Proof First observe that the sets $\pi (g^{[0,\infty ]} \mathfrak {v}) = \mathfrak {v}^+ \cup \{v(t) : v \in \mathfrak {v} \text { and } t \ge 0\}$ and $\pi (g^{[-\infty ,0]} \mathfrak {v}) = \mathfrak {v}^- \cup \{v(t) : v \in \mathfrak {v} \text { and } t \le 0\}$ are closed in $\overline {X}$ because $ \mathfrak {v}$ is compact.

For each $i \in \mathbb {N}$ , let $v_i \in \mathfrak {v} \cap \gamma _i g^{-t_i} \mathfrak {v}$ . Passing to a subsequence if necessary, we may assume the sequence $(v_i)$ converges to some $v_0 \in \mathfrak {v}$ , and $(\gamma _i^{-1} g^{t_i} v_i)$ converges to some $w_0 \in \mathfrak {v}$ . Let $x_0 = v_0(0)$ and $y_0 = w_0(0)$ . Recall that $\gamma _i y_0 \to \xi \in \partial X$ . We may assume the sequence $(\gamma _i^{-1} x_0)$ converges to some $\eta \in \partial X$ .

We know $d(\gamma _i w_0, g^{t_i} v_i) \to 0$ , so $d(\gamma _i y_0, v_i(t_i)) \to 0$ . Since $\pi (g^{[0,\infty ]} \mathfrak {v})$ is closed, we may conclude $\xi = \lim v_i(t_i) \in \mathfrak {v}^+$ . Now for each $i \in \mathbb {N}$ let $w_i = \gamma _i^{-1} g^{t_i} v_i$ . Then $d(\gamma _i^{-1} v_0, g^{-t_i} w_i) = d(\gamma _i^{-1} v_0, \gamma _i^{-1} v_i) \to 0$ , and so $d(\gamma _i^{-1} x_0, w_i(-t_i)) \to 0$ . Since each $w_i \in \mathfrak {v}$ and $\pi (g^{[-\infty ,0]} \mathfrak {v})$ is closed, we see that $\eta = \lim w_i(-t_i) \in \mathfrak {v}^-$ .

Thus $\gamma _i x_0 \to \xi \in \mathfrak {v}^+$ and $\gamma _i^{-1} x_0 \to \eta \in \mathfrak {v}^-$ . Apply Theorem 3.1.

Proof Suppose not. Then for each $i \in \mathbb {N}$ there exist $\gamma _i \in \Gamma $ and $t_i \to \infty $ such that $v_i \in \mathfrak {v} \cap \gamma _i g^{-t_i} \mathfrak {v}$ but $\gamma _i \mathfrak {v}^+ \nsubseteq U_\lambda $ for all $i, \lambda $ . Since $(\gamma _i)$ escapes to infinity, we may assume $\gamma _i x \to \xi \in \partial X$ for some $\xi \in \partial X$ and $x \in X$ . This contradicts Lemma 3.3. Therefore, the theorem must hold.

Proof Suppose, by way of contradiction, there exist $\alpha> 0$ and sequences of $\delta _n> 0$ and $v_n, w_n \in \widetilde {\mathfrak {v}}_{\varepsilon , \delta _n}$ such that $\delta _n \to 0$ but $d(v_n, w_n) \ge 4\varepsilon + \operatorname {\mathrm {diam}} CS(v_0) + \alpha $ for all n. For each n find $s_n, t_n \in [-\varepsilon , 2\varepsilon ]$ such that $g^{-s_n} v_n, g^{-t_n} w_n \in \mathfrak {v}_{0,\delta }$ . By the triangle inequality, $d(g^{-s_n} v_n, g^{-t_n} w_n) \ge \operatorname {\mathrm {diam}} CS(v_0) + \alpha $ for all n. We may assume $g^{-s_n} v_n \to v$ and $g^{-t_n} w_n \to w$ for some $v, w \in \bigcap _{\delta> 0} \mathfrak {v}_{0,\delta }$ . Thus $v,w \in CS(v_0)$ , hence $d(v, w) \le \operatorname {\mathrm {diam}} CS(v_0)$ , contradicting $g^{-s_n} v_n \to v$ and $g^{-t_n} w_n \to w$ . Therefore, the statement of the lemma must hold.

Proof Let $p = v_0(0)$ . Because $\Gamma $ acts freely and properly discontinuously on X, there is some $r_0> 0$ such that $d(p, \gamma p) \ge r_0$ for all non-trivial $\gamma \in \Gamma $ . Let $\varepsilon _0 = {r_0} / {30}$ , and let $\delta _0> 0$ be small enough that $\operatorname {\mathrm {diam}} \widetilde {\mathfrak {v}}_{3\varepsilon _0,\delta _0} < 15\varepsilon _0$ . This implies $\widetilde {\mathfrak {v}}_{3\varepsilon _0,\delta _0} \cap \gamma \widetilde {\mathfrak {v}}_{3\varepsilon _0,\delta _0} = \varnothing $ for all $\gamma \neq \operatorname {\mathrm {id}}$ by the triangle inequality.

Now let $\varepsilon \in [0,\varepsilon _0]$ and $\delta \in (0,\delta _0]$ . Let $\gamma , \gamma ' \in \Gamma $ be such that $\operatorname {\mathrm {E}}(\widetilde {\mathfrak {w}}^{\gamma }) \cap \operatorname {\mathrm {E}}(\widetilde {\mathfrak {w}}^{\gamma '})$ is non-empty. By definition of $\widetilde {\mathfrak {v}}$ , there exist $t' \in \mathbb {R}$ and $w \in GX$ such that $w \in g^{t} \widetilde {\mathfrak {w}}^{\gamma } \cap g^{t'} \widetilde {\mathfrak {w}}^{\gamma '}$ . Then

$$ \begin{align*} w &\in (g^{t} \widetilde{\mathfrak{v}} \cap \gamma \widetilde{\mathfrak{v}}) \cap (g^{t'} \widetilde{\mathfrak{v}} \cap \gamma' g^{t' - t} \widetilde{\mathfrak{v}})= (g^{t} \widetilde{\mathfrak{v}} \cap g^{t'} \widetilde{\mathfrak{v}}) \cap (\gamma \widetilde{\mathfrak{v}} \cap \gamma' g^{t' - t} \widetilde{\mathfrak{v}}). \end{align*} $$

So $w \in g^{t} \widetilde {\mathfrak {v}} \cap g^{t'} \widetilde {\mathfrak {v}}$ , hence $| t' - t | \le 3\varepsilon $ by definition of $\widetilde {\mathfrak {v}}$ . Then also

$$ \begin{align*} \gamma^{-1} w &\in \widetilde{\mathfrak{v}}_{\varepsilon,\delta} \cap \gamma^{-1} \gamma' g^{t' - t} \widetilde{\mathfrak{v}}_{\varepsilon,\delta} \subset g^{-\varepsilon_0} \widetilde{\mathfrak{v}}_{3\varepsilon_0,\delta_0} \cap \gamma^{-1} \gamma' g^{-\varepsilon_0} \widetilde{\mathfrak{v}}_{3\varepsilon_0,\delta_0}, \end{align*} $$

which is empty by the previous paragraph unless $\gamma ^{-1} \gamma ' = \operatorname {\mathrm {id}}$ . Therefore $\gamma = \gamma '$ .

Proof Property (1) follows from continuity of $\pi _p$ , (2) and (4) from Lemma 2.1, (3) from the definitions, and (5) from Lemma 4.2.

Proof Since $m( \mathfrak {v}_{\varepsilon ,\delta })> 0$ , we have $ N^{\mathrm{unclipped}} = 0$ if and only if $\mu (\operatorname {\mathrm {E}}( \widehat {\mathfrak {w}}^{\mathrm{unclipped}})) = 0$ . Thus the lemma is trivial if $\mu (\operatorname {\mathrm {E}}( \widehat {\mathfrak {w}}^{\mathrm{unclipped}})) = 0$ . So assume $\mu (\operatorname {\mathrm {E}}( \widehat {\mathfrak {w}}^{\mathrm{unclipped}}))> 0$ . Start with the identity

$$ \begin{align*} N^{\mathrm{unclipped}} = \sum_{\gamma \in I^{\mathrm{unclipped}}} 1 = \sum_{\gamma \in I^{\mathrm{unclipped}}} \frac{1}{\mu(\operatorname{\mathrm{E}}(\widehat{\mathfrak{w}}^\gamma))} \int_{\operatorname{\mathrm{E}}(\widehat{\mathfrak{w}}^\gamma)}\, d\mu. \end{align*} $$

By Corollary 6.9,

$$ \begin{align*} \frac{1}{ C_{\varepsilon,\delta}} \le e^{-ht} \frac{\mu(\operatorname{\mathrm{E}}( \mathfrak{v}))}{\mu(\operatorname{\mathrm{E}}(\widehat{\mathfrak{w}}^\gamma))} \le C_{\varepsilon,\delta} \end{align*} $$

for $\gamma $ unclipped, so

$$ \begin{align*} \frac{1}{ C_{\varepsilon,\delta}} \mu(\operatorname{\mathrm{E}}( \widehat{\mathfrak{w}}^{\mathrm{unclipped}})) &= \! \sum_{\gamma \in I^{\mathrm{unclipped}}} \frac{1}{ C_{\varepsilon,\delta}} \int_{\operatorname{\mathrm{E}}(\widehat{\mathfrak{w}}^\gamma)}\, d\mu \\ &\le e^{-ht} \mu(\operatorname{\mathrm{E}}( \mathfrak{v})) N^{\mathrm{unclipped}} \\ &\le \! \sum_{\gamma \in I^{\mathrm{unclipped}}} C_{\varepsilon,\delta} \int_{\operatorname{\mathrm{E}}(\widehat{\mathfrak{w}}^\gamma)}\, d\mu \\ &= C_{\varepsilon,\delta} \, \mu(\operatorname{\mathrm{E}}( \widehat{\mathfrak{w}}^{\mathrm{unclipped}})).\\[-37pt] \end{align*} $$

Proof Let $\alpha = \delta - r> 0$ . By Corollary 3.5, there exists $t_1 \ge 0$ such that for all $t \ge t_1$ and $\gamma \in I_{r,t}$ (that is, $ \mathfrak {v}_r \cap \gamma g^{-t} \mathfrak {v}_r \neq \varnothing $ ), $\gamma \mathfrak {v}_\delta ^+ \subseteq {B_{\rho }} ( \mathfrak {v}_r^+, \alpha ) = \mathfrak {v}_\delta ^+$ . Similarly, there exists $t_2 \ge 0$ such that for all $t \ge t_2$ and $\gamma \in I_{r,t}$ (that is, $ \mathfrak {v}_r \cap \gamma ^{-1} g^t \mathfrak {v}_r \neq \varnothing $ ), $\gamma ^{-1} \mathfrak {v}_\delta ^- \subseteq {B_{\rho }} ( \mathfrak {v}_r^-, \alpha ) = \mathfrak {v}_\delta ^-$ . So for all $t \ge t_0 = \max \{t_1, t_2\}$ , if $\gamma \in I_{r,t}$ then $\gamma \in I^{\text {unclipped}}_{\delta ,t}$ .

Proof By Corollary 5.17, $\lim _{t \to \infty } \mu (\operatorname {\mathrm {E}}( \mathfrak {w}_{r,t})) = ({2}/{\varepsilon }) m( \mathfrak {v}_r)^2$ for all $r \in (0, \delta _0]$ . Hence $\delta $ is a point of continuity of the function $f(r) = \lim _{t \to \infty } \mu (\operatorname {\mathrm {E}}( \mathfrak {w}_{r,t}))$ . So by Corollary 8.2,

$$ \begin{align*} &\lim_{t \to \infty} \mu(\operatorname{\mathrm{E}}( \mathfrak{w}^{\mathrm{unclipped}}_{\delta,t})) = \lim_{t \to \infty} \mu(\operatorname{\mathrm{E}}( \mathfrak{w}_{\delta,t}))\\ &\quad= \frac{2}{\varepsilon} m( \mathfrak{v}_{\delta})^2 \le \lim_{t \to \infty} \mu(\operatorname{\mathrm{E}}( \widehat{\mathfrak{w}}^{\mathrm{unclipped}}_{\delta,t})) \le \lim_{t \to \infty} \mu(\operatorname{\mathrm{E}}(\widehat{\mathfrak{w}}_{\delta,t}))\\ &\quad= \frac{2}{\widehat{\varepsilon}} m(\widehat{\mathfrak{v}}_{\delta})^2 = \frac{2\widehat{\varepsilon}}{\varepsilon^2} m( \mathfrak{v}_{\delta})^2. \end{align*} $$

But now

$$ \begin{align*} \frac{1}{ C_{\varepsilon,\delta}} \le {\widetilde{\lim_{t \to \infty}}} \frac{ N^{\mathrm{unclipped}}_{\delta,t}}{2 e^{ht} m( \mathfrak{v}_{\delta})} \le \frac{\widehat{\varepsilon}}{\varepsilon} C_{\varepsilon,\delta} \end{align*} $$

by Lemma 7.1 because $\mu (\operatorname {\mathrm {E}}( \mathfrak {v})) = {m( \mathfrak {v})}/{\varepsilon }$ .

Proof Whenever $\delta ' \in (\delta , \delta _0]$ , we find $ N^{\mathrm{unclipped}}_{\delta ,t} \le N_{\delta ,t} \le N^{\mathrm{unclipped}}_{\delta '\!,t} \le N_{\delta '\!,t}$ for all t sufficiently large by Lemma 8.1, hence

$$ \begin{align*}\phi(r) = \limsup_{t \to \infty} e^{-ht} N^{\mathrm{unclipped}}_{r,t} \quad \text{and} \quad \psi(r) = \limsup_{t \to \infty} e^{-ht} N_{r,t}\end{align*} $$

satisfy $\phi (\delta ) \le \psi (\delta ) \le \phi (\delta ') \le \psi (\delta ')$ . Taking a decreasing sequence $\delta ^{\prime }_n \to \delta $ such that each $\delta ^{\prime }_n> \delta $ is a point of continuity of $r \mapsto m( \mathfrak {v}_r)$ , we find by Lemma 8.3 that

$$ \begin{align*}\frac{2 m( \mathfrak{v}_{\delta})}{ C_{\varepsilon,\delta}} \le \liminf_{t \to \infty} e^{-ht} N^{\mathrm{unclipped}}_{\delta,t} \le \liminf_{t \to \infty} e^{-ht} N_{\delta,t}\end{align*} $$

and

$$ \begin{align*}\psi(\delta) \le \liminf_{n \to \infty} \phi(\delta^{\prime}_n) \le \liminf_{n \to \infty} 2 m(\widehat{\mathfrak{v}}_{\delta^{\prime}_n}) C_{\varepsilon,\delta^{\prime}_n} = 2 m(\widehat{\mathfrak{v}}_{\delta}) C_{\varepsilon,\delta}. \\[-40pt] \end{align*} $$

Proof Let $t> 0$ and $\gamma \in I_{\varepsilon ,\delta ,t}^{\mathrm{unclipped}}$ . Since $\gamma \mathfrak {v}^+ \subseteq \mathfrak {v}^+$ , the nested intersection $\bigcap _{n \in \mathbb {N}} \gamma ^n \mathfrak {v}^+$ of compact sets must contain a point $\xi \in \partial X$ . Similarly, the nested intersection $\bigcap _{n \in \mathbb {N}} \gamma ^{-n} \mathfrak {v}^-$ must contain a point $\eta \in \partial X$ . Since $\operatorname {\mathrm {E}}( \mathfrak {v}) = \mathfrak {v}^- \times \mathfrak {v}^+$ , there is some geodesic $v \in \mathfrak {v}$ with endpoints $\operatorname {\mathrm {E}}(v) = (\eta , \xi )$ . We may assume v is the central geodesic in its parallel set—that is, $v(0)$ is the circumcenter of $CS(v)$ —so then $\gamma $ must stabilize $g^{\mathbb {R}} v$ (as a set). Thus $\gamma $ must act on v by $\gamma v = g^{t'} v$ for some $t' \in \mathbb {R}$ . By Lemma 6.4, $| t' - t | \le \varepsilon + 3\operatorname {\mathrm {diam}} \pi ( \mathfrak {v}_{0,\delta })$ . It follows from Lemma 4.2 that $t'> 0$ . Thus $ \mathfrak {v}$ contains an axis for $\gamma $ .

Proof Let $v \in \mathfrak {v}_{\varepsilon ,\delta }$ be an axis for $\gamma \in \Gamma $ , that is, $\gamma v = g^{| \gamma |} v$ , and observe that $t + \tau _\gamma (v) = | \gamma |$ . By Lemma 6.4 we see that if then $\tau _\gamma (v) \in [-\varepsilon , \varepsilon ]$ ; this proves the first inclusion. For the second, it suffices to show $\tau _\gamma (v) \in [-\varepsilon , \varepsilon ]$ implies $\gamma \in I_{\varepsilon ,\delta ,t}$ . So assume $\tau _\gamma (v) \in [-\varepsilon , \varepsilon ]$ . We may also assume $s(v) = 0$ . If $\tau _{\gamma }(v) \in [-\varepsilon , 0]$ , then $\gamma ^{-1} g^t v = g^{-\tau _{\gamma }(v)} v \in \mathfrak {v}_{\varepsilon ,\delta }$ , and therefore $v \in \mathfrak {v} \cap g^{-t} \gamma \mathfrak {v}$ . If $\tau _{\gamma }(v) \in [0, \varepsilon ]$ , then for $w = g^{\tau _{\gamma }(v)} v \in \mathfrak {v}_{\varepsilon ,\delta }$ we find $\gamma ^{-1} g^t w = g^{-\tau _{\gamma }(v)} w = v$ , and therefore $w \in \mathfrak {v} \cap g^{-t} \gamma \mathfrak {v}$ . Thus in either case $ \mathfrak {v} \cap g^{-t} \gamma \mathfrak {v}$ is not empty, hence $\tau _{\gamma }(v) \in [-\varepsilon , \varepsilon ]$ implies $\gamma \in I_{\varepsilon ,\delta ,t}$ .

Proof By Lemma 9.2, $ N_{\varepsilon ,\delta ,t}^{\mathrm{unclipped}} \le N_{\varepsilon ,\delta ,t}^{\mathrm{periodic}} \le N_{\varepsilon ,\delta ,t}$ for all sufficiently large t, hence

$$ \begin{align*}\liminf_{t \to \infty} \frac{ N_{\varepsilon,\delta,t}^{\mathrm{unclipped}}}{2e^{ht} m( \mathfrak{v}_{\varepsilon,\delta})} \le {\widetilde{\lim_{t \to \infty}}} \frac{ N_{\varepsilon,\delta,t}^{\mathrm{periodic}}}{2e^{ht} m( \mathfrak{v}_{\varepsilon,\delta})} \le \limsup_{t \to \infty} \frac{ N_{\varepsilon,\delta,t}}{2e^{ht} m( \mathfrak{v}_{\varepsilon,\delta})}.\end{align*} $$

Now apply the bounds from Lemma 8.3 and Lemma 8.5.

Proof The intersection segments for v with $ \mathfrak {v}_{\varepsilon ,\delta }$ are each of length $\varepsilon $ , and they are pairwise disjoint. Hence $\lambda _{\Gamma }^{v} (\operatorname {\mathrm {pr}} \mathfrak {v}_{\varepsilon ,\delta }) = \varepsilon \cdot \operatorname {\mathrm {S}}^{ \mathfrak {v}_{\varepsilon ,\delta }}(v)$ .

Proof Lemma 11.1 gives us

$$ \begin{align*} \sum_{[\gamma] \in \mathfrak C_\Gamma^{U} (t - \varepsilon, t + \varepsilon)} \operatorname{\mathrm{S}}^{ \mathfrak{v}_{\varepsilon,\delta}}({\mathfrak a}[\gamma]) = \frac{t}{\varepsilon} {\mathrm{Conj}}_\Gamma^{U} (t - \varepsilon, t + \varepsilon) \tilde{\lambda}_{{\mathfrak a},t,\varepsilon}^{{\mathrm{mult}}, U} (\operatorname{\mathrm{pr}} \mathfrak{v}_{\varepsilon,\delta}), \end{align*} $$

and we apply Lemma 10.1.

Proof Combine Proposition 9.4 and Lemma 11.2, and observe that

Proof By definition,

$$ \begin{align*} \lambda_{{\mathfrak a}, t + \alpha, 2 \alpha}^{{\mathrm{mult}}, U} = \frac{1}{{\mathrm{Conj}}_\Gamma^{U} (t - \alpha, t + \alpha)} \sum_{[\gamma] \in \mathfrak C_\Gamma^{U} (t - \alpha, t + \alpha)} \frac{1}{| \lambda_{\Gamma}^{{\mathfrak a}[\gamma]} |} \lambda_{\Gamma}^{{\mathfrak a}[\gamma]} \end{align*} $$

and

$$ \begin{align*} \tilde{\lambda}_{{\mathfrak a},t,\alpha}^{{\mathrm{mult}}, U} = \frac{1}{{\mathrm{Conj}}_\Gamma^{U} (t - \alpha, t + \alpha)} \sum_{[\gamma] \in \mathfrak C_\Gamma^{U} (t - \alpha, t + \alpha)} \frac{1}{t} \lambda_{\Gamma}^{{\mathfrak a}[\gamma]}. \end{align*} $$

Since $t - \alpha \le | \lambda _{\Gamma }^{{\mathfrak a}[\gamma ]} | \le t + \alpha $ for all $\lambda _{\Gamma }^{{\mathfrak a}[\gamma ]} \in \mathfrak C_\Gamma ^{U} (t - \alpha , t + \alpha )$ , we see that for $t> \alpha $ ,

$$ \begin{align*} \frac{t}{t + \alpha} \tilde{\lambda}_{{\mathfrak a},t,\alpha}^{{\mathrm{mult}}, U} (V) \le \lambda_{{\mathfrak a}, t + \alpha, 2 \alpha}^{{\mathrm{mult}}, U} (V) \le \frac{t}{t - \alpha} \tilde{\lambda}_{{\mathfrak a},t,\alpha}^{{\mathrm{mult}}, U} (V) \end{align*} $$

for all Borel sets $V \subseteq GX$ . The conclusion of the lemma follows.

Proof By Corollary 11.3 and Lemma 11.4,

The fact that $\lambda _{{\mathfrak a}, t + \varepsilon , 2 \varepsilon }^{{\mathrm{mult}}, \mathfrak {v}_{\varepsilon ,\delta }}$ is a probability measure gives us the desired inequality.

Proof By Proposition 5.2, there is some zero-width $v_0 \in V$ . By Lemma 2.1, there exist $\delta> 0$ and $\varepsilon> 0$ such that $ \mathfrak {v}_{\varepsilon ,\delta } = \mathfrak {v}(v_0,\varepsilon ,\delta )$ is completely contained in $\mathcal R \cap V$ . We may assume $\varepsilon \le \min \{\alpha , \varepsilon _0\}$ and that $\delta \in (0, \delta _0]$ is chosen such that

by Corollary 11.5. Thus there exist $C> 0$ and $t_0> 0$ such that for all $t \ge t_0$ ,

$$ \begin{align*}C \frac{e^{ht}}{t} \le {\mathrm{Conj}}_\Gamma^{ \mathfrak{v}_{\varepsilon,\delta}} (t - \varepsilon, t + \varepsilon) \le {\mathrm{Conj}}_\Gamma^{U} (t - \varepsilon, t + \varepsilon) \le {\mathrm{Conj}}_\Gamma^{U} (t - \alpha, t + \alpha).\\[-40pt] \end{align*} $$

Proof By Lemma 11.7, there exist $C'> 0$ and $t^{\prime }_0> 0$ such that for all $t \ge t^{\prime }_0$ ,

$$ \begin{align*}{\mathrm{Conj}}_\Gamma^{U} \bigg(t - \frac{\alpha}{2}, t + \frac{\alpha}{2}\bigg) \ge C' \cdot \frac{e^{ht}}{t}.\end{align*} $$

So let $C = C' \cdot e^{{-h \alpha } / {2}}$ and $t_0 = t^{\prime }_0 + {\alpha }/{2}$ . Then for all $t \ge t_0$ ,

$$ \begin{align*}{\mathrm{Conj}}_\Gamma^{U} (t - \alpha, t) \ge C' \cdot \frac{e^{h (t - {\alpha}/{2})}}{t - {\alpha}/{2}} = C \cdot \frac{t}{t - {\alpha}/{2}} \cdot \frac{e^{ht}}{t} \ge C \frac{e^{ht}}{t}. \\[-43pt] \end{align*} $$

Proof Consider that for $\gamma \in \Gamma $ with an axis in K, we know $d(\gamma p, p) \le | \gamma | + 2 \operatorname {\mathrm {diam}} \pi (K)$ , and therefore for all $h'> h$ ,

$$ \begin{align*} \sum_{t> 0} e^{-h't} {\mathrm{Conj}}_\Gamma^{K}(t,t) &= \sum_{\substack{[\gamma] \in \mathfrak C(\Gamma) \\ \text{with an axis in }K}} e^{-h' | \gamma |} \le \sum_{\substack{\gamma \in \Gamma \\ \text{with an axis in }K}} e^{-h' | \gamma |} \\ &\le \sum_{\substack{\gamma \in \Gamma \\ \text{with an axis in }K}} e^{-h'\, d(\gamma p, p) + 2 h' \operatorname{\mathrm{diam}} \pi(K)} \\ &\le e^{2 h' \operatorname{\mathrm{diam}} \pi(K)} \sum_{\substack{\gamma \in \Gamma \\ \text{with an axis in }K}} e^{-h'\, d(\gamma p, p)} \end{align*} $$

converges because h is the critical exponent of the Poincaré series for Patterson’s construction. It follows that $\lim _{t \to \infty } e^{-h't} {\mathrm{Conj}}_\Gamma ^{K} (0,t) = 0$ .

Proof Let $\alpha> 0$ . By Corollary 11.8, there exist $C> 0$ and $t^{\prime }_0> 0$ such that

$$ \begin{align*}{\mathrm{Conj}}_\Gamma^{U} (t - \alpha, t) \ge C \frac{e^{ht}}{t}\end{align*} $$

for all $t \ge t^{\prime }_0$ . Now by definition, ${\mathrm{Conj}}_\Gamma ^{K} (0,{t}/{2}) = {\mathrm{Conj}}_\Gamma ^{\Gamma K} (0,{t}/{2}) \ge {\mathrm{Conj}}_\Gamma ^{U} (0,{t}/{2})$ . Since $\lim _{t \to \infty } 2t e^{-({3}/{2}) ht} {\mathrm{Conj}}_\Gamma ^{K} (0,t) = 0$ by Lemma 12.1, there exists $t_0 \ge t^{\prime }_0$ such that

$$ \begin{align*}{\mathrm{Conj}}_\Gamma^{U} \bigg(0,\frac{t}{2}\bigg) \le {\mathrm{Conj}}_\Gamma^{K} \bigg(0,\frac{t}{2}\bigg) \le C \cdot \frac{e^{({3}/{4}) ht}}{t}\end{align*} $$

for all $t \ge t_0$ . Since every imprimitive $[\gamma ] \in \mathfrak C_\Gamma ^{U} (t - \alpha , t) ~\backslash~ \mathfrak C_\Gamma ^{{\mathrm{prime}}, U} (t - \alpha , t)$ is a multiple of some $[\phi ] \in \mathfrak C_\Gamma ^{U} (0, {t}/{2})$ , we see that

$$ \begin{align*}0 \le {\mathrm{Conj}}_\Gamma^{U} (t - \alpha, t) - {\mathrm{Conj}}_\Gamma^{{\mathrm{prime}}, U} (t - \alpha, t) \le {\mathrm{Conj}}_\Gamma^{U} (0,{t}/{2}).\end{align*} $$

Thus

$$ \begin{align*}{\mathrm{Conj}}_\Gamma^{U} (t - \alpha, t) \ge {\mathrm{Conj}}_\Gamma^{{\mathrm{prime}}, U} (t - \alpha, t) \ge {\mathrm{Conj}}_\Gamma^{U} (t - \alpha, t) - {\mathrm{Conj}}_\Gamma^{U} (0,{t}/{2})\end{align*} $$

and therefore

$$ \begin{align*}1 \ge \frac{{\mathrm{Conj}}_\Gamma^{{\mathrm{prime}}, U} (t - \alpha, t)}{{\mathrm{Conj}}_\Gamma^{U} (t - \alpha, t)} \ge 1 - \frac{{\mathrm{Conj}}_\Gamma^{U} (0,{t}/{2})}{{\mathrm{Conj}}_\Gamma^{U} (t - \alpha, t)} \ge 1 - e^{-({1}/{4}) ht}. \\[-44pt] \end{align*} $$

Proof Let W be a Borel subset of $GX$ . By the definitions,

$$ \begin{align*} \lambda_{{\mathfrak a},t,\alpha}^{{\mathrm{mult}}, GX}(W) = \frac{1}{{\mathrm{Conj}}_\Gamma^{U} (t - \alpha, t)} \sum_{[\gamma] \in \mathfrak C_\Gamma (t - \alpha, t)} \frac{1}{\| \lambda_{\Gamma}^{{\mathfrak a}[\gamma]} \|} \lambda_{\Gamma}^{{\mathfrak a}[\gamma]}(W) \end{align*} $$

and

$$ \begin{align*} \lambda_{{\mathfrak a},t,\alpha}^{{\mathrm{prime}}, GX}(W) = \frac{1}{{\mathrm{Conj}}_\Gamma^{{\mathrm{prime}}, U} (t - \alpha, t)} \sum_{[\gamma] \in \mathfrak C_\Gamma^{\mathrm{prime}} (t - \alpha, t)} \frac{1}{\| \lambda_{\Gamma}^{{\mathfrak a}[\gamma]} \|} \lambda_{\Gamma}^{{\mathfrak a}[\gamma]}(W). \end{align*} $$

The outside coefficients are asymptotically equal (and non-zero), and the difference in the sums is at most ${\mathrm{Conj}}_\Gamma ^{U} (t - \alpha , t) - {\mathrm{Conj}}_\Gamma ^{{\mathrm{prime}}, U} (t - \alpha , t)$ , which is asymptotically zero compared to ${\mathrm{Conj}}_\Gamma ^{U} (t - \alpha , t)$ by Lemma 12.2.

Proof For any fixed $a \in \mathbb {R}$ and $m \in \mathbb {Z}$ ,

$$ \begin{align*} \lim\limits_{t \to \infty} \frac{\int_{0}^{t} F(x) \; dx}{\int_{a}^{t} F(x) \; dx} = 1 \qquad \text{and} \qquad \lim\limits_{t \to \infty} \frac{\sum_{k = 0}^{\lfloor {t}/{2\varepsilon} \rfloor} 2\varepsilon F(t - (2k+1) \varepsilon)} {\sum_{k = 0}^{\lfloor {t}/{2\varepsilon} \rfloor - m} 2\varepsilon F(t - (2k+1) \varepsilon)} = 1, \end{align*} $$

so without loss of generality we may assume F is positive and non-decreasing on $[0, \infty )$ . We may similarly assume, for $\alpha> 0$ fixed, that $1 \le ({F(x + \varepsilon )}/{F(x)}) \le C + \alpha $ for all $x> -2\varepsilon $ . Let $t> 0$ and put $n = \lfloor {t}/{2\varepsilon } \rfloor $ . For each $k = 0, 1, 2, \dotsc , n$ , we have

$$ \begin{align*} \frac{1}{C + \alpha} F(t - (2k+1)\varepsilon) \le F(x) \le (C + \alpha) F(t - (2k+1)\varepsilon) \end{align*} $$

for all $x \in [t - (2k+2)\varepsilon , t - 2k\varepsilon ]$ . Thus

$$ \begin{align*} \frac{1}{C + \alpha} 2\varepsilon F(t - (2k+1)\varepsilon) \le \int_{t - (2k+2)\varepsilon}^{t - 2k\varepsilon} F(x) \; dx \le (C + \alpha) 2\varepsilon F(t - (2k+1)\varepsilon) \end{align*} $$

for each $k = 0, 1, 2, \dotsc , n$ , and therefore

$$ \begin{align*} \frac{1}{C + \alpha} \sum_{k = 0}^{n-1} 2\varepsilon F(t - (2k+1)\varepsilon) &\le \int_{2\varepsilon}^{t} F(x) \; dx \le \int_{0}^{t} F(x) \; dx \\ &\le \int_{-2\varepsilon}^{t} F(x) \; dx \le (C + \alpha) \sum_{k = 0}^{n} 2\varepsilon F(t - (2k+1)\varepsilon). \end{align*} $$

But

$$ \begin{align*} \lim_{t \to \infty} \frac {\sum_{k = 0}^{\lfloor {t}/{2\varepsilon} \rfloor} 2\varepsilon F(t - (2k+1)\varepsilon)} {\sum_{k = 0}^{\lfloor {t}/{2\varepsilon} \rfloor-1} 2\varepsilon F(t - (2k+1)\varepsilon)} = 1, \end{align*} $$

so

$$ \begin{align*} \frac{1}{C + \alpha} &\le {\widetilde{\lim_{t \to \infty}}} \frac {\int_{0}^{t} F(x) \; dx} {\sum_{k = 0}^{\lfloor {t}/{2\varepsilon} \rfloor} 2\varepsilon F(t - (2k+1)\varepsilon)} \le C + \alpha. \end{align*} $$

As $\alpha> 0$ was arbitrary, we find

$$ \begin{align*} \frac{1}{C} \le {\widetilde{\lim_{t \to \infty}}} \frac{\int_{0}^{t} F(x) \; dx}{\sum_{k = 0}^{n} 2\varepsilon F(t - (2k+1)\varepsilon)} \le C. \\[-44pt] \end{align*} $$

Proof It is a standard fact that for any fixed $t_0> 0$ ,

(1) $$ \begin{align} \lim\limits_{t \to \infty} {\int_{t_0}^{t} \frac{e^{hx}}{x} \; dx} \bigg/ { }{\frac{e^{ht}}{ht}} = 1. \end{align} $$

This comes from the calculation

$$ \begin{align*} \int_{t_0}^{t} \frac{e^{hx}}{x} \; dx = \frac{e^{hx}}{hx} \bigg \vert_{t_0}^{t} + \int_{t_0}^{t} \frac{e^{hx}}{hx^2} \; dx = \frac{e^{ht}}{ht} - \frac{e^{h t_0}}{h t_0} + \int_{t_0}^{t} \frac{e^{hx}}{hx^2} \; dx; \end{align*} $$

the second term of the last expression tends to zero relative to ${e^{ht}} / {ht}$ because it is constant, the third because $\lim \nolimits _{x \to \infty } {({e^{hx}}/{hx^2})} / {({e^{hx}}/{x})} = 0$ . On the other hand, for all $\varepsilon> 0$ , Lemma 13.1 gives us

$$ \begin{align*} e^{-h \varepsilon} \le {\widetilde{\lim_{t \to \infty}}} {\sum_{k = 0}^{\lfloor {t}/{2\varepsilon} \rfloor} 2\varepsilon \frac{e^{h(t - (2k+1) \varepsilon)}}{t - (2k+1) \varepsilon}} \bigg/ { }{\int_{t_0}^{t} \frac{e^{hx}}{x} \; dx} \le e^{h \varepsilon} \end{align*} $$

and therefore

$$ \begin{align*} e^{-h \varepsilon} \le {\widetilde{\lim_{t \to \infty}}} {\sum_{k = 0}^{\lfloor {t}/{2\varepsilon} \rfloor} 2\varepsilon \frac{e^{h(t - (2k+1) \varepsilon)}}{t - (2k+1) \varepsilon}} \bigg/ { }{\frac{e^{ht}}{ht}} \le e^{h \varepsilon} \end{align*} $$

from (1).

Proof Let ${\mathfrak a}$ be a choice of axis, and let $0 < \alpha < e^{-{1}/{2}} \operatorname {\mathrm {injrad}}(\Gamma \backslash X)$ . Let $\gamma _1, \gamma _2 \in \Gamma $ represent distinct conjugacy classes $[\gamma _1], [\gamma _2] \in P$ . Let $v = {\mathfrak a}[\gamma _1]$ and $w = {\mathfrak a}[\gamma _2]$ , and write $\bar v = \operatorname {\mathrm {pr}} v$ and $\bar w = \operatorname {\mathrm {pr}} w$ . We may assume, replacing w by $\gamma w$ and $\gamma _2$ by $\gamma \gamma _2 \gamma ^{-1}$ (for some $\gamma \in \Gamma $ ) if necessary, that $d(\bar v, \bar w) = d(v,w)$ .

Write $n = \lceil t_0 \rceil $ . Suppose, by way of contradiction, that $d(g_\Gamma ^k \bar v, g_\Gamma ^k \bar w) \le \alpha $ for all $k = 0, 1, 2, \dotsc , n$ . For each $k \in \mathbb {Z}$ , let $\varphi _k \in \Gamma $ satisfy $d(g^k v, \varphi _k g^k w) = d(g_\Gamma ^k \bar v, g_\Gamma ^k \bar w)$ . (Note we assumed above that $\varphi _0 = \operatorname {\mathrm {id}}$ .) Then

$$ \begin{align*}e^{-{1}/{2}} d(v(k \pm \tfrac{1}{2}), \varphi_k w(k \pm \tfrac{1}{2})) \le d(g^k v, \varphi_k g^k w) = d(g_\Gamma^k \bar{v}, g_\Gamma^k \bar{w}) \le \alpha\end{align*} $$

for all $k = 0, 1, 2, \dotsc , n$ , hence $d(v(k \pm \tfrac {1}{2}), \varphi _k w(k \pm \tfrac {1}{2})) \le e^{{1}/{2}} \alpha $ . So

$$ \begin{align*}d(\varphi_k w(k + \tfrac{1}{2}), \varphi_{k + 1} w(k + \tfrac{1}{2})) \le 2e^{{1}/{2}} \alpha < 2\operatorname{\mathrm{injrad}}(\Gamma \backslash X),\end{align*} $$

and therefore $\varphi _n = \varphi _{n - 1} = \dotsb = \varphi _{0} = \operatorname {\mathrm {id}}$ . It follows that $d(v(k), w(k)) = d(\bar v(k), \bar w(k)) \le \alpha $ for all $k = 0, 1, 2, \dotsc , n$ . Thus $d(v(t), w(t)) \le \alpha $ for all $t \in [0,t_0]$ by convexity.

Find $t_1, t_2 \in [t_0 - \alpha , t_0]$ such that $\gamma _1 v = g^{t_1} v$ and $\gamma _2 w = g^{t_2} w$ . Then

$$ \begin{align*}d(\gamma_2^{-1} \gamma_1 v(0), w(0)) = d(\gamma_2^{-1} v(t_1), \gamma_2^{-1} w(t_2)) = d(v(t_1), w(t_2)) \le 2\alpha.\end{align*} $$

Hence $d(\gamma _2^{-1} \gamma _1 v(0), v(0)) \le 3\alpha < 2\operatorname {\mathrm {injrad}}(\Gamma \backslash X)$ , which is only possible if $\gamma _2^{-1} \gamma _1$ is trivial. This contradicts our hypothesis that $[\gamma _1]$ and $[\gamma _2]$ are distinct. Therefore, there must be some $k \in \{0, 1, 2, \dotsc , n\}$ such that $d(g_\Gamma ^k \bar v, g_\Gamma ^k \bar w)> \alpha $ , and thus we see that $\operatorname {\mathrm {pr}}({\mathfrak a}(P))$ is $(n,\alpha )$ -separated.

Proof By compactness of the space of $g^t$ -invariant Borel probability measures on $\Gamma \backslash GX$ under the weak* topology, every subsequence $(\nu _{k_j})$ has at least one weak* accumulation point $\nu $ of $\{\nu _k\}$ . By uniqueness of the measure of maximal entropy, it suffices to prove that every such $\nu $ is a measure of maximal entropy for $g_\Gamma ^t$ .

Let $\nu $ be a weak* accumulation point of $\{\nu _k\}$ ; passing to a subsequence if necessary, we may assume $\nu _k \to \nu $ in the weak* topology. Fix a measurable partition ${\mathcal A} = \{A_1, \dotsc , A_m\}$ of $\Gamma \backslash GX$ such that $\delta := \operatorname {\mathrm {diam}} {\mathcal A} < \varepsilon $ and $\nu (\partial A_i) = 0$ . Let $n_k = \lceil t_k \rceil $ and $\phi = g^1$ . Since the closed geodesics in $\operatorname {\mathrm {pr}}({\mathfrak a}(P_k))$ are $(n_k,\varepsilon )$ -separated by Lemma 14.2, they are also $(n_k,\delta )$ -separated. But by construction, no two geodesics in any one $\alpha \in {\mathcal A}_\phi ^{(n_k + 1)}$ are $(n_k,\delta )$ -separated, hence each $\alpha \in {\mathcal A}_\phi ^{(n_k + 1)}$ touches at most one geodesic from $\operatorname {\mathrm {pr}}({\mathfrak a}(P_k))$ . But Lemma 14.2 holds for arbitrary choice of axis, including flowing each geodesic in ${\mathfrak a}(P_k)$ by a different amount; hence each $\alpha \in {\mathcal A}_\phi ^{(n_k + 1)}$ touches at most one orbit from $\operatorname {\mathrm {pr}}({\mathfrak a}(P_k))$ . Thus $\nu _k(\alpha ) \le {1}/{\# {P_k}}$ . Therefore the entropy

$$ \begin{align*} H_{\nu_k}({\mathcal A}_\phi^{(n_k + 1)}) = \sum_{\alpha \in {\mathcal A}_\phi^{(n_k + 1)}} - \nu_k(\alpha) \log \nu_k(\alpha) \ge \sum_{\alpha \in {\mathcal A}_\phi^{(n_k + 1)}} \nu_k(\alpha) \log \# {P_k} = \log \# {P_k}. \end{align*} $$

Since $\nu (\partial A_i) = 0$ for all $A_i \in {\mathcal A}$ , we have $H_{\nu _k}({\mathcal A}_\phi ^{(q)}) \to H_{\nu }({\mathcal A}_\phi ^{(q)})$ and thus

$$ \begin{align*} h_{\nu}(\phi) \ge h_{\nu}(\phi, {\mathcal A}) = \lim_{q \to \infty} \frac{H_{\nu} ({\mathcal A}_\phi^{(q)})}{q} = \lim_{q \to \infty} \lim_{k \to \infty} \frac{H_{\nu_k} ({\mathcal A}_\phi^{(q)})}{q}. \end{align*} $$

By Lemma 14.1 and the inequality $H_{\nu _k}({\mathcal A}_\phi ^{(n_k + 1)}) \ge \log \# {P_k}$ from above,

$$ \begin{align*} \lim_{q \to \infty} \lim_{k \to \infty} \frac{H_{\nu_k} ({\mathcal A}_\phi^{(q)})}{q} \ge {\widetilde{\lim_{k \to \infty}}} \frac{H_{\nu_k} ({\mathcal A}_\phi^{(n_k + 1)})}{n_k} \ge \lim_{k \to \infty} \frac{\log \# {P_k}}{t_k} = h. \end{align*} $$

Therefore $h_{\nu }(\phi ) \ge h$ , which shows that $\nu $ is a measure of maximal entropy. Because $m_\Gamma $ is the unique such probability measure by Proposition 5.3, we have $\nu = m_\Gamma $ .