2 Preliminary results

Let $\mu $ be a probability measure on a countable group G. We assume that $\mu $ is generating: that is, the semigroup generated by the support of $\mu $ equals G. We define the step space as $(G^{\mathbb {N}}, \mu ^{\mathbb {N}})$ and the map $\pi : G^{\mathbb {N}} \to G^{\mathbb {N}}$ as $\pi ((g_n)_{n \in \mathbb {N}}) := (w_n)_{n \in \mathbb {N}}$ , with for any n

$$ \begin{align*}w_n := g_1 g_2 \dots g_n.\end{align*} $$

The target space of $\pi $ is denoted by $\Omega $ and called the path space; as a set, it equals $G^{\mathbb {N}}$ and is equipped with the measure $\mathbb {P}_\mu := \pi _\star (\mu ^{\mathbb {N}})$ .

Then we define the first-passage function $F_\mu (x, y)$ as

$$ \begin{align*}F_\mu(x, y) := \mathbb{P}_\mu(\exists n \ : \ w_n x = y)\end{align*} $$

for any $x, y \in G$ , and the Green metric $d_\mu $ on G, introduced in [Reference Blachère and Brofferio6], as

$$ \begin{align*}d_\mu(x, y) := - \log F_\mu(x, y).\end{align*} $$

The following fact is well-known.

Lemma 2.1. Let $p : G \to H$ be a group homomorphism, let $\mu $ be a probability measure on G, and let $\overline {\mu } := p_\star \mu $ . Then for any $x, y \in G$ ,

$$ \begin{align*}d_{\overline{\mu}}(p(x), p(y)) \leq d_\mu(x, y).\end{align*} $$

Proof. Since p induces a map from paths in G to paths in H, we have $\overline {\mu }^n(p(g)) \geq \mu ^n(g)$ for any $g \in G$ , any $n \geq 0$ . Hence

$$ \begin{align*}\mathbb{P}_{\overline{\mu}}(p(x), p(y)) \geq \mathbb{P}_\mu(x, y)\end{align*} $$

for any $x, y \in G$ , from which the claim follows.

We shall use the following criterion, which relates the absolute continuity of the hitting measure to the fundamental inequality. Recall that a group action is geometric if it is isometric, properly discontinuous and cocompact.

For each $g \in G$ , let $\ell (g)$ denote its translation length, namely

$$ \begin{align*}\ell(g) := \lim_{n \to \infty} \frac{d_{\mathbb{H}}(o, g^n o)}{n}.\end{align*} $$

Equivalently, $\ell (g)$ is the length of the corresponding closed geodesic on the quotient surface. The mechanism to utilise Theorem 2.2 is through the following lemma, similar to the one from [Reference Kosenko32].

Lemma 2.3. Suppose that the hitting measure is absolutely continuous. Then for any $g \in G$ , we have

$$ \begin{align*}\ell(g) \leq d_\mu(1, g).\end{align*} $$

Proof. If not, then $\ell (g)> d_\mu (1, g) \geq 0$ , and hence g is loxodromic. Let us pick some $o \in \mathbb {H}^2$ , which lies on the axis of g, so that $d_{\mathbb {H}}(o, g^k o) = \ell (g^k) = k \ell (g)$ for any k. Moreover, by the triangle inequality for the Green metric, one has $d_\mu (1, g^k) \leq k d_\mu (1, g)$ , and hence

$$ \begin{align*}d_{\mathbb{H}}(o, g^k o) - d_\mu(1, g^k) \geq k \ell(g) - k d_\mu(1, g) = k (\ell(g) - d_\mu(1, g));\end{align*} $$

thus, since $\ell (g) - d_\mu (1, g)> 0$ ,

$$ \begin{align*}\sup_{k \in \mathbb{N}} \left|d_{\mathbb{H}}(o, g^k o) - d_\mu(1, g^k) \right| = + \infty,\end{align*} $$

which contradicts Theorem 2.2.

Let F be a free group, freely generated by a finite set S. Recall that the (hyperbolic) boundary $\partial F$ of F is the set of infinite, reduced words in the alphabet $S \cup S^{-1}$ . Given a finite, reduced word g, we denote as $C(g) \subseteq \partial F$ the cylinder determined by g, namely the set of infinite, reduced words that start with g.

Lemma 2.4. Consider a random walk on the free group

$$ \begin{align*}F_m = \left\langle s_1^{\pm 1}, \dots, s_m^{\pm 1} \right\rangle,\end{align*} $$

defined by a probability measure $\mu $ on the generators. If we denote $x_i := F_\mu (1, s_i)$ , $\check {x}_i := F_\mu (1, s_i^{-1})$ and the hitting measure on the boundary of $F_m$ by $\nu $ , then

$$ \begin{align*} \nu(C(s_i)) = \frac{x_i (1 - \check{x}_{i})}{1 - x_i \check{x}_{i} }. \end{align*} $$

A similar lemma is stated in [Reference Lalley34, Exercise 5.14].

Proof. For any infinite word $w = s_{j_1} s_{j_2} s_{j_3} \dots $ , there exist two possibilities:

1. 1. There exists a subword $s_{j_1} \dots s_{j_k}$ such that it equals $s_i$ in $F_m$ .

2. 2. No subword $s_{j_1} \dots s_{j_k}$ equals $s_i$ , so it belongs to the set of paths that never hit $s_i$ .

In the first case, we denote this subword by $w_1$ , and we consider $w_1^{-1} w$ ; we apply the same procedure, but replacing $s_i$ with $s_i^{-1}$ at each subsequent step. This procedure yields the equality

$$ \begin{align*} \begin{aligned} \nu(C(s_i)) &= \mathbb{P}(1 \rightarrow s_i \nrightarrow 1) + \mathbb{P}(1 \rightarrow s_i \rightarrow 1 \rightarrow s_i \nrightarrow 1) + \dots = \\ &= \sum_{n=0}^{\infty} F_\mu(1, s_i)^{n+1} F_\mu(1, s_i^{-1})^{n} (1 - F_\mu(1, s_i^{-1})) \\ & = F_\mu(1, s_i)(1 - F_\mu(1, s_i^{-1})) \sum_{n=0}^{\infty} \left(F_\mu(1, s_i)F_\mu(1, s_i^{-1})\right)^n = \frac{x_i (1 - \check{x}_{i})}{1 - x_i \check{x}_{i} }. \end{aligned}\\[-36pt] \end{align*} $$

3 A criterion for singularity

Theorem 3.1. Let $\mu $ be a finitely supported measure on a cocompact Fuchsian group, and let S be the support of $\mu $ . Suppose that

(3.1) $$ \begin{align} \sum_{g \in S \cup S^{-1}} \frac{1}{1 + e^{\ell(g)}} < 1. \end{align} $$

Then the hitting measure $\nu $ on $\partial \mathbb {D}$ is singular with respect to the Lebesgue measure.

Proof. Denote as $(g_i^{\pm })_{i = 1}^{m}$ the elements of $S \cup S^{-1}$ , let F be a free group of rank m with generators $(h_i)_{i =1}^{m}$ , and let $\widetilde {\mu }$ be a measure on F with $\widetilde {\mu }(h_i^\pm ) = \mu (g_i^\pm )$ . Moreover, let us denote

$$ \begin{align*} x_i &:= F_{\widetilde{\mu}}(1, h_i) = \mathbb{P}_{\widetilde{\mu}}( \exists n \ : \ w_n = h_i) \\ \check{x}_i & := F_{\widetilde{\mu}}(1, h_i^{-1}). \end{align*} $$

Then we have

(3.2) $$ \begin{align} \sum_{i =1}^{m} \left( \frac{x_i (1 - \check{x}_{i})}{1 - x_i \check{x}_{i} } + \frac{\check{x}_i (1 - x_{i})}{1 - x_i \check{x}_{i} } \right) = 1. \end{align} $$

Indeed, if $\widetilde {\nu }$ is the hitting measure on $\partial F$ , by Lemma 2.4, the measure of the cylinder $C(h_i)$ starting with $h_i$ is

$$ \begin{align*}\widetilde{\nu}(C(h_i)) = \frac{x_i (1 - \check{x}_{i})}{1 - x_i \check{x}_{i} }, \qquad \widetilde{\nu}(C(h_i^{-1})) = \frac{\check{x}_i (1 - x_{i})}{1 - x_i \check{x}_{i} },\end{align*} $$

from which, since the cylinders are disjoint and cover the boundary, equation (3.2) follows.

Then by equation (3.1), there exists an index i such that

$$ \begin{align*}\frac{2}{1 + e^{\ell(g_i)}} < \frac{x_i (1 - \check{x}_{i})}{1 - x_i \check{x}_{i} } + \frac{\check{x}_i (1 - x_{i})}{1 - x_i \check{x}_{i} },\end{align*} $$

which is equivalent to

$$ \begin{align*}e^{\ell(g_i)}> \frac{2 - x_i - \check{x}_i}{x_i + \check{x}_i - 2 x_i \check{x}_i}.\end{align*} $$

Finally, an algebraic computation yields

$$ \begin{align*}\frac{2 - x_i - \check{x}_i}{x_i + \check{x}_i - 2 x_i \check{x}_i} \geq \min \left\{ \frac{1}{x_i}, \frac{1}{\check{x}_i} \right\},\end{align*} $$

and thus we obtain

(3.3) $$ \begin{align} \ell(g_i)> \inf \{ - \log x_i, - \log \check{x}_i \}. \end{align} $$

If the hitting measure $\nu $ on $S^1 = \partial \mathbb {D}$ is absolutely continuous, then by Lemma 2.3 and Lemma 2.1, we get

$$ \begin{align*}\ell(g_i) \leq d_{\mu}(1, g_i) \leq d_{\widetilde{\mu}}(1, h_i) = - \log x_i\end{align*} $$

for any i. If we apply the same inequality to $g_i^{-1}$ , we also have

$$ \begin{align*}\ell(g_i) = \ell(g_i^{-1}) \leq d_{\mu}(1, g_i^{-1}) \leq d_{\widetilde{\mu}}(1, h_i^{-1}) = - \log \check{x}_i,\end{align*} $$

and hence

$$ \begin{align*}\ell(g_i) \leq \inf \{ -\log x_i, -\log \check{x}_i \},\end{align*} $$

which contradicts equation (3.3), showing that $\nu $ is singular with respect to Lebesgue measure.

4 Parametrisation of the space of polygons

Let P be a convex, compact polygon in the hyperbolic disk $\mathbb {D}$ , with $2m$ sides and interior angles $\{ \gamma _1, \dots , \gamma _{2m} \}$ .

We say that P is centrally symmetric if there exists a point $o \in \mathbb {D}$ so that P is invariant under symmetry with respect to the point O. This clearly implies that opposite sides have equal length and opposite angles are equal.

Poincaré’s theorem provides conditions to ensure that the group generated by side pairings is discrete (see [Reference Maskit39]). In particular, one needs a condition on the angles, which in our setting can be formulated as follows.

Definition 4.1. A centrally symmetric polygon P satisfies the cycle condition if there exists an integer $k \geq 1$ such that

$$ \begin{align*}\sum_{i = 1}^{m} \gamma_{2 i} = \sum_{i = 1}^m \gamma_{2i - 1} = \frac{2 \pi}{k}.\end{align*} $$

Let $S := \{ g_1, \dots , g_{2 m} \}$ be the set of hyperbolic translations identifying opposite sides of P, with $g_{i+m} = g_i^{-1}$ . By Poincaré’s theorem [Reference Maskit39], if the polygon P satisfies the cycle condition, then the group G generated by S is discrete.Footnote ¹

More precisely, denote as $g_i$ the hyperbolic translation mapping the ith side of P to its $(i+m)$ th side. If m is even, there is only one equivalence class of vertices, and it is fixed by the transformation

$$ \begin{align*}b := g_{m} g_{m-1}^{-1} \dots g_2 g_1^{-1} g_{m}^{-1} g_{m-1} \dots g_2^{-1} g_1,\end{align*} $$

which is called the cycle transformation in the language of [Reference Maskit39]. If m is odd, then there are two equivalence classes of vertices, fixed, respectively, by the transformations

$$ \begin{align*}b_1 := g_m g_{m-1}^{-1} \dots g_2^{-1} g_1 \qquad \mathrm{and} \qquad b_2 := g_m^{-1} g_{m-1} \dots g_2 g_1^{-1}.\end{align*} $$

Thus the presentations defining the group G are

$$ \begin{align*}\begin{array}{ll} \langle g_1, \dots, g_m \ : \ b^k = 1 \rangle& \mathrm{if}\ m\ \mathrm{is\ even} \\ \langle g_1, \dots, g_m \ : \ (b_1)^k = (b_2)^k = 1 \rangle & \mathrm{if\ }m\mathrm{\ is\ odd.} \end{array}\end{align*} $$

The following is our main geometric inequality.

Theorem 4.2. Let P be a centrally symmetric, hyperbolic polygon satisfying the cycle condition, with $2m$ sides, and let $S := \{ g_1, \dots , g_{2 m} \}$ be the set of hyperbolic translations identifying opposite sides of P. Then we have

(4.1) $$ \begin{align} \sum_{g \in S} \frac{1}{1 + e^{\ell(g)}} < 1. \end{align} $$

Remarks. The inequality (4.1) has the same form as the main inequality in [Reference Anderson, Canary, Culler and Shalen1] and [Reference Culler and Shalen14] for free Kleinian groups; more recently, a stronger version for free Fuchsian groups has been obtained in [Reference He28], while generalisations in variable curvature (and any dimension) are due to [Reference Hou29], [Reference Balacheff and Merlin3].

Equation (4.1) is also reminiscent of McShane’s identity [Reference McShane40], where one obtains the equality by taking the infinite sum over all group elements of a punctured torus group. Our inequality, however, does not follow from any of them; in fact, it is in a way stronger than these, as a cocompact surface group can be deformed to a finite covolume group and then to a Schottky (hence free) group by increasing the translation lengths of the generators.

It is interesting to point out that the above inequalities have an interpretation in terms of hitting measures of stochastic processes (see, e.g., [Reference Labourie and Tan33]). Here, we go along the opposite route: we prove the geometric inequality (4.1), and then we use it to conclude properties about the hitting measure.

Finally, there are generating sets of G for which equation (4.1) fails. Indeed, the mechanism behind the inequality is that since all curves corresponding to $(g_i)_{i = 1}^{m}$ intersect each other, by the collar lemma, at most one of them can be short. In general, on a surface of genus g one can choose a configuration of $3 g - 3$ short curves and construct a Dirichlet domain for which the corresponding side pairing does not satisfy equation (4.1).

The proof of this inequality will take up most of the paper until Section 6. To begin with, let us note that a way to parametrise the space of all symmetric hyperbolic polygons is to write, by [Reference Buser10, Example 2.2.7],

(4.2) $$ \begin{align} \cos(\gamma_i) = - \cosh(a_i) \cosh(a_{i+1}) \cos(\alpha_i) + \sinh(a_i) \sinh(a_{i+1}) \end{align} $$

with $i = 1, \dots , m$ , where $(a_i)$ are the distances between the base point and the ith side, $(\alpha _i)$ are the angles at the origin and $(\gamma _i)$ are the angles at the vertices (see Figure 2). Since $\ell (g_i) \geq 2 a_i$ , it is enough to show

$$ \begin{align*}\sum_{i = 1}^m \frac{1}{1 + e^{2 a_i}} < \frac{1}{2}\end{align*} $$

under the constraints $\sum _{i=1}^m \alpha _i = \pi $ and $\sum _{i=1}^m \gamma _i = \frac {2 \pi }{k}$ .

Figure 2 Angles at the centre and vertices of a symmetric hyperbolic octagon.

The fundamental geometric idea in our approach to Theorem 4.2 is that two intersecting curves cannot both be short, as a consequence of the collar lemma [Reference Buser9]. For instance, we get:

Proof. From the collar lemma [Reference Buser9], we have

$$ \begin{align*}\sinh(a_i) \sinh(a_j) \geq 1\end{align*} $$

for all $i \neq j$ . Recall that

$$ \begin{align*}\frac{2}{1 + e^{2a}} = 1 - \tanh(a),\end{align*} $$

and hence, if we set $s := \sinh (a_1)$ , we obtain for $i \neq 1$ that $\sinh (a_i) \geq \frac {1}{s}$ ; thus

$$ \begin{align*}\tanh(a_i) = \frac{\sinh(a_i)}{\sqrt{1 + \sinh(a_i)^2} } = \frac{1}{\sqrt{1+\frac{1}{\sinh(a_i)^2}}} \geq \frac{1}{\sqrt{1 + s^2}},\end{align*} $$

and hence

$$ \begin{align*}\sum_{i = 1}^m \tanh(a_i) \geq \frac{s}{\sqrt{1 + s^2}} + \frac{m-1}{\sqrt{1 + s^2}}> m -1\end{align*} $$

if and only if $s < \frac {2 (m-1)}{m (m-2) }$ .

To actually prove Theorem 4.2, however, we need an improvement on the previous estimate. Let us rewrite equation (4.2) above as

$$ \begin{align*}\cos(\alpha_i) = \tanh(a_i) \tanh(a_{i+1}) - \frac{\cos(\gamma_i)}{\cosh(a_i) \cosh(a_{i+1})};\end{align*} $$

and recalling that

$$ \begin{align*}\tanh^2(x) + \frac{1}{\cosh^2(x)} = 1,\end{align*} $$

we obtain, by setting $z_i = \tanh (a_i)$ ,

(4.3) $$ \begin{align} \cos(\alpha_i) = z_i z_{i+1} - \cos(\gamma_i) \sqrt{1 - z_i^2} \sqrt{1 - z_{i+1}^2} \end{align} $$

with $0 \leq z_i \leq 1$ . Finally, we want to show

$$ \begin{align*}\sum_{i = 1}^m \frac{1}{1 + e^{2 a_i}} = \sum_{i = 1}^m \frac{1 - z_i}{2} \overset{?}{<} \frac{1}{2},\end{align*} $$

which is equivalent to

(4.4) $$ \begin{align} \sum_{i = 1}^m z_i \overset{?}{>} m - 1. \end{align} $$

Now, let us first assume that $\gamma _i \leq \pi /2$ for all $1 \leq i \leq m$ . Then equation (4.3) yields

$$ \begin{align*}\cos(\alpha_i) \leq z_i z_{i+1},\end{align*} $$

and hence the constraint becomes

(4.5) $$ \begin{align} \sum_{i = 1}^m \arccos(z_i z_{i+1}) \leq \pi. \end{align} $$

Note that $z_1 \to 0$ implies $\cos \alpha _1 \leq z_1 z_2 \to 0$ and thus $\alpha _1 \to \frac {\pi }{2}$ , and $\cos \alpha _m \leq z_m z_1 \to 0$ and thus $\alpha _m \to \frac {\pi }{2}$ , and hence also $\alpha _2, \alpha _3, \dots , \alpha _{m-1} \to 0$ , which implies $z_2, z_3, \dots , z_m \to 1$ .

5 An optimisation problem

By the above discussion, if we set $x_i = 1 - z_i$ , we reduce the proof of Theorem 4.2 (at least in the case all angles of P are acute) to the following optimisation problem (see also Figure 3).

Theorem 5.1. Let $m \geq 3$ and $0 \leq x_i \leq 1$ with $\sum _{i=1}^m x_i = 1$ . Then

$$ \begin{align*}\sum_{i = 1}^m \arccos((1-x_i) (1-x_{i+1})) \geq \pi.\end{align*} $$

Moreover, equality holds if and only if there exists an index i such that $x_i = 1$ and $x_j = 0$ for all $j \neq i$ .

Figure 3 The graph of $f(x) := \sum _{i = 1}^3 \mathrm {arccos}((1-x_i)(1-x_{i+1}))$ subject to the constraint $\sum _{i = 1}^3 x_i = 1$ , compared with the constant function at height $\pi $ . The lack of convexity (or concavity) of f makes the proof of Theorem 5.1 trickier.

In the statement of Theorem 5.1 and elsewhere from now on, all indices i are meant modulo m. The next is the main technical lemma.

Lemma 5.2. Let $m \geq 3$ and $0 \leq x_i \leq 1$ with $\sum _{i=1}^m x_i = 1$ . Then

$$ \begin{align*}\sum_{i = 1}^m \sqrt{x_i + x_{i+1} - x_i x_{i+1}} \geq \sqrt{ 4 + 3 \sum_{i =1}^m x_i x_{i+1}}.\end{align*} $$

Proof. Set $\Delta _i := x_i + x_{i+1} - x_i x_{i+1}$ . Note that

$$ \begin{align*}\Delta_i \geq \max\{ x_i, x_{i+1}\},\end{align*} $$

and hence

(5.1) $$ \begin{align} \sqrt{\Delta_i}\sqrt{\Delta_{i+1}} \geq x_{i+1}. \end{align} $$

Moreover, since $m \geq 2$ , we have $x_{i+1} + x_{i+2} \leq \sum _{i =1}^m x_i = 1$ , and hence if we multiply by $(x_{i+1} + x_{i+2})$ , we obtain

$$ \begin{align*} \Delta_i & = x_i + x_{i+1} - x_i x_{i+1} \\ & \geq (x_i + x_{i+1}) (x_{i+1} + x_{i+2}) - x_i x_{i+1} \\ & \geq x_{i+1}^2 + x_{i+1} x_{i+2}. \end{align*} $$

Similarly, we obtain

$$ \begin{align*} \Delta_{i+2} & = x_{i+2} + x_{i+3} - x_{i+2} x_{i+3} \\ & \geq (x_{i+2} + x_{i+3})(x_{i+1} + x_{i+2}) - x_{i+2} x_{i+3} \\ & \geq x_{i+2}^2 + x_{i+1} x_{i+2}. \end{align*} $$

Thus Cauchy-Schwarz yields

(5.2) $$ \begin{align} \sqrt{\Delta_i}\sqrt{\Delta_{i+2}} \geq \sqrt{ x_{i+1}^2 + x_{i+1} x_{i+2}} \sqrt{ x_{i+2}^2 + x_{i+1} x_{i+2}} \geq 2 x_{i+1} x_{i+2}. \end{align} $$

By squaring both sides, our desired inequality is equivalent to

$$ \begin{align*}\sum_{i =1}^m \Delta_i + 2 \sum_{1 \leq i < j \leq m} \sqrt{\Delta_i} \sqrt{\Delta_j} \geq 4 + 3 \sum_{i =1}^m x_i x_{i+1};\end{align*} $$

thus, using $\sum _{i =1}^m \Delta _i = 2 - \sum _{i = 1}^m x_i x_{i+1}$ , it is enough to prove

(5.3) $$ \begin{align} \sum_{1 \leq i < j \leq m} \sqrt{\Delta_i} \sqrt{\Delta_j} \geq 1 + 2 \sum_{i = 1}^m x_i x_{i+1}. \end{align} $$

Now note that

$$ \begin{align*}\sum_{1 \leq i < j \leq m} \sqrt{\Delta_i} \sqrt{\Delta_j} = \sum_{i = 1}^m \sqrt{\Delta_i} \sqrt{\Delta_{i+1}} + M\end{align*} $$

with

(5.4) $$ \begin{align} \hphantom{\hspace{-0pt}}M & = 0 \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\mathrm{if\ } m = 3\phantom{.} \end{align} $$

(5.5) $$ \begin{align} M & = \sum_{i = 1}^2 \sqrt{\Delta_i} \sqrt{\Delta_{i+2}} & \mathrm{if\ }m = 4\phantom{.} \end{align} $$

(5.6) $$ \begin{align} M & \geq \sum_{i=1}^m \sqrt{\Delta_i} \sqrt{\Delta_{i+2}} & \mathrm{if\ }m \geq 5. \end{align} $$

Thus for $m \geq 5$ , we have, using equations (5.6), (5.1) and (5.2),

$$ \begin{align*} \sum_{1 \leq i < j \leq m} \sqrt{\Delta_i} \sqrt{\Delta_j} & \geq \sum_{i=1}^m \sqrt{\Delta_i} \sqrt{\Delta_{i+1}} + \sum_{i=1}^m \sqrt{\Delta_i} \sqrt{\Delta_{i+2}} \\ & \geq \sum_{i=1}^m x_{i+1} + 2 \sum_{i=1}^m x_{i+1} x_{i+2} \\ & \geq 1 + 2\sum_{i=1}^m x_{i+1} x_{i+2}, \end{align*} $$

which yields equation (5.3) and hence completes our proof. The cases $m = 3$ and $m = 4$ need to be dealt with separately. If $m = 3$ , we obtain, by multiplying by $\sum _{i = 1}^3 x_i = 1$ ,

$$ \begin{align*}\Delta_i = x_i^2 + x_{i+1}^2 + \sum_{i = 1}^3 x_i x_{i+1},\end{align*} $$

so by Cauchy-Schwarz, we get

$$ \begin{align*}\sqrt{\Delta_i} \sqrt{\Delta_{i+1}} \geq x_{i+1}^2 + x_i x_{i+2} + \sum_{i = 1}^3 x_i x_{i+1},\end{align*} $$

and hence

$$ \begin{align*} \sum_{i=1}^3 \sqrt{\Delta_i} \sqrt{\Delta_{i+1}} & \geq \sum_{i=1}^3 x_i^2 + 4 \sum_{i=1}^3 x_i x_{i+1} \\ & = \left( \sum_{i=1}^3 x_i \right)^2 + 2 \sum_{i = 1}^3 x_i x_{i+1} \\ & = 1 + 2 \sum_{i = 1}^3 x_i x_{i+1}, \end{align*} $$

which yields equation (5.3), as desired. Finally, if $m = 4$ , then we note

$$ \begin{align*}\sum_{1 \leq i < j \leq 4} \sqrt{ \Delta_i } \sqrt{\Delta_j} = \sum_{i = 1}^4 \sqrt{\Delta_i} \sqrt{\Delta_{i+1}} + \sum_{i = 1}^2 \sqrt{\Delta_i} \sqrt{\Delta_{i+2}}\end{align*} $$

and, again by Cauchy-Schwarz,

$$ \begin{align*}\sqrt{\Delta_1} \sqrt{\Delta_3} \geq \sqrt{x_1^ 2 + x_2^2 + x_1 x_4 + x_2 x_3} \sqrt{x_3^ 2 + x_4^2 + x_1 x_4 + x_2 x_3} \geq 2 x_1 x_4 + 2 x_2 x_3\end{align*} $$

and similarly

$$ \begin{align*}\sqrt{\Delta_2} \sqrt{\Delta_4} \geq 2 x_1 x_2 + 2 x_3 x_4;\end{align*} $$

thus, using equation (5.1),

$$ \begin{align*}\sum_{1 \leq i < j \leq 4} \sqrt{ \Delta_i } \sqrt{\Delta_j} \geq \sum_{i = 1}^4 x_i + 2 \sum_{i = 1}^4 x_i x_{i+1} = 1 + 2 \sum_{i = 1}^4 x_i x_{i+1},\end{align*} $$

which is again equation (5.3). This completes the proof.

Proof of Theorem 5.1

By setting $f(x) := \frac {2}{\pi } \arccos (1-x)$ , our claim is equivalent to

$$ \begin{align*}\sum_{i = 1}^m f(x_i + x_{i+1} - x_i x_{i+1}) \geq 2\end{align*} $$

under the constraint $\sum _{i=1}^m x_i = 1$ , with $m \geq 3$ and $0 \leq x_i \leq 1$ .

Let us set $\Delta _i := x_i + x_{i+1} - x_i x_{i+1}$ and $\sigma := \sum _{i =1}^m x_i x_{i+1}$ . Observe that $2 \sigma \leq (\sum _{i=1}^m x_i)^2 = 1$ . Then we have, by Lemma 5.3,

$$ \begin{align*} \sum_{i = 1}^m f(\Delta_i) & \geq \frac{2}{3} \sum_{i=1}^m \sqrt{\Delta_i} + \frac{1}{3} \sum_{i=1}^m \Delta_i \end{align*} $$

and using Lemma 5.2 and the fact $\sum _{i=1}^m \Delta _i = 2 - \sigma $ , we obtain

$$ \begin{align*} &\geq \frac{2}{3} \sqrt{4 + 3 \sigma} + \frac{1}{3} (2- \sigma) \geq 2, \end{align*} $$

where in the last step, we apply Lemma 5.3 (2). This completes the proof of the inequality. By Lemma 5.3 (1), equality implies that $\Delta _i = 0, 1$ for every i, which in turn implies that $x_i = 0, 1$ for all i. Since $\sum _{i = 1}^m x_i = 1$ , this can only happen if $x_i = 1$ for exactly one index i.

6 The obtuse angle case

The proof in the previous section works as long as all angles $\gamma _i$ are less than or equal to $\pi /2$ . If one of them is obtuse, we have a geometric argument to reduce ourselves to that case.

6.1 Neutralising pairs

We call a neutralising pair for P a pair $\{\gamma _i, \gamma _{i+1}\}$ of adjacent interior angles of P with $\gamma _i + \gamma _{i+1} \leq \pi $ . Whenever we have a neutralising pair, we can apply the following lemma (see Figure 4).

Lemma 6.1. Let $ABCDE$ be a hyperbolic pentagon with right angles $\widehat {B}$ and $\widehat {E}$ , and suppose that $\widehat {C} < \pi /2$ and $\widehat {C} + \widehat {D} \leq \pi $ . Then there exist points F on the line $\overline {BC}$ and G on the line $\overline {ED}$ such that the hyperbolic pentagon $ABFGE$ satisfies $\widehat {F} \leq \pi /2$ and $\widehat {G} \leq \pi /2$ .

Figure 4 The hyperbolic pentagon of Lemma 6.1.

Proof. Let M be the midpoint of $\overline {CD}$ , and let F be the foot of the orthogonal projection of M to $\overline {BC}$ . If the intersection, which we denote as G, of the lines $\overline {FM}$ and $\overline {ED}$ exists, then we claim that the angle $\delta = D \widehat {G} F$ satisfies $\delta \leq \pi /2$ .

To see this, let $F'$ be the symmetric point to F with respect to M. Then $CFM$ and $DMF'$ are equal triangles. Hence $E \widehat {D} F' = E \widehat {D} M + M \widehat {D} F' = E \widehat {D} C + B \widehat {C} D \leq \pi $ , and hence $F'$ lies on the segment $\overline {M G}$ . Moreover, $D \widehat {F'} M = C \widehat {F} M = \pi /2$ , and hence $\delta = D \widehat {G} M \leq \pi /2$ .

If the lines $\overline {FM}$ and $\overline {ED}$ do not intersect, we take as G the foot of the perpendicular from M to the line $\overline {ED}$ , and we define as $F"$ the intersection of $\overline {GM}$ with the line $\overline {BC}$ . Then the hyperbolic pentagon $ABF'GE$ satisfies $\widehat {F''} < \pi /2$ and $\widehat {G} = \pi /2$ , so the claim follows by replacing F by $F''$ .

We say that P has disjoint neutralising pairs if there is a set $\Pi := \{ \{ \gamma _{i_1}, \gamma _{i_1 + 1} \}, \dots , \{ \gamma _{i_k}, \gamma _{i_k +1} \} \}$ of neutralising pairs such that each obtuse angle of P belongs to some pair in $\Pi $ , and the pairs are disjoint: that is, $\{ \gamma _{i_r}, \gamma _{i_r +1} \} \cap \{ \gamma _{i_s}, \gamma _{i_s +1} \} = \emptyset $ for any $r \neq s$ .

Let us use the notation

$$ \begin{align*}\varphi(x_1, x_2, \dots, x_m) := \sum_{i = 1}^m \frac{1}{1 + e^{2 x_i}}.\end{align*} $$

Proposition 6.2. Let P be a centrally symmetric hyperbolic polygon with $2m$ sides and centre o, and let $\ell _1, \dots , \ell _m$ be the distances between o and the midpoints of the sides. If P has disjoint neutralising pairs, there exists a centrally symmetric hyperbolic $2m$ -gon $P'$ with no obtuse angles and such that

$$ \begin{align*}\varphi(\ell_1, \ell_2, \dots, \ell_m) \leq \varphi(d^{\prime}_1, d^{\prime}_2, \dots, d^{\prime}_m),\end{align*} $$

where $d_i'$ is the distance between o and the ith side of $P'$ .

Proof. Let us denote as $d_i$ the distance between o and the ith side of P. Note that by definition, $d_i \leq \ell _i$ for all i.

If the polygon P only has acute angles, we take $P = P'$ and note that by definition, $d_i' = d_i \leq \ell _i$ , which yields the claim.

Suppose now that the hyperbolic polygon P has one obtuse angle, say $\gamma _1$ , which belongs to a neutralising pair, and let $\ell _1$ correspond to the side adjacent to the obtuse angle and the other angle, say $\gamma _2$ , in the neutralising pair. Consistently with this choice, let us denote as $s_1, s_2, \dots , s_{2m}$ the sides of P.

Let us now consider the hyperbolic pentagon delimited by $s_{2m}, s_1, s_2$ and the orthogonal projections from o to $s_2$ and $s_{2m}$ . Let us call this pentagon $ABCDE$ , where $o = A$ , the side $s_1$ is denoted $\overline {DC}$ , the orthogonal projection from o to $s_2$ is B and the orthogonal projection from o to $s_{2m}$ is E.

Using Lemma 6.1, let us replace P by a new polygon $P'$ obtained by substituting the pentagon $ABCDE$ by the pentagon $ABFGE$ , which satisfies $\widehat {F} = \pi /2 $ and $\widehat {G} \leq \pi /2$ . If we denote by $d_1'$ the distance between $o = A$ and $\overline {FG}$ , then we have

$$ \begin{align*}d_1' = d(A, \overline{FG}) \leq \ell_1.\end{align*} $$

On the other hand, note that for $i = 2, \dots , m$ , the distance between o and the ith side is the same for P and $P'$ . That is, $d_i = d_i'$ for $i = 2, \dots , m$ . Hence,

$$ \begin{align*}\varphi(\ell_1, \ell_2, \dots, \ell_m) \leq \varphi(\ell_1, d_2, \dots, d_m) \leq \varphi(d_1', d_2, \dots, d_m) = \varphi(d_1', d_2', \dots, d_m').\end{align*} $$

If there is more than one neutralising pair (by symmetry, the number of neutralising pairs is even), we can analogously replace each side adjacent to the pair by rotating it around its midpoint. This proves the claim.

6.2 The general case

Let $(p_i)_{i =1}^{2m}$ denote the vertices of P and $(q_i)_{i =1}^{2m}$ denote the midpoints of the sides, indexed so that $q_i$ lies between $p_{i-1}$ and $p_{i}$ . Let o denote the centre of symmetry of P. Let $\alpha _i = q_i \widehat {o} q_{i+1}$ be the angles at the origin and $\gamma _i = q_i \widehat {p}_i q_{i+1}$ the angles at the vertices of P. By the cycle condition and symmetry, we have

$$ \begin{align*}\sum_{i = 1}^m \alpha_i = \pi, \qquad \sum_{i = 1}^{m} \gamma_i = \frac{2 \pi}{k},\end{align*} $$

where $k \geq 1$ is an integer. Note that if $k \geq 2$ , at most one of the $\gamma _i$ is obtuse, and hence P has disjoint neutralising pairs. However, if $k = 1$ , P need not have disjoint neutralising pairs; in particular, it may have three consecutive obtuse angles. To deal with this case, we need the notion of a dual polygon.

6.3 Dual polygons

Given a centrally symmetric polygon P with centre o and sum of angles $4 \pi $ , we construct its dual polygon $\widehat {P}$ as follows.

Let $Q_i$ be the quadrilateral delimited by $o, q_i, p_i, q_{i+1}$ . As in Figure 5, we can cut and rearrange the $Q_i$ s with $1 \leq i \leq m$ as follows. Glue all vertices $(p_i)_{i = 1}^m$ to a single point, which we now denote as v, so that the $Q_i$ s with $1 \leq i \leq m$ lie in counterclockwise order around v. Since $\sum _{i=1}^{m} \gamma _i = 2 \pi $ , the copies of $Q_i$ fit together, creating a new polygon $\widehat {P}$ with m sides. By construction, the sides of $\widehat {P}$ have lengths $2 \ell _1, \dots , 2 \ell _m$ , with $\ell _i = d(o, q_i)$ . Also by construction, the angles of $\widehat {P}$ are $\alpha _1, \dots , \alpha _m$ , and hence their sum is $\sum _{i = 1}^m \alpha _i = \pi $ .

Figure 5 The construction of the dual polygon. Top: the polygon P, in blue. Bottom: the dual polygon $\widehat {P}$ , in red. The angles $\alpha _i$ at the origin in P become the angles at the vertices of $\widehat {P}$ ; on the other hand, the angles $\gamma _i$ at the vertices of P become the angles at the point $\widehat {v}$ in the interior of $\widehat {P}$ . Quadrilaterals of the same colour are congruent. The point $\widehat {v}$ is the common intersection of the four coloured regions in the bottom picture.

We define the pair $(\widehat {P}, v)$ to be the dual polygon to $(P, o)$ .

The duality relation

$$ \begin{align*}(P, o) \leftrightarrow (\widehat{P}, v)\end{align*} $$

defines a bijective correspondence between centrally symmetric $2m$ -gons with sum of angles $4 \pi $ and m-gons with the sum of angles $\pi $ together with a choice of a point inside them.

To see that this is a bijection, let us construct the inverse map as follows: given an m-gon $\widehat {P}$ and a point v inside it, denote as $v_1, \dots , v_m$ its vertices. Decompose $\widehat {P}$ as the union of m quadrilaterals $R_1, \dots , R_m$ by drawing the segments joining v and the midpoints of the sides of $\widehat {P}$ ; then take copies of the polygons $R_1, \dots , R_m, R_1, \dots , R_m$ and glue them in this order by identifying all $ v_i$ s to a point, which we call o. This will create a polygon P whose sum of angles is twice the sum of internal angles of $\widehat {P}$ , and hence $4 \pi $ . Moreover, by construction, this polygon is centrally symmetric about the point o.

Given a polygon P with $2m$ sides and a point o inside P, we define

$$ \begin{align*}\Sigma(P) := \sum_{i =1}^{m} \frac{1}{1 + e^{2 \ell_i}},\end{align*} $$

where $\ell _i$ are the segments connecting o and the midpoint of the $ith$ side. Let us also define

$$ \begin{align*}\widehat{\Sigma}(P) := \sum_{i = 1}^m \frac{1}{1 + e^{s_i}},\end{align*} $$

where $s_i$ are the lengths of the sides of P. Then note that we have

$$ \begin{align*}\Sigma(P) = \widehat{\Sigma}(\widehat{P}).\end{align*} $$

In particular, $\Sigma (P)$ does not depend on v but only on $\widehat {P}$ .

Lemma 6.3. Let P be a centrally symmetric hyperbolic polygon with $2m$ sides and the total sum of its interior angles $4 \pi $ . Then there exists a centrally symmetric hyperbolic polygon $P'$ with the same number of sides so that $\Sigma (P) = \Sigma (P')$ so that $P'$ has at most four obtuse angles, which belong to disjoint neutralising pairs.

Proof. Let $\widehat {P}$ be the dual polygon to P, as defined above. We claim that we can pick another point $v'$ inside $\widehat {P}$ so that at most two of the angles at $v'$ are obtuse.

To do this, let $(r_i)_{i = 1}^m$ be the midpoints of the sides of $\widehat {P}$ ; join $r_1$ and $r_3$ by a segment, and pick any point $v'$ in the interior of that segment. Then consider the angles $\gamma _i' := r_i \widehat {v'} r_{i+1}$ with $1 \leq i \leq m$ . Since $\sum _{i = 1}^2 \gamma _i' = \pi $ and $\sum _{i = 3}^m \gamma _i' = \pi $ , at most two of the angles $\gamma _i'$ with $1 \leq i \leq m$ can be obtuse.

Then we define $P'$ to be the dual of $(\widehat {P}, v')$ . Since P and $P'$ have the same dual, we have $\Sigma (P) = \Sigma (P')$ . By our previous choice of $v'$ , in $P'$ , there are at most $4$ obtuse angles $\gamma _i'$ , and there are no three consecutive obtuse angles; hence, for all of them there exists another adjacent angle $\gamma ^{\prime }_{i\pm 1}$ so that $\gamma ^{\prime }_i + \gamma ^{\prime }_{i\pm 1} < \pi $ . Hence, $P'$ has neutralising pairs.

By putting together these reductions, we can complete the proof of Theorem 4.2. Let us see the details.

Proof of Theorem 4.2

Let us first suppose that $\gamma _i \leq \pi /2$ for all i. We know by equation (4.5) that $\sum _{i = 1}^m \arccos (z_i z_{i+1}) \leq \pi $ with $0 < z_i < 1$ . Then we need to show that $\sum _{i = 1}^m z_i> m- 1$ . Suppose not; then there exists $z_i$ with $\sum _{i = 1}^m z_i \leq m-1$ . Then there exists $(z_i')_{i = 1}^m$ with $0 \leq z_i \leq z_i' \leq 1$ for all i, so that $\sum z_i' = m - 1$ . Then we have, by Theorem 5.1, $\pi \leq \sum _{i = 1}^m \arccos ( z_i' z_{i+1}') \leq \sum _{i = 1}^m \arccos ( z_i z_{i+1}) \leq \pi $ , and hence $\sum _{i = 1}^m \arccos ( z_i' z_{i+1}') = \pi $ , which by the second part of Theorem 5.1 implies $z_i' = 0$ for some i, and hence also $z_i = 0$ , which is a contradiction.

In the general case, we first apply Lemma 6.3 to reduce to the case where P has disjoint neutralising pairs. Then by applying Proposition 6.2, we reduce to the case of P having no obtuse angles, which we can deal with as above. This completes the proof.

Proof of Theorem 1.2

Theorem 4.2 shows that the criterion of Theorem 3.1 holds, proving the singularity of hitting measure.

7 Coxeter groups

Let P be a centrally symmetric convex polygon with $2 m$ sides in $\mathbb {H}^2$ , with each angle $\gamma _i$ at the vertices being equal to $\frac {\pi }{k_i}$ for some natural $k_i> 1$ , for $1 \leq i \leq 2m$ . Then due to [Reference Davis15, Theorem 6.4.3], the group of isometries generated by hyperbolic reflections $R := \{ r_1, \dots , r_{2 m} \}$ with respect to the sides of P acts geometrically on $\mathbb {H}^2$ . Therefore, it is a hyperbolic group, so Theorem 2.2 can be applied to it. Such groups are referred to as hyperbolic Coxeter groups.

Below, we will show that Theorem 1.2 can be quickly generalised to hyperbolic Coxeter groups.

Lemma 7.1. Let $m> 1$ . Consider a random walk on the free product of $2m$ copies of $\mathbb {Z} / 2 \mathbb {Z}$

$$ \begin{align*}F^{\prime}_{2m} = \left\langle s_1, \dots, s_{2m} \ | \ s_i^2 = 1 \right\rangle,\end{align*} $$

defined by a probability measure $\mu $ on the generators. If we denote $x_i := F_\mu (1, s_i)$ for $1 \le i \le 2m$ and the hitting measure on the boundary of $F^{\prime }_{2m}$ by $\nu $ , then

$$ \begin{align*} \nu(C(s_i)) = \frac{x_i}{1+x_i}. \end{align*} $$

Proof. The proof of this lemma can be obtained similarly to the proof of Lemma 2.4 for $F_m$ because the Cayley graphs for $F_m$ and $F^{\prime }_{2m}$ are isometric.

More precisely, a sample path converges to the boundary of the cylinder $C(s_i)$ if and only if it crosses the edge $s_i$ an odd number of times. This leads to the following computation:

$$ \begin{align*} \begin{aligned} \nu(C(s_i)) &= \mathbb{P}(1 \rightarrow s_i \nrightarrow 1) + \mathbb{P}(1 \rightarrow s_i \rightarrow 1 \rightarrow s_i \nrightarrow 1) + \dots = \\ &= \sum_{n=0}^{\infty} F_\mu(1, s_i)^{2 n+1} (1 - F_\mu(1, s_i)) \\ & = \sum_{k = 1}^\infty (-1)^{k+1} x_i^k = \frac{x_i}{1 + x_i}.\\[-35pt] \end{aligned} \end{align*} $$

A measure $\mu $ on the set $R = \{ r_1, \dots , r_{2 m} \}$ of reflections through the sides of P is called geometrically symmetric if $\mu (r_i) = \mu (r_{i+m})$ for each $1 \leq i \leq m$ .

Theorem 7.2. Let $\mu $ denote a geometrically symmetric measure supported on the generators $R = \{ r_1, \dots , r_{2 m} \}$ of a hyperbolic Coxeter group. Suppose that

(7.1) $$ \begin{align} \sum_{i=1}^m \dfrac{1}{1 + e^{\ell(r_i r_{i+m})/2}} < \dfrac{1}{2}. \end{align} $$

Then the hitting measure $\nu $ in $\partial \mathbb {D}$ is singular with respect to the Lebesgue measure.

Proof. The proof of this theorem is quite similar to the proof of Theorem 3.1. We consider a measure $\tilde {\mu }$ on a free product $\left \langle h_1, \dots , h_{2m} \ | \ h_i^2 = 1 \right \rangle $ of $2m$ copies of $\mathbb {Z} / 2\mathbb {Z}$ uniquely defined by $\tilde {\mu }(h_i) = \mu (r_i)$ .

If $\nu $ were to be absolutely continuous, then a similar argument would yield that

$$ \begin{align*} \begin{aligned} \ell(r_i r_{i+m}) &\le d_{\mu}(1, r_i r_{i+m}) \le d_{\mu}(1, r_i) + d_{\mu}(1, r_{i+m}) \\ &\le d_{\tilde{\mu}}(1, h_i) + d_{\tilde{\mu}}(1, h_{i+m}) = 2 d_{\tilde{\mu}}(1, h_i) = -2\log \, x_i. \end{aligned} \end{align*} $$

Keep in mind that $d_{\tilde {\mu }}(1, h_i) = d_{\tilde {\mu }}(1, h_{i+m})$ due to $\tilde {\mu }$ being geometrically symmetric as well. Therefore,

$$ \begin{align*} \dfrac{x_i}{1+x_i} \le \dfrac{1}{1 + e^{\ell(r_i r_{i+m})/2}}, \end{align*} $$

and due to Lemma 7.1, we obtain

$$ \begin{align*} 1 = \sum_{i=1}^{2m} \frac{x_i}{1 + x_i} \leq 2 \sum_{i=1}^{m} \dfrac{1}{1 + e^{\ell(r_i r_{i+m})/2}} < 1, \end{align*} $$

which delivers a contradiction.

Proof. Let us recall that $(g_i)^m_{i =1}$ denotes the translations identifying the opposite sides of P. It is easily seen that $\ell (r_i r_{i+m}) = 2\ell (g_i) = 2\ell (g_{i+m})$ for every $1 \le i \le m$ . However, we can apply Theorem 4.2, because there are no obtuse angles, to get

$$ \begin{align*} \sum_{i = 1}^m \frac{2}{1 + e^{\ell(r_i r_{i+m})/2}} = \sum_{g \in S} \frac{1}{1 + e^{\ell(g)}} < 1. \end{align*} $$

We conclude the proof by applying Theorem 7.2.