# Small Systole Sets and Coxeter Groups

Olivier Mathieu, Ingrid Irmer

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Olivier Mathieu, Ingrid Irmer. Small Systole Sets and Coxeter Groups. 2023. hal-04255464

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Distributed under a Creative Commons CC BY 4.0 - Attribution - International License
# SMALL SYSTOLE SETS AND COXETER GROUPS

INGRID IRMER AND OLIVIER MATHIEU

**ABSTRACT.** The systoles of a hyperbolic surface $\Sigma$ are the shortest closed geodesics. We say that the systoles *fill* the surface if the set $\text{Syst}(\Sigma)$ of all systoles cuts $\Sigma$ into polygons.

We refine an idea of Schmutz [15] to construct closed hyperbolic surfaces $\Sigma$ of arbitrarily large genus with a small set $\text{Syst}(\Sigma)$ that fills. In fact, for the surfaces $\Sigma$ considered, the cardinality of $\text{Syst}(\Sigma)$ is in $o(g/\sqrt{\ln g})$, where $g$ is the genus of $\Sigma$. The proof is based on the theory Coxeter groups, combined with some elementary number theory.

## CONTENTS

<table><tr><td>1. Introduction</td><td>1</td></tr><tr><td>2. The 2-dimensional Cayley Complex Cay<sup>+</sup> W</td><td>4</td></tr><tr><td>3. Uniformization of 2k-Regular Tesselations</td><td>9</td></tr><tr><td>4. The subgroup H(k) of W(k)</td><td>17</td></tr><tr><td>5. Bounds on Fill(g)</td><td>19</td></tr><tr><td>References</td><td>21</td></tr></table>

**Keywords** Coxeter groups, Tits representation, hyperbolic surfaces, systoles, Thurston spine.

## 1. INTRODUCTION

### 1.1 General introduction

In this paper, surfaces are assumed to be oriented. A *systole* of a hyperbolic surface $\Sigma$ is an essential closed geodesic of minimal length. Let $\text{Syst}(\Sigma)$ be the set of all systoles of $\Sigma$. We say that $\text{Syst}(\Sigma)$ *fills* the surface if it cuts $\Sigma$ into polygons.

For $g \ge 2$, the Teichmüller space $\mathcal{T}_g$ is a manifold used for parameterising closed hyperbolic surfaces of genus $g$. In [16], Thurston defined a remarkable subspace $\mathcal{P}_g \subset \mathcal{T}_g$, called the *Thurston spine*. It consists of all surfaces $\Sigma \in \mathcal{T}_g$ for which $\text{Syst}(\Sigma)$ fills.

Since $\mathcal{P}_g$ is nonempty by [16], one can meaningfully define the integer $\text{Fill}(g)$ as the smallest cardinality of $\text{Syst}(\Sigma)$ when $\Sigma$ varies over $\mathcal{P}_g$. Trying to understand the dimension of $\mathcal{P}_g$ leads to the question of finding an upper bound for $\text{Fill}(g)$. In this direction, we prove

**Theorem 25.** There exists an infinite set $A$ of integers $g \ge 2$ such that

$$ \text{Fill}(g) \le \frac{57}{\sqrt{\ln \ln \ln g}} \frac{g}{\sqrt{\ln g}} $$

for any $g \in A$.
In fact, the theorem 25 proved in the main body is slightly stronger than the previous statement. Since $57/\sqrt{\ln \ln \ln g}$ belongs to $o(1)$, it implies the result stated in the abstract. From our main result, we will deduce, in a forthcoming preprint [12], a related bound for the codimension of the Thurston spine $\mathcal{P}_g$.

## 1.2 Previous works for Fill(g)

The idea of studying examples for which the systoles cuts the surface into regular right-angled polygons goes back to [15], and this idea has been used extensively in the subsequent works [1][14][2]. Schmutz's paper [15] seems to have been motivated by the observation that many critical points of mapping class group-equivariant Morse functions are of this type. Classical examples are the Bolza surface and the Klein quartic. This investigation also led to the study of upper bounds of Fill(g). Hyperbolic surfaces with 2g systoles have been found in [15],[1] and [14]. The recent result of [2] can be reformulated as follows.

**Theorem.** [2] There is infinite set B of integers $g \ge 2$ and an increasing function $\psi$ (discussed below) such that

$$ \text{Fill}(g) \le g/\psi(g) \text{ for any } g \in B. $$

The function $\psi$ is only implicitly defined in loc. cit., but a rough estimate is given in a footnote. It is clear that $\psi(g)$ is in $o(\lg^* g)$, where $\lg^*$ is, essentially, the inverse of the Ackerman function $n \mapsto F(4, n)$ (see [4], Section 3.2, pp. 58-59). In particular $\lg^*$ is smaller than the $m$th-iterate $\lg^{(m)} = \lg \circ \lg \circ \dots \lg$, of the base-2 logarithmic function $\lg$, for any $m \ge 1$. Therefore $\psi(g)$ is much smaller than the factor $\sqrt{\ln g}$ in our denominator.

Conversely, a rough lower bound $\sim \pi g / \ln g$ has been found in [1]. According to loc. cit., this lower bound seems difficult to obtain. Intuitively, the difficulty comes from the fact that a small number $N$ of filling systoles implies a relatively large systole length, at least $O(g/N)$. However, as a loose general rule, the number of systoles increases with the systole length.

## 1.3 Main ideas of the proof and organisation of the paper

The surfaces studied in this paper were motivated by the examples in Theorem 36 of [15]. In the examples of loc. cit. the systoles cut the surface into regular right-angled polygons. However, here we will use the refined notion of standard tesselations. In order to explain this, we first need to give some definitions.

For simplicity, we will only consider hexagonal tesselations. A decoration of the regular right-angled hexagon $P$ is a cyclic indexing of the edges by $\mathbb{Z}/6\mathbb{Z}$. Since a cyclic indexing of the edges defines an orientation of the hexagon, $P$ admits exactly two decorations, up to orientation preserving isometry. By definition, a standard tesselation of an oriented hyperbolic closed surface $\Sigma$ is a tesselation by decorated regular right-angled hexagons. By definition, the curves of a standard tesselation $\tau$ are the maximal geodesic components of the 1-skeleton $\tau_1$ of $\tau$. The standard tesselation $\tau$ is called $2k$-regular if all its curves consist of exactly $2k$ edges. Since each side of a regular right-angled hexagon has length arcosh 2, all curves of a $2k$-regular standard tesselation are closed geodesics of length $2k$ arcosh 2. We also define the Coxeter group $W(k)$ by the presentation

$$ \langle (s_i)_{i \in \mathbb{Z}/6\mathbb{Z}} | s_i^2 = 1, (s_i s_{i+1})^2 = 1, \text{ and } (s_i s_{i+2})^k = 1, \forall i \in \mathbb{Z}/6\mathbb{Z} \rangle $$
Let $\epsilon: W(k) \to \mathbb{Z}/2\mathbb{Z}$ be the sign homomorphism, defined by $\epsilon(s_i) = -1$, for any $i \in \mathbb{Z}/6\mathbb{Z}$, and set $W(k)^+ = \text{Ker} \, \epsilon$.

Before going to the core of the proof, we will explain the connection between the Coxeter group $W(k)$ and the use of decorated tiles. The tesselations considered here are, somehow, “doubly regular”: the tiles are regular hexagons, and all curves have the same length. These two properties appear in [15] and the subsequent papers [14][2]. The advantage of adding a decoration to the tiles is explained by the following observation, proved in Section 3.

**Theorem12.** *There is a one-to-one correspondence between*

(1) the $2k$-regular standard tesselations $\tau$ of closed oriented surfaces, and

(2) the finite index subgroups $H \subset W(k)^+$ satisfying (11.1) and (11.2).

The assertions (11.1) and (11.2) are described in Lemma 11 of Section 3. These conditions, which are usually satisfied, are easy to check.

We can now describe the main ideas of the proof. Using the previous statement, obtaining a bound on $\text{Fill}(g)$ is reduced to the theory of Coxeter groups. The theory of right-angled Coxeter groups is a classical tool to investigate the tesselations of the Poincare half-plane $\mathbb{H}$ [5]. Our construction is similar, but the Coxeter groups $W(k)$ are not right-angled. The proof contains three parts.

In the first part, namely in Sections 2 and 3, we look at the delicate question - are the set of curves of a $2k$-regular standard tesselation $\tau$ and the set of systoles of the corresponding surface identical? A partial answer is provided at the end of Section 3.

**Criterion 18.** *Assume that $k \ge 4$ is even. Let $H$ be a subgroup of $W(k)^+$, any conjugate of which intersects $B_{4k}$ trivially, where $B_{4k}$ is the ball of radius $4k$ in $W(k)$. Then the set of curves of the corresponding tesselation $\tau$ is the set of systoles.*

The proof of the criterion is quite long, and it uses a new result on the combinatorics of Coxeter groups $W$, namely the Theorem 9 proved in Section 2. We define the *Cayley complex* $\text{Cay}^+ W$ of $W$ by attaching some $2$-cells to its Cayley graph $\text{Cay} W$. Theorem 9 involves Coxeter groups endowed with right-angled partition. It shows that some “relatively short” loops of $\text{Cay} W$ are null-homotopic in $\text{Cay}^+ W$.

In the second part of the proof, i.e. Section 4, we find an upper bound for the index of a subgroup $H$ satisfying Criterion 18:

**Proposition 22.** *For any $k \ge 3$, there exists a normal subgroup $H(k)$ of $W(k)$ satisfying the criterion 18 with*

$$[W(k) : H(k)] \le 3^{72k\phi(2k)},$$

where $\phi$ is Euler totient function.

Its proof uses the *Tit's representation* $\rho: W(k) \to \text{GL}_6(K)$ [17], where $K$ is the number field $\mathbb{Q}(\cos \pi/k)$. We have $H(k) = \{w \in W(k) \mid \rho(w) \in \Gamma\}$, where $\Gamma$ is a suitable congruence subgroup of $\text{GL}_6(K)$.

The last part of the proof, in Section 5, explains the factor $\frac{57}{\sqrt{\ln \ln \ln g}}$. It is based on Landau's Theorem [10][9] about the asymptotics of $\phi(k)/k$, which is a corollary of the classical prime number theorem [6][8].
## 2. THE 2-DIMENSIONAL CAYLEY COMPLEX $Cay^+ W$

Given a Coxeter system $(W, S)$, the *Tits combinatorics* [18] describes the loops in its Cayley graph $Cay W$.

In this section, we define a Cayley complex $Cay^+ W$ obtained by attaching a collection of 2-cells to $Cay W$, for each commutative rank two parabolic subgroup of $W$. This square complex $Cay^+ W$ is unrelated with the well-known simplicial complexes of [5], like the Coxeter's complex and the Davis's complex.

We also define the notion of right-angled partition of a Coxeter group $W$. Theorem 9, proved in this section, states that for a Coxeter group $W$ endowed with a right-angled partition, the “relatively short” loops of $Cay W$ are null-homotopic in $Cay^+ W$. This result will be the main ingredient of the proof of Criterion 18 in Section 3.

### 2.1 Coxeter Groups

Let $S$ be a set. A square matrix $M = (m_{s,t})_{s,t \in S}$ is a *Coxeter matrix* if it satisfies the following conditions:

(1) $m_{s,s} = 1$,

(2) for $s \neq t$, $m_{s,t}$ belongs to $\mathbb{Z}_{\ge 2} \cup \{\infty\}$,

(3) $m_{s,t} = m_{t,s}$.

The group $W$ defined by the presentation

$$ \langle s \in S \mid (st)^{m_{s,t}} = 1, \forall s, t \in S \text{ with } m_{s,t} \neq \infty \rangle $$

is called the *Coxeter group* associated with the Coxeter matrix $M$. Unless stated otherwise, it will be assumed that the set $S$ is finite. Its cardinality is called the *rank* of $W$. The pair $(W, S)$ is called a *Coxeter system*.

Let $\epsilon: W \to \{\pm 1\}$ be the group homomorphism uniquely defined by the property that $\epsilon(s) = -1$ for any $s \in S$. This is called the *signature homorphism*.

Denote by $W_S$ the free monoid generated by $S$. An element $w$ of $W_S$ is a word $w = s_1 \dots s_n$ where $s_1, \dots, s_n$ belong to the alphabet $S$. By definition $l(w) := n$ is the length of $w$. There is a natural monoid homomorphism $W_S \to W, w \mapsto \bar{w}$ whose restriction to $S$ is the identity. The element $w$ is called a *word representative* of $\bar{w}$. The *Bruhat length* of an element $u \in W$, denoted by $l(u)$, is the minimal length of a word representative of $u$. A word $w \in W_S$ is called *reduced* if $l(w) = l(\bar{w})$.

### 2.2 The *Tits word combinatorics*

For any distinct $s, t \in S$ with $m_{s,t} < \infty$, let $w(s, t)$ be the word of length $m_{s,t}$ starting with $s$ and alternating the letters $s$ and $t$. The relation $(st)^{m_{s,t}} = 1$ is in fact equivalent to $\overline{w(s, t)} = \overline{w(t, s)}$.

The *subwords* of a word $w = s_1s_2 \dots s_n \in W_S$ are the words $s_{i_1}s_{i_2} \dots s_{i_k}$, where $1 \le i_1 < i_2 < \dots < i_k \le n$ and the *substrings* of $w$ are the subwords $s_p s_{p+1} \dots s_q$ where $1 \le p \le q \le n$.

An *elementary reduction* [3] is a pair of words $(w, w')$ such that $w'$ can be obtained from $w$ by one of the following reductions:

* *Reduction of first type:* deleting two consecutive identical letters in $w$, or

* *Reduction of second type:* replacing in $w$ a substring $w(s, t)$ by $w(t, s)$.
Consequently, we have $l(w') = l(w) - 2$ for a reduction of the first type, and $l(w') = l(w)$ otherwise.

**Theorem 1.** (Tits) Let $w, w'$ be two words with $\bar{w} = \bar{w}'$. If $w'$ is reduced, one can transform $w$ into $w'$ by a sequence of elementary reductions.

Besides the original reference in French [18], the reader can consult Davis's book [5], section 3.4. (The elementary reductions are called elementary $M$-operations in loc.cit..) Our presentation of the Tits Theorem is close to Casselman's webpage [3].

### 2.3 Girth of W

Set $\gamma(W) = 2 \min_{s \neq t} m_{s,t}$. The integer $\gamma(W)$, possibly infinite, is the girth of the Cayley graph of $W$, see [11, Lemma 2.1]. It will also be called the girth of $W$. For any distinct $s, t \in S$, we have $l(w(s,t)) \ge \gamma(W)/2$. Therefore any word $w$ with $l(w) < \gamma(W)/2$ can be reduced only by reductions of the first type. A consequence of Theorem 1 is the following

**Corollary 2.** Let $w, w'$ be two words with $\bar{w} = \bar{w}'$. Assume that $l(w) < \gamma(W)/2$.

(1) if $w$ and $w'$ are reduced, we have $w = w'$,

(2) if $w$ is not reduced, it contains a substring $ss$ for some $s \in S$.

### 2.4 Loops

By definition, a *cyclic word* on the alphabet $S$ of length $n$ is a word $w = s_1 \dots s_n$ in $W_S$ modulo a cyclic permutation. For example the cyclic words $s_1s_2s_3$ and $s_3s_1s_2$ are equal.

For a cyclic word $w = s_1 \dots s_n$, it will be convenient to assume that the indices belong to $\mathbb{Z}/n\mathbb{Z}$. The substrings of length $l \le n$ of the cyclic word $w = s_1s_2\dots s_n$ are the words $u = s_i s_{i+1} \dots s_{i+l-1}$, for some $i \in \mathbb{Z}/n\mathbb{Z}$. For example $s_4s_1$ is a substring of the cyclic word $s_1s_2s_3s_4$.

A word $w = s_1 \dots s_n$ in $W_S$ of length $n > 0$ is called a *loop* if $\bar{w} = 1$. For a cyclic word $w = s_1 \dots s_n$, the condition $\bar{w} = 1$ is independent of its representatives in $W_S$. It follows that $w$ is called a *cyclic loop* if any of its representatives in $W_S$ is a loop. Since $\epsilon(s) = -1$ for any $s \in S$, the length of any loop or cyclic loop is even.

**Corollary 3.** Let $w = s_1 \dots s_{2n}$ be a cyclic loop, with $2n < \gamma(W)$. Then there are two distinct indices $i, j \in \mathbb{Z}/2n\mathbb{Z}$ such that $s_i = s_{i+1}$ and $s_j = s_{j+1}$.

*Proof.* Set $u = s_1s_2\dots s_n$ and $v = s_{2n}s_{2n-1}s_{n+1}$. We have $\bar{u} = \bar{v}$ and $l(u) = l(v) = n < \gamma(W)/2$. It follows that $u$ is reduced iff $v$ is reduced.

If both $u$ and $v$ are reduced, it follows from the first assertion of Corollary 2 that $u = v$, hence we have $s_n = s_{n+1}$ and $s_{2n} = s_{2n+1}$.

Otherwise, we can transform $u$ and $v$ into reduced words by a sequence of elementary reductions of the first type. By the second assertion of Corollary 2, there exist $i, j$ with $1 \le i < n$ and $n \le j < 2n$ such that $s_i = s_{i+1}$ and $s_j = s_{j+1}$.

In both cases, the corollary is proved. $\square$

### 2.5 Parabolic subgroups

For a subset $I$ of $S$, let $W_I \subset W$ be the subgroup generated by $I$, let $W_I$ be the set of words on the alphabet $I$. The subgroups $W_I$ are called the *parabolic subgroups* of $W$. It
is well-known that $(W_I, I)$ is a Coxeter system, with Coxeter matrix $(m_{s,t})_{s,t \in I}$, see e.g. [5] Section 4.1.

It is clear from Theorem 1 that any reduced expression of an element $w \in W_I$ is in $W_I$. It follows that $W_I \cap W_J = W_{I \cap J}$ for any two subsets $I, J$ of $S$. Given $w \in W$, the smallest subset $I$ such that $w \in W_I$ is called the *support* of $w$.

For a subset $I$ in $S$ and $t \in S$, set

$$I(t) := \{s \in I \mid m_{s,t} = 2\} = \{s \in I \mid s \neq t \text{ and } st = ts\}, \text{ and} \\ t^{W_I} = \{t^w \mid w \in W_I\},$$

where, as usual, $t^w := wtw^{-1}$.

**Lemma 4.** Let $I$ be a subset of $S$.

(1) For any $t \in S \setminus I$, the centraliser of $t$ in $W_I$ is $W_{I(t)}$. In particular, we have $t^{W_I} \simeq W_I/W_{I(t)}$.

(2) For elements $t \neq t'$ of $S \setminus I$, we have $t^{W_I} \cap t'^{W_I} = \emptyset$.

This lemma appears to be well-known. Since we did not find an exact reference, we provide a quick proof.

**Proof of Assertion (1).** Set

$$W_I^{I(t)} = \{w \in W_I \mid llearners > l(w) \forall s \in I(t)\}.$$

Let $w \in W_I^{I(t)}$ with $w \neq 1$ and let $w = s_1 \dots s_n$ be any reduced expression for $w$. Since $t$ is not in the support of $w$, the word $s_1 \dots s_n t$ is reduced. Since $s_n \notin I(t)$, no reduction of second type involves $t$. Therefore any reduced expression of $wt$ ends with the letter $t$, therefore

$$wt \neq tw.$$

By Section 4.5 of [5], any $w \in W_I$ can be uniquely written as $w = uv$, where $u \in W_I^{I(t)}$ and $u \in W_{I(t)}$. Thus the previous statement is equivalent to Assertion (1).

**Proof of Assertion (2)** Let $t \neq t'$ be elements in $S \setminus I$. The support of any element in $w \in t^{W_I}$ (resp. $w \in t'^{W_I}$) contains $t$ but not $t'$ (resp. contains $t'$ but not $t$). Therefore $t^{W_I}$ and $t'^{W_I}$ are disjoint. $\square$

## 2.6 The Cayley complex $Cay^+ W$

By definition, the *Cayley graph* of $W$, denoted $Cay W$, is the graph whose vertices are the elements $v \in W$ and whose edges are the pairs $(v, vs)$, for $v \in W$ and $s \in S$. We will now define the *Cayley complex* $Cay^+ W$ by attaching some 2-cells to $Cay W$.

Let $\mathfrak{P}$ be the set of pairs $I = \{s, t\}$ of commuting elements of $S$. For $I \in \mathfrak{P}$, any $W_I$-coset $vW_I$ is a subgraph of $Cay W$ consisting of a 4-cycle, which can be seen as the boundary of a plain square $c(v, I)$. An example is the square $c_1$ shown in Figure 1. Therefore we can attach the 2-cell $c(v, I)$ along $\partial c(v, I)$ to the Cayley graph $Cay W$. By definition, the *Cayley complex* $Cay^+ W$ is the 2-dimensional complex obtained by attaching the 2-cells $c(v, I)$, where $I$ varies over $\mathfrak{P}$ and $v$ varies over a set of representatives of $W/W_I$.

It remains to add one remark to complete the definition of $Cay^+ W$. The group $W_I$ acts (by the right action) of the coset $vW_I$, and this action can be extended to the cell $c(v, I)$. The two generators $s$ and $t$ of $W_I$ are the median reflections of the square $c(v, I)$. We require
<figure><img src="image_2.png" /><figcaption>FIGURE 1. The rectangle $c_1 \cup c_2 \dots \cup c_n$ in Cay$^+$ W.</figcaption></figure>

that the 2-cells $c(v, I)$ are glued compatibly with the $W_I$-action. It follows that the $W$-action on Cay $W$ extends naturally to Cay$^+$ $W$.

Note that any word $w = s_1s_2\dots s_n$ in $W_S$ defines a path, denoted by $|w|$, in the Cayley graph Cay $W$. The $n$ successive edges of $|w|$ are

$$ (1, s_1), (s_1, s_1s_2), \dots, (s_1s_2, \dots, s_{n-1}, s_1s_2, \dots, s_n). $$

Let $v \in W$. By definition the path $|w|$ is based at 1 and the path $v|w|$ is based at $v$. By $W$-equivariance, it is clear that

* for any $w_1, w_2 \in W_S$, the paths $|w_1|$ and $|w_2|$ are homotopic in Cay$^+$ $W$ iff $v|w_1|$ and $v|w_2|$ are homotopic, and

* for any loop $w \in W_S$, $|w|$ is null-homotopic in Cay$^+$ $W$ iff $v|w|$ is.

Therefore the next statement and its proof only involve loops based at 1.

For $s \in S$ and $I \subset S$, set

$$ I(s) = \{t \in I \mid m_{s,t} = 2\}. $$

**Lemma 5.** Let $u \in W_{I(s)}$ be a reduced word. Then the paths $|sus|$ and $|u|$ are homotopic in Cay$^+$ $W$.

*Proof.* By definition, we have $u = t_1 \dots t_n$ where all $t_i$ commute with $s$ and $n = l(u)$. For each integer $i$ with $1 \le i \le n$, let $I_i = \{s, t_i\}$ and $c_i = c(t_1t_2 \dots t_{i-1}, I_i)$. Then $c_1 \cup c_2 \dots \cup c_n$ is a rectangle of Cay$^+$ $W$. As shown in Figure 1, the lower side of the rectangle is the path $|t_1t_2 \dots t_n|$ and the three other sides represent the path $|st_1t_2 \dots t_n s|$. It follows that inside Cay$^+$ $W$, the path $|sus|$ and $|u|$ are homotopic. $\square$

## 2.7 Gal's Theorem

Following [7], a partition $S = R \sqcup B$ of $S$ is called a Gal's partition if $m_{s,t}$ is an even integer or is infinite for any $s \in R$ and $t \in B$. For a Gal's partition $S = R \sqcup B$, there are group homorphisms $\mu_R : W \to W_R$ and $\mu_B : W \to W_B$ uniquely defined by

$$ \mu_R(s) = s \text{ if } s \in R \text{ and } \mu_R(s) = 1 \text{ if } s \notin R, $$

$$ \mu_B(s) = 1 \text{ if } s \in B \text{ and } \mu_B(s) = s \text{ if } s \notin B. $$

Set

$$ \mathrm{Ker} \mu_R := \overline{W}_B. $$

The notation $\overline{W}_B$ is intended to emphasise that $\mathrm{Ker} \mu_R$ is the normal closure of $W_B$ [7]. Set $\overline{B} = \cup_{t \in B} t^{W_R}$.

**Theorem 6.** (Gal) Let $S = R \sqcup B$ be a Gal's partition.
Then the pair $(\overline{W}_B, \overline{B})$ is a Coxeter system, possibly of infinite rank. Moreover, its Coxeter matrix $(m_{\sigma,\tau})_{\sigma,\tau \in \overline{B}}$ is defined by

(1) If $\sigma = s^w$ and $\tau = t^w$ for some $s, t \in B$ and $w \in W_R$, then $m_{\sigma,\tau} = m_{s,t}$,

(2) If $\sigma = t^w$ and $\tau = t^{ws}$ for some $t \in B$, $s \in R$ and $w \in W_R$, then $m_{\sigma,\tau} = m_{s,t}/2$, and

(3) otherwise, we have $m_{\sigma,\tau} = \infty$.

For the proof, see Proposition 2.1 and Corollary 3.1 of [7].

## 2.8 Right-angled partitions

In order to use Gal's Theorem, we will restrict ourselves to a certain type of Gal's partitions. Recall that a Coxeter group is called right-angled if the non-diagonal entries of its Coxeter matrix are 2 or $\infty$. By analogy, a partition $S = R \cup B$ of $S$ will be called a right-angled partition if $m_{s,t} = 2$ or $\infty$, for any $s \in R$ and $t \in B$.

**Lemma 7.** Let $S = R \cup B$ be a right-angled partition. Then we have

$$ \gamma(\overline{W}_B) = \gamma(W_B). $$

*Proof.* Let $\sigma, \tau \in \overline{B}$. If $\sigma = t^w$, $\tau = t^{ws}$ for some $t \in B$, $s \in R$ and $w \in W_R$ such that $m_{s,t}$ is finite, we have $m_{s,t} = 2$ and $\sigma = \tau$. Hence $m_{\sigma,\tau}$ is a diagonal entry of the Coxeter matrix of $\overline{W}_B$. Otherwise, by Theorem 6, we have $m_{\sigma,\tau} = m_{s,t}$ for some $s, t \in B$ or $m_{\sigma,\tau} = \infty$. It follows that $\gamma(\overline{W}_B) = \gamma(W_B)$. $\square$

## 2.9 Loops for Coxeter groups with a right-angled partition

Let $I$ be a subset of $S$. For any word or cyclic word $w = s_1s_2...s_n \in W_S$, let $l_I(w)$ be the number of its letters in $I$. Therefore for any partition $S = R \cup B$ of $S$, we have

$$ l(w) = l_R(w) + l_B(w). $$

**Lemma 8.** Let $S = R \cup B$ be a right-angled partition. Let $w$ be a cyclic loop on the alphabet $S$ such that $l_R(w) < \gamma(W_R)$ and $l_B(w) < \gamma(W_B)$. Then one of the following assertions holds

(1) $w$ contains a substring $sus$, where $s \in B$ and $u$ is a reduced word on the alphabet $B(s)$, or

(2) $w$ contains a substring $tt$, where $t \in R$.

*Proof.* If $l_B(w) = 0$, then Assertion (2) holds by Corollary 3.

From now on, let us assume that $l_B(w) > 0$. As $w$ is a cyclic word, we can write it as

$$ w = u_1s_1u_2s_2...u_ks_k, $$

where $k = l_B(w)$, $s_1, s_2...s_k$ are in $B$ and $u_1, ..., u_k$ are in $W_R$. (It is not excluded that some words $u_i$ have length zero.) As usual, the indices $1, 2...k$ are viewed as elements of $\mathbb{Z}/k\mathbb{Z}$.

Set $v_1 := \overline{u_1}$, $v_2 := \overline{u_1u_2}$, $v_3 := \overline{u_1u_2u_3}$... For any index $i \in \mathbb{Z}/k\mathbb{Z}$, set $\sigma_i = s_i^{v_i}$. Since $\mu_R(\overline{w}) = 1$, it follows that $v_n := \overline{u_1u_2...u_n} = 1$. Therefore the identity $\overline{w} = 1$ is equivalent to

$$ \sigma_1\sigma_2...\sigma_k = 1 $$

By definition, each $\sigma_i$ belongs to $\overline{B}$. Moreover, as a word on the alphabet $\overline{B}$, the word $\sigma_1\sigma_2...\sigma_k$ is a loop.

By Lemma 7 we have
$$ \gamma(\overline{W}_B) = \gamma(W_B). $$

Since $k < \gamma(\overline{W}_B)$, it follows from Corollary 3 that there are two indices $i, j \in \mathbb{Z}/k\mathbb{Z}$ such that $\sigma_i = \sigma_{i+1}$ and $\sigma_j = \sigma_{j+1}$. We can choose $i$ and $j$ in such a way that $l(u_{i+1}) \le l(u_{j+1})$. Since we have $l(u_{i+1}) + l(u_{j+1}) \le l_R(w) < \gamma(W_R)$, we have

$$ l(u_{i+1}) < \gamma(W_R)/2. $$

Set $s = s_i$, $s' = s_{i+1}$ and $u = u_{i+1}$. The equality $\sigma_i = \sigma_{i+1}$ is equivalent to

$$ s = s^{\overline{u}}. $$

By Assertion (2) of Lemma 4, we have $s = s'$. Therefore $w$ contains the substring

$sus,$

where $l(u) < \gamma(W_R)/2$.

To finish the proof, let us consider two cases.

*Case 1:* $u$ is not reduced. By Lemma 2, the word $u$ contains the substring $tt$ for some $t \in I$, therefore Assertion (2) holds.

*Case 2:* $u$ is reduced. Since $s = s^{\overline{u}}$, it follows from the Assertion (1) of Lemma 4 that $u$ belongs to $W_R(s)$. Therefore $u$ is a word on the alphabet $R(s)$, and Assertion (1) holds. $\square$

## 2.10 Homotopically trivial paths in Cay$^+$W

**Theorem 9.** Let $S = R \sqcup B$ be a right-angled partition. Let $w$ be a cyclic loop on the alphabet $S$ such that $l_R(w) < \gamma(W_R)$ and $l_B(w) < \gamma(W_B)$.

Then $|w|$ is null-homotopic in Cay$^+$W.

*Proof.* If $w$ contains a substring $tt$, $|w|$ is homotopic in Cay$^+$W to the loop $w'$ obtained by deleting this substring.

Otherwise, by Lemma 8, $w$ contains a substring $sus$, where $s \in B$ and $u$ is a reduced word on the alphabet $R(s)$. Hence by Lemma 5, the loop $|w|$ is homotopic to the loop $w'$ obtained by replacing the substring $sus$ by $u$.

In both cases, we have $l(w') = l(w) - 2$, while $l_R(w') < \gamma(W_R)$ and $l_B(w') < \gamma(W_B)$. Therefore by induction, $|w|$ is null-homotopic in Cay$^+$W. $\square$

## 3. UNIFORMIZATION OF 2k-REGULAR TESSELATIONS

Let $k$ be an integer. For simplicity, it will be assumed that $k \ge 3$ in the whole section. For $k = 1$ or 2, the theory is not difficult, but some details are slightly different.

We define the notion of 2k-regular standard hexagonal tesselation of a hyperbolic surface $\Sigma$. Except where stated otherwise, all tesselations considered in the paper have hexagonal tiles, so we will skip the term hexagonal in what follows.

We will first show a formal uniformization theorem for these tesselations. There is a universal surface $\mathcal{H}$, endowed with such a tesselation, on which a certain Coxeter group $W(k)$ acts, such that any closed surface with a 2k-regular standard tesselation is isometric to $\mathcal{H}/H$ for some finite index subgroup of $W(k)$. Conversely, a finite index subgroup $H \subset W(k)$ for which the tesselation on $\mathcal{H}/H$ is 2k-regular can be readily characterised.

By definition, the curves of a tesselation are the maximal geodesics containing the sides of the tiles. All curves of a 2k-regular tesselation have length $2k$ arcosh 2. This leads to the
question - are the set of systoles and the set of curves of the tesselation of $\mathcal{H}/H$ identical?
This is partly answered by the main result of Section 3, namely the Criterion 18.

The surface $\mathcal{H}$ has two realizations. The first one, denoted by $\mathcal{H}(k)$, is a quotient of the Poincaré half-plane $\mathbb{H}$. The second realization is the Coxeter complex Cay$^+W(k)$. The proof of Criterion 18 uses these two realizations of $\mathcal{H}$, and it combines standard hyperbolic trigonometry and Theorem 9 of the previous section.

### 3.1 The decorated hexagons $\mathcal{P}$ and $\bar{\mathcal{P}}$

Let $\mathbb{H}$ be the Poincaré half-plane, endowed with its hyperbolic metric. Recall that a hexagon $P \subset \mathbb{H}$ is *regular* if its automorphism group is flag-transitive, i.e. if Aut $P$ acts transitively on the pairs $(e, v)$, where $e$ is a side and $v \in e$ is a vertex of $P$. The following lemma follows readily from hyperbolic trigonometry, e.g. [13] pp.90-96.

**Lemma 10.** *Up to isometry, there exists a unique hyperbolic right-angled hexagon $P$ whose side lengths are all equal. Moreover $P$ is regular and the common length of its sides is $L := \arccosh 2$.*

By definition, the *decorated hexagon* $\mathcal{P}$ is the oriented hexagon $P$, whose sides, $S_1, S_2, \dots, S_6$ are indexed by $\mathbb{Z}/6\mathbb{Z}$ in an anti-clockwise direction. (The orientation of $P$ induces an orientation of its sides.) Let $\bar{\mathcal{P}}$ be the same hexagon with opposite orientation. By definition the *red sides* are $S_1, S_3$ and $S_5$, and the other three are called the *blue sides*.

### 3.2 The $2k$-regular standard tesselations

Let $\Sigma$ be an oriented hyperbolic surface, finite or infinite, and let $\tau$ be a tesselation of $\Sigma$ whose tiles are the decorated hexagons $\mathcal{P}$ or $\bar{\mathcal{P}}$. The tesselation $\tau$ is called a *standard tesselation* if it satisfies the following axioms

(AX1) The tiles are glued along sides of the same index,

(AX2) Each vertex of the tesselation has valence four.

In this definition, it should be understood that the tiles of a standard tesselation *are always the decorated right-angled regular hexagons $\mathcal{P}$ and $\bar{\mathcal{P}}$*. The second axiom implies that the sum of the four angles at each vertex is $2\pi$, so it is equivalent to the fact that $\Sigma$ has no boundary.

Let $\tau$ be a standard tesselation. By definition, a *curve of the tesselation $\tau$* is a maximal geodesic in the 1-skeleton of $\tau$. Since the angle between two adjacent edges of same index is $\pi$, each curve $C$ is a maximal geodesic of $\Sigma$ consisting of a union of adjacent edges of the same index $i$. This common index $i$ is called the *index* of the curve $C$.

Given a positive integer $k$, a standard tesselation $\tau$ is called *$k$-regular* if it satisfies the following additional requirement

(AX3) Each curve $C$ of the tesselation is a closed geodesic of length $kL$.

For a standard tesselation $\tau$ of $\Sigma$ satisfying (AX3), each curve $C$ of index $i$ alternately meets curves of index $i+1$ and curves of index $i-1$, so $k$ is an even integer. From now on, we will only speak about $2k$-regular tesselations to emphasise the fact that $2k$ is an even integer.

Let $Tess(\Sigma, 2k)$ be the set of all $2k$-regular standard tesselations of $\Sigma$. Once again, it is tacitly assumed that the tiles are the hexagons $\mathcal{P}$ and $\bar{\mathcal{P}}$.
### 3.3 The universal tesselated surface $\mathcal{H}(k)$

Let $\mathbb{H}$ be the Poincaré half-plane. By Lemma 10, there is a unique right-angled regular hexagon. Hence, by the Poincaré polygon Theorem, there exists a unique (up to isometry) standard tesselation $\tau_\infty$ of $\mathbb{H}$. Let us choose one tile $\tilde{T}$ of the tesselation $\tau_\infty$ and let $\tilde{S}_i$ be its side of index $i$. The tile $\tilde{T}$ will be called the distinguished tile of $\tau_\infty$. For $i \in \mathbb{Z}/6\mathbb{Z}$, let $\Delta_i$ be the line containing the side $\tilde{S}_i$ and let $s_i$ be the reflection across the line $\Delta_i$. The subgroup of $\mathrm{PGL}_2(\mathbb{R})$ generated by the six reflections $(s_i)_{i \in \mathbb{Z}/6\mathbb{Z}}$ is the right-angled Coxeter group $W(\infty)$ with presentation

$$ \langle (s_i)_{i \in \mathbb{Z}/6\mathbb{Z}} | s_i^2 = 1 \text{ and } (s_i s_{i+1})^2 = 1, \forall i \in \mathbb{Z}/6\mathbb{Z} \rangle. $$

Let $k \ge 2$ and let $N(k)$ be the normal subgroup of $W(\infty)$ generated by the elements $(s_is_{i+2})^k$, for $i \in \mathbb{Z}/6\mathbb{Z}$, and all their conjugates. The group $W(k) := W(\infty)/N(k)$ is the Coxeter group with presentation

$$ \langle (s_i)_{i \in \mathbb{Z}/6\mathbb{Z}} | s_i^2 = 1, (s_i s_{i+1})^2 = 1, \text{ and } (s_i s_{i+2})^k = 1, \forall i \in \mathbb{Z}/6\mathbb{Z} \rangle $$

Set $W(\infty)^+ := W(\infty) \cap \mathrm{PSL}_2(\mathbb{R}) = \{w \in W(\infty) | \epsilon(w) = 1\}$. It follows from the Poincaré Theorem that $W(\infty)$ acts freely and transitively on the set of tiles of $\tau_\infty$. So a nontrivial element $w \in W(\infty)^+$ is elliptic iff it is conjugate to $s_is_{i+1}$ for some $i \in \mathbb{Z}/6\mathbb{Z}$. Since $k \ge 2$, the subgroup $N(k)$ acts freely on $\mathbb{H}$.

Set $\mathcal{H}(k) = \mathbb{H}/N(k)$ and let $\tau_k$ be the standard tesselation of $\mathcal{H}(k)$ induced by $\tau_\infty$. Note that $s_{i-1}s_{i+1}(\Delta_i) = \Delta_i$ and its restriction to the line $\Delta_i$ is a translation of length $2L$. It follows that $\tau_k$ is a $2k$-regular tesselation. The distinguished tile of $\mathcal{H}(k)$, denoted by $T$, is the image of $\tilde{T}$ in $\mathcal{H}(k)$.

Set $W(k)^+ = \{w \in W(k) | \epsilon(w) = 1\}$ and set $t_i = s_{i-1}s_{i+1}$ for any $i \in \mathbb{Z}/6\mathbb{Z}$. For any element $t \in W(k)$, let $t^{W(k)}$ be its conjugacy class. When a subgroup $H \subset W(k)^+$ acts freely on $\mathcal{H}(k)$, let $\tau_H$ be the tesselation induced by $\tau_k$ on the surface $\mathcal{H}(k)/H$. Note that the condition $H \subset W(k)^+$ ensures that $\mathcal{H}(k)/H$ is orientable.

**Lemma 11.** A subgroup $H \subset W(k)^+$ acts freely on $\mathcal{H}(k)$ iff

$$ (11.1) \qquad H \cap (s_i s_{i+1})^{W(k)} = \emptyset \text{ for all } i \in \mathbb{Z}/6\mathbb{Z}. $$

Moreover assume that $H$ satisfies the condition (11.1). Then the tesselation $\tau_H$ is $2k$-regular iff the following condition holds

$$ (11.2) \qquad H \cap (t_i^l)^{W(k)} = \emptyset \text{ for all } i \in \mathbb{Z}/6\mathbb{Z} \text{ and } 1 \le l < k. $$

*Proof.* Let $w \in H$. Since $W(k)$ is tile-transitive, $w$ has a fixed point in $\mathcal{H}(k)$ iff $w^v$ has a fixed point in $T$, for some $v \in W(k)$. By hypothesis, $w^v$ is not a reflection therefore $w^v = s_is_{i+1}$ for some $i$ in $\mathbb{Z}/6\mathbb{Z}$, which proves the first assertion.

Assume that $\tau_H$ is not $2k$-regular. By assumption, there is a curve $C$ of the tesselation $\tau_k$ whose image in $\mathcal{H}(k)/H$ has length less than $kL$. Since $W(k)$ is tile-transitive, we can assume that $C$ contains the side $S_i$ of the distinguished tile $T$, for some $i \in \mathbb{Z}/6\mathbb{Z}$. We have $h(C) = C$, for some nontrivial $h \in H$. Since it has no fixed points, $h|_C$ is a rotation. It follows that $h = t_i^l$ for some $l \in \mathbb{Z}/6\mathbb{Z}$, which proves the second assertion. $\square$

Set

$$ \mathcal{Tess}(*, 2k) = \cup_{\Sigma} \mathcal{Tess}(\Sigma, 2k), $$
where $\Sigma$ varies over all oriented closed hyperbolic surfaces of genus $g \ge 2$.

In what follows, we will only use the previous Lemma 11. However we would like to briefly
explain that $\mathcal{H}(k)$ is the universal cover of all $2k$-regular standard tesselations, as will now
be shown.

**Theorem 12.** The map $H \mapsto \tau_H$ is a one-to-one correspondence between

(1) the finite index subgroups $H \subset W(k)^+$ satisfying (11.1) and (11.2), and

(2) the $2k$-regular standard tesselations $\tau$ of closed oriented surfaces.

In the previous statement, it should be understood that the word “subgroups” refers to
conjugacy classes of subgroups and “tesselations” refers to isometry classes of tesselations.

*Proof.* The Lemma 11 shows that this map is well-defined.

We will now define the inverse map. Let $\tau \in \mathcal{T}ess(*, 2k)$. By definition, $\tau$ is a tesselation of some oriented closed surface $\Sigma$. The tesselation $\tau$ induces a standard tesselation of its universal cover $\mathbb{H}$. Since the induced tesselation is isometric to $\tau_\infty$, there is an embedding $\pi_1(\Sigma, p) \subset W(\infty)$, where $p \in \Sigma$ is a base point. Since $\tau$ is $2k$-regular, it follows that $\pi_1(\Sigma, p) \supset N(k)$, and therefore $\tau = \tau_H$, where $H = \pi_1(\Sigma, p)/N(k)$. $\square$

it Remark. The universal tesselated surfaces $\mathcal{H}(k)$ can be defined for $k = 1$ and $2$. In fact $\mathcal{H}(1)$ is the genus 2 surface of [15], which was the starting point of our paper.

### 3.4 The homeomorphism $Cay^+W(k) \simeq \mathcal{H}(k)$

Since $W(k)$ acts freely and transitively on the set of tiles, the Cayley graph $Cay W(k)$ can
be identified with the dual graph of the tesselation $\tau_k$.

Let us recall more precisely the definition of the dual graph $\tau_k^*$ of $\tau_k$. Let $T$ be the distinguished tile of $\mathcal{H}(k)$ and $X$ be its center. For $w \in W(k)$, set $T(w) = w.T$, $X(w) = w.X$ and, for any $i \in \mathbb{Z}/6\mathbb{Z}$, let $S_i(w)$ be the side of $T(w)$ of index $i$.

By definition,

$$V(\tau_k^*) := \{X(w) \mid w \in W(k)\}$$

is the set of vertices of $\tau_k^*$. For $w \in W(k)$ and $i \in \mathbb{Z}/6\mathbb{Z}$,

$$O := T(w) \cup T(\text{ws}_i)$$

is a right-angled octagon with two sides of length $2L$ and all others of length $L$. Let $\mathbf{e}(w, \text{ws}_i)$ be the geodesic arc joining $X(w)$ and $X(\text{ws}_i)$ in $O$. In this instance, the arcs are not oriented, so $\mathbf{e}(w, \text{ws}_i) = \mathbf{e}(\text{ws}_i, w)$. By definition

$$E(\tau_k^*) := \{\mathbf{e}(w, \text{ws}_i) \mid w \in W(k) \text{ and } i \in \mathbb{Z}/6\mathbb{Z}\}$$

is the set of edges of $\tau_k^*$. The duality property means that each tile $T(w)$ contains exactly one vertex of $\tau_k^*$, namely $X(w)$, and each of its sides $S_i(w)$ meets exactly one edge of $\tau_k^*$, namely $\mathbf{e}(w, \text{ws}_i)$. The natural homeomorphisms between an edge $|(w, \text{ws}_i)|$ of $Cay W(k)$ and an edge $\mathbf{e}(w, \text{ws}_i)$ of $\tau_k^*$ provide a natural homeomorphism

$$\mathrm{Cay} W(k) \simeq \tau_k^*.$$

It follows that the topological graph $Cay W(k)$ is embedded in $Cay^+W(k)$ and in $\mathcal{H}(k)$.
In fact, the two embeddings of $Cay W(k)$ are the same, as it is shown in the next

**Lemma 13.** The embedding $Cay W(k) \subset \mathcal{H}(k)$ extends to a homeomorphism

$$\mathrm{Cay}^+ W(k) \simeq \mathcal{H}(k).$$
<figure><img src="image_3.png" /><figcaption>FIGURE 2. The quadrilateral from the proof of Lemma 13.</figcaption></figure>

*Proof.* By definition, Cay$^+$W(k) is tesselated by quadrilaterals and H(k) is tesselated by hexagons. Roughly speaking, it will be shown that these two tesselations are dual to each other, see Figure 2.

Since $k \ge 3$, the rank-two commutative parabolic subgroups of W(k) are the subgroups $W_{I(i)}$, where $I(i) = \{s_i, s_{i+1}\}$ and $i \in \mathbb{Z}/6\mathbb{Z}$. For $v \in W(k)$, it is clear that

$$ \mathbf{H} = T(v) \cup T(v s_i) \cup T(v s_{i+1}) \cup T(v s_i s_{i+1}) = W(k)_{I(i)} \cdot T(v) $$

is a right-angled 12-gon with four sides of length $2L$ and all others of length $L$. Let $\mathbf{Q}(v, i)$ be the quadrilateral contained in $\mathbf{H}$ whose set of vertices is

$$ W(k)_{I(i)} \cdot X(v) = \{X(v), X(v s_i), X(v s_{i+1}), X(v s_i s_{i+1})\} $$

as shown in Figure 2. Consequently, we have $\mathbf{Q}(v, i) = \mathbf{Q}(v', i)$ if $v = v' \mod W(k)_{I(i)}$.

Set $A_i(v) = S_i(v) \cap S_{i+1}(v)$ and let $M_i(v)$ be the midpoint $S_i(v)$ and $\Omega_i(v) \subset T(v)$ be the convex quadrilateral with vertices given by $M_i(v)$, $A_i(v)$, $M_{i+1}(v)$ and $X(v)$. The tile $T(v)$ is tesselated by the six quadrilaterals $\Omega_i(v)$. Since

$$ \mathbf{Q}(v, i) \cap T(v) = \Omega_i(v) $$

the set of quadrilaterals

$$ \{\mathbf{Q}(v, i) \mid i \in \mathbb{Z}/6\mathbb{Z} \text{ and } v \in W(k)\} $$

tesselates $\mathcal{H}(k)$.

Since Cay W(k) has been identified with the graph $\tau_k^* \subset \mathcal{H}(k)$, there is an equality

$$ \partial \mathbf{Q}(v, i) = \partial \mathbf{c}(v, I_i). $$

which can be extended to a homeomorphism

$$ \mathbf{Q}(v, i) \simeq \mathbf{c}(v, I_i). $$

Therefore Cay$^+$W(k) is homeomorphic to

$$ \cup_{i,v} \mathbf{Q}(v, i) $$

where $i$ varies over $\mathbb{Z}/6\mathbb{Z}$ and $v$ over $W(k)$. It follows that Cay$^+$W(k) is homeomorphic to $\mathcal{H}(k)$. $\square$
### 3.5 The combinatorial datum $\omega(\gamma)$ associated to arcs in $\mathcal{H}(k)$

We will now start to investigate the length of the closed geodesics of the surfaces $\mathcal{H}(k)/H$. To do so, we will first look at the lengths of the arcs $\gamma$ in $\mathcal{H}(k)$. In this section, we will associate a word $\omega(\gamma)$ over the letters $(s_i)_{i \in \mathbb{Z}/6\mathbb{Z}}$. Its length will be called the combinatorial length of an arc. Then, we will find a lower bound for the combinatorial length of the closed geodesic. The relation between the hyperbolic lengths and the hyperbolic lengths will be examined in the next subsection.

We will now provide the precise definition of the combinatorial datum $\omega(\gamma)$ associated to some geodesic paths $\gamma$. Let $\mathrm{Arc}_T(\mathcal{H}(k))$ be the set of geodesic arcs $\gamma: [0, l] \to \mathcal{H}(k)$ of length $l$ such that

(1) $\gamma(t)$ lies in $T^0$ for small $t \neq 0$

(2) $\gamma(0) \in C$ and $\gamma(l) \in C'$ for some curves $C, C'$ of $\tau_k$,

where $T^0$ is the interior of the distinguished cell $T$ of $\tau_k$. Note that the first condition implies that $\gamma$ is not an arc of a curve of $\tau_k$. Of course, the previous conditions do not ensure that $\gamma$ necessarily lifts a closed geodesic in some $\mathcal{H}(k)/H$. It is the case only if the indices of $C$ and $C'$ are equal together with some position and angle conditions for $\gamma(0)$ and $\gamma(l)$. In the notation $\mathrm{Arc}_T(\mathcal{H}(k))$ the index $T$ emphasizes that $\gamma$ originates in $T$.

Now we are going to define a word

$$ \omega(\gamma) = s_{i_1}s_{i_2}\dots s_{i_N}, $$

associated with the path $\gamma$. First assume that $\gamma$ does not meet any vertex of the tesselation. Then $i_1, \dots, i_{N-1}$ are the indices of the curves successively crossed by $\gamma$ and $i_N$ is the index of the curve $C'$ passing through $\gamma(l)$. Note that $\omega(\gamma)$ does not encode the index of the initial curve $C$.

When $\gamma$ does meet some vertices $v$, we need a convention to precise the order of the curves that $\gamma$ meets. When $\gamma$ crosses a vertex at the intersection of two curves of indices $i$ and $i+1$, we consider that $\gamma$ first crosses the curve of index $i$ and then the curve of index $i+1$. Similarly, if $\gamma(l)$ terminates at such point, we consider that $\gamma$ first crosses the curve of index $i$ and then terminates on a curve of index $i+1$. As before, the datum $\omega(\gamma)$ does not encode any information about $\gamma(0)$.

The integer $N = l(\omega(\gamma))$ is called the *combinatorial length* of $\gamma$.

**Lemma 14.** Let $\gamma \in \mathrm{Arc}_T(\mathcal{H}(k))$ be a closed geodesic. Then $\gamma$ is freely homotopic to the loop $|\omega(\gamma)|$ in $\mathrm{Cay} W(k)$.

*Proof.* Set $\omega(\gamma) = s_{i_1}s_{i_2}\dots s_{i_N}$, where $N$ is the combinatorial length of $\gamma$. Let $0 < t_1 \le \dots \le t_n \dots \le t_{N-1} \le l$ be the successive time at which $\gamma(t_n)$ crosses a curve of index $i_n$. Also set $t_0 = 0$ and $t_N = l$. For any $n$ with $1 \le n \le N$, let $\gamma_n$ be the restriction of $\gamma$ to $[t_{n-1}, t_n]$, so we have

$$ \gamma = \gamma_1 * \gamma_2 * \dots * \gamma_N, $$

where the $*$ denotes the concatenation of paths.

Recall that, for $v \in W(k)$, $T(v)$ denotes the tile $v.T$ and $X(v)$ is its center. Let $v_0, v_1, \dots, v_N$ be the elements of $W(k)$ defined by $v_0 = 1$ and $v_n = \overline{s_{i_1}s_{i_2}\dots s_{i_n}}$ for $n \ge 1$. By definition,
<figure><img src="image_4.png" /><figcaption>FIGURE 3. The concatenation of arcs from the proof of Lemma 14.</figcaption></figure>

$\gamma(t_n)$ belongs to the side $T(v_n) \cap T(v_{n-1})$ for any $0 \le n \le N$. Let $\delta_n \subset T(v_n)$ be the oriented geodesic arc from $\gamma(t_n)$ to $X(v_n)$, and let $\bar{\delta}_n$ be the same arc with the opposite orientation.

Since $\gamma$ is a loop, we have $v_N = 1$ and therefore $\delta_N = \bar{\delta}_0$. It follows that $\gamma$ is freely homotopic to

$$ \bar{\delta}_0 * \gamma_1 * \delta_1 * \bar{\delta}_1 * \gamma_1 * \dots * \gamma_N * \delta_N = \tilde{\gamma}_1 * \tilde{\gamma}_2 * \dots * \tilde{\gamma}_N, $$

where $\tilde{\gamma}_n = \bar{\delta}_{n-1} * \gamma_n * \delta_n$ for any $n \ge 1$. This is illustrated in Figure ?? . By definition $\tilde{\gamma}_n$ is a path joining $X(v_{n-1})$ to $X(v_n)$. Since $\tilde{\gamma}_n$ lies in the convex octagon $T(v_{n-1} \cup T(v_n))$, it is homotopic to the geodesic arc $\mathbf{e}(v_{n-1}, v_n)$ going from $X(v_{n-1})$ to $X(v_n)$.

Hence $\gamma$ is freely homotopic to

$$ \mathbf{e}(v_0, v_1) * \mathbf{e}(v_1, v_2) * \dots * \mathbf{e}(v_{N-1}, v_N), $$

which is precisely the loop $|\omega(\gamma)|$ in Cay $W(k)$. $\square$

Set $S = \{s_i \mid i \in \mathbb{Z}/6\mathbb{Z}\}$, $R = \{s_1, s_3, s_5\}$ and $B = \{s_2, s_4, s_6\}$. The elements of $R$, respectively of $B$, are called the red letters, respectively the blue letters. For any word $w = s_{i_1}s_{i_2}\dots s_{i_N} \in W_S$, let $l_R(w)$ (respectively $l_R(w))$ be the number of red (respectively blue) letters in $w$.

**Corollary 15.** Assume that $k$ is even. Let $\gamma \in \mathrm{Arc}_T(\mathcal{H}(k))$ be a closed geodesic. Then we have

$$ l_R(\omega(\gamma)) \ge 2k \text{ or } l_B(\omega(\gamma)) \ge 2k. $$

*Proof.* Assume otherwise, i.e. $l_R(\omega(\gamma)) < 2k$ and $l_B(\omega(\gamma)) < 2k$. Since $k$ is even, the partition $S = R \cup B$ is right-angled. Therefore by Theorem 9, the path $|\omega(\gamma)|$ is null-homotopic in Cay$^+W(k)$. By Lemma 13 Cay$^+W(k)$ is homeomorphic to $\mathcal{H}(k)$ and by Lemma 14 $\gamma$ is freely homotopic to $|\omega(\gamma)|$. This contradicts the fact that no closed geodesic is null-homotopic on a hyperbolic surface. $\square$

### 3.6 Combinatorial length versus geometric length

Let $\gamma \in \mathrm{Arc}_T(\mathcal{H}(k))$ be a geodesic arc of $\mathcal{H}(k)$. In this section, we will compare the metric length $L(\gamma)$ of $\gamma$ with its combinatorial length $l(\omega(\gamma))$.

**Lemma 16.** Let $\gamma \in \mathrm{Arc}_T(\mathcal{H}(k))$ be a geodesics arc. We have $L(\gamma) > L$ whenever one of the following conditions is satisfied

$$ (1) \ l(\omega(\gamma)) = 1 \text{ and } \gamma \text{ joins two non-consecutive sides of a tile, or} $$
$$ (2) l(\omega(\gamma)) = 2. $$

*Proof.* The first statement is well known, see e.g. [15][2]. Let $\gamma \in \text{Arc}_T(\mathcal{H}(k))$ be an arc and let $\gamma = \gamma_1 * \gamma_2$ be its factorization into arcs of combinatorial length 1. If $\gamma_1$ or $\gamma_2$ join two non-consecutive sides of a tile, then it is already proved that $L(\gamma) > L$.

Otherwise, after applying some isometry, we can assume that $\gamma(0)$ belongs to $S_1(1)$, then $\gamma$ crosses $S_2(1)$ and ends on $S_3(s_2)$, where $S_i(v)$ denotes the side of index $i$ of the tile $T(v)$. Since $T(1) \cup T(s_2)$ is a convex octagon, we can lift it to the Poincaré half plane. Let $\Delta_1$ and $\Delta_3$ be the line containing the arcs $S_1(1)$ and $S_3(s_2)$. The lift of $\gamma$ joins $\Delta_1$ and $\Delta_3$. Since $S_2(1) = S_2(s_2)$ is the common perpendicular to $\Delta_1$ and $\Delta_3$, we have $L(\gamma) > L(S_2(1)) = L$. $\square$

**Lemma 17.** Let $\gamma \in \text{Arc}_T(\mathcal{H}(k))$ be a closed geodesic. If $l_R(\omega(\gamma)) \ge 2k$ or $l_B(\omega(\gamma)) \ge 2k$, then $\gamma$ has length $> 2kL$.

*Proof.* We can assume that $l_R(\omega(\gamma)) \ge 2k$, and write $\gamma = \gamma_1 * \gamma_2 * \cdots * \gamma_N$, where $N = l_R(\omega(\gamma))$ and each $\gamma_n$ is an arc joining a red side to another red side.

If $l(\gamma_n) = 1$, then $\gamma$ joins two red sides of the same tile. Since these sides are not consecutive, we have $L(\gamma_n) > L$ by the part (1) of Lemma 16. Otherwise, we have $l(\gamma_n) \ge 2$ and $L(\gamma_n) > L$ by the part (2) of Lemma 16. Therefore each arc $\gamma_n$ has length $> L$. It follows that

$$ L(\gamma) = \sum_{1 \le n \le N} L(\gamma_n) > NL \ge 2kL. \quad \square $$

### 3.7 Systoles of $\mathcal{H}(k)/H$

Assume now that $k$ is even. It follows from Corollary 15 and Lemma 17 that the systoles of $\mathcal{H}(k)$ are the curves of the tesselation. Let $H \subset W(k)^+$ be a subgroup satisfying the conditions (11.1) and (11.2). Then all curves of the tessellation in $\mathcal{H}(k)/H$ have length $2kL$. Determining when these curves of $\tau_H$ are the systoles is a delicate question. The next criterion provides a partial answer.

For $w \in W(k)$, set $H^w = wHw^{-1}$. Also set

$$ B_{4k} := \{w \in W(k) \mid l(w) \le 4k\}. $$

**Criterion 18.** Let $H \subset W(k)^+$ be a subgroup such that $H^w \cap B_{4k} = \{1\}$ for all $w \in W(k)$.

Then $H$ acts freely on $\mathcal{H}(k)$ and the tesselation $\tau_H$ is 2k-regular. Moreover, the set of systoles of $\mathcal{H}(k)/H$ is exactly the set of curves of $\tau_H$.

*Proof.* By Lemma 11, $H$ acts freely on $\mathcal{H}(k)$ and the tesselation $\tau_H$ is 2k-regular. The curves of the tesselation $\tau_H$ have length $2kL$. Hence, it remains to prove that any closed geodesic $\bar{\gamma}$ of $\mathcal{H}(k)/H$, which is not a curve, has length $L(\bar{\gamma}) > 2kL$.

The choice of a distinguished tile $\bar{T}$ was arbitrary, so we can assume that $\bar{\gamma}$ intersects the interior of $\bar{T}$. Hence there is a geodesic arc $\gamma \in \text{Arc}_T(\mathcal{H}(k))$ which lifts $\bar{\gamma}$. Since $L(\bar{\gamma}) = L(\gamma)$, it is enough to show that $L(\gamma) > 2kL$.

If $l(\omega(\gamma)) > 4k$, then by Lemma 16, we have $L(\gamma) > 2kL$. Assume now that $l(\omega(\gamma)) \le 4k$. Since we have

$$ H^w \cap B_{4k} = \{1\} \text{ for all } w \in W(k), $$
it follows that $\gamma$ is a closed geodesic of $\mathcal{H}(k)$. By Corollary 15, we have
$$l_R(\omega(\gamma)) \geq 2k \text{ or } l_B(\omega(\gamma)) \geq 2k,$$
hence, by Lemma 16, we have $L(\gamma) > 2kL$. $\square$

### 4. THE SUBGROUP $H(k)$ OF $W(k)$

For simplicity, we will assume that the integer $k$ is $\ge 3$. Set $K := \mathbb{Q}(\cos \pi/k)$ and let $\mathcal{O}$ be the ring of integers of the field $K$.

In this section, we use the *Tits representation* $\rho : W(k) \to GL_6(K)$ to find a subgroup $H(k)$ of $W(k)$ satisfying the hypothesis of Criterion 18 with index $[W(k) : H(k)] \le 3^{72k\phi(2k)}$, see Proposition 22.

#### 4.1 The *Tits representation*

In [17], *Tits* defined a faithful representation of any Coxeter group. References are [17] or [5] Appendix D. Here we will describe his result for the groups $W(k)$.

Let $(\alpha_i)_{i \in \mathbb{Z}/6\mathbb{Z}}$ be a basis of the six-dimensional vector space $K^6$. There is a symmetric bilinear form $B$ on $K^6$ given by

$$ (1) \quad B(\alpha_i, \alpha_i) = 2 $$

$$ (2) \quad B(\alpha_i, \alpha_j) = 0, \text{ if } i - j = \pm 1, $$

$$ (3) \quad B(\alpha_i, \alpha_j) = -2 \cos(\pi/k), \quad i - j = \pm 2, \text{ and} $$

$$ (4) \quad B(\alpha_i, \alpha_j) = -2, \text{ if } i - j = \pm 3. $$

For any $\alpha \in K^6$ with $B(\alpha, \alpha) = 2$, let $s_\alpha$ be the hyperplane reflection, defined by
$$ s_\alpha(\lambda) = \lambda - B(\alpha, \lambda)\alpha, $$

for any $\lambda \in K^6$. The *Tits representation* of $W(k)$ is the group homomorphism
$$ \rho : W(k) \to GL_6(K) $$

defined on the generators by $\rho(s_i) = s_{\alpha_i}$, for any $i \in \mathbb{Z}/6\mathbb{Z}$.

**Theorem 19.** (Tits [17][5]) *The representation $\rho$ is faithful.*

#### 4.2 The $l_\infty$-norms of $K$ and $\text{End}(K^6)$

Let $\mathcal{F}$ be the set of field embeddings $v : K \to \mathbb{R}, x \mapsto x_v$. The field $K$ is totally real, and its degree is $\phi(2k)/2$, where $\phi$ is the Euler totient function. It follows that $\text{Card} \, \mathcal{F} = \phi(2k)/2$.
For $x \in K$, let $\|x\|$ be its $l_\infty$-norm defined by

$$ \|x\| = \text{Max}_{v \in \mathcal{F}} |x_v|. $$

We have $\|xy\| \le \|x\|\|y\|$. This norm should not be confused with the usual norm $N_{K/\mathbb{Q}}(x) := \prod_{v \in \mathcal{F}} x_v$, which is a determinant. Let $x \in \mathcal{O} \setminus \{0\}$. Since $|N_{K/\mathbb{Q}}(x)|$ is a positive integer, we have

$$ \|x\| \ge 1 \text{ if } x \in \mathcal{O} \setminus \{0\}. $$

For $i, j \in \mathbb{Z}/6\mathbb{Z}$, let $E_{i,j} \in \text{End}(K^6)$ be the linear map

$$ E_{i,j} : v \in K^6 \mapsto \langle \alpha_j | v \rangle \alpha_i. $$

Since $B(\alpha_i, \alpha_j)$ is a circulant matrix, an easy computation shows that

$$ \det B(\alpha_i, \alpha_j) = 64(3 \cos^4(\pi/k) - 4 \cos^3(\pi/k)) $$

where $\mu_6$ is the group of $6^{\text{th}}$ roots of unity. Since $k \ge 3$, the bilinear form $B$ is nondegenerate and the set

$$ \{E_{i,j} | i, j \in \mathbb{Z}/6\mathbb{Z}\} $$
is a basis of $\mathrm{End}(K^6)$. Therefore any element $A \in \mathrm{End}(K^6)$ can be written as $A = \sum_{i,j \in \mathbb{Z}/6\mathbb{Z}} a_{i,j} E_{i,j}$, where $a_{i,j} \in K$. Its $l_\infty$-norm $\|A\|$ is defined by

$$ \|A\| = \mathrm{Max}_{i,j \in \mathbb{Z}/6\mathbb{Z}} \|a_{i,j}\|. $$

For each $i \in \mathbb{Z}/6\mathbb{Z}$, set $F_i = E_{i,i}$ and for any word $w = i_1 \dots i_n$ on the alphabet $\mathbb{Z}/6\mathbb{Z}$, set

$$ F_w = F_{i_1} \dots F_{i_n}. $$

The $l_\infty$-norms of $\mathrm{End}(K^6)$ is not multiplicative, i.e. $\|AB\|$ is not necessarily $\le \|A\|\|B\|$. Nevertheless, $\|F_w\|$ can still be estimated, as shown in the next lemma.

**Lemma 20.** Let $w$ be a word of length $n$ over the alphabet $\mathbb{Z}/6\mathbb{Z}$. We have

$$ \|F_w\| \le 2^n. $$

*Proof.* The Galois conjugates of $\cos(m\pi/k)$ are the numbers $\cos(m\pi/k)$, for $m$ prime to $k$, hence we have $\|B(\alpha_i, \alpha_j)\| \le 2$ for any $i, j \in \mathbb{Z}/6\mathbb{Z}$. Since $E_{i_1,j_1}E_{i_2,j_2} = B(\alpha_{j_1}, \alpha_{j_2})E_{i_1,j_2}$, for any $i_1, j_1, i_2, j_2 \in \mathbb{Z}/6\mathbb{Z}$, it can be proven by induction over $n$ that $F_w = a_w E_{i_1,i_2}$ for some $a_w \in \mathcal{O}$ with $\|a_w\| \le 2^{n-1} < 2^n$, from which the lemma follows. $\square$

### 4.3 The $l_\infty$-norm of the *Tits representation*

Recall that $\rho : W(k) \to \mathrm{GL}_6(K)$ denotes the *Tits representation*.

**Lemma 21.** For any $w \in W(k)$, we have

$$ \|\rho(w) - 1\| < 3^{l(w)}. $$

*Proof.* Set $n = l(w)$ and let $w = s_{i_1} \dots s_{i_n}$ be a reduced decomposition of $w$. Let $\mathcal{V}$ be the collection of all nonempty subwords of the word $s_{i_1} \dots s_{i_n}$. For $l > 0$, set $\mathcal{V}_l = \{v \in \mathcal{V} \mid l(v) = l\}$. Since some subwords appear more than once, $\mathcal{V}$ and $\mathcal{V}_l$ are sets with multiplicity. For example, if $w = s_1s_2s_3s_2$, the subword $s_1s_2$ appears twice.

We have $\rho(s_i) = 1 - F_i$, for any $i \in \mathbb{Z}/6\mathbb{Z}$. Hence we obtain

$$ \rho(w) - 1 = \sum_{v \in \mathcal{V}} (-1)^{l(v)} F_v, \text{ and} \\ \|\rho(w) - 1\| \le \sum_{v \in \mathcal{V}} \|F_v\|. $$

By Lemma 20, we have

$$ \|F_v\| \le 2^{l(v)}. $$

Since $\mathrm{Card} \, \mathcal{V}_l = \binom{n}{l}$, we obtain

$$ \|\rho(w) - 1\| \le \sum_{1 \le l \le n} \binom{n}{l} 2^l = (3^n - 1) < 3^n. $$

$\square$

### 4.4 Finite quotients of $W(k)$

Set $\mathcal{R} := \bigoplus_{i,j \in \mathbb{Z}/6\mathbb{Z}} \mathcal{O}E_{i,j}$. Since $E_{i_1,j_1}E_{i_2,j_2} = B(\alpha_{j_1}, \alpha_{j_2})E_{i_1,j_2}$, $\mathcal{R}$ is a subring of $\mathrm{End}(K^6)$. Since $s_{\alpha_i} = 1 - E_{i,i}$, we have $\rho(W(k)) \subset \mathcal{R}^*$, where $\mathcal{R}^*$ is the group of invertible elements of $\mathcal{R}$. Set

$$ H(k) = \{w \in W(k) \mid \rho(w) - 1 \in 3^k \mathcal{R}\}. $$

For any integer $n > 0$, set

$$ B_n = \{w \in W(k) \mid l(w) \le n\}. $$

**Proposition 22.** The subgroup $H(k)$ is normal and lies inside $W(k)^+$. Moreover we have

(1) $B_{4k} \cap H(k) = \{1\}$, and

(2) $\mathrm{Card} W(k)/H(k) \le 3^{72k\phi(2k)}$.
*Proof.* The group $\mathcal{R}(k) := \mathcal{R}^* \cap (1+3^k\mathcal{R})$ is a normal subgroup of $\mathcal{R}^*$, hence $H(k)$ is normal. Since we have $\det \rho(h) \equiv 1 \mod 3$, for any $h \in H(k)$, the group $H(k)$ lies in $W(k)^+$. It remains to prove Assertions (1) and (2).

1. *Proof that* $B_{4k} \cap H(k) = \{1\}$. Let $w \in B_{4k} \cap H(k)$. By definition, we have $\rho(w) - 1 = 3^k A$, where $A$ belongs to $\mathcal{R}$. By Lemma 20, we have

$$ \|3^k A\| = \|\rho(w) - 1\| < 3^{4k} = N, $$

therefore $\|A\| < 1$. Since for any $a \in \mathcal{O}$, we have $a = 0$ or $\|a\| \ge 1$ and since $\|A\| = \text{Max}\|a_{i,j}\|$, we have $A = 0$. By Theorem 19, it follows that $w = 1$.

2. *Proof that* $\text{Card } W(k)/H(k) \le 3^{72k\phi(k)}$.

By definition, there is an embedding

$$ W(k)/H(k) \subset \mathcal{R}/N\mathcal{R}. $$

As an abelian group, $\mathcal{O}$ is isomorphic to $\mathbb{Z}^{\phi(k)/2}$, hence $\mathcal{R}/N\mathcal{R} \simeq \mathbb{Z}/N\mathbb{Z}^{18\phi(k)}$. It follows that

$$ \text{Card } W(k)/H(k) \le N^{18\phi(k)} = 3^{72k\phi(k)}. $$

□

5. BOUNDS ON Fill(g)

The last step of the proof of the bound on Fill(g) stated in the Introductin involves the factor $57/\sqrt{\ln \ln \ln g}$. This will now be shown to be a consequence of a 1904 result of E. Landau.

5.1 The set B and Landau's Theorem

Let $p_1 < p_2 < \dots$ be the ordered list of all odd prime numbers. For any $n \ge 1$, set $q_n = 2 \prod_{k=1}^n p_i$, and set $B := \{q_1, q_2, \dots\} = \{6, 30, 210, \dots\}$. The following classical theorem is an improvement of the prime number Theorem of de la Vallée Poussin [6] and Hadamard [8].

**Theorem 23.** (Landau [10]) While $k$ varies over $B$, we have

$$ \phi(k) \sim e^{-\gamma} \frac{k}{\ln \ln k}, $$

where $\gamma = 0.577\dots$ is Euler's constant.

For the proof, see [9], Theorem 328 p. 352.

5.2 The constant $\delta = 12\sqrt{e^{-\gamma}\ln 3} = 9.42\dots$

Let $k \ge 3$. By Proposition 22, there is a subgroup $H(k) \subset W(k)^+$ such that

(1) $B_{4k} \cap H(k) = \{1\}$, and

(2) $\text{Card } W(k)/H(k) \le 3^{72k\phi(2k)}$.

By Lemma 18, $H(k)$ acts freely on $\mathcal{H}(k)$. Let $g_k$ be the genus of the closed surface $\Sigma(k) := \mathcal{H}(k)/H(k)$ and let $\text{Sys}(\Sigma(k))$ be the set of systoles of $\Sigma(k)$. Set

$$ \delta = 12\sqrt{e^{-\gamma}\ln 3}. $$

**Lemma 24.** (1) We have

$$ \lim_{k \to \infty} g_k = \infty $$
(2) Let $\delta^+ > \delta$. For almost all $k \in B$, we have

$$ \mathrm{Card} \ \mathrm{Sys} (\Sigma(k)) \leq \frac{6\delta^+}{\sqrt{\ln \ln \ln g_k}} \frac{g_k}{\sqrt{\ln g_k}}. $$

*Proof.* By Lemma 18, $\mathrm{Sys}(\Sigma(k))$ is exactly the set of curves of the tesselation $\tau_{H(k)}$. Since the systole length of $\Sigma(k)$ tends to $\infty$ as $k$ tends to $\infty$, so is its genus $g_k$, which proves the first assertion.

In the proof of the second assertion, the integer $k$ varies over $B$. Set

$$ \delta(k) = 3^{72k\phi(2k)} = 3^{144k\phi(k)}. $$

By Theorem 23, we have $\ln \delta(k) \sim 144e^{-\gamma} \ln 3k^2 / \ln \ln k$, hence

$$ (24.1) \qquad (\ln \ln k) \ln \delta(k) \sim \delta^2 k^2. $$

It follows that

$$ (24.2) \qquad k > \sqrt{\ln \delta(k)}, \text{ for } k \gg 0. $$

When $k$ tends to infinity, we have $\ln \ln \ln \delta(k) \sim \ln \ln \ln \sqrt{\delta(k)}$. Hence Equation (24.2) implies that

$$ (24.3) \qquad (\delta^+)^2 \ln \ln k > \delta^2 \ln \ln \ln \delta(k) \text{ for } k \gg 0. $$

Combining Equations (24.1) and (24.3) we get that

$$ (\delta^+ k)^2 > \ln \delta(k) \ln \ln \ln \delta(k) \text{ for } k \gg 0, $$

thus we have

$$ (24.4) \qquad \frac{1}{k} < \frac{\delta^+}{\sqrt{\ln \ln \ln \delta(k)}} \frac{1}{\sqrt{\ln \delta(k)}}, \text{ for } k \gg 0. $$

Let $f_0, f_1$ and $f_2$ be the number of vertices, edges and tiles of the tesselation $\tau_{H(k)}$. Since it is a hexagonal tesselation and each vertex has valence 4, we have $f_1 = 3f_2$ and $f_0 = f_1/2$. Since $2(g_k - 1) = f_1 - f_2 - f_0$, we have $2(g_k - 1) = f_2/2 = [W(k) : H(k)]/2$, hence we have $g_k \le \delta(k)$. It follows that

$$ (24.5) \qquad \frac{1}{k} < \frac{\delta^+}{\sqrt{\ln \ln \ln g_k}} \frac{1}{\sqrt{\ln g_k}}, \text{ for } k \gg 0. $$

The number of systoles is $f_1/2k = 6(g_k - 1)/k < 6g_k/k$. It follows from equation (24.5) that

$$ (24.6) \qquad \mathrm{Card} \ \mathrm{Syst}(\Sigma(k)) < \frac{6\delta^+}{\sqrt{\ln \ln \ln g_k}} \frac{g_k}{\sqrt{\ln g_k}}, \text{ for } k \gg 0. \qquad \square $$

## 5.3 The bound for Fill(g)

The following statement is a stronger form of the theorem stated in the introduction.

**Theorem 25.** There exists an infinite set $A$ of integers $g \ge 2$ and, for any $g \in A$, a closed oriented hyperbolic surface $\Sigma_g$ of genus $g$, endowed with a standard hexagonal tesselation $\tau_g$, satisfying the following assertions

(1) the set of curves of $\tau_g$ is the set of systoles of $\Sigma_g$, and

(2) we have

$$ \mathrm{Card} \ \mathrm{Syst}(\Sigma_g) \leq \frac{57}{\sqrt{\ln \ln \ln g}} \frac{g}{\sqrt{\ln g}}. $$

*Proof.* Let us use the notations of Subsection 5.2 and set $\delta^+ = 9.5 = 57/6$.

By the first assertion of Lemma 24, there is an infinite subset $B' \subset B$ such that the map $k \in B' \mapsto g_k \in \mathbb{Z}$ is injective, Set $A := \{g_k | k \in B'\}$ and, for $g \in A$ set $\Sigma_g = \Sigma(k)$ and $\tau_g = \tau_{H(k)}$ where $k \in B'$ is uniquely defined by $g_k = g$.

It follows from the second assertion of Lemma 24 that
$$ \text{Card Sys}(\Sigma_g) \le \frac{57}{\sqrt{\ln \ln \ln g}} \frac{g}{\sqrt{\ln g}}, \text{ for any } g \in A. \quad \square $$

## 5.4 Final remark

The constant 57 in the theorem can be replaced by any real number $a > 6\delta = 56.547...$. This constant can be improved in two ways.

First, one can use the fact that the Tits representations lies inside the orthogonal group $\O_6(K, q)$, where $q$ is the quadratic form defined by $B$. Second, the results concerning hexagon tesselations clearly extend to $2p$-gon tesselations, for any $p \ge 3$. Using octagons instead of hexagons provides a marginally better bound. We have restricted ourselves to this version in order to keep the paper as elementary as possible.

The paper [1] and our result suggests that $\text{Fill}(g)$ should be of “order of magnitude” $g/(\ln g)^\alpha$ for some $\alpha$ with $1/2 \le \alpha \le 1$, but we cannot formulate a precise conjecture at this stage.

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DEPARTMENT OF MATHEMATICS, SOUTHERN UNIVERSITY OF SCIENCE AND TECHNOLOGY, SHENZHEN, CHINA
SUSTECH INTERNATIONAL CENTER FOR MATHEMATICS, SOUTHERN UNIVERSITY OF SCIENCE AND
TECHNOLOGY, SHENZHEN, CHINA

INSTITUT CAMILLE JORDAN DU CNRS, UCBL, 69622 VILLEURBANNE CEDEX, FRANCE

Email address: ingridmary@sustech.edu.cn

Email address: mathieu@math.univ-lyon1.fr