The Cheeger constants of random Belyi surfaces

We briefly introduce the idea on the proof of Theorem 1 here. By definition of the Cheeger constant, it suffices to show that for a generic surface $S^{C}(\Gamma,\mathcal{O})$, there exists a colletion of curves whose union separates $S^{C}(\Gamma,\mathcal{O})$ into two parts such that the ratio quantity in $h(S^{C}(\Gamma,\mathcal{O}))$ can be bounded from above by the desired upper bound in Theorem 1. By the construction of hyperbolic surfaces in [4], we know that

where each $D_{i}$ is a horoball in $S^{O}(\Gamma,\mathcal{O})$ together with its infinity point, and each small triangle has three boundary curves of lengths $\leq 1$ (for example see Figure 8). For large $n>0$ and a generic surface $S^{C}(\Gamma,\mathcal{O})$, we first split as $\bigcup\limits_{i\in\mathcal{I}}D_{i}=\left(\bigcup\limits_{i\in\mathcal{I}_{1}}D_{i}\right)\bigcup\left(\bigcup\limits_{i\in\mathcal{I}_{2}}D_{i}\right)$ where each $D_{i}$ with $i\in\mathcal{I}$ contains a hyperbolic disk of large radius (see Section 3). Next we construct certain simple curves $\{\eta_{i}^{j}\}$ of lengths at most $O(\log n)$ separating each $D_{i}$ $(i\in\mathcal{I}_{1})$ into certain subdomains $\{D(\eta_{i}^{j},\eta_{i}^{k})\}$, each of which has area at most $O\left(\frac{n}{(\log n)^{2}}\right)$ (see Lemma 6). Rewrite the decomposition as

and

where $\delta\in(0,\frac{1}{2})$ is arbitrary. Recall that for any $J$,

Then Theorem 1 follows by letting $n\to\infty$ and $\delta\to 0$ since

Section 2 provides some necessary background from [4] and basic properties on two-dimensional hyperbolic geometry. In Section 3, based on [3, 4] we provide several bounds on hyperbolic lengths and areas, and also give a special division for certain disks from a decomposition of $S^{C}(\Gamma,\mathcal{O})$ which is important in the proof of Theorem 1 (see Lemma 6). We prove our main result Theorem 1 in Section 4.

The authors would like to thank the anonymous referees for their careful reading and valuable comments, and especially would like to thank one referee for sharing his/her idea on how to obtain the current upper bound $\frac{3}{2\pi}$ in Theorem 1, improving our former one $\frac{2}{3}$. They significantly improve the quality of this paper. We also would like to thank Yuhao Xue for helpful discussions, and thank Bram Petri and Zeev Rudnick for their interests and comments on this project. The second named author is partially supported by the NSFC grant No. $12171263$.

Let $\mathbb{H}=\{x+y\textbf{i};\ y>0\}$ be the upper half-plane endowed with the standard hyperbolic metric $ds^{2}=\frac{dx^{2}+dy^{2}}{y^{2}}.$

For any $0<a<b$, set

Then their hyperbolic lengths satisfy

We also set

Then its hyperbolic area satisfies

Now we recall the construction of Belyi surfaces in [4], i.e., the surface $S^{C}(\Gamma,\mathcal{O})$ associated with an oriented graph $(\Gamma,\mathcal{O})$. Let $\Gamma$ be a finite $3-$regular graph and $V(\Gamma)$ be the set of all its vertices. An orientation $\mathcal{O}$ on $\Gamma$ is an assignment, for each vertex $v\in V(\Gamma)$, of a cyclic ordering for the three edges emanating from $v$. Set

Up to a Möbius transformation, one may assume that an ideal triangle $T$ always has vertices $0,1$ and $\infty$ (see Figure 2). The solid segments are geodesics joining the point $\frac{1+\sqrt{3}\textbf{i}}{2}$ and the points in $\left\{\textbf{i},\ 1+\textbf{i},\ \frac{1+\textbf{i}}{2}\right\}$. The dotted segments are horocycles joining pairs of points in $\left\{\textbf{i},\ 1+\textbf{i},\ \frac{1+\textbf{i}}{2}\right\}$. The ideal triangle $T$ has a natural clockwise orientation

In this article, similar as in [4], the points $\left\{\textbf{i},\ 1+\textbf{i},\ \frac{1+\textbf{i}}{2}\right\}$ are called the mid-points of the three sides of $T$, even each side has infinite length. And a dotted segment (see Figure 2) joining two mid-points of an ideal triangle is always called a short horocycle segment. It is clear that each short horocycle segment has hyperbolic length equal to $1$.

Given an element $(\Gamma,\mathcal{O})\in\mathcal{F}^{\star}_{n}$, we replace each vertex $v\in V(\Gamma)$ by a copy of $T$, such that the natural clockwise orientation of $T$ coincides with the orientation of $\Gamma$ at the vertex $v$. If two vertices of $\Gamma$ are joined by an edge, we glue the two copies of $T$ along the corresponding sides subject to the following conditions:

1. (i)

   the mid-points of two sides are glued together;

2. (ii)

   the gluing preserves the orientations of two copies of $T$.

As in [4], the surface $S^{O}(\Gamma,\mathcal{O})$ is uniquely determined by the two conditions above, and it is a complete hyperbolic surface with area equal to $2\pi n$.

For any points $p,q,r\in\mathbb{R}\cup\{\infty\}$, denote by $\mathfrak{L}(p,q)$ the hyperbolic geodesic line joining $p$ and $q$, and denote by $\Delta(p,q,r)$ the ideal triangle with vertices $p,q$ and $r$. Let $L$ and $R$ be the matrices as follows:

They represent the following two automorphisms of $\mathbb{H}$:

Then $L$ and $R$ generate a subgroup $G$ of $\mathrm{PSL(2,\mathbb{Z})}$. Set

Then the set $G(T)$ consists of ideal triangles forming a partition of the upper half-plane $\mathbb{H}$ as in Figure 3.

There exists a fundamental domain of $S^{O}(\Gamma,\mathcal{O})$ in $\mathbb{H}$ such that it is a finite union of ideal triangles in $G(T)$. For example, let $v_{1}$ and $v_{2}$ be two vertices of a $3-$regular graph $\Gamma$ as shown in Figure 4, and the orientations $\mathcal{O}_{1}$ at $v_{1}$ and $v_{2}$ are $(1,2,3)$ and $(1,3,2)$ respectively. From the construction of $S^{O}(\Gamma,\mathcal{O}_{1})$, it is not hard to see that the union

is a fundamental domain of $S^{O}(\Gamma,\mathcal{O}_{1})$ where the boundary sides $\mathfrak{L}(0,1)$ and $\mathfrak{L}(1,2)$ are identified, and the boundary sides $\mathfrak{L}(0,\infty)$ and $\mathfrak{L}(2,\infty)$ are identified. Then $S^{O}(\Gamma,\mathcal{O}_{1})$ is a three punctured sphere (see Figure 4). Using the same $3-$regular graph $\Gamma$, consider the orientation $\mathcal{O}_{2}$ such that the orientations at the two vertices are both $(1,2,3)$ as shown in Figure 4, then $S^{O}(\Gamma,\mathcal{O}_{2})$ is a torus with one puncture.

Recall that [4, Definition 4.1] says that a left-hand-turn path on $(\Gamma,\mathcal{O})$ is a closed path on $\Gamma$ such that at each vertex, the path turns left in the orientation $\mathcal{O}$. Every left-hand-turn path of $(\Gamma,\mathcal{O})$ corresponds to a horocycle loop in $S^{O}(\Gamma,\mathcal{O})$ which encloses a puncture point and consists of certain short horocycle segments as described in subsection 2.2. One may denote such loops by canonical horocycle loops. For any two canonical horocycle loops, they are either disjoint or tangent to each other. We remark here that a canonical horocycle loop may also be tangent to itself. In the remaining, a tangency point always means either a tangency point between two canonical horocycle loops or a self-tangency point of one canonical horocycle loop.

For a random element $(\Gamma,\mathcal{O})\in\mathcal{F}^{\star}_{n}$, denote by $\mathrm{LHT(\Gamma,\mathcal{O})}$ the number of left-hand-turn paths in $(\Gamma,\mathcal{O})$ and by $\mathbb{E}_{n}[\mathrm{LHT(\Gamma,\mathcal{O})}]$ the expected value of $\mathrm{LHT(\Gamma,\mathcal{O})}$ over $\mathcal{F}^{\star}_{n}$. Then $\mathrm{LHT(\Gamma,\mathcal{O})}$ is also the number of punctures of $S^{O}(\Gamma,\mathcal{O})$, and the genus of $S^{O}(\Gamma,\mathcal{O})$ is equal to $1+\frac{n-\mathrm{LHT(\Gamma,\mathcal{O})}}{2}$. We enclose this section by the following estimate which is a direct consequence of [4, Theorem 2.3].

The following estimate is well-known to experts. We prove it for completeness here.

Given $(\Gamma,\mathcal{O})\in\mathcal{F}_{n}^{\star}$ that is defined in (4), we let $\{L_{i}\}_{i\in\mathcal{I}(\Gamma,\mathcal{O})}$ denote the set of all canonical horocycle loops of $S^{O}(\Gamma,\mathcal{O})$. Write the index set $\mathcal{I}(\Gamma,\mathcal{O})$ as $\mathcal{I}$ for simplicity. For each $i\in\mathcal{I}$, the canonical horocycle loop $L_{i}$ bounds a punctured disk $N_{i}\subset S^{O}(\Gamma,\mathcal{O})$ with a puncture $p_{i}$. Moreover $D_{i}=N_{i}\cup\{p_{i}\}$ is an open topological disk in $S^{C}(\Gamma,\mathcal{O})$.

Now we recall the so-called large cusps condition defined in [3, 4]. First for simplicity, we write $S^{O}(\Gamma,\mathcal{O})$ and $S^{C}(\Gamma,\mathcal{O})$ as $S^{O}$ and $S^{C}$ respectively. For any $i\in\mathcal{I}$, up to a conjugation one may lift the puncture $p_{i}$ to $\infty$ in the boundary $\partial\mathbb{H}$ of the upper half plane $\mathbb{H}$. Let $\pi_{i}:\mathbb{H}\to S^{O}$ be the covering map with $\pi_{i}(\infty)=p_{i}$ and assume that the Mobius transformation through doing one turn around $p_{i}$ corresponds to the parabolic isometry $z\mapsto z+1$. For any $l>0$, denote

and

Then for suitable $l>0$, the set $C_{i}(l)$ is a cusp around $p_{i}$ whose boundary curve $h_{i}(l)$ is the projection of a horocycle. We also denote by $C^{l}$ the standard cusp

endowed with the hyperbolic metric. Now we recall [3, Definition 2.1] or [4, Definition 3.1] saying that.

Brooks and Makover in [4] proved that for any $l>0$, as $n\to\infty$ a generic hyperbolic surface $S^{O}(\Gamma,\mathcal{O})$ has cusps of length $\geq l$. More precisely

Let $S^{C}$ be the conformal compactification of $S^{O}$ endowed with the hyperbolic metric $ds_{S^{C}}^{2}$, i.e., $S^{C}=S^{O}\bigcup\{p_{i}\}_{i\in\mathcal{I}}$. For any point $p\in S^{C}$ and $r>0$, denote

According to the proof of [3, Theorem 2.1], the following theorem holds.

Let $(\Gamma,\mathcal{O})\in\mathcal{F}^{\star}_{n}(c)$ that is defined in (8). For each $i\in\mathcal{I}$, we assume that the canonical horocycle loop $L_{i}$ around $p_{i}$ consists of $d_{i}$ short horocycle segments. So we have

where $N_{i}\subset S^{O}(\Gamma,\mathcal{O})$ is the punctured disk enclosed by the canonical horocycle loop $L_{i}$ around $p_{i}$ in $S^{O}(\Gamma,\mathcal{O})$. Recall that each $S^{O}(\Gamma,\mathcal{O})$ consists of $2n$ ideal triangles each of which contains three short horocycle segments. So we have

Divide $\mathcal{I}$ into the following two subsets

If $i\in\mathcal{I}_{1}$, then for $n$ large enough we have

Recall that for every $i\in\mathcal{I}$, $D_{i}=N_{i}\cup\{p_{i}\}$ is a topological disk in $S^{C}(\Gamma,\mathcal{O})$. For any $i\in\mathcal{I}_{1}$, as in Figure 5, there are four closed loops around the puncture point $p_{i}$ as follows:

1. (a)

   $L_{i}$ is the canonical horocycle loop under the hyperbolic metric $ds_{S^{O}}^{2}$;

2. (b)

   $h_{i}(l(a))$ is the horocycle loop with length $l(a)$ under the hyperbolic metric $ds_{S^{O}}^{2}$;

3. (c)

   $h_{i}(l(1))$ is the horocycle loop with length $l(1)$ under the hyperbolic metric $ds_{S^{O}}^{2}$;

4. (d)

   the loop $\partial\mathbb{B}(p_{i},2\sqrt{2}r(1))$ is the boundary of the geodesic ball $\mathbb{B}(p_{i},2\sqrt{2}r(1))$ under the hyperbolic metric $ds_{S^{C}}^{2}$.

Then combining with (7), (14) and Part $(2)$ of Theorem 5 we have

which in particular tells that the four loops $L_{i},h_{i}(l(a)),h_{i}(l(1))$ and $\partial\mathbb{B}(p_{i},2\sqrt{2}r(1))$ are pairwisely disjoint.

Assume that the set of all tangency points on $L_{i}$ is

arranged in the anti-clockwise direction. For each $1\leq j\leq d_{i}$, we let $p_{i}A_{j}$ be the geodesic ray joining $p_{i}$ and $A_{j}$ under the hyperbolic metric $ds^{2}_{S^{O}}$. We remark here that there may exist several intersection points between $p_{i}A_{j}$ and the loop $\partial\mathbb{B}(p_{i},2\sqrt{2}r(1))$. Now define $B_{j}$ to be the last intersection point between them on the direction from $p_{i}$ to $A_{j}$, i.e., $B_{j}$ is the unique point on the ray $p_{i}A_{j}$ such that

where $B_{j}A_{j}$ is the subsegment of $p_{i}A_{j}$ joining $B_{j}$ and $A_{j}$. By (15) we know that $\mathbb{B}(p_{i},2\sqrt{2}r(1))\subset C_{i}(l(a))\cup\{p_{i}\}$ where $C_{i}(l(a))\cup\{p_{i}\}$ is topologically a disk. Then for any point $Q\in\mathbb{B}(p_{i},2\sqrt{2}r(1))$, there exists a unique shortest geodesic $\overline{p_{i}Q}$ joining $p_{i}$ and $Q$ under the hyperbolic metric $ds_{S^{C}}^{2}$. In particular, we let $\overline{p_{i}B_{j}}$ be the unique shortest geodesic joining $p_{i}$ and $B_{j}$. Now consider the concatenation (see Figure 5 for an illustration)

By construction, it is not hard to see that

1. (1)

   for $1\leq j\neq k\leq d_{i}$, $\eta_{i}^{j}\cap\eta_{i}^{k}=\{p_{i}\}$;

2. (2)

   the curve $\eta_{i}^{j}$ joins $p_{i}$ and $A_{j}$ with $\eta_{i}^{j}\subset D_{i}$.

Moreover, the length $\ell_{C}(\eta_{i}^{j})$ can be effectively bounded from above. More precisely, it follows by (2), (12) and Lemma 3 that for each $1\leq j\leq d_{i}$ and $n$ large enough,

Assume that $\alpha,\beta\subset D_{i}$ are two simple curves joining $p_{i}$ and certain points on $L_{i}$ such that $\alpha\cap\beta=\{p_{i}\}$ and both $\alpha$ and $\beta$ only intersect with $L_{i}$ at their endpoints. Denote by $D(\alpha,\beta)$ the domain enclosed by $\alpha,\ \beta$ and $L_{i}$ in the anti-clockwise direction from $\alpha$ to $\beta$ (see Figure 6 for an illustration).

For any $1\leq j\neq k\leq d_{i}$, for simplicity we denote