Proof of Proposition 4.2

Since $d(N z_1,N z_2)\leq d(z_1,z_2)$ for any $z_1,z_2\in \mathbb {D}$ , we have $\|gN\|\leq \|g\|$ for all $g\in G$ . Thus, on the one hand, we have

On the other hand, we take $g^{(k)}\in G$ satisfying $\chi (g^{(k)})=k$ and $d(N 0,N g^{(k)}(0))=d(0,g^{(k)}(0))$ for each positive integer k. From (4.1), we have $\|g^{(k)}\|\leq d(0,g^{(k)}0)$ for every positive integer k. Hence, we have

Therefore, $\unicode{x3bb} _{G|N}=\inf _{g\in G/N}\|g\|/|\chi (g)|$ .

Next, we prove the last claim in Proposition 4.2. Suppose that $\Lambda _p(G)=\Lambda _p(N)$ . Note that $g^{(m)}\neq g^{(n)}$ for distinct positive integers m and n because they are in disjoint cosets of N. As G contains no elliptic elements, we have $g^{(m)}(0)\neq g^{(n)}(0)$ . Since the G-orbit of $0$ is a discrete subset of $\mathbb {D}$ , we thus deduce that is unbounded. By Proposition 4.1, there exists a constant C such that $\|g^{(n)}N\|\geq d(N 0,N g^{(n)}0)-C=d(0,g^{(n)}(0))-C$ for every positive integer n. Consequently,

(4.4) $$ \begin{align} \sup_{n\geq1} \|g^{(n)}N\|\geq \sup_{n\geq1} d(0,g^{(n)}(0))-C=+\infty. \end{align} $$

For each positive integer n, we have $\|g^{(n)}N\|=\unicode{x3bb} _{G|N}\cdot |\chi (g^{(n)})|=n\unicode{x3bb} _{G|N}$ . From this fact and (4.4), we deduce that $\unicode{x3bb} _{G|N}>0$ when $\Lambda _p(G)=\Lambda _p(N)$ .

If contains at least one parabolic element, say $g_p$ , then we have $\chi (g_p)\neq 0$ and $\|g_p\|=0$ . Thus, in this case,

It follows that $\unicode{x3bb} _{G|N}=0$ if contains at least one parabolic element.

Proof of Lemma 4.3

For any $a\in G_0$ , note that is by construction a closed subset of $\partial R_a$ and bounded away from both of the two endpoints of the geodesic $\partial R_a$ . Therefore, is compact for any $a\in G_0$ . Since $G_0$ is finite, is compact as well.

Let $g\in G$ . Suppose that $d(D,gD)>0$ and $a_pa_p$ is not an initial block of the reduced word representing g for any $a_p\in \Pi _0$ . We show . Note that $d(D,gD)=d(\partial D,g(\partial D))$ , $\partial D=\bigcup _{a\in G_0}\partial R_a$ , and $G_0$ is finite, so there are ${a,b\in G_0}$ such that $d(\partial R_a, g(\partial R_b))=d(D,gD)>0$ . Thus, there are further $z_1\in \partial R_a$ and ${z_2\in \partial R_b}$ such that

(4.6) $$ \begin{align} d(z_1,g(z_2))= d(\partial R_a, g(\partial R_b)); \end{align} $$

indeed, it is known that the geodesic segment $[z_1,g(z_2)]$ is perpendicular to both $\partial R_a$ and $g(\partial R_b)$ [Reference Beardon1, Ch. 7]. We claim that . As $z_1\in \partial R_a$ , we only need to show that $z_1\notin \bigcup _{x\in \mathfrak {E}_a}O^{\prime }_{a,x}$ . Take $x\in \mathfrak {E}_a$ arbitrarily. First, we consider the case where $x\notin \mathfrak {E}_{a'}$ for any . Since $z_2\in \partial R_b$ and $d(\partial R_a,g(\partial R_b))>0$ , by our construction, we have that $g(z_2)\notin O_{a,x}$ , which hence implies that $g(z_2)\notin O^{\prime }_{a,x}$ . Since the geodesic segment $[z_1,g(z_2)]$ and the geodesic $L_{a,x}$ are both perpendicular to $\partial R_a$ , we have $[z_1,g(z_2)]$ is disjoint from $L_{a,x}$ , showing that $z_1$ and $g(z_2)$ are on the same side of the geodesic $L_{a,x}$ . As we have seen that $g(z_2)\notin O^{\prime }_{a,x}$ , we deduce that $z_1\notin O^{\prime }_{a,x}$ .

If $x\in \mathfrak {E}_{a'}$ for some , then a is parabolic and x is the fixed point of a. Recall that for any parabolic $a_p\in G_0$ , $a_pa_p$ is assumed not to be an initial block of the reduced word representing g. Since $z_2\in \partial R_b$ , we have $g(z_2)\notin O_{a,x}$ . Repeating the same arguments we made when handling the previous case, we have that $z_1\notin O^{\prime }_{a,x}$ still holds when $x\in \mathfrak {E}_{a'}$ for some . Hence, holds in any case. As g is an isometry acting on $\mathbb {D}$ , (4.6) gives

$$ \begin{align*} d(z_2,g^{-1}(z_1))= d(\partial R_b, g^{-1}(\partial R_a)), \end{align*} $$

which further equals $d(D,g^{-1}D)>0$ . Hence, by symmetry, we also have that . Now that we have shown that and , we have

Since , we also have . Therefore, , provided that $d(D,gD)>0$ and $a_pa_p$ is not an initial block of the reduced word representing g for any $a_p\in \Pi _0$ .

Finally, we show the last claim. Fix an arbitrary $g\in G$ and pick $g'\in gN$ satisfying that

$$ \begin{align*} d(D,g'D) = d(ND,N gD). \end{align*} $$

If $d(ND,N gD)=0$ , then $d(D,g'D)=0$ , so $g'$ must be the identity or a side-pairing transformation in $G_0$ or a power of some parabolic $a_p\in G_0$ . If $g'$ is the identity or a power of some parabolic element of $G_0$ , then using our assumption that $\Lambda _p(G)=\Lambda _p(N)$ , we can show that $g'\in N$ , which implies that $g\in N$ as well. Hence, , so our last claim holds in this case. The remaining case is that $g'\in G_0$ is hyperbolic. Let $z_+$ be the point at which $\partial R_{g'}$ intersects the axis of $g'$ and $z_-$ be the point at which $\partial R_{\overline {g'}}$ intersects the axis of $g'$ . Then, we have $g'(z_-)=z_+$ . Since we took so that $z_-$ and $z_+$ lie in , we thus have

Therefore, we still have when g is a hyperbolic element of $G_0$ . Hence, our last claim holds when $d(ND,N gD)=0$ .

Suppose that $d(D,g'D)=d(ND,N gD)>0$ . Since $g'\in gN$ , we also have $d(ND,\, N g'D)=d(ND,N gD)$ . Therefore, by Lemma 4.4, $a_pa_p$ is not a subword, and hence not an initial block, of the reduced word representing $g'$ for any parabolic $a_p\in G_0$ . By Lemma 4.3(1), we have . Consequently, on the one hand, we obtain

On the other hand, because . We can thus conclude that also holds when $d(ND,N gD)>0$ . Our proof is now complete.

Proof. Fix an arbitrary $g\in G$ . We shall define a constant $C_0\geq 0$ independent of the choice of g and show for this $C_0$ that $n\mapsto d(ND, N g^nD)-C_0$ is superadditive. For any positive integer m, take $g^{(m)}\in g^mN$ such that

(4.7) $$ \begin{align} d(D, g^{(m)}D)=d(ND, N g^mD). \end{align} $$

Take positive integers k and l arbitrarily. Then, there exists some group element $g^*(k,l)$ represented by an initial block of the reduced word representing $g^{(k+l)}$ satisfying that

(4.8) $$ \begin{align} |\chi(g^*(k,l))-k\cdot\chi(g)|\leq\max|\chi(G_0)|. \end{align} $$

Indeed, one may take $g^*(k,l)$ to be the longest initial block of $g^{(k+l)}$ such that $\chi (g^*(k,l))\leq k\chi (g)$ if $\chi (g)\geq 0$ and take $g^*(k,l)$ to be the longest initial block of $g^{(k+l)}$ such that $\chi (g^*(k,l))\geq k\chi (g)$ if $\chi (g)<0$ .

By definition, the reduced word representing $g^*(k,l)$ is an initial block of the reduced word representing $g^{(k+l)}$ . This implies that for any $z_1,z_2\in D$ , the geodesic segment $[z_1,g^{(k+l)}z_2]$ is bound to intersect $g^*(k,l)D$ . Take $z_3\in g^*(k,l)D \cap [z_1,g^{(k+l)}z_2]$ . Then, we have

$$ \begin{align*} d(z_1, g^{(k+l)}z_2) &=d(z_1,z_3)+d(z_3,g^{(k+l)}z_2)\\ &\geq d(z_1,g^*(k,l)D)+d(g^*(k,l)D, g^{(k+l)}z_2). \end{align*} $$

It thus follows that $d(D, g^{(k+l)}D)\geq d(D, g^*(k,l)D)+d(g^*(k,l)D, g^{(k+l)}D)$ . Combining this with (4.7) and the fact that $g^{(k+l)}\in g^{k+l}N$ , we deduce that

$$ \begin{align*} d(ND, N g^{k+l}D) &\geq d(D, g^*(k,l)D)+d(g^*(k,l)D, g^{(k+l)}D)\\ &\geq d(ND, N g^*(k,l)D)+ d(N g^*(k,l)D, N g^{k+l}D). \end{align*} $$

Lemma 4.3 shows that we can replace all D appearing above by , so we have

By the triangle inequality for the metric on the quotient surface $\mathbb {D}/N$ , we further have

From (4.8), we know that $|\chi (g^{-k}g^*(k,l))|\leq \max |\chi (G_0)|$ . Thus, we obtain

(4.9)

where we define

Note that the maximum in the defining equation of $C_0$ exists because is completely determined by the value of $\chi (h)$ for $h\in G$ . Since is compact, it has finite diameter, which in turn implies that $C_0$ is finite. Subtracting $C_0$ from both sides of (4.9), we can conclude that is superadditive.

Proof of Proposition 4.1

Fix an arbitrary $g\in G$ . Since

$$ \begin{align*} \|gN\|=\inf_{n\geq 1}\frac{d(N 0,N g^n(0))}{n}, \end{align*} $$

the second half of (4.2) is straightforward, so it only remains to show the first half. Take $C_0$ satisfying the condition in Lemma 4.5. Then, by Fekete’s lemma, we have

Set . Clearly, C is finite. Hence, by the triangle inequality, we have

Therefore, the first inequality in (4.2) is true as well.

Proof. We first show that q is real-valued and real-analytic. When $\Pi _0=\varnothing $ , $\tilde {f}=f:\Lambda _r(G)\rightarrow \Lambda _r(G)$ is uniformly expanding and conjugate to a topological Markov shift with finite alphabet. For this case, we refer to [Reference Pesin and Weiss29] for a proof. Now, we only need to treat the case where $\Pi _0\neq \varnothing $ . Since $\tilde {\psi }$ is bounded, we have $\mathcal {P}(-q\tilde {\zeta }+\beta \tilde {\psi })<\infty $ if and only if $\mathcal {P}(-q\tilde {\zeta })<\infty $ . Since $\tilde \zeta $ is Hölder continuous by Proposition 6.1, it follows from [Reference Mauldin and Urbański26, Proposition 2.19] that $\mathcal {P}(-q\tilde {\zeta })<\infty $ if and only if ${\sum _{\omega \in \tilde {I}}\exp (-q\sup \tilde {\zeta }|_{[\omega ]})<\infty }$ . By well-known estimates from hyperbolic geometry [Reference Bowen and Series5], we have for each $q\in \mathbb {R}$ , $\sum _{{\omega \in \tilde {I}}}\exp (-q\sup \tilde {\zeta }|_{[\omega ]})\asymp \sum _{\ell \in \mathbb {N}}\ell ^{-2q}$ . Hence, $\mathcal {P}(-q\tilde {\zeta }+\beta \tilde {\psi })<\infty $ if and only if ${q>1/2}$ . It follows that for every $\beta \in \mathbb {R}$ ,

$$ \begin{align*} \lim_{q\searrow1/2}\mathcal{P}(-q\tilde{\zeta}+\beta\tilde{\psi})=+\infty. \end{align*} $$

For each $\beta \in \mathbb {R}$ and $q_{1}<q_{2}$ , we have

$$ \begin{align*} \mathcal{P}(-q_{2}\tilde{\zeta}+\beta\tilde{\psi})-\mathcal{P}(-q_{1}\tilde{\zeta}+\beta\tilde{\psi})\le(q_{2}-q_{1})(-\inf\tilde{\zeta}). \end{align*} $$

Since $\inf \tilde {\zeta }>0$ by Proposition 6.1, it follows that ${\lim _{q\rightarrow +\infty }\mathcal {P}(-q\tilde {\zeta }+\beta \tilde {\psi })=-\infty }$ . Since $q\mapsto \mathcal {P}(-q\tilde {\zeta }+\beta \tilde {\psi })$ is convex and finite on $(1/2,+\infty )$ , the intermediate value theorem implies the existence of a unique $q(\beta )\in \mathbb {R}$ satisfying $\mathcal {P}(-q(\beta )\tilde {\zeta }+\beta \tilde {\psi })=0$ for each ${\beta \in \mathbb {R}}$ , so q is real-valued. Since $\mathcal {P}$ is real-analytic and $(\partial /\partial q)\mathcal {P}(-q\tilde {\zeta }+\beta \tilde {\psi })\le -\inf \tilde {\zeta }<0$ [Reference Mauldin and Urbański26], it follows from the implicit function theorem that q is real-analytic.

Next, we show that $\mathcal {P}(-q\tilde {\zeta }+\beta \tilde {\psi })=\mathcal {P}(-q\tilde {\zeta }-\beta \tilde {\psi })$ for all $q,\beta \in \mathbb {R}$ . Note that for any $\tilde {f}$ -periodic $x\in \Lambda _r(G)$ , there is some hyperbolic $g\in G$ such that x is one of its two fixed points. Based on this fact, we define $\iota (x)$ as the unique point in $\Lambda _r(G)$ satisfying that there is a hyperbolic $g\in G$ such that $\{x,\iota (x)\}$ is the set of fixed points of g. This definition is valid because all group elements of G fixing x form a cyclic subgroup of G, and thus share common fixed points. Clearly, $\iota ^2(x)=x$ holds for any $\tilde {f}$ -periodic $x\in \Lambda _r(G)$ . Moreover, for any positive integer n and any $\tilde {f}$ -periodic $x\in \Lambda _r(G)$ satisfying $\tilde {f}^n(x)=x$ , we have $\tilde {f}^n(\iota (x))=\iota (x)$ , $\rho _n(x)=\rho _n(\iota (x))$ , and

$$ \begin{align*} \iota(x)_0\cdots\iota(x)_{\rho_n(x)-1}=(x_0\cdots x_{\rho_n(x)-1})^{-1}, \end{align*} $$

where we identify a reduced word over $G_0$ with a group element of G. Since $d(0,g0)=d(0,g^{-1}0)$ and $\chi (g)=-\chi (g^{-1})$ for any $g\in G$ , we deduce from (6.3) and (6.4) that

$$ \begin{align*} \sup_{n\in\mathbb{N}}\sup_{\tilde{f}^n(x)=x}|\tilde{S}_n\tilde{\zeta}(x)-\tilde{S}_n\tilde{\zeta}(\iota(x))|&<+\infty,\\ \sup_{n\in\mathbb{N}}\sup_{\tilde{f}^n(x)=x}|\tilde{S}_n\tilde{\psi}(x)+\tilde{S}_n\tilde{\psi}(\iota(x))|&=0. \end{align*} $$

Consequently, for any $q,\beta \in \mathbb {R}$ ,

(6.6) $$ \begin{align} \sum_{\tilde{f}^{n}x=x}\exp(-q\tilde S_{n}\tilde{\zeta}(\iota(x))+\beta \tilde S_{n}\tilde{\psi}(\iota(x)))\asymp\sum_{\tilde{f}^{n}x=x}\exp(-q\tilde S_{n}\tilde{\zeta}(x)-\beta \tilde S_{n}\tilde{\psi}(x)). \end{align} $$

It is shown in [Reference Jaerisch, Kesseböhmer and Lamei16] that for any $\phi :\Lambda _r(G)\rightarrow \mathbb {R}$ , we have

$$ \begin{align*}\mathcal{P}(\phi)=\limsup_{n\rightarrow \infty}\frac{1}{n}\log \sum_{\tilde{f}^{n}x=x}\exp(\tilde S_{n}\phi(x)).\end{align*} $$

Hence, using the facts about $\iota $ we pointed out, we have for any $q,\beta \in \mathbb {R}$ ,

$$ \begin{align*} \mathcal{P}(-q\tilde{\zeta}+\beta\tilde{\psi})&=\limsup_{n\rightarrow+\infty}\frac{1}{n}\log \sum_{\tilde{f}^{n}x=x}\exp(-q \tilde S_{n}\tilde{\zeta}(\iota(x))+\beta \tilde S_{n}\tilde{\psi}(\iota(x)))\\ &=\limsup_{n\rightarrow+\infty}\frac{1}{n}\log \sum_{\tilde{f}^{n}x=x}\exp(-q \tilde S_{n}\tilde{\zeta}(x)-\beta \tilde S_{n}\tilde{\psi}(x))=\mathcal{P}(-q\tilde{\zeta}+\beta\tilde{\psi}). \end{align*} $$

Thus, $q(\beta )=q(-\beta )$ for any $\beta \in \mathbb {R}$ . From this and the convexity of q, it follows that q attains its minimum at $0$ . It is well known that q is either strictly convex or affine. Suppose for a contradiction that q is affine. Then, by symmetry, $q(\beta )=q(0)$ for all $\beta \in \mathbb {R}$ . This contradicts the fact that $\mathcal {P}(\beta \tilde {\psi })\rightarrow +\infty $ as $\beta \rightarrow +\infty $ .

Now that we have shown that q is even and real-analytic, it follows that $\alpha _-=-\alpha _+$ . The properties of the convex conjugate $q^*$ are immediate consequences of the corresponding properties of q.

Proof. By [Reference Jaerisch, Kesseböhmer and Lamei16, Corollary 2.2, Remark 2.4, and Corollary 3.5], we have for any $\beta \in \mathbb {R}$ ,

$$ \begin{align*} q(\beta)=\inf\bigg\{q \in \mathbb{R} \,\bigg|\sum_{n=1}^\infty \sum_{\tilde{f}^{n}x=x} \exp (-q \tilde S_n \tilde{\zeta}(x)+\beta \tilde S_n \tilde \psi(x))<+\infty\bigg\}. \end{align*} $$

For $\beta \geq 0$ , on the one hand, we have

$$ \begin{align*} \sum_{\tilde{f}^{n}x=x}\exp (-q \tilde S_n \tilde{\zeta}(x)+\beta \tilde S_n \tilde \psi(x) )\leq \sum_{\tilde{f}^{n}x=x} \exp (-q \tilde S_n \tilde{\zeta}(x)+\beta |\tilde S_n \tilde \psi(x)| ), \end{align*} $$

while on the other hand, by (6.6), we also have

$$ \begin{align*} \sum_{\tilde{f}^{n}x=x} \exp (-q \tilde S_n \tilde{\zeta}(x)+\beta |\tilde S_n \tilde \psi(x)| )&\leq\sum_{\tilde{f}^{n}x=x}\exp (-q \tilde S_n \tilde{\zeta}(x)+\beta \tilde S_n \tilde \psi(x) )\\ &\quad +\sum_{\tilde{f}^{n}x=x}\exp (-q \tilde S_n \tilde{\zeta}(x)-\beta \tilde S_n \tilde \psi(x) )\\ &\asymp\sum_{\tilde{f}^{n}x=x}\exp (-q \tilde S_n \tilde{\zeta}(x)+\beta \tilde S_n \tilde \psi(x)). \end{align*} $$

It thus follows that for $\beta \ge 0$ , we have

$$ \begin{align*} q(\beta)=\inf\bigg\{q\geq 0 \,\bigg|\sum_{n=1}^\infty \sum_{\tilde{f}^{n}x=x} \exp (-q \tilde S_n \tilde{\zeta}(x)+\beta |\tilde S_n \tilde \psi(x)| )<+\infty\bigg\}. \end{align*} $$

Here, we note that the range of q can be restricted to non-negative numbers because ${q(\beta )\geq q(0)=\dim _{H}(\Lambda (G))\geq 0}$ for all $\beta \in \mathbb {R}$ . By [Reference Jaerisch, Kesseböhmer and Lamei16, Corollary 2.2, Remark 2.4, and Corollary 3.5], we have

$$ \begin{align*} q(\beta)=\inf\bigg\{q \in \mathbb{R} \,\bigg|\sum_{n=1}^\infty \sum_{\omega \in \tilde{\Sigma}^n} \exp \sup (-q \tilde S_n \tilde{\zeta}+\beta |\tilde S_n \tilde \psi|)|_{[\omega]}<+\infty\bigg\}. \end{align*} $$

By (6.3) and (6.4), we have

$$ \begin{align*} q(\beta)=\inf\bigg\{q \in \mathbb{R} \,\bigg|\sum_{n=1}^\infty \sum_{\omega \in \tilde{\Sigma}^n} \exp(-qd(0,\omega 0)+\beta d(N 0,N \omega 0))<+\infty\bigg\}. \end{align*} $$

It is not difficult to see that

$$ \begin{align*} \sum_{n=1}^\infty \sum_{\omega \in \tilde{\Sigma}^n} \exp(-qd(0,\omega0)+\beta d(N 0,N \omega 0))\asymp\sum_{g\in G}\exp(-qd(0,g0)+\beta d(N 0,N g 0)). \end{align*} $$

Hence, for each $\beta \ge 0$ , we have $q(\beta )=s(\beta )$ , which proves the first assertion.

To prove the second assertion, let $\alpha \ge 0$ . Since q is even by Proposition 6.3, we have for $\beta <0$ , $q(\beta )-\alpha \beta \leq q(-\beta )+\alpha \beta $ . Therefore, $-q^{*}(\alpha )=\inf _{\beta \in \mathbb {R}}\{ q(\beta )- \alpha \beta \} =\inf _{\beta \ge 0}\{ q(\beta )- \alpha \beta \}$ . Hence, by the first assertion, $-q^{*}(\alpha )=\inf _{\beta \ge 0}\{ s(\beta )-\alpha \beta \}$ . The proof is complete.

Proof. The second inclusion is clear. To show the first, take $x\in \Lambda _r(G)$ from the left-hand side of the first inclusion arbitrarily. By (6.1), we have for any $\alpha \geq 0$ ,

$$ \begin{align*} \sup_{n\geq 1}||S_{\rho_n(x)}\psi(x)|-\alpha \tilde S_{n}\tilde{\zeta}(x)|=\sup_{n\geq 1}||\tilde S_{n}\tilde{\psi}(x)|-\alpha \tilde S_{n}\tilde{\zeta}(x)|<+\infty. \end{align*} $$

Combining this with (6.3), (6.4), and Proposition 6.2(3), we have

(6.7) $$ \begin{align} C_\alpha:=\sup_{n\ge1}|d(N 0,N \gamma_x(\tau_{\rho_n(x)}(x)))-\alpha d( 0, \gamma_x(\tau_{\rho_n(x)}(x)))|<+\infty. \end{align} $$

Also notice that the induced sequence $\tilde {\pi }^{-1}(x)\in \tilde {\Sigma }$ is in $F^{\mathbb {N}}$ for some finite $F\subseteq \tilde {I}$ , which implies that $\sup _{n\in \mathbb {N}}\{\rho _{n+1}(x)-\rho _n(x)\}<+\infty $ . Since $\zeta $ is bounded,

$$ \begin{align*} \sup_{n\in\mathbb{N}}\{\tau_{\rho_{n+1}(x)}(x)-\tau_{\rho_n(x)}(x)\}=\sup_{n\in\mathbb{N}}\{\tilde{S}_{n+1}\tilde{\zeta}(x)-\tilde{S}_n\tilde{\zeta}(x)\}<+\infty. \end{align*} $$

Further note that $\tau _m(x)$ is monotonically increasing in m and diverges to infinity as ${m\rightarrow +\infty }$ , so $\sup _{t\geq 0}\inf _{n\in \mathbb {N}}|t-\tau _{\rho _n(x)}(x)|<+\infty $ . Combining this fact with (6.7) and the fact that $d(N z,N w)\leq d(z,w)$ for all $z,w\in \mathbb {D}$ , we obtain

$$ \begin{align*} \sup_{t\geq 0}|d(N 0,N \gamma_x(t))-\alpha d( 0, \gamma_x(t))|&\leq C_\alpha+(1+\alpha)\sup_{t\geq 0}\inf_{n\in\mathbb{N}}|t-\tau_{\rho_n(x)}(x)|<+\infty, \end{align*} $$

by using the triangle inequality. This shows the first inclusion we claim.

Now, it only remains to prove the third inclusion in Lemma 6.5. For $\alpha =0$ , our assumption $\Lambda _p(N)=\Lambda _p(G)$ implies

Now, let $\alpha>0$ and . Since $y\in \Lambda _r(G)$ , we have that $\tau _{\rho _n(y)}(y)=\tilde {S}_n\tilde {\zeta }(y)$ tends to infinity as $n\rightarrow +\infty $ . Hence,

$$ \begin{align*} \liminf_{n\rightarrow+\infty} \frac{d(N 0,N\gamma_y(\tau_{\rho_n(y)}(y)))}{\tilde{S}_n\tilde{\zeta}(y)}=\frac{d(N 0,N\gamma_{{y}}(\tau_{\rho_n(y)}(y)))}{\tau_{\rho_n(y)}(y)}\ge \alpha. \end{align*} $$

By (6.1) and (6.4), this yields

$$ \begin{align*} \liminf_{n\rightarrow+\infty}\frac{|\tilde{S}_n\tilde{\psi}(y)|}{\tilde{S}_n\tilde{\zeta}(y)}=\liminf_{n\rightarrow+\infty}\frac{|S_{\rho_n(y)}\psi(y)|}{\tilde{S}_n\tilde{\zeta}(y)}\ge \alpha. \end{align*} $$

Since $\alpha> 0$ , this in particular shows that either $\tilde {S}_n\tilde {\psi }(y)>0$ for sufficiently large $n\in \mathbb {N}$ or $\tilde {S}_n\tilde {\psi }(y)<0$ for sufficiently large $n\in \mathbb {N}$ . For the former case, we have $y\in \mathcal {L}^{\geq \alpha }_{\tilde {\psi },\tilde {\zeta }}$ ; for the latter case, we have $y\in \mathcal {L}^{\leq -\alpha }_{\tilde {\psi },\tilde {\zeta }}$ .

Proof of Theorem 1.1

Proposition 6.3 states that q is real-analytic and strictly convex, so its derivative $q'$ exists and is monotonically increasing. Define $\alpha _\pm =\lim _{\beta \rightarrow \pm \infty }q'(\beta )$ . By Proposition 6.3, we have $\alpha _+=-\alpha _->0$ . Let $\alpha \in [0,\alpha _+)$ . Our goal is to show that

(6.8) $$ \begin{align} \dim_{H}\mathcal{UE}_{N}^{\alpha}=\dim_{H}\mathcal{LE}_{N}^{\alpha}=\dim_H (\mathcal{L}^{\ge \alpha}_{\tilde{\psi},\tilde{\zeta}}\cup\mathcal{L}^{\le -\alpha}_{\tilde{\psi},\tilde{\zeta}})=-q^*(\alpha)>0. \end{align} $$

Note that this gives the desired formula in Theorem 1.1 because we have $-q^*(\alpha )=\inf _{\beta \ge 0} \{s(\beta )-\alpha \beta \}$ by Lemma 6.4 and $\mathcal {UE}^\alpha _N\subseteq \mathcal {E}^\alpha _N\subseteq \mathcal {LE}^\alpha _N$ .

We now turn to the proof of (6.8). By Lemma 6.5 and Theorem 5.1, we have $\dim _{H}\mathcal {UE}_{N}^{\alpha }\ge -q^*(\alpha )$ . Here, we have used that the metric $d_{\tilde \zeta }$ on $\tilde {\Sigma }$ is equivalent to the Euclidean metric on $\Lambda _r(G)$ . Moreover, by Lemma 6.5, we have

$$ \begin{align*}\dim_{H}\mathcal{LE}_{N}^{\alpha}\le \dim_H (\mathcal{L}^{\ge \alpha}_{\tilde{\psi},\tilde{\zeta}}\cup\mathcal{L}^{\le -\alpha}_{\tilde{\psi},\tilde{\zeta}} ). \end{align*} $$

Hence, to complete the proof of (6.8), it suffices to show $\dim _H (\mathcal {L}^{\ge \alpha }_{\tilde {\psi },\tilde {\zeta }}\cup \mathcal {L}^{\le -\alpha }_{\tilde {\psi },\tilde {\zeta }})\le -q^*(\alpha )$ . Since $q'(0)=0$ by Proposition 6.3, it follows from Theorem 5.1 and Remark 5.2 that

$$ \begin{align*} \dim_H (\mathcal{L}^{\ge \alpha}_{\tilde{\psi},\tilde{\zeta}}\cup\mathcal{L}^{\le -\alpha}_{\tilde{\psi},\tilde{\zeta}})&\le \max \{ \dim_H (\mathcal{L}^{\#\alpha}_{\tilde{\psi},\tilde{\zeta}}), \dim_H \big(\mathcal{L}^{\#(-\alpha)}_{\tilde{\psi},\tilde{\zeta}})\}\\ &=\max \{-q^*(\alpha), -q^*(-\alpha)\}. \end{align*} $$

By Proposition 6.3, we have $\max \{-q^*(\alpha ), -q^*(-\alpha ) \}=-q^*(\alpha )$ . The proof of (6.8) is thus complete.

Next, we show $\alpha _+=1$ . By (6.8), we have $[0,\alpha _+)\subseteq \mathrm {Ran}\,\mathcal {UE}_N$ . Since $\mathrm {Ran}\,\mathcal {UE}_N\subseteq [0,1]$ , it follows that $\alpha _+ \le 1$ . Combining Theorem 5.1, Lemma 6.5, and Lemma 6.6, we also have $\alpha _+\geq 1$ . Hence, $\alpha _+=1$ . It then follows that $[0,1)\subseteq \mathrm {Ran}\,\mathcal {UE}_N\subseteq \mathrm {Ran}\,\mathcal {LE}_N\subseteq [0,1]$ .

The analytic properties of the escape rate spectrum for $\alpha \in [0,1)$ follow from Proposition 6.3. The proof of the main theorem is complete except for the boundary point $\alpha =1$ , which will be investigated in the next subsection.

Proof. Since $\alpha _{+}=1$ by the proof of Theorem 1.1, we have

$$ \begin{align*} \lim_{\beta\rightarrow +\infty}\frac{\int\tilde{\psi}\,{d}\tilde{\mu}_{\beta}}{\int\tilde{\zeta}\,{d}\tilde{\mu}_{\beta}}=1. \end{align*} $$

Hence, since $\tilde {\psi }$ is bounded, it follows that

$$ \begin{align*} \limsup_{\beta\rightarrow+\infty}\int\tilde{\zeta}\,{d}\tilde{\mu}_{\beta}<+\infty. \end{align*} $$

By [Reference Kesseböhmer and Urbański22, Lemma 3.2], we conclude that $\{ \tilde {\mu }_{\beta }\mid \beta>0\}$ is tight.

Proof. The first two assertions are shown in [Reference Kesseböhmer and Urbański22, Proposition 4.3]. To show the last two, first note that $\tilde {\psi }$ is a bounded continuous function, so by the weak* convergence of $(\tilde {\mu }_{\beta _m})_{m\geq 1}$ to $\tilde {\mu }$ , we have $\lim _{m\rightarrow +\infty }\int _{\tilde {\Sigma }}\tilde {\psi }\,{d}\tilde {\mu }_{\beta _m}=\int _{\tilde {\Sigma }}\tilde {\psi }\,{d}\tilde {\mu }$ , which is finite due to the boundedness of $\tilde {\psi }$ . Since we also have

(6.9) $$ \begin{align} \lim_{m\rightarrow+\infty}\frac{\int_{\tilde{\Sigma}}\tilde{\psi}\,{d}\tilde{\mu}_{\beta_m}}{\int_{\tilde{\Sigma}}\tilde{\zeta}\,{d}\tilde{\mu}_{\beta_m}}=1, \end{align} $$

we deduce that

$$ \begin{align*} 0<\inf\tilde{\zeta}\leq\lim_{m\rightarrow +\infty}\int_{\tilde{\Sigma}}\tilde{\zeta}\,{d}\tilde{\mu}_{\beta_m}=\int_{\tilde{\Sigma}}\tilde{\zeta}\,{d}\tilde{\mu}<+\infty. \end{align*} $$

Combining this with (6.9), we have $\int _{\tilde {\Sigma }}\tilde {\psi }\,{d}\tilde {\mu }>0$ . The proof is thus complete.

Proof of Theorem 1.1 (continued)

We are now in the position to prove the lower bound $\dim _{H}(\mathcal {UE}^1_N)\ge -q^*(1)$ . Let $\alpha _m=q'(\beta _m)$ for all and note that

$$ \begin{align*} -q^*(\alpha_m)=q(\beta_m)-\alpha_m\beta_m=\frac{h(\tilde{\mu}_{\beta_m},\tilde{\sigma})}{\int_{\tilde{\Sigma}}\tilde{\zeta}\,{d}\tilde{\mu}_{\beta_m}}. \end{align*} $$

Since $\lim _{m\to \infty }\alpha _m=1$ , it follows from Lemma 6.8(2) and (3) that

$$ \begin{align*} \frac{h(\tilde{\mu},\tilde{\sigma})}{\int_{\tilde{\Sigma}}\tilde{\zeta}\,{d}\tilde{\mu}}\geq \limsup_{m\rightarrow+\infty}\frac{h(\tilde{\mu}_{\beta_m},\tilde{\sigma})}{\int_{\tilde{\Sigma}}\tilde{\zeta}\,{d}\tilde{\mu}_{\beta_m}}=-q^*(1). \end{align*} $$

By employing standard arguments involving the mass distribution principle (see for instance [Reference Przytycki and Urbański30, Theorem 9.1.11]), we only need to show that $\tilde {\mu }(\mathcal {UE}^1_N)=1$ . Since $\limsup _{n\rightarrow +\infty } \tilde {S}_n\tilde {\psi }/\tilde {S}_n\tilde {\zeta }\le \alpha _+=1$ , we conclude that for any $\tilde {x}\in \tilde {\Sigma }$ ,

$$ \begin{align*} \liminf_{n\rightarrow+\infty}\frac{1}{n}(\tilde{S}_n\tilde{\psi}(\tilde{x})-\tilde{S}_n\tilde{\zeta}(\tilde{x})) \leq 0. \end{align*} $$

Let $F \subseteq \tilde {I}$ be a finite set satisfying $\tilde \mu (\tilde {\Sigma } \cap F^{\mathbb {N}} )=1$ whose existence is guaranteed by Lemma 6.8(1). Since $\tilde {\psi }$ and $\tilde {\zeta }$ are Hölder continuous, from the tools in ergodic optimization [Reference Jenkinson19], we have that $(\tilde {\psi }-\tilde {\zeta })|_{\tilde {\Sigma } \cap F^{\mathbb {N}}}$ is cohomologous to a non-positive function, which means that there exists a Hölder continuous function $\phi :\tilde {\Sigma } \cap F^{\mathbb {N}}\rightarrow \mathbb {R}$ such that

(6.10) $$ \begin{align} (\tilde{\psi}-\tilde{\zeta})|_{\tilde{\Sigma} \cap F^{\mathbb{N}}}+\phi\circ\tilde{\sigma}-\phi \leq 0. \end{align} $$

By the $\tilde {\sigma }$ -invariance of $\tilde {\mu }$ , we have $\int _{\tilde {\Sigma }\cap F^{\mathbb {N}}} (\phi \circ \tilde {\sigma }-\phi ) \,{d}\tilde {\mu }=0$ . This observation, along with Lemma 6.8 (3), (4), and the equality (6.9), implies that

$$ \begin{align*} \int_{\tilde{\Sigma}\cap F^{\mathbb{N}}} (\tilde{\psi}-\tilde{\zeta}+\phi\circ\tilde{\sigma}-\phi)\,{d}\tilde{\mu}&=\int_{\tilde{\Sigma}\cap F^{\mathbb{N}}}(\tilde{\psi}-\tilde{\zeta})\,{d}\tilde{\mu}\\ &=\int_{\tilde{\Sigma}}(\tilde{\psi}-\tilde{\zeta})\,{d}\tilde{\mu}=\lim_{m\rightarrow+\infty}\int_{\tilde{\Sigma}}(\tilde{\psi}-\tilde{\zeta})\,{d}\tilde{\mu}_{\beta_m}=0, \end{align*} $$

where the second equality holds because $\tilde {\mu }(\tilde {\Sigma }\cap F^{\mathbb {N}})=1$ . Combining this with (6.10), we have

$$ \begin{align*} \tilde{\mu} (((\tilde{\psi}-\tilde{\zeta})|_{\tilde{\Sigma} \cap F^{\mathbb{N}}}+\phi\circ\tilde{\sigma}-\phi)^{-1}(\{0\}))=1. \end{align*} $$

Since $\tilde {\sigma }$ preserves $\tilde {\mu }$ , we further have

$$ \begin{align*} \tilde{\mu} \bigg(\bigcap_{n=0}^\infty\tilde{\sigma}^{-n}((\tilde{\psi}-\tilde{\zeta})|_{\tilde{\Sigma} \cap F^{\mathbb{N}}}+\phi\circ\tilde{\sigma}-\phi)^{-1}(\{0\})\bigg)=1. \end{align*} $$

Since $\phi $ is bounded, we have

$$ \begin{align*} \bigcap_{n=0}^\infty\tilde{\sigma}^{-n}(((\tilde{\psi}-\tilde{\zeta})|_{\tilde{\Sigma} \cap F^{\mathbb{N}}}+\phi\circ\tilde{\sigma}-\phi)^{-1}(\{0\}))\subseteq \mathcal{UL}^1_{\tilde{\psi},\tilde{\zeta}}\subseteq \mathcal{UE}^1_N. \end{align*} $$

Hence, $\tilde {\mu }(\mathcal {UE}^1_N)=1$ . This shows that $\mathcal {UE}^1_N$ is non-empty and has Hausdorff dimension no less than $-q^*(1)$ . Let us turn to the upper bound. By a standard covering argument, we obtain

$$ \begin{align*} \dim_{H}(\mathcal{L}^{\ge 1}_{\tilde{\psi},\tilde{\zeta}})\leq-q^*(1)\quad \textrm{and}\quad \dim_{H}(\mathcal{L}^{\le -1}_{\tilde{\psi},\tilde{\zeta}})\leq-q^*(-1). \end{align*} $$

By Proposition 6.3, we have $-q^*(1)=-q^*(-1)$ . Hence, by Lemma 6.5, we have $\dim _{H}(\mathcal {LE}_{N}^1)\le -q^*(1)$ . The proof is complete.

Proof. The first claim is precisely (7.2), so we only need to prove the second claim.

For , let $\omega $ be the reduced word over $G_0$ representing g. For the sake of convenience, we set $\omega _{-1}=a$ and $\omega _{|\omega |}=\bar {a}$ . Since , there are ${k,l\in \{0,\ldots ,|\omega |-1\}}$ such that and $\{\omega _{k-1},\omega _{l+1}\}\subseteq \{\bar {a},a\}$ . Apparently, $k\neq l+1$ . Moreover, note that every geodesic segment joining $\partial R_{\bar {a}}$ and $\omega \partial R_{\bar {a}}$ must intersect $\omega _0\cdots \omega _{j}\partial R_{\overline {\omega _{j}}}$ for all $j\in \{-1,\ldots ,|\omega |\}$ . Therefore,

$$ \begin{align*} d(\partial R_{\bar{a}},g\partial R_{\bar{a}})&\geq d(\omega_1\cdots\omega_{k-1}\partial R_{\overline{\omega_{k-1}}},\omega_1\cdots\omega_k\partial R_{\overline{\omega_k}})\\ &\quad+ d(\omega_1\cdots\omega_l\partial R_{\overline{\omega_l}},\omega_1\cdots\omega_{l+1}\partial R_{\overline{\omega_{l+1}}})\\ &= d(\partial R_{\overline{\omega_{k-1}}},\partial R_{\omega_k})+d(\partial R_{\overline{\omega_l}},\partial R_{\omega_{l+1}})\geq \|a\|+2c_0>\|a\|, \end{align*} $$

showing the second claim.

Proof of Proposition 7.3

Let z be the point at which the geodesic $\partial R_{\bar {a}}$ and $\mathcal {X}_a$ intersect. We claim that

(7.3) $$ \begin{align} d(N z,N a^n z)=d(z,a^nz) \end{align} $$

for any $n\in \mathbb {Z}$ . This claim apparently holds when $n=0$ . Also note that (7.3) holds if and only if $d(N z,N a^{-n} z)=d(z,a^{-n}z)$ , since $N a^{-n}:\mathbb {D}/N\rightarrow \mathbb {D}/N$ and $a^{-n}:\mathbb {D}\rightarrow \mathbb {D}$ are isometries. Therefore, it suffices to prove (7.3) for positive n.

Given an arbitrary positive integer n, let $\omega $ be a reduced word over $G_0$ which, considered as a group element of G, is in $a^nN$ and satisfies $d(z,\omega z)=d(N z,N a^nz)$ . Note that there exist reduced words $\omega ^{(1)},\ldots ,\omega ^{(n)}$ over $G_0$ such that:

1. (1) $\omega ^{(j)}$ , considered as a group element of G, is in $aN$ for any $j\in \{1,\ldots ,n\}$ ;

2. (2) the last symbol appearing in $\omega ^{(j)}$ is a for all $j\in \{1,\ldots ,n\}$ ;

3. (3) their concatenation $\omega ^{(1)}\cdots \omega ^{(n)}$ is reduced as well and is an initial block of the reduced word $\omega $ .

Indeed, we can define $\omega ^{(1)},\ldots ,\omega ^{(n)}$ by letting $\omega ^{(1)}\cdots \omega ^{(j)}$ be the shortest initial block of $\omega $ which is in $a^jN$ for all $j\in \{1,\ldots , n\}$ . Then, as we argued in the proof of Lemma 7.4, we can show that

$$ \begin{align*} d(\partial R_{\bar{a}},\omega\partial R_{\bar{a}})&\geq d(\partial R_{\bar{a}},\omega^{(1)}\cdots\omega^{(n)}\partial R_{\bar{a}})\\ &\geq\sum_{j=1}^nd(\omega^{(1)}\cdots\omega^{(j-1)}\partial R_{\bar{a}},\omega^{(1)}\cdots\omega^{(j)}\partial R_{\bar{a}})=\sum_{j=1}^nd(\partial R_{\bar{a}},\omega^{(j)}\partial R_{\bar{a}}). \end{align*} $$

Since $\omega ^{(j)}\in aN$ for all $j\in \{1,\ldots ,n\}$ , we deduce from Lemma 7.4 that

(7.4) $$ \begin{align} d(\partial R_{\bar{a}},\omega\partial R_{\bar{a}})\geq n\|a\|=d(z,a^nz), \end{align} $$

where the equality above holds because z is on the axis $\mathcal {X}_a$ of a. Since $z\in \partial R_{\bar {a}}$ , on the one hand, we have

$$ \begin{align*} d(N z,N a^nz)=d(z,\omega z)\geq d(\partial R_{\bar{a}},\omega\partial R_{\bar{a}})\geq d(z,a^nz). \end{align*} $$

Since $d(N w_1,N w_2)\leq d(w_1,w_2)$ for all $w_1,w_2\in \mathbb {D}$ , we also have, on the other hand, $d(N z,N a^nz)\leq d(z,a^n z)$ . Consequently, (7.3) holds for any positive integer n. Thus, as we argued in the beginning of this proof, (7.3) holds for all $n\in \mathbb {Z}$ .

Now, we can show Proposition 7.3. Let $z_1$ and $z_2$ be two arbitrary points on the axis $\mathcal {X}_a$ of a. Then, there exist distinct $n_1,n_2\in \mathbb {Z}$ such that $z_1$ and $z_2$ lie on the geodesic segment joining $a^{n_1}z$ and $a^{n_2}z$ . Without loss of generality, we assume that $d(z_1,a^{n_1}z)\leq d(z_2,a^{n_1}z)$ . Then, on the one hand, we have

(7.5) $$ \begin{align} d(N z_1,N z_2)\leq d(z_1,z_2)=d(a^{n_1}z,a^{n_2}z)-d(a^{n_1}z,z_1)-d(z_2,a^{n_2}z_2). \end{align} $$

On the other hand, using the fact that (7.3) holds for any $n\in \mathbb {Z}$ , we obtain

$$ \begin{align*} d(a^{n_1}z,a^{n_2}z)=d(z,a^{n_2-n_1}z)=d(N z,N a^{n_2-n_1}z)=d(N a^{n_1}z,N a^{n_2}z). \end{align*} $$

Thus, by the triangle inequality and the fact that $d(N w_1,N w_2)\leq d(w_1,w_2)$ for all $w_1,w_2\in \mathbb {D}$ , we have

$$ \begin{align*} d(N z_1,N z_2)&\geq d(N a^{n_1}z,N a^{n_2}z)-d(N a^{n_1}z,N z_1)-d(N z_2,N a^{n_2}z_2)\\ &\geq d(a^{n_1}z,a^{n_2}z)-d(a^{n_1}z,z_1)-d(z_2,a^{n_2}z_2). \end{align*} $$

Combining this with (7.5), we see that $d(N z_1,N z_2)=d(z_1,z_2)$ .

Proof of Proposition 7.2

Let z be the point at which $\partial R_{\bar {a}}$ and the axis $\mathcal {X}_a$ of a intersect. For each $x\in \Lambda (G)=\Lambda (N)$ , $\gamma _{x,z}:[0,+\infty )\rightarrow \mathbb {D}$ is defined in the same way as in §2.

We begin with showing $\mathcal {UE}^1_N\subseteq \Lambda _p(N)\cup G\cdot \Lambda (a)$ . Take an arbitrary x from . Then, $\chi (x_0\cdots x_{n-1})=\sum _{j=0}^{n-1}\chi (x_j)$ diverges to $+\infty $ or $-\infty $ as $n\rightarrow +\infty $ . For every positive integer k, define

$$ \begin{align*} n(k)=\min\bigg\{n\in\mathbb{N}\,\bigg|\sum_{j=0}^{n-1}\chi(x_j)=sk\bigg\}=\min\{n\in\mathbb{N}|x_0\cdots x_{n-1}\in a^{sk}N\}, \end{align*} $$

where $s=1$ if $\sum _{j=0}^\infty \chi (x_j)=+\infty $ and $s=-1$ otherwise. Since $\max _{b\in G_0}|\chi (b)|=1$ for our example, $n(k)$ is well defined. It also follows that $x_{{n(k)-1}}=a$ if $\sum _{j=1}^\infty \chi (x_j)=+\infty $ , and $x_{{n(k)-1}}=\bar {a}$ otherwise. Further for each positive integer k, define $t_k$ as the unique positive number satisfying

$$ \begin{align*} \gamma_{x,z}(t_k)\in {x_0\cdots x_{n(k)-1}\partial R_{\overline{x_{n(k)-1}}}}. \end{align*} $$

Also, set $t_0=0$ for convenience. Then, $t_k\leq t_{k+1}$ for any positive integer k. Moreover, for any integer $k\geq 2$ ,

$$ \begin{align*} t_k=d(z,\gamma_{x,z}(t_k))&=\sum_{j=0}^{k-1}d(\gamma_{x,z}(t_j),\gamma_{x,z}(t_{j+1}))\geq \sum_{j=1}^{k-1}d(\gamma_{x,z}(t_j),\gamma_{x,z}(t_{j+1}))\\ &\geq \sum_{j=1}^{k-1}d(\partial R_{{\overline{x_{n(j)-1}}}},{x_{n(j)}\cdots x_{n(j+1)-1}}\partial R_{{\overline{x_{n(j+1)-1}}}}). \end{align*} $$

If $\sum _{j=1}^\infty \chi (x_j)=+\infty $ , note that $x_{{n(j)-1}}=a$ and ${x_{n(j)}\cdots x_{n(j+1)-1}}\in aN$ for any positive integer j. Hence, by Lemma 7.4, we have for any integer $k\geq 2$ ,