Proof. First assume that $\gamma $ has an infinite cutting sequence $(g_n)_{n=0}^\infty $ . Since cutting sequences are shortest by Proposition 2.1, $(g_0\cdots g_n)_{n=0}^\infty \subset G$ are pairwise distinct. Since R is locally finite, $\mathrm {diam}( g_0\cdots g_nR) \rightarrow 0$ as $n\rightarrow \infty $ . Hence, $g_0\cdots g_n0\rightarrow \gamma ^+$ and therefore, $\gamma ^+\in \Lambda $ . To prove that $\gamma ^+\in \Lambda _c$ , we assume for the sake of contradiction that $\gamma ^+$ is fixed by some parabolic element of G. By [Reference Beardon3, Corollary 9.2.9], $\gamma ^+$ is the G-image of some cusp of R. This implies that $\gamma $ has a finite cutting sequence and gives the desired contradiction. Hence, $\gamma ^+\in \Lambda _c$ .

Figure 4 A fundamental domain R of a finitely generated Fuchsian group of the first kind with eight sides: v is a cusp, $e_1$ and $e_8$ , $e_2$ and $e_6$ , $e_3$ and $e_7$ , $e_4$ and $e_5$ are identified in pairs, which yields a hyperbolic surface of genus $2$ . The bidirectional arrows in the inner (respectively outer) circle indicate the elements of the partition of $\mathbb S^1$ defined by W in equation (3.4) (respectively $W'$ in equation (3.6)). In the coarser partition defined by $W'$ , the sets $L_i(v)\ (i=2,3,\ldots )$ are combined into a single set $L(v)$ . Similarly, $R_i(v)\ (i=3,4,\ldots )$ are combined into $R(v)$ .

Conversely, assume that $\gamma \in \mathscr {R}$ with $\gamma ^+\in \Lambda $ has no infinite cutting sequence. In this case, $\gamma ^+$ belongs to the Euclidean boundary of some image of R under G. Hence, by [Reference Beardon3, Theorem 9.3.8], $\gamma ^+$ is fixed by some parabolic element of G and therefore, $\gamma ^+ \notin \Lambda _c$ . The proof is complete.

Proof. Clearly, $a_0\cdots a_{n}0 \notin a_0 \cdots a_n H(\overline a_{n+1})$ and $a_0\cdots a_{n+1}0 \in H(\overline a_{n+1})$ . Since by Proposition 2.3 the elements of $E(\Sigma ^+)$ are shortest, $a_0\cdots a_{k}0 \in H(\overline a_{n+1})$ for $k>n$ . Hence, the lemma follows.

Proof. Recall that each $a_0\cdots a_{n-1} \in E^n(\Sigma ^+)$ has word length n by Proposition 2.3. For convenience, we work in the upper half-plane $\mathbb {H}$ . We choose a conjugacy which maps a point in the complement of the Euclidean closure of the arc cut off by $H(\overline a_0)$ in $\mathbb {S}^1$ to infinity. Put $r_n=\max _{a_0\cdots a_{n-1}\in E^n(\Sigma ^+)} \mathrm {diam}( a_0\cdots a_{n-1}R)$ . Since R is locally finite, we have $r_n\to 0$ as $n\to \infty $ .

If $C(\overline a_n)$ is a free side of R, then $|\partial \mathbb {H}\cap a_0\cdots a_{n-1}H(\overline a_n) | \le r_n$ . If $C(\overline a_n)$ is not a free side, we assume for simplicity that the side s of $a_0 \cdots a_{n-1} R$ contained in $a_0 \cdots a_{n-1} C(\overline a_n)$ has one vertex v at infinity, and one vertex $v'$ in $\mathbb {H}$ . Denote the other side of $a_0 \cdots a_{n-1} R$ emanating from the vertex $v'$ by $s'$ , and the endpoint of $s'$ not equal to $v'$ by $v"$ . Denote by $s"$ the side of $a_0 \cdots a_{n-1} R$ emanating from the vertex $v"$ not equal to $s'$ . Since the circular arcs containing s and $s"$ are disjoint [Reference Bowen and Series7, Lemma 2.2], it is easy to see that $|\partial \mathbb {H}\cap a_0\cdots a_{n-1}H(\overline a_n)| $ is bounded from above by the Euclidean distance of v and $v"$ . Since v and $v"$ are vertices of the fundamental domain $a_0 \cdots a_{n-1} R$ , we conclude $|\partial \mathbb {H}\cap a_0\cdots a_{n-1}H(\overline a_n)|\le r_n $ . The remaining cases can be treated in a similar fashion. The proof of the first assertion is complete. The second assertion follows from the first one because $\Theta (a_0\cdots a_{n})$ is contained in the Euclidean closure of $a_0\cdots a_{n-1}H(\overline a_n)$ .

Proof. By Lemma 2.4, we have $a_0\cdots a_{n-1}0\in a_0\cdots a_{n-2}H(\overline a_{n-1})$ . By Lemma 2.5, $\textrm {diam}(a_0\cdots a_{n-2}H(\overline a_{n-1}))$ tends to zero uniformly, as $n\rightarrow \infty $ . Since $\xi $ has the f-expansion $(a_n)_{n=0}^{\infty }$ , it belongs to the Euclidean closure of $a_0\cdots a_{n-2}H(\overline a_{n-1})$ for each n. The lemma follows.

Proof. By Lemmas 2.2 and 2.6, the cutting orbit $(g_0\cdots g_n0)_{n=0}^\infty $ and the BS orbit $(a_0\cdots a_{n}0)_{n=0}^\infty $ converge to the same point $\gamma ^+$ . Hence, the conclusion is a consequence of [Reference Series38, Proposition 3.2] and [Reference Series38, Theorem 3.1].

Proof. Let $n\geq 2$ and let $a_0\cdots a_{n-1}\in E^n(\Sigma ^+)$ . By the chain rule and the mean value theorem for $\log |f'|$ , for $\xi ,\eta \in \Theta (a_0\cdots a_{n-1})$ , we have

$$ \begin{align*}\begin{split}\log\frac{|(f^{n})'\xi|}{|(f^{n})'\eta|}&\ll \sum_{j=0}^{n-1}|f^j(\Theta(a_0\cdots a_{n-1}))| \ll \sum_{j=0}^{n-2}\Theta_{\max,n-j}+2\pi,\end{split}\end{align*} $$

which is $o(n)$ by Lemma 2.5.

Proof. By Lemma 2.7, the copies $g_0\cdots g_{n-1}R$ and $a_0\cdots a_{n-1}R$ of R share a common side of N, or else share a common vertex of N in $\mathbb D$ . The triangle inequality yields

$$ \begin{align*}|t_n(\gamma)-d(0,a_0\cdots a_{n-1}0)|\leq n(R)\max\{d(0,g0)\colon g\in G_R\} \ll1.\end{align*} $$

Hence, it suffices to show that for all $n\geq 1$ and $a_0\cdots a_{n-1}\in E^n(\Sigma ^+)$ ,

(2.4) $$ \begin{align}1 \ll\frac{|\Theta(a_0\cdots a_{n-1})|}{\exp(-d(0,a_0\cdots a_{n-1}0))}\ll D_n.\end{align} $$

Let $n\geq 1$ and let $a_0\cdots a_{n-1}\in E^n(\Sigma ^+)$ . Let $\xi _+$ and $\xi _-$ denote the boundary points of $\Theta (a_0\cdots a_{n-1})$ . Let $\theta>0$ denote the angle between the geodesic arcs joining $a_0\cdots a_{n-1}0$ to $\xi _+$ and $\xi _-$ . Since all $a_0,\ldots , a_{n-1}$ are Möbius transformations, equation (2.3) gives

(2.5) $$ \begin{align}\theta_{0}\leq \theta\leq 2\pi .\end{align} $$

Split $\Theta (a_0\cdots a_{n-1})$ into three disjoint arcs $\Theta _+$ , $\Theta _0$ , $\Theta _-$ so that $\xi _+\in \Theta _+$ , $\xi _-\in \Theta _-$ and the $\overline {a_{0}\cdots a_{n-1}}$ -images of the three arcs have the same Euclidean lengths. We use $\Theta _+$ and $\Theta _-$ as a buffer, and estimate $|\Theta _0|$ rather than $|\Theta (a_0\cdots a_{n-1})|$ itself. The mean value theorem gives

(2.6) $$ \begin{align}\frac{1 }{3D_n}\leq\frac{|\Theta_0 |}{|\Theta(a_0\cdots a_{n-1})|}\leq\min\bigg\{\frac{D_n}{3},1\bigg\}.\end{align} $$

Let $d(0,a_0\cdots a_{n-1}0)=r$ . Rotate the Poincaré disk so that $a_0\cdots a_{n-1}0$ is placed on the negative part of the real axis. By Lemma 2.6, there exists $n_0\geq 1$ such that if $n\geq n_0$ , then for any $a_0\cdots a_{n-1}\in E^{n}(\Sigma ^+)$ , $\Theta (a_0\cdots a_{n-1})$ is contained in the Euclidean open ball of radius $1/100$ about $-1$ . In particular, $\Theta (a_0\cdots a_{n-1})$ does not contain $1$ . We apply the Möbius transformation $T\colon \mathbb P^1\to \mathbb P^1$ given by

$$ \begin{align*}T(z)=\frac{\cosh({r}/{2})z+\sinh({r}/{2})}{\sinh({r}/{2})z+\cosh({r}/{2})}.\end{align*} $$

This carries the four geodesics through $a_0\cdots a_{n-1}0$ separating $\Theta _+$ , $\Theta _0$ , $\Theta _-$ to rays through $0$ at an equal angle $\theta /3$ . Since $1\notin \Theta (a_0\cdots a_{n-1})$ and $T(1)=1$ , $1\notin T(\Theta (a_0\cdots a_{n-1}))$ holds. Therefore, $T(\Theta _0)$ lies in the complement of the domain $\{z\in \mathbb D\colon |\mathrm {arg}(z)|\leq \theta _0/3\}$ . A calculation shows

$$ \begin{align*}|(T^{-1})'z|=\bigg|\kern-3pt\cosh\frac{r}{2}-\mathrm{Re}(z)\sinh\frac{r}{2}-\sqrt{-1}\mathrm{Im}(z)\sinh\frac{r}{2}\bigg|^{-2}.\end{align*} $$

Since $T(\Theta _0)$ is uniformly bounded away from $1$ in the Euclidean distance, we have $|(T^{-1})'z|\asymp e^{-r}$ . Since $|\Theta _0|=\int _{T(\Theta _0)}|(T^{-1})'z||{d}z|$ , this yields

(2.7) $$ \begin{align}|\Theta_0|\asymp \theta e^{-r}.\end{align} $$

Combining equations (2.5), (2.6) and (2.7), we obtain equation (2.4).

Proof. Let $(a_n)_{n=0}^{\infty }$ denote the f-expansion of $\gamma ^+$ . By the mean value theorem, there exists $\xi \in \Theta (a_0\cdots a_{n-1})$ such that $|(f^n)'\xi ||\Theta (a_0\cdots a_{n-1})|=|f^n(\Theta (a_0\cdots a_{n-1}))|$ . By equation (2.3), we have $|f^n(\Theta (a_0\cdots a_{n-1}))|\in [\theta _0,2\pi ]$ , and

$$ \begin{align*}\theta_0D_n^{-1}|\Theta(a_0\cdots a_{n-1})|^{-1}\leq|(f^n)'\gamma^+|\leq 2\pi D_n|\Theta(a_0\cdots a_{n-1})|^{-1}.\end{align*} $$

This together with Proposition 2.9 yields the desired double inequalities. The rest of the assertions follow from Proposition 2.8.

Proof. If G has no parabolic element, some power of f is uniformly expanding [Reference Series37, Theorem 5.1]. Hence, we have $\alpha _->0$ by Proposition 2.10. If G has a parabolic element, then R has a cusp, which is a neutral periodic point of f. By [Reference Jaerisch and Takahasi14], the set ${\{x\in \Lambda \colon \overline \chi (x)=\underline {\chi }(x)=0\}}$ has positive Hausdorff dimension, while $\Lambda \setminus \Lambda _c$ is a countable set. Hence, we have $\mathscr {L}(0)\neq \emptyset $ and $\mathscr {H}(0)\neq \emptyset $ by Proposition 2.10. Hence, we obtain $\alpha _-=0$ .

Proof. By [Reference Bowen and Series7, Lemmas 2.3 and 2.5], $f|_{\Lambda }$ is a transitive Markov map with respect to the partition of $\varDelta $ into arcs with endpoints given by two consecutive points in W. By Lemma 3.2 below, the proof of which is similar to that of [Reference Bowen and Series7, Lemma 2.3], f is also a transitive Markov map with respect to the finite Markov partition $(\varDelta (a))_{a\in S}$ .

Let $P \in W'$ . Since f clearly preserves cusps and improper vertices of R, we may assume in the following that P is neither a cusp nor an improper vertex of R.

First suppose that there exists a vertex $v\in V\cap \mathbb D$ such that $P\in W'(v)$ . Let $i\in \mathbb Z$ satisfy $v\in C(\bar e_{i-1})\cap C(\bar e_{i})$ . We distinguish three cases. (i) If $P\in [P_{i-1},P_i)$ , then $f(P) \in \bar e_{i-1}(P) \in W(\bar e_{i-1}(v))=W'(\bar e_{i-1}(v))$ . (ii) If $P\in [P_{i},P_{i+1})$ , then $f(P) \in \bar e_{i+1}(P) \in W(\bar e_{i+1}(v))=W'(\bar e_{i+1}(v))$ . (iii) If $P\in [P_{i+1},P_{i-1})$ , then, by [Reference Bowen and Series7, Lemma 2.2], we have $P=Q_{i+1}$ . Since P is neither a cusp nor an improper vertex, it follows that $P\in W'(v')$ for the vertex $v'\in V\cap \mathbb {D}$ of R next to v in anticlockwise order. Hence, by case (i) with v replaced by $v'$ , we obtain $f(P)\in W'$ .

Next we suppose that there exists a cusp $v'\in V_c$ such that $P \in W'(v')$ . Let $j\in \mathbb Z$ satisfy $v'\in C(\bar e_{j-1})\cap C(\bar e_{j})$ . Then we have $P\in \partial L_1(v')$ or $P\in \partial R_2(v')$ . First suppose that $P\in \partial R_2(v')$ . By equation (3.5), we have $f(P)\in \partial R_1(\bar e_{j-1}(v'))\subset W'(\bar e_{j-1}(v'))$ . Now suppose that $P\in \partial L_1(v')$ . If $P\in \partial L_1(v')\cap \partial L_2(v)$ , then we have $f(P)\in \partial L_1(\bar e_{j}(v'))\subset W'(\bar e_{j}(v'))$ by equation (3.5). Otherwise, we have $P=Q_{j+1}\in \partial L_1(v')\cap \partial R_1(v')$ . Since P is neither a cusp nor an improper vertex, it follows that $P\in W'(v")$ for the vertex $v"\in V\cap \mathbb {D}$ of R next to $v'$ in anticlockwise order. Hence, by case (i) with v replaced by $v"$ , we obtain $f(P)\in W'$ . This completes the proof of $f(W')\subset W'$ .

It remains to show $\lim _{\xi \nearrow U_{i+1}} f(\xi )\in W'$ for any $i\in \mathbb Z$ , where $U_{i+1}=P_{i+1}$ if $C(\bar e_i)\cap C(\bar e_{i+1})\neq \emptyset $ and $U_{i+1}=Q_{i+1}$ otherwise (see §2.2). In the former case, with $v\in C(\bar e_i)\cap C(\bar e_{i+1})$ , we have $\lim _{\xi \nearrow U_{i+1}} f(\xi )=\lim _{\xi \nearrow P_{i+1}} \bar e_i (\xi ) \in W'(\bar e_i(v))$ . In the latter case, we have $\lim _{\xi \nearrow U_{i+1}} f(\xi )= \bar e_i(Q_{i+1})\in W'$ , because $e_i(Q_{i+1})$ is an improper vertex of R. This completes the proof of the lemma and that of Proposition 3.1.

Proof. If G is of the first kind, then clearly all the three sets are equal to $\mathbb S^1$ . Suppose G is of the second kind. Then $\varDelta $ is the union of finitely many arcs. Points in $\partial \varDelta \setminus \varDelta $ are improper vertices of R. Since each improper vertex of R is paired with another, it is easy to see that improper vertices of R are not limit points and, in particular, $\partial \varDelta \setminus \varDelta $ is not contained in $\Lambda $ . Interior points of the complement of $\varDelta $ are not limit points, because no copy of R can accumulate at such a point. We have verified that $\Lambda \subset \varDelta $ .

Since $\Lambda $ is G-invariant and $\Lambda \subset \varDelta $ , we obtain $\Lambda \subset \bigcap _{n=0}^{\infty } f^{-n}(\varDelta )$ . To prove the equalities in the proposition, we first show the next lemma.

To complete the proof of Proposition 3.3, it remains to show $\pi (X)\subset \Lambda $ . Since $f|_{\Lambda }$ is transitive by Proposition 3.1, the periodic points of $\sigma $ are dense in X. Since $\pi $ is continuous, it suffices to show that for any $k\geq 1$ and any fixed point $x=(x_n)_{n=0}^{\infty }\in X$ of $\sigma ^k$ , $\pi (x)\in \Lambda $ holds. Observe that the Möbius transformation $\overline {x_0\cdots x_{k-1}}\in G$ satisfies $\overline {x_0\cdots x_{k-1}}(\pi (x))=\pi (x)$ . Since $x_0\cdots x_{k-1}$ is not the identity in G by Proposition 2.3, and since $\Lambda $ contains all fixed points of elements of $G\setminus \{1 \}$ in $\mathbb S^1$ , we obtain $\pi (x)\in \Lambda $ .

Proof. The coding map $\pi $ is one-to-one except on the preimage of the countable set $B=\bigcup _{n=0}^{\infty } f^{-n}(\bigcup _{a\in S}\partial \varDelta (a))$ . Since $\mathbb S^1$ is one-dimensional, $\pi $ is at most two-to-one on B. Since f preserves boundary points of the elements of the Markov partition, $f^{-1}(B)=B$ and so $\sigma ^{-1}(\pi ^{-1}(B))=\pi ^{-1}(B).$

We have $f\circ \pi =\pi \circ \sigma $ , and the restriction of $\pi $ to $X\setminus \pi ^{-1}(B)$ has a continuous inverse. Hence, $\pi $ induces a measurable bijection between $X\setminus \pi ^{-1}(B)$ and $\pi (X)\setminus B$ . This and $\Lambda \subset \pi (X)$ in Proposition 3.3 imply that for any $\mu \in \mathcal M(\Lambda ,f)$ with $\mu (B)=0$ , there exists $\nu \in \mathcal M(X,\sigma )$ such that $\mu =\nu \circ \pi ^{-1}$ and $h(\mu )=h(\nu )$ .

If $\mu \in \mathcal M(\Lambda ,f)$ and $\mu (B)>0$ , there exist $\rho \in (0,1]$ and $\mu _1,\mu _2\in \mathcal M(\Lambda ,f)$ such that $\mu _1(B)=0$ , $\mu _2(B)=1$ and $\mu =(1-\rho )\mu _1+\rho \mu _2$ . Since B is a countable set, $\mu _2$ is supported on a periodic orbit of f. By Proposition 3.3, there exists $\nu _2\in \mathcal M(X,\sigma )$ that is supported on a periodic orbit of $\sigma $ and satisfies $\mu _2=\nu _2\circ \pi ^{-1}$ . By the previous paragraph, there exists $\nu _1\in \mathcal M(X,\sigma )$ with $\mu _1=\nu _1\circ \pi ^{-1}$ . Set $\nu =(1-\rho )\nu _1+\rho \nu _2$ . Then $\mu =\nu \circ \pi ^{-1}$ and $h(\mu )=(1-\rho )h(\mu _1)=(1-\rho )h(\nu _2)=h(\nu )$ , as required in the first assertion of the lemma. A proof of the second one is analogous.

Proof. Using Proposition 2.8 and the irreducibility of the finite Markov shift X in Proposition 3.1, we can construct a measure supported on periodic points whose Lyapunov exponent is arbitrarily close to $\alpha _+$ . Hence, $\sup \{\overline {\chi }(\xi )\colon \xi \in \Lambda _c\}\leq \sup \{\chi (\mu )\colon \mu \in \mathcal M(\Lambda ,f)\}$ holds. The reverse inequality follows from Birkhoff’s ergodic theorem. Combining this equality with $\alpha _+=\sup \{\overline {\chi }(\xi )\colon \xi \in \Lambda _c\}$ which follows from Proposition 2.10, we obtain the first equality in the lemma. A proof of the second one is analogous.

Proof. From Lemma 3.6 and $\alpha _-\geq 0.$

Proof. Let $g=e_{i_1}\cdots e_{i_{|g|}}$ be a shortest representation of g. We replace all anticlockwise half-cycles in this representation by the corresponding clockwise half-cycles, and obtain (possibly) another shortest representation $g=e_{j_1}\cdots e_{j_{|g|}}$ that contains no anticlockwise half-cycle. By Proposition 2.3, $e_{j_1}\cdots e_{j_{|g|}}\in E(\Sigma ^+)$ holds.

Let $g=e_{j_1}\cdots e_{j_{|g|}}$ , $g=e_{k_1}\cdots e_{k_{|g|}}$ be two admissible shortest representations of g. Suppose $e_{j_1}\neq e_{k_1}$ . Then we have a relation $\bar e_{k_{|g|}}\cdots \bar e_{k_1}e_{j_1}\cdots e_{j_{|g|}}=1$ . Since the vertex cycles give a complete set of the relations of G and both representations of g are shortest, $\bar e_{k_{|g|}}\cdots \bar e_{k_1}e_{j_1}\cdots e_{j_{|g|}}$ contains a cycle that contains $\bar e_{k_1}e_{j_1}$ . It follows that one of the two representations of g contains an anticlockwise half cycle, and this yields a contradiction since both representations are admissible. Hence, we obtain $e_{j_1}=e_{k_1}$ . Repeating this argument, we obtain $e_{j_i}\neq e_{k_i}$ for $1\leq i\leq |g|$ .

Proof of Proposition 3.8

For $n\geq 1$ and $a_0\cdots a_{n-1}\in E^n(\Sigma ^+)$ , let $E^n(X,a_0\cdots a_{n-1})$ denote the set of elements of $\omega $ in $E^{n}(X)$ such that $\varDelta (\omega )\subset \Theta (a_0\cdots a_{n-1})$ . Clearly,

(3.8) $$ \begin{align}1\leq\#E^n(X,a_0\cdots a_{n-1})\leq 2n(R).\end{align} $$

By equation (2.3), for $x\in \Theta (a_0\cdots a_{n-1})$ ,

(3.9) $$ \begin{align}\theta_0D_n^{-1}\leq\frac{|\Theta(a_0\cdots a_{n-1})|}{|(f^{n})'x|^{-1}}\leq 2\pi D_n.\end{align} $$

By Proposition 2.9, and equations (3.8) and (3.9), there exists a constant $C\geq 1$ such that for all $\beta ,t\in \mathbb R$ , we have

(3.10) $$ \begin{align} C^{-\beta}D_n^{-2\beta}(2\pi)^{-\beta}&\leq\frac{\sum_{\omega\in E^n(X,a_0\cdots a_{n-1})}\sup_{[\omega]}\exp(\beta\sum_{k=0}^{n-1}\varphi\circ\sigma^k)e^{-tn}}{\exp(-\beta d(0,a_0\cdots a_{n-1}0))e^{-tn}} \nonumber\\[-15pt]& \\[-3pt]&\leq 2n(R) \cdot C^\beta D_n^{2\beta}\theta_0^{-\beta}.\nonumber \end{align} $$

By Lemma 3.9 and Proposition 2.3, there is a one-to-one correspondence between $G\setminus \{1\}$ and $E(\Sigma ^+)$ . Therefore, rearranging the double inequalities in equation (3.10), summing the result over all words in $ E^{n}(\Sigma ^+)$ , then summing the result over all $n\geq 1$ and then using [Reference Mauldin and Urbański21, Theorem 2.1.3], we obtain

$$ \begin{align*}P(\beta\varphi,\sigma)\leq\inf\bigg\{t\in\mathbb R\colon\sum_{n=1}^\infty D_n^{2\beta}\sum_{g\in G,|g|=n}\exp(-\beta d(0,g0)-t|g|)<+\infty\bigg\},\end{align*} $$

and $P(\beta \varphi ,\sigma )\leq P(\beta )$ . A similar reasoning shows the reverse inequality. Lemma 3.5 implies $P(\beta \varphi ,\sigma )=P(\beta \phi ,f)$ . This completes the proof of Proposition 3.8.

Proof. Let $C_0>0$ be so large that the hyperbolic disk around $0$ of radius $C_0$ covers all of the intersection of R and the Nielsen region of G [Reference Beardon3, §8.5] except small neighbourhoods of the cusps. Let $M_0\ge 1$ be a large number to be determined later.

Suppose for a contradiction the assertion of the lemma fails. Then there exists $\gamma \in \mathscr {R}$ with infinite cutting sequence $(g_n)_{n=0}^\infty $ (see Lemma 2.2) and there exists $k\ge M_0$ such that $f^k(\gamma ^+)\in \tilde \varDelta $ and, for every $n\in \{k-M_0,\ldots , k+M_0\}$ ,

$$ \begin{align*}d(g_0\cdots g_n0,\gamma\cap g_0\cdots g_nR)> C_0.\end{align*} $$

This means that $\gamma $ performs a deep cusp excursion between the $(k-M_0)$ th and the $(k+M_0)$ th crossing of fundamental domains and therefore, the cutting symbols $g_{k-M_0},\ldots , g_{k+M_0}$ of $\gamma $ are given by the periodic sequence of labels of sides ending at one of the cusps of R, say $v_0\in V_c$ . We conclude by Lemmas 2.4 and 2.7 that the partial BS orbit $(a_{0}\cdots a_{n}0)_{n\colon |n-k|\leq M_0-2}$ appears in the same order in the partial cutting orbit $(g_0\cdots g_n0)_{n\colon |n-k|\leq M_0}$ . This implies that the first $M_0 -2$ symbols of the f-expansion of $f^k(\gamma ^+)$ are given by the periodic sequence of sides ending at the cusp $v_0$ . If $M_0$ is large enough depending on the prime periods of the cusps, this implies $f^k(\gamma ^+)\in L(v_0)\cup R(v_0)$ contradicting $f^k(\gamma ^+) \in \tilde \varDelta $ .

Proof. Let $C_0$ and $M_{0}$ denote the constants in Lemma 4.1. Let $n> M_0$ . There exists $k\geq n$ and $a_1\cdots a_k{\kern-0.5pt}\in{\kern-0.5pt} E^k(\Sigma ^+)$ such that $\tilde {\omega }_{0}\cdots \tilde {\omega }_{n-1} {\kern-0.5pt}={\kern-0.5pt} a_1 \cdots a_k $ and ${f^{k}(\gamma ^{+}){\kern-0.5pt}={\kern-0.5pt}(\tilde {f})^{n}(\gamma ^{+}){\kern-0.5pt}\in{\kern-0.5pt} \tilde {\varDelta }}$ . By Lemma 4.1, there exists $n_{*}\in \{k-M_{0},\ldots ,k+M_{0}\}$ such that $d(g_{0}\cdots g_{n_{*}}0,\gamma \cap g_{0}\cdots g_{n_{*}}R)\leq C_0$ . By the triangle inequality, we have

$$ \begin{align*} d(a_1\cdots a_k0,\gamma) \le 2M_{0}\max_{g\in G_{R}}d(0,g0) + d(a_{1}\cdots a_{n_{*}}0,g_0\cdots g_{n_{*}}0) + C_0. \end{align*} $$

Since the second term of the right-hand side does not exceed $n(R)\max _{g\in G_R}\{d(0,g0)\}$ by Lemma 2.7, the proposition follows.

Proof. Let $C_0$ and $M_{0}$ denote the constants in Lemma 4.1. Let $n> M_0$ and $\tilde \omega _0\cdots \tilde \omega _{n-1}\in E^{n}(\tilde X)$ . Let $\gamma \in \mathscr {R}$ such that $\gamma ^+ \in \Lambda _c\cap \tilde \varDelta (\tilde \omega _0\cdots \tilde \omega _{n-1})$ with the cutting sequence $(g_j)_{j=0}^{\infty }$ . By Lemma 4.1, for each $k\in \{M_0,\ldots ,n-1\}$ , we fix an integer $j(k)$ such that $|j(k)-|\tilde \omega _0\cdots \tilde \omega _k||\leq M_0$ and

(4.1) $$ \begin{align} d(g_0 \cdots g_{j(k)}0,\gamma\cap g_0 \cdots g_{j(k)}R)\leq C_0. \end{align} $$

We write all the distinct elements of the sequence $j(M_0),\ldots ,j(n-1)$ as $j_1,j_2,\ldots ,j_q$ in the increasing order with some $q\ge (n-1-M_0)/(2M_0)$ . For each $1\le k \le q$ , there exists $p_k\in \gamma \cap g_{0}\cdots g_{j_k}R$ such that $d(g_{0}\cdots g_{j_k}0,p_k)\le C_0.$ Using equation (4.1) and Lemma 2.7, we derive the existence of a uniform constant $C'>0$ such that

$$ \begin{align*} d(0,\tilde\omega_0\cdots \tilde\omega_{n-1}0) \ge d(p_1,p_q) -C'. \end{align*} $$

Divide the geodesic segment from $p_1$ to $p_q$ into segments of hyperbolic length $C_0$ , with one shorter segment, say $\gamma _1,\ldots , \gamma _N$ , for some $N\ge 1$ . By equation (4.1), for each ${1\leq \ell \leq q}$ , the orbit point $g_{0}\cdots g_{j_\ell }0$ is within the hyperbolic distance $C_0$ of one of the geodesic segments $\gamma _1,\ldots , \gamma _N$ .

Partition the set $\{g_{0}\cdots g_{j_\ell }0\colon 1\leq \ell \leq q\}$ into subsets $O_1,\ldots , O_N$ so that $d(O_k, \gamma _k) \le C_0$ for $1\le k\le N$ . Since G acts properly discontinuously on $\mathbb D$ , there exists an integer $M\geq 1$ such that $\# O_k\le M$ for $1\le k\le N$ . Hence, $N\ge q/M$ . Combining this with equation (4.1) yields

$$ \begin{align*}d(0,\tilde\omega_0\cdots \tilde\omega_{n-1}0)\geq C_0\bigg(\frac{q}{M}-1\bigg)-C'\geq C_0\bigg(\frac{n-1-M_0}{2M_0M}-1\bigg)-C'.\end{align*} $$

Hence, the lemma follows for $\alpha _0=C_0/(3M_0M)$ and sufficiently large n.

Proof. Let $C>0$ denote the constant in Proposition 4.2. Let $n\geq 1$ and $\tilde \omega _0\cdots \tilde \omega _{n-1}\in E^{n}(\tilde X)$ and let $\xi \in \tilde \varDelta (\tilde {\omega }_{0}\cdots \tilde {\omega }_{n-1})\cap \Lambda _c$ . Let $\gamma \in \mathscr {R}$ be the ray through zero with $\gamma ^{+}=\xi $ . By Proposition 4.2, for all sufficiently large $n\geq 1$ , we have

(4.2) $$ \begin{align} d(\tilde{\omega}_{0}\cdots\tilde{\omega}_{n-1}0,\gamma)\le C. \end{align} $$

Since $\tilde {f}^{n}(\gamma ^{+})=(\tilde {\omega }_{0}\cdots \tilde {\omega }_{n-1})^{-1}\gamma ^{+}$ , it follows from the well-known properties of the Poisson kernel [Reference Beardon3] that

$$ \begin{align*}\log|(\tilde{f}^{n})'\gamma^{+}|=d(0,p),\end{align*} $$

where $p\in \mathbb D$ denotes the point of intersection between $\gamma $ and the horocircle at $\gamma ^{+}$ through $\tilde {\omega }_{0}\cdots \tilde {\omega }_{n-1}0$ . By equation (4.2), we have $ d(p,\tilde {\omega }_{0}\cdots \tilde {\omega }_{n-1}0)\le 2C $ and thus,

$$ \begin{align*} |\log|(\tilde{f}^{n})'\gamma^{+}|-d(0,\tilde{\omega}_{0}\cdots\tilde{\omega}_{n-1}0)|\le2C. \end{align*} $$

The first assertion of the proposition now follows from Lemma 4.3.

To prove the second assertion, first note that the estimate in equation (4.2) remains intact if $\gamma \in \mathscr {R}$ is a ray through zero whose endpoint $\gamma ^+$ is in between two points in $\tilde \varDelta (\tilde {\omega }_{0}\cdots \tilde {\omega }_{n-1})\cap \Lambda _c$ . Consequently, the first assertion of the proposition also holds if $\xi $ is taken from the smallest arc in $\mathbb {S}^{1}$ containing $\tilde \varDelta (\tilde {\omega }_{0}\cdots \tilde {\omega }_{n-1})\cap \Lambda _c$ . From this and the mean value theorem, the second assertion of the proposition follows.

Proof. If G has a parabolic element, then $\alpha _-=0$ by Lemma 2.12, and $\alpha _+>0$ since the induced Markov interval map $\tilde f$ is uniformly expanding by Proposition 4.4. If G has no parabolic element, it follows from [Reference Lalley19, Corollary 11.3] that the function $\varphi $ in equation (3.7) is not cohomologous to a constant. Since f is piecewise $C^2$ and some iterate of f is uniformly expanding, $\varphi $ is Hölder continuous. By a standard argument [Reference Bowen6, Proposition 4.5], we conclude that $\alpha _-<\alpha _+$ .

Proof. The case where G has no parabolic element has already been explained. Suppose G has a parabolic element. If equation (1.1) holds, then f is a non-uniformly expanding, finitely irreducible Markov map in the sense of [Reference Jaerisch and Takahasi14]. Since $\Lambda \setminus \bigcup _{n=0}^\infty f^{-n}(V_c)$ is contained in $\bigcup _{n=0}^\infty f^{-n}(\tilde \Lambda )$ and $V_c$ is countable, we have $\dim _{\mathrm {H}}(\Lambda ) = \dim _{\mathrm {H}}(\tilde \Lambda )$ . Hence, f is saturated by [Reference Jaerisch and Takahasi14, Proposition 5.2(c)]. Even if equation (1.1) does not hold, we have shown in Proposition 4.4 that some power of the induced Markov map $\tilde f$ is uniformly expanding. Hence, the argument in the proof of [Reference Jaerisch and Takahasi14, Proposition 5.2(c)] works almost verbatim to conclude that f is saturated.

Proof. Proposition 2.10 gives $\mathscr {H}(\alpha )=\mathscr {L}(\alpha )$ , and so $b(\alpha )=\dim _{\mathrm {H}}\mathscr {L}(\alpha )$ . If G has no parabolic element, then some power of f is uniformly expanding [Reference Series37, Theorem 5.1], and so the result is well known, see for example [Reference Olsen24, Reference Pesin26, Reference Pesin and Weiss28, Reference Pesin and Weiss29, Reference Weiss42].

Suppose G has a parabolic element. The assertion in item (a) follows from [Reference Jaerisch and Takahasi14, Main Theorem(a)]. To derive the desired formula in item (b), we aim to apply [Reference Jaerisch and Takahasi14, Main Theorem(b)]. By Proposition 3.1, f is a finitely irreducible Markov map. By Proposition 2.8, f has mild distortion, and by Proposition 5.2, f is saturated. In addition to these conditions, in [Reference Jaerisch and Takahasi14, Main Theorem(b)], it is assumed that the map satisfies a non-contracting condition as in equation (1.1). However, the non-contracting condition was used in [Reference Jaerisch and Takahasi14] only to ensure the non-existence of points with negative pointwise Lyapunov exponent. Although we do not assume the Bowen–Series map f satisfies equation (1.1), the formulae in [Reference Jaerisch and Takahasi14, Main Theorem(b)] remain intact for the level sets $\mathscr L(\alpha )$ since all points in these sets have non-negative pointwise Lyapunov exponents. This proves the desired formula in item (b).

Let $\alpha \in [\alpha _-,\alpha _+]\setminus \{0\}$ . To remove the limit $\varepsilon \to 0$ in equation (5.2), we use Lemma 3.5 to transfer the problem to $\mathcal M(X,\sigma )$ . Since the function $\varphi :X\rightarrow \mathbb {R}$ is continuous and the entropy function is upper semicontinuous on the compact space $\mathcal M(X,\sigma )$ , we can choose a convergent sequence $\{\mu _n\}_{n=1}^{\infty }$ in $\mathcal M(X,\sigma )$ with positive entropy such that its weak* limit point $\mu _{\infty }$ is an expanding measure satisfying $\dim (\mu _{\infty }\circ \pi ^{-1})=b(\alpha ).$ This yields the desired formula in item (c).

Proof. Put $\delta _0=\sup \{\dim (\mu )\colon \mu \in \mathcal M(\Lambda ,f)\}$ . Since f is saturated by Proposition 5.2, we have $\delta _0=\delta _G.$ Set $\delta _1=\mathrm {\min }\{\beta \geq 0\colon P(\beta )=0\}.$ It suffices to show $\delta _0=\delta _1$ . By definition, $\delta _0\geq \dim (\mu )$ holds for any expanding measure $\mu \in \mathcal M(\Lambda ,f)$ . Hence, $P(\delta _0)\leq 0$ and so $\delta _0\geq \delta _1$ . Suppose for a contradiction that $\delta _0>\delta _1$ . Then there exists $\varepsilon>0$ such that $\delta _0>\delta _1+\varepsilon $ and an expanding measure $\mu $ such that $\dim (\mu )>\delta _1+\varepsilon $ , and so $P(\delta _1+\varepsilon )>0$ . However, by the definition of $\delta _1$ and the monotonicity of pressure, we have $P(\delta _1+\varepsilon ) \leq P(\delta _1+\varepsilon /2)\leq 0$ , and a contradiction arises. Therefore, $\delta _0=\delta _1$ holds.

Proof. By Proposition 5.1(a), we have $\beta _+<\infty .$ Suppose for a contradiction that $\lim _{\beta \nearrow \beta _+} P'(\beta )<0.$ Take a sequence $\{\beta _n\}_{n=1}^{\infty }$ with $\beta _n\nearrow \beta _+$ as $n\to \infty $ and ${\lim _{n\to \infty }P'(\beta _n)<0}.$ Let $\mu _{\beta _n}$ be the equilibrium state for the potential $-\beta _n\log |f'|$ and let $\mu $ be an weak* accumulation point of $\{\mu _{\beta _n}\}_{n=1}^{\infty }$ . Recall that X is a finite Markov shift where the entropy function is upper semicontinuous, and the function $\varphi \colon X\to \mathbb R$ in equation (3.7) is continuous. Hence, $\mu $ is an equilibrium state for the potential $-\beta _+\log |f'|$ , namely, ${h(\mu )-\beta _+ \chi (\mu )=0}$ . Since $P'(\beta _n)=-\chi (\mu _{\beta _n})$ by Proposition 5.1(c), we have $\chi (\mu )=\lim _{n\to \infty }\chi (\mu _{\beta _n})=-\lim _{n\to \infty }P'(\beta _n)>0.$ By Proposition 5.4, $\mu $ is a measure of maximal dimension for G, which is a contradiction to Proposition 5.5.

Proof. Lemma 2.12 gives $\alpha _-=0$ . By the definition of $\beta _+$ , the pressure is equal to zero on $[\beta _+,+\infty )$ . By Proposition 5.4, we have $\beta _+=\delta _G$ . By Proposition 5.1(b), P is analytic on $(-\infty ,\beta _+)$ . The continuous differentiability of P at $\beta =\delta _G$ is a consequence of Lemma 5.6 and the convexity of P. This shows that P is $C^1$ on $\mathbb R$ . Since P is analytic, convex and non-increasing on $(-\infty , \delta _G)$ , an elementary inductive argument on the power series expansion of P shows that either P is affine on $(-\infty ,\delta _G)$ or strictly convex on $(-\infty ,\delta _G)$ . The first case is ruled out by Lemma 5.6 and the assumption that G is non-elementary, which gives $\delta _G>0$ .

Proof. We have $P(\beta (\alpha ))+\alpha \beta (\alpha )= h(\mu _{\beta (\alpha )})-\beta (\alpha )\alpha +\alpha \beta (\alpha ) =h(\mu _{\beta (\alpha )})\leq \alpha b(\alpha )$ , where the last inequality follows from Proposition 5.3(c). Again by Proposition 5.3(c), there exists an expanding measure $\mu \in \mathcal M(\Lambda ,f)$ such that $\chi (\mu )=\alpha $ and $\dim (\mu )=b(\alpha )$ . Then, $\alpha b(\alpha )=h(\mu )= h(\mu )-\beta (\alpha )\alpha +\alpha \beta (\alpha ) \leq P(\beta (\alpha ))+\alpha \beta (\alpha )$ , and so the first equality in equation (5.4) holds.

The second equality in equation (5.4) follows from the first. Since P and $\beta $ are analytic, the $\mathscr {H}$ -spectrum is analytic on $(\alpha _-,\alpha _+)$ by the first formula in equation (5.4). Differentiating the first equality in equation (5.4), and combining with the first equality in equation (5.4) and the fact that $P'(\beta (\alpha ))=-\alpha $ , yields the equality in equation (5.5).

Proof. Recall that $\beta (\alpha )$ is the unique solution of the equation $P'(\beta )+\alpha =0$ . By the implicit function theorem, $\alpha \in (\alpha _-,\alpha _+)\mapsto \beta (\alpha )$ is differentiable and

(5.7) $$ \begin{align}\beta'(\alpha)=-\frac{1}{P"(\beta(\alpha))}.\end{align} $$

We apply l’Hôpital’s rule to equation (5.5) together with equation (5.7) to obtain $\lim _{\alpha \searrow \alpha _-}b'(\alpha )=-\lim _{\beta \nearrow \delta _G}1/(2P"(\beta ))$ when one of the two limits exists. Since $P"(\beta )>0$ for $\beta <\delta _G$ , we obtain $\lim _{\alpha \searrow \alpha _-}b'(\alpha )=-\infty $ . Since $\lim _{\alpha \nearrow \alpha _+}b'(\alpha )=-\infty $ by the Main Theorem, we must have $b"(\alpha _*)=0$ for some $\alpha _*\in (\alpha _-,\alpha _+)$ .