ so that

$\partial^{\pm}_{e'}(\Delta\times (\mathrm{interval}))\subseteq\Delta$

for any $e'\in E(\Xi')$. For each $v'\in V(\Xi')$ let $\Delta_{v'}=\Delta\cap X_{v'}$ and let ${f'_i}=\{$ incident edge map of infinite degree $\}$.

Applying lemma 29 to $\Gamma^H=X_{v'}$, for some large $d$, set $X_{\hat{v}}=\Gamma_d$. (Here we use the fact that vertex groups of $\mathscr{X}'$ are finitely generated)

Define $\mathscr{X}^+$ as follows:

$\bullet \Xi^+=\Xi^-$

$\bullet$ For each $v^+\in V(\Xi^+)$, the edge space is the $X_{v^+}$ that the lemma produced from the corresponding $v'$.

Now, by construction, $\bigcup_{v^+}X_{v^+}$ can be completed to a graph of spaces $\mathscr{X}^+$ so that the map

$X_{\mathscr{X}'}\longrightarrow X_{\mathscr{X}}$

factors through $X_{\mathscr{X}'}\longrightarrow X_{\mathscr{X}}$ and $\Delta$ embeds. Let $\mathscr{X}^-$ be identical to $\mathscr{X}^+$ except with +’s and -‘s exchanged. Clearly $\mathscr{X}^+\cup\mathscr{X}^-$ satisfies Stallings condition, as required. $\Box$

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,

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

Also

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

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

$\Gamma_A=\mathbb{Z}^2\rtimes_A\mathbb{Z}$

is not hyperbolic for any Anosov $A$.