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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


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))


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.


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