# Small eigenvalues of closed Riemann surfaces for large genus

Yunhui Wu  Tsinghua University, Beijing, China<EMAIL_ADDRESS>and  Yuhao Xue  Tsinghua University, Beijing, China
<EMAIL_ADDRESS>
###### Abstract.

In this article we study the asymptotic behavior of small eigenvalues of
Riemann surfaces for large genus. We show that for any positive integer $k$,
as the genus $g$ goes to infinity, the smallest $k$-th eigenvalue of Riemann
surfaces in any thick part of moduli space of Riemann surfaces of genus $g$ is
uniformly comparable to $\frac{1}{g^{2}}$ in $g$.

In the proof of the upper bound, for any constant $\varepsilon>0$, we will
construct a closed Riemann surface of genus $g$ in any $\varepsilon$-thick
part of moduli space such that it admits a pants decomposition whose boundary
curves all have length equal to $\varepsilon$, and the number of separating
systole curves in this surface is uniformly comparable to $g$.

##  1\. Introduction

For a closed hyperbolic Riemann surface $X_{g}$ of genus $g\geq 2$, the
spectrum of the Laplacian on $X_{g}$ is a fascinating topic in a number of
mathematical fields for a long time. It is well-known that the spectrum of
$X_{g}$ is a discrete subset in $\mathbb{R}^{\geq 0}$ and consists of
eigenvalues with finite multiplicity. We enumerate them, counted with
multiplicity, in the following increasing order

$0=\lambda_{0}(X_{g})<\lambda_{1}(X_{g})\leq\lambda_{2}(X_{g})\leq\cdots.$

Buser [6] showed that for any constant $\varepsilon>0$, there exists a
hyperbolic surface $X_{g}$ of genus $g\geq 2$ such that
$\lambda_{2g-3}(X_{g})<\varepsilon$. Schoen, Wolpert and Yau [23] showed that
for any integer $i\in[1,2g-3]$, the $i$-th eigenvalue $\lambda_{i}(X_{g})$ is
comparable to a quantity $L_{i}$ on $X_{g}$. Here $L_{i}$ is the minimal
possible sum of the lengths of simple closed geodesics in $X_{g}$ which cut
$X_{g}$ into $i+1$ pieces. Clearly the quantity $L_{i}\ (1\leq i\leq 2g-3)$
can be arbitrary close to $0$ for certain Riemann surfaces. And they [23] also
showed that $\lambda_{2g-2}(X_{g})>c(g)$ where $c(g)>0$ is a constant only
depending on $g$. Then Buser [8, Theorem 8.1.4] showed that the constant
$c(g)$ can be chosen to be uniform (independent of $g$). Otal and Rosas [20]
showed that if the metric is analytic, then the constant $c(g)$ can be exactly
chosen to be $\frac{1}{4}$. Ballmann, Matthiesen and Mondal [1] showed that
the assumption on the real analyticity of the hyperbolic metric in [20] can be
removed.

Let $\mathcal{M}_{g}$ be the moduli space of Riemann surfaces of genus $g$.
Fix any integer $k\geq 1$, as introduced above, by Schoen-Wolpert-Yau [23] the
$k$-th eigenvalue $\lambda_{k}$ can be arbitrarily closed to $0$ near certain
boundary of $\mathcal{M}_{g}$ for large enough $g$. For any
$X=X_{g}\in\mathcal{M}_{g}$, let $h(X)$ be the Cheeger constant of
$X\in\mathcal{M}_{g}$ (one may see [12] or Section 3 for the definition).
Mirzakhani [17, Theorem 4.8] showed that the following limit probability
holds:

$\lim_{g\to\infty}\frac{\mathop{\rm Vol}_{wp}(\\{X\in\mathcal{M}_{g};\
h(X)\geq\frac{\log 2}{2\pi+\log 2}\\})}{\mathop{\rm
Vol}_{wp}(\mathcal{M}_{g})}=1$

where $\mathop{\rm Vol}_{wp}$ is the Weil-Petersson measure induced from the
classical Weil-Petersson metric on $\mathcal{M}_{g}$.

Recall that the Cheeger inequality [12] tells that the first eigenvalue
$\lambda_{1}(X)\geq\frac{1}{4}h^{2}(X)$. Hence, Mirzakhani’s equation above
gives that in a probabilistic sense the first eigenvalue of Riemann surfaces
is uniform positive as $g\to\infty$. More precisely,

$\lim_{g\to\infty}\frac{\mathop{\rm Vol}_{wp}(\\{X\in\mathcal{M}_{g};\
\lambda_{1}(X)\geq\frac{1}{4}\cdot(\frac{\log 2}{2\pi+\log
2})^{2}\\})}{\mathop{\rm Vol}_{wp}(\mathcal{M}_{g})}=1.$

We refer to the length of a shortest closed geodesic in $X$ as the _systole_
of $X$ and denote it by $\mathop{\rm sys}(X)$. The systole function
$\mathop{\rm sys}(\cdot):\mathcal{M}_{g}\to\mathbb{R}^{+}$ is continuous, but
not smooth as corners appear when it is realized by multiple essential
isotropy classes of simple closed curves. Many results on the geometry of
moduli space $\mathcal{M}_{g}$ especially for large genus $g$ can be stated in
terms of the systole function. One may see [3, 10, 13, 16, 22, 25] for recent
related topics.

For any constant $\varepsilon>0$, let $\mathcal{M}_{g}^{\geq\varepsilon}$ be
the $\varepsilon$-thick part of the moduli space $\mathcal{M}_{g}$. That is,

$\mathcal{M}_{g}^{\geq\varepsilon}:=\\{X\in\mathcal{M}_{g};\ \mathop{\rm
sys}(X)\geq\varepsilon\\}.$

It is known that the set $\mathcal{M}_{g}^{\geq\varepsilon}$ is compact for
all $\varepsilon>0$, which is due to Mumford [19]. And it is not hard to see
that $\mathcal{M}_{g}^{\geq\varepsilon}$ is always nonempty for large enough
$g$. Actually Buser-Sarnak [5] showed that
$\max_{X\in\mathcal{M}_{g}}\mathop{\rm sys}(X)$ is uniformly comparable to
$\log{g}$ as $g\to\infty$.

For any fixed positive integer $k$, the $k$-th eigenvalue

$\lambda_{k}(\cdot):\mathcal{M}_{g}\to\mathbb{R}^{+}$

defines a continuous and bounded function [8]. As introduced above we know
that for large enough $g$,

$\inf_{X\in\mathcal{M}_{g}}\lambda_{k}(X)=0.$

Restricted on the thick part, there exists a constant $c(g,k,\epsilon)>0$,
depending on $g,k$ and $\varepsilon$, such that

$\inf_{X\in\mathcal{M}_{g}^{\geq\varepsilon}}\lambda_{k}(X)\geq
c(g,k,\epsilon)>0.$

In this paper, our main goal is to study the asymptotic behavior of this
quantity $\inf_{X\in\mathcal{M}_{g}^{\geq\varepsilon}}\lambda_{k}(X)$ as
$g\to\infty$.

Notation. In this paper, for two positive functions $f_{1}$ and $f_{2}$ of
genus $g$ we say

$f_{1}\succ_{g}f_{2}$

if there exists a universal constant $C>0$, independent of $g$, such that

$\liminf_{g\to\infty}\frac{f_{1}}{f_{2}}\geq C.$

Similarly, we say

$f_{1}\prec_{g}f_{2}$

if

$\limsup_{g\to\infty}\frac{f_{1}}{f_{2}}\leq C.$

We say

$f_{1}\asymp_{g}f_{2}$

if

$f_{1}\succ_{g}f_{2}\quad\text{and}\quad f_{1}\prec_{g}f_{2}.$

Now we are ready to state our result.

######  Theorem 1.

For any integer $k\geq 1$ and constant $\varepsilon>0$, we have

$\inf_{X\in\mathcal{M}_{g}^{\geq\varepsilon}}\lambda_{k}(X)\asymp_{g}\frac{1}{g^{2}}.$

More precisely, there exist two positive constants $\alpha(\varepsilon)$ and
$\beta(\varepsilon)$, only depending on $\varepsilon$, such that for $g$ large
enough,

$\frac{\alpha(\varepsilon)}{g^{2}}\leq\min_{X\in\mathcal{M}_{g}^{\geq\varepsilon}}\lambda_{k}(X)\leq\frac{\beta(\varepsilon)\cdot
k^{2}}{g^{2}}.$

######  Remark 2.

For small positive constant $\varepsilon<2\ln(1+\sqrt{2})$, the constant
$\beta(\varepsilon)$ above can be chosen to be $c\cdot\varepsilon$ where $c>0$
is a universal constant independent of $\varepsilon$. One may see Theorem 17
for details.

The lower bound $\frac{1}{g^{2}}$ for the growth rate will follow by the
Cheeger inequality. The largest part of our proof of the theorem above is to
show that this growth rate $\frac{1}{g^{2}}$ is optimal as $g\to\infty$. We
will construct a Riemann surface in $\mathcal{M}_{g}^{\geq\varepsilon}$ and
show that the $k$-th eigenvalue of this surface achieves this growth rate
$\frac{1}{g^{2}}$ for large $g$. The existence of such a Riemann surface is
independently interesting. We formulate it as follows. A curve $\gamma$ is
said to be _separating_ on $X$ if the complement $X\setminus\gamma$ has two
components.

######  Proposition 3.

For any constant $\varepsilon>0$, there exists an integer $g(\varepsilon)>0$,
depending on $\varepsilon$, such that for all $g\geq g(\varepsilon)$, there
exists a Riemann surface $\mathcal{X}_{g}$ of genus $g$ satisfying that

$(a)$. $\mathcal{X}_{g}\in\mathcal{M}_{g}^{\geq\varepsilon}$.

$(b)$. There exists a pants decomposition $P$ of $\mathcal{X}_{g}$ such that
for every boundary curve $\gamma$ in the boundary $\partial P$ of $P$, the
length $\ell(\gamma)$ of $\gamma$ is equal to $\varepsilon$.

$(c)$. The number $\\#\\{\gamma\in\partial P;\ \text{$\gamma$ is separating on
$\mathcal{X}_{g}$}\\}\asymp_{g}g$.

For small enough constant $\varepsilon>0$, by using the Collar Lemma it is not
hard to construct such a surface. Petri [21, Proposition 6.2] constructed a
closed surface satisfying (b) and all other simple curves that are not part of
the given pants decomposition are “long enough”. One may also see [14] for
related constructions (in more complicated settings). The new insights of
Proposition 3 are that the constant $\varepsilon$ can be arbitrarily large and
the number of separating systolic curves on $\mathcal{X}_{g}$ is uniformly
comparable to $g$. By construction in Section 4, the twist parameters on the
boundary curves of $P$ can be chosen to be arbitrary. So actually we construct
a subset in $\mathcal{M}_{g}^{\geq\varepsilon}$ of dimension $(3g-3)$
consisting of closed Riemann surfaces of genus $g$ satisfying Proposition 3.

######  Remark 4.

It is interesting to study the asymptotic behavior of the quantity
$\sup_{X\in\mathcal{M}_{g}}\lambda_{k}(X)$ as $g\to\infty$. It is known that
$\limsup_{g\to\infty}\sup_{X\in\mathcal{M}_{g}}\lambda_{1}(X)\leq\frac{1}{4}$.
One may see [4, 7, 9, 18] for related details. One conjecture is as follows.

######  Conjecture 5.

$\lim_{g\to\infty}\sup_{X\in\mathcal{M}_{g}}\lambda_{1}(X)=\frac{1}{4}$.

### Plan of the paper.

Section 2 will provide some necessary background and basic properties on
hyperbolic Riemann surfaces. In Section 3 we will provide a proof for the
lower bound of Theorem 1. The construction of Riemann surface will be
discussed in Section 4, that is to show Proposition 3. In Section 5 we will
complete the proof of Theorem 1, that is to prove the upper bound of Theorem
1.

### Acknowledgement

The authors would like to greatly thank Hugo Parlier and Kasra Rafi for the
invaluable discussions on Proposition 3. The first named author is partially
supported by China’s Recruit Program for Global Young Experts.

##  2\. Preliminaries

In this section we will set up the notations and provide some necessary
background on two-dimensional hyperbolic geometry, regular graphs and spectrum
theory of the Laplace operator.

###  2.1. Hyperbolic surfaces

Let $X_{g}$ be a closed hyperbolic surface of genus $g\geq 2$ and
$\gamma\subset X_{g}$ be a non-trivial loop. There always exists a unique
closed geodesic, still denoted by $\gamma$, representing this loop. The Collar
Lemma says that it has a tubular neighborhood which is a topological cylinder
with a standard hyperbolic metric. And the width of this cylinder, only
depending on the length of $\gamma$, goes to infinity as the length of
$\gamma$ goes to $0$. This was first observed by Keen in [15] and then
improved by many people. We use the following version [8, Theorem 4.1.1].

######  Lemma 6 (Collar lemma).

Let $\gamma_{1},\gamma_{2},...,\gamma_{m}$ be disjoint simple closed geodesics
on a closed hyperbolic Riemann surface $X_{g}$, and $\ell(\gamma_{i})$ be the
length of $\gamma_{i}$. Then $m\leq 3g-3$ and we can define the collar of
$\gamma_{i}$ by

$K(\gamma_{i})=\\{x\in X_{g};\ \mathop{\rm dist}(x,\gamma_{i})\leq
w(\gamma_{i})\\}$

where

$w(\gamma_{i})=\mathop{\rm arcsinh}\frac{1}{\sinh\frac{1}{2}l(\gamma_{i})}$

is the half width of the collar.

Then the collars are pairwise disjoint for $i=1,...,m$. Each $K(\gamma_{i})$
is isomorphic to a cylinder
$(\rho,t)\in[-w(\gamma_{i}),w(\gamma_{i})]\times\mathbb{S}^{1}$, where
$\mathbb{S}^{1}=\mathbb{R}/\mathbb{Z}$, with the metric

$ds^{2}=d\rho^{2}+\ell(\gamma_{i})^{2}\cosh^{2}\rho dt^{2}.$

And for a point $(\rho,t)$, the point $(0,t)$ is its projection on the
geodesic $\gamma_{i}$, $\lvert\rho\rvert$ is the distance to $\gamma_{i}$, $t$
is the coordinate on $\gamma_{i}\cong\mathbb{S}^{1}$.

As the length $\ell(\gamma)$ of the central closed geodesic goes to $0$, the
width $w$ tends to infinity; and if the length $\ell(\gamma)$ goes to
infinity, the width $w$ tends to zero. In this paper, we mainly deal with the
case that $\ell(\gamma)$ is large.

###  2.2. Pants

_A pair of pants_ $\mathcal{P}$ is a compact Riemann surface of $0$ genus with
$3$ boundary closed geodesics. The complex structure is uniquely determined by
the lengths of the three boundary closed geodesics.

For any constant $\varepsilon>0$, let $\mathcal{P}_{\varepsilon}$ be the pair
of pants whose boundary curves all have length equal to $\varepsilon$. Let $d$
be a shortest path between two different boundary curves. We denote its length
also by $d$. The curve $d$ is perpendicular to the boundary curves. Let $\tau$
be a shortest path with both end points on a given boundary curve and assume
that $\tau$ is not homotopic a piece of the boundary curve (with end points
fixed). Then this curve $\tau$ is clearly also perpendicular to the given
boundary curve at both end points. We denote the length of $\tau$ also by
$\tau$.

$\varepsilon$$\varepsilon$$\varepsilon$$d$$d$$d$$\tau$ Figure 1.

The three geodesics $d$ divide the pants into two equal right-angled hexagons.
Let $h$ be the shortest path between a pair of opposite sides in the right-
angled hexagon. We denote the length of $h$ also by $h$. By symmetry, clearly
$h$ intersects the both boundary curves at the midpoints. And

$\tau=2h.$

$\varepsilon/2$$\varepsilon/2$$\varepsilon/2$$d$$d$$d$$h$ Figure 2.

For right-angled pentagons and hexagons, by formulas in [8, Page 454] we have

$\cosh\frac{\varepsilon}{2}=\sinh^{2}{\frac{\varepsilon}{2}}\cosh{d}-\cosh^{2}\frac{\varepsilon}{2}$

and

$\cosh\frac{\varepsilon}{2}=\sinh h\sinh\frac{\varepsilon}{4}.$

Thus,

(1) $\displaystyle d=2\mathop{\rm
arcsinh}\frac{1}{2\sinh\frac{\varepsilon}{4}}$

and

(2) $\displaystyle\tau=2h=2\mathop{\rm
arcsinh}\frac{\cosh\frac{\varepsilon}{2}}{\sinh\frac{\varepsilon}{4}}.$

Thus, for large $\varepsilon$, we have

(3)
$\displaystyle\lim_{\varepsilon\to\infty}\frac{d}{2e^{-\frac{\varepsilon}{4}}}=1.$

And we always have

(4) $\displaystyle\tau>\frac{1}{2}\varepsilon.$

###  2.3. Trivalent graphs

A _trivalent graph_ is a finite $3$-regular connected graph. That is, every
vertex has three emanating edges. Every compact Riemann surface of genus
$g\geq 2$ has a pants decomposition of $(2g-2)$ pairs of pants. If we
associate each pair of pants to be a vertex, and an edge between two vertices
if two pairs of pants have a common boundary curve, then this associates each
compact Riemann surface of genus $g\geq 2$ to a trivalent graph with $(2g-2)$
vertices of $(3g-3)$ edges. Actually the opposite way is also true. One may
see [8, Section 3.5] for more details.

Recall that the _girth_ of a connected graph is the length of the shortest
non-trivial cycle in this graph where each edge has length equal to $1$.
Actually more general, an $n$-regular graph is a graph such that all vertices
have degree $n$. A trivalent graph is a $3$-regular graph. Let $U(n,E,w)$ be
the number of unlabeled $n$-regular graphs with $E$ edges with girth at least
$w$.

We say two functions $f_{1}$ and $f_{2}$ of variable $E$ $f_{1}\sim_{E}f_{2}$
if $\lim_{E\to\infty}\frac{f_{1}(E)}{f_{2}(E)}=1$.

By Bollob$\acute{\text{a}}$s [2] and Wormald [24] we know that,

######  Theorem 7 (Bollob$\acute{\text{a}}$s).

For $n\geq 3$, we have

$U(n,E,3)\sim_{E}\exp(-\sum_{i=1}^{2}\frac{(n-1)^{i}}{2i})\frac{(2E)!}{2^{E}E!V!(n!)^{V}}$

where $V=2E/n$ is the number of vertices.

######  Theorem 8 (Wormald).

For $n\geq 3,w\geq 3$, we have

$U(n,E,w)\sim_{E}\exp(-\sum_{i=3}^{w-1}\frac{(n-1)^{i}}{2i})U(n,E,3)$

where $V=2E/n$ is the number of vertices.

Thus,

$U(n,E,w)\sim_{E}\exp(-\sum_{i=1}^{w-1}\frac{(n-1)^{i}}{2i})\frac{(2E)!}{2^{E}E!V!(n!)^{V}}.$

This goes to infinity as $E\rightarrow\infty$ for any given positive numbers
$n$ and $w$.

In particular for any given (large) girth $w$, $n=3$ and $E=(3g-3)$ we have

(5) $\displaystyle\lim_{g\to\infty}U(3,3g-3,w)=\infty.$

So for large enough $g$, there always exist trivalent graphs with edges
$(3g-3)$ of arbitrary large girth. In Section 4 we will use this fact to
construct the Riemann surfaces in Proposition 3.

###  2.4. Spectrum

Let $X_{g}$ be a closed Riemann surface of genus $g\geq 2$ which corresponding
to a hyperbolic metric on $X_{g}$. Let $\Delta$ be the Laplacian with respect
to this metric. A number $\lambda$ is called an _eigenvalue_ if $\Delta
f+\lambda\cdot f=0$ on $X_{g}$. And the corresponding function $f$ is called
an _eigenfunction_. It is known that the set of eigenvalues is an infinite
sequence of non-negative numbers

$0=\lambda_{0}(X_{g})<\lambda_{1}(X_{g})\leq\lambda_{2}(X_{g})\leq\cdots.$

Let $\\{f_{i}\\}_{i\geq 0}$ be its corresponding orthonormal sequence of
eigenfunctions. Clearly $f_{0}$ is the constant function
$\frac{1}{4\pi(g-1)}$. The mini-max principle tells that for any integer
$k\geq 0$,

(6) $\displaystyle\lambda_{k}(X_{g})=\inf\\{\frac{\int_{X_{g}}|\nabla
f|^{2}}{\int_{X_{g}}f^{2}};\ f\in H^{1}(X_{g})\ \text{and}\ \int_{X_{g}}f\cdot
f_{i}=0\ \forall i\in[0,k-1]\\}$

where $H^{1}(X_{g})$ is the completion of the space of smooth functions on
$X_{g}$. One may see [11] for details.

##  3\. Lower Bound

In this section, we will use the Cheeger inequality in [12] to give the lower
bound in Theorem 1. More precisely, we will show that

######  Proposition 9.

For any constant $\varepsilon>0$ there exists a constant
$\alpha(\varepsilon)>0$, only depending on $\varepsilon$, such that for large
enough $g$,

$\min_{X\in\mathcal{M}_{g}^{\geq\varepsilon}}\lambda_{1}(X)\geq\frac{\alpha(\varepsilon)}{g^{2}}.$

First we recall the Cheeger inequality. Let $X_{g}$ be a closed Riemann
surface of genus $g\ (g\geq 2)$ and $\Omega\subset X_{g}$ be a domain with
smooth boundary. The _first eigenvalue $\sigma_{1}(\Omega)$ of Neumann type_
on $\Omega$ is defined as

$\sigma_{1}(\Omega):=\inf_{\int_{\Omega}f=0}\frac{\int_{\Omega}\lvert\nabla
f\rvert^{2}}{\int_{\Omega}f^{2}}$

where $\nabla$ is the hyperbolic gradient in the sense of the hyperbolic
metric corresponding to the complex structure of $X_{g}$.

The _Cheeger isoperimetric constant_ $h(\Omega)$ is defined as

$h(\Omega):=\inf\frac{\mathop{\rm length}(\Gamma)}{\min\\{\mathop{\rm
Area}(A_{1}),\mathop{\rm Area}(A_{2})\\}}$

where the infimum is taken over all smooth curves $\Gamma$ which divide
$\Omega$ into two pieces $A_{1}$ and $A_{2}$.

######  Lemma 10 (Cheeger inequality, [12]).

Then

$\sigma_{1}(\Omega)\geq\frac{1}{4}h^{2}(\Omega).$

We will apply this Cheeger inequality to prove Proposition 9.

The following two elementary isoperimetric inequalities will also be applied
later.

######  Lemma 11.

Assume that $\Omega\subset X_{g}$ is an open disk. Then

$\mathop{\rm Area}(\Omega)\leq\mathop{\rm length}(\partial\Omega).$

###### Proof.

Since $\Omega\subset X_{g}$ is an open disk, one may regard $\Omega$ as a disk
in the upper half plane $\mathbb{H}$ endowed with the standard hyperbolic
metric $ds^{2}=\frac{dx^{2}+dy^{2}}{y^{2}}$. For a smooth function $f(x,y)$ on
$\Omega$, direct computations give that

$\nabla f=y^{2}\frac{\partial f}{\partial x}\frac{\partial}{\partial
x}+y^{2}\frac{\partial f}{\partial y}\frac{\partial}{\partial y},$ $|\nabla
f|^{2}=y^{2}(|\frac{\partial f}{\partial x}|^{2}+|\frac{\partial f}{\partial
y}|^{2}),$ $\Delta f=y^{2}(\frac{\partial^{2}f}{\partial
x^{2}}+\frac{\partial^{2}f}{\partial y^{2}})$

By Stokes’ Theorem and the Cauchy-Schwarz inequality, we have

$\displaystyle\int_{\Omega}\Delta f$ $\displaystyle=$
$\displaystyle\int_{\partial\Omega}\nabla f\cdot\vec{n}$ $\displaystyle\leq$
$\displaystyle\int_{\partial\Omega}|\nabla f|.$

Set $f(x,y)=-\ln y$. It is clear that

$\Delta f=1\quad\emph{and}\quad|\nabla f|=1.$

Plug the two equations above into (3), we get

$\int_{\Omega}dV\leq\int_{\partial\Omega}ds.$

That is,

$\mathop{\rm Area}(\Omega)\leq\mathop{\rm length}(\partial\Omega).$

The proof is complete. ∎

######  Lemma 12.

Assume that a domain $\Omega\subset X_{g}$ is topologically a cylinder. Then

$\mathop{\rm Area}(\Omega)\leq\mathop{\rm length}(\partial\Omega).$

###### Proof.

Since $\Omega\subset X_{g}$ is topologically a cylinder, the boundary
$\partial\Omega$ consists of two homotopic simple closed loops. We split the
proof into two cases.

Case-1. The two boundary closed loops of $\Omega$ are homotopic to a point on
$X_{g}$.

For this case, one may assume that one boundary closed loop bounds a disk $D$
such that

$\Omega\subset D\quad\text{and}\quad\partial D\subset\partial\Omega.$

Then by Lemma 11 we have,

$\displaystyle\mathop{\rm Area}(\Omega)$ $\displaystyle\leq$
$\displaystyle\mathop{\rm Area}(D)$ $\displaystyle\leq$
$\displaystyle\mathop{\rm length}(\partial D)$ $\displaystyle\leq$
$\displaystyle\mathop{\rm length}(\partial\Omega).$

Case-2. The two boundary closed loops of $\Omega$ are not homotopic to a
point.

For this case, one may assume that $\gamma$ is the unique closed geodesic
representing a component of $\partial\Omega$. Set

$l=\mathop{\rm length}(\gamma).$

Let $\mathbb{R}\times\mathbb{S}^{1}$ be the infinite cylinder of parameters
$(\rho,t)$ endowed with the hyperbolic metric

$ds^{2}=d\rho^{2}+l^{2}\cosh^{2}\rho dt^{2}.$

This is a hyperbolic cylinder with infinite width whose unique closed geodesic
has length equal to $l$. By assumption one may assume that $\Omega$ is a
subset of this infinite hyperbolic cylinder $\mathbb{R}\times\mathbb{S}^{1}$.
For a smooth function $f(\rho,t)$ on $\Omega$, direct computations show that

$\nabla f=\frac{\partial
f}{\partial\rho}\frac{\partial}{\partial\rho}+\frac{1}{l^{2}\cosh^{2}\rho}\frac{\partial
f}{\partial t}\frac{\partial}{\partial t},$ $|\nabla f|^{2}=|\frac{\partial
f}{\partial\rho}|^{2}+\frac{1}{l^{2}\cosh^{2}\rho}|\frac{\partial f}{\partial
t}|^{2},$ $\Delta
f=\frac{\partial^{2}f}{\partial\rho^{2}}+\tanh{\rho}\cdot\frac{\partial
f}{\partial\rho}+\frac{1}{l^{2}\cosh^{2}\rho}\frac{\partial^{2}f}{\partial
t^{2}}.$

Similar as in the proof of Lemma 11 we have

$\displaystyle\int_{\Omega}\Delta f$ $\displaystyle=$
$\displaystyle\int_{\partial\Omega}\nabla f\cdot\vec{n}$ $\displaystyle\leq$
$\displaystyle\int_{\partial\Omega}|\nabla f|.$

Set $f(\rho,t)=\ln\cosh\rho$. It is clear that

$\Delta f=1\quad\emph{and}\quad|\nabla f|=|\tanh\rho|.$

Plug these two equations above into (3), we get

$\displaystyle\int_{\Omega}dV$ $\displaystyle\leq$
$\displaystyle\int_{\partial\Omega}|\tanh\rho|ds$ $\displaystyle\leq$
$\displaystyle\int_{\partial\Omega}ds.$

That is,

$\mathop{\rm Area}(\Omega)\leq\mathop{\rm length}(\partial\Omega).$

The proof is complete. ∎

Now we are ready to prove Proposition 9.

###### Proof of Proposition 9.

Let $X_{g}\in\mathcal{M}_{g}^{\geq\varepsilon}$. By Cheeger’s inequality it
suffices to provide a lower bound for the Cheeger isoperimetric constant
$h(X_{g})$.

Let $\Gamma$ be a set of smooth curves dividing $X_{g}$ into two disjoint
pieces $A_{1}$ and $A_{2}$. Then $\Gamma$ must be one of the following three
cases:

$(a)$. $\Gamma$ contains a simple closed curve bounding a disk $D$ in $X_{g}$.

$(b)$. $\Gamma$ contains two simple closed curves $\tau$ and $\gamma$ which
are homotopic to each other in $X_{g}$.

$(c)$. $\Gamma$ is not of type (a) and (b). That is, no two pairwise simple
closed curves in $\Gamma$ are homotopic, and no simple closed curve in
$\Gamma$ is homotopically trivial. In particular, $\Gamma$ contains at least
one nontrivial closed curve.

If $\Gamma$ is of type (a), by Lemma 11 we have

$\mathop{\rm length}(\Gamma)\geq\mathop{\rm Area}(D).$

Thus,

$\displaystyle\frac{\mathop{\rm length}(\Gamma)}{\min\\{\mathop{\rm
Area}(A_{1}),\mathop{\rm Area}(A_{2})\\}}\geq\frac{\mathop{\rm
length}(\Gamma)}{\mathop{\rm Area}(D)}\geq 1.$

If $\Gamma$ is of type (b), then the two curves $\tau$ and $\gamma$ bounds a
cylinder $\Omega$ in $X_{g}$. By Lemma 12 we have

$\displaystyle\mathop{\rm length}(\Gamma)$ $\displaystyle\geq$
$\displaystyle\mathop{\rm length}(\gamma)+\mathop{\rm length}(\tau)$
$\displaystyle=$ $\displaystyle\mathop{\rm length}(\partial\Omega)$
$\displaystyle\geq$ $\displaystyle\mathop{\rm Area}(\Omega).$

Thus,

$\displaystyle\frac{\mathop{\rm length}(\Gamma)}{\min\\{\mathop{\rm
Area}(A_{1}),\mathop{\rm Area}(A_{2})\\}}\geq\frac{\mathop{\rm
length}(\Gamma)}{\mathop{\rm Area}(D)}\geq 1.$

If $\Gamma$ is of type (c), since $\Gamma$ contains at least one nontrivial
closed curve,

$\mathop{\rm length}(\Gamma)\geq\mathop{\rm sys}(X_{g})\geq\varepsilon.$

It is clear that

$\displaystyle\min\\{\mathop{\rm Area}(A_{1}),\mathop{\rm Area}(A_{2})\\}$
$\displaystyle\leq$ $\displaystyle\frac{1}{2}\mathop{\rm Area}(X_{g})$
$\displaystyle=$ $\displaystyle 2\pi(g-1).$

Thus,

$\frac{\mathop{\rm length}(\Gamma)}{\min\\{\mathop{\rm
Area}(A_{1}),\mathop{\rm Area}(A_{2})\\}}\geq\frac{\varepsilon}{2\pi(g-1)}.$

From these three cases above, the Cheeger isoperimetric constant $h(X_{g})$
satisfies that

$\displaystyle h(X_{g})$ $\displaystyle=$ $\displaystyle\inf\frac{\mathop{\rm
length}(\Gamma)}{\min\\{\mathop{\rm Area}(A_{1}),\mathop{\rm Area}(A_{2})\\}}$
$\displaystyle\geq$ $\displaystyle\min\\{1,\frac{\varepsilon}{2\pi(g-1)}\\}.$

By Lemma 10, we have

$\displaystyle\lambda_{1}(X_{g})$ $\displaystyle\geq$
$\displaystyle\frac{h^{2}(X_{g})}{4}$ $\displaystyle\geq$
$\displaystyle\min\\{\frac{1}{4},\frac{\varepsilon^{2}}{16\pi^{2}(g-1)^{2}}\\}.$

For large enough $g$, clearly we get

$\lambda_{1}(X_{g})\geq\frac{\varepsilon^{2}}{16\pi^{2}}\cdot\frac{1}{g^{2}}.$

Then conclusion follows by choosing

$\alpha(\varepsilon)=\frac{\varepsilon^{2}}{16\pi^{2}}.$

∎

##  4\. Proof of Proposition 3

In this section we will construct a Riemann surface satisfying Proposition 3.

For a trivalent graph $G$ with $V$ vertices of $E$ edges. It is clear that

$3V=2E.$

So every trivalent graph must have even vertices. For any $\varepsilon>0$, we
let

$d(\varepsilon)=2\mathop{\rm arcsinh}\frac{1}{2\sinh\frac{\varepsilon}{4}}$

be the constant in (1). One may assume $W(\varepsilon)>0$ is a constant, only
depending on $\varepsilon$, such that

(9) $\displaystyle W(\varepsilon)\cdot d(\varepsilon)\geq 2\varepsilon.$

Clearly as $\varepsilon$ goes to infinity, $W(\varepsilon)$ also goes to
infinity. By (5) we know that there exists an even integer
$V_{0}=V_{0}(\varepsilon)>0$, only depending on $\varepsilon$, such that for
all even integer $V\geq V_{0}$, there exists a trivalent graph with $V$
vertices, denoted by $G(V)$, such that the girth $\mathop{\rm girth}(G(V))$ of
$G(V)$ satisfies that

(10) $\displaystyle\mathop{\rm girth}(G(V))\geq W(\varepsilon).$

In particular,

$\mathop{\rm girth}(G(V_{0}))\geq W(\varepsilon).$

For large enough $g\ (g>>V_{0})$, there exists two numbers $g_{0}$ and $V_{1}$
such that

(11) $\displaystyle 2g-2=g_{0}(V_{0}+2)+(V_{1}+2)-2$

where $V_{1}$ satisfies

(12) $\displaystyle V_{0}\leq V_{1}\leq 2V_{0}+2.$

In particular, by (10) there exists a trivalent graph $G(V_{1})$ with vertices
$V_{1}$ such that

$\mathop{\rm girth}(G(V_{1}))\geq W(\varepsilon).$

We put $g_{0}$ graphs $G(V_{0})$ and a graph $G(V_{1})$ from left to right and
add one edge between any two consecutive graphs. Then we get a new trivalent
graph, denoted by $G_{g}$. One may see this graph $G_{g}$ as in figure 3.

$G(V_{0})$$G(V_{0})$$G(V_{0})$$G(V_{0})$$G(V_{1})$ Figure 3. Graph $G_{g}$

The number of vertices of $G_{g}$ is $g_{0}(V_{0}+2)+(V_{1}+2)-2$. By (11) we
know that this graph $G_{g}$ has $(2g-2)$ vertices. And it is clear that any
non-trivial simple cycle in $G_{g}$ can only happen in either one $G(V_{0})$
or $G(V_{1})$. Thus, the girth satisfies

(13) $\displaystyle\mathop{\rm girth}(G_{g})\geq W(\varepsilon).$

######  Remark 13.

If $\varepsilon$ is small enough, we may choose a special graph $G_{g}$ as
shown in figure 4. In this case, the girth $W(\varepsilon)=1$.

Figure 4.

The surface $\mathcal{X}_{g}$ of genus $g$ in Proposition 3 is constructed as
follows.

Recall that $\mathcal{P}_{\varepsilon}$ is a pair of pants whose boundary
closed geodesics all have length equal to $\varepsilon$. We replace each
vertex in the trivalent graph $G_{g}$ by $\mathcal{P}_{\varepsilon}$ and glue
two pair of pants along one boundary loop if they are connected by one edge.
In this way, we get a Riemann surface $\mathcal{X}_{g}$ of genus $g$. For
examples: for large $\varepsilon$ and graph as showed in figure 3, the surface
$\mathcal{X}_{g}$ looks like figure 5; and for small $\varepsilon$ and the
graph as showed in figure 4, the surface $\mathcal{X}_{g}$ looks like figure
6.

$\Sigma(G(V_{0}))$$\Sigma(G(V_{0}))$$\Sigma(G(V_{1}))$ Figure 5. Surface
$\mathcal{X}_{g}$ Figure 6. Surface $\mathcal{X}_{g}$

The graph $G_{g}$ has $(3g-3)$ edges. The corresponding closed loops in
$\mathcal{X}_{g}$ form a pants decomposition, denoted by $P$, of
$\mathcal{X}_{g}$.

In the Fenchel-Nielsen coordinate with respect to this pants decomposition
$P$, the surface $\mathcal{X}_{g}$ has $(3g-3)$ length parameters being
$\varepsilon$ and arbitrary twist parameters. So we not only just construct
one surface, actually we construct a subset of dimension $(3g-3)$ in moduli
space $\mathcal{M}_{g}$ in some sense. In this paper, we only use one such a
surface to get the upper bound in Theorem 1.

The (thick) closed geodesics in figures 5 and 6 are separating. They are
$g_{0}$ ones in total. So in particular, we have the number

(14) $\displaystyle g_{0}\leq\\#\\{\gamma\in\partial P;\ \text{$\gamma$ is
separating on $\mathcal{X}_{g}$}\\}\leq(3g-3).$

By (11) we know that

(15) $\displaystyle\\#\\{\gamma\in\partial P;\ \text{$\gamma$ is separating on
$\mathcal{X}_{g}$}\\}\asymp_{g}g.$

######  Remark 14.

These $g_{0}$ separating loops separate the surface $\mathcal{X}_{g}$ into
$(g_{0}+1)$ components. Each component has similar area. More precisely, the
largest area of the components is not bigger than twice that of the smallest
one. Moreover, these components can be sorted from left to right. These are
important in proving that the $k$-th eigenvalue of this surface
$\mathcal{X}_{g}$ roughly realizes the rate of growth $\frac{1}{g^{2}}$ for
large $g$. In Section 5 we will see how these work in our proof.

Now we are ready to prove this surface $\mathcal{X}_{g}$ satisfies all those
three properties in Proposition 3.

###### Proof of Proposition 3.

Let $\mathcal{X}_{g}$ be the closed Riemann surface of genus $g$ as
constructed above.

First by the construction above, the obvious pants decomposition $P$ clearly
satisfies part $(b)$.

By (15) we also know that part $(c)$ holds.

So it suffices to show part $(a)$. We will show that

$\mathop{\rm sys}(\mathcal{X}_{g})=\varepsilon.$

Let $\gamma$ be a shortest non-trivial loop on $\mathcal{X}_{g}$. Since every
loop in the boundary $\partial P$ has length equal to $\varepsilon$, we have

$\ell(\gamma)\leq\varepsilon.$

If $\gamma\in\partial P$, then we are done.

Now we assume that $\gamma\notin\partial P$. Since $P$ is a pants
decomposition of $\mathcal{X}_{g}$, one may assume that $\gamma$
(transversely) intersects with certain boundary closed geodesics in $P$. This
in particular induces a closed cycle, denoted by $\Gamma$, on the
corresponding trivalent graph $G_{g}$. One may see it as follows: as $\gamma$
goes from one pair of pants to its neighboring one through a boundary curve,
$\Gamma$ goes from the corresponding vertex to the corresponding neighboring
vertex through the corresponding edge, and finally $\Gamma$ will come back to
its original vertex.

We will split the remaining proof into two cases.

Case-1: The closed cycle $\Gamma\subset G_{g}$ is non-trivial.

For this case, $\Gamma$ contains a non-trivial simple cycle in $G_{g}$. By
(13) we know that

$\mathop{\rm girth}(G_{g})\geq W(\varepsilon).$

So we have,

$\mathop{\rm length}{\Gamma}\geq W(\varepsilon).$

Recall that on each pair of pant $\mathcal{P}_{\varepsilon}$, the distance
between two boundary curves is $d(\varepsilon)$ (as shown in figure 1). So we
have

$\ell(\gamma)\geq W(\varepsilon)\cdot d(\varepsilon).$

Then by (9) we have

$\ell(\gamma)\geq 2\varepsilon.$

Which is a contradiction.

Case-2: The closed cycle $\Gamma\subset G_{g}$ is trivial.

For this case, the closed cycle $\Gamma$ must turn back twice in the sense
that it goes through one edge and then immediately comes back through this
edge. For turning back, the corresponding arc $\tau^{\prime}$ in $\gamma$ will
be contained in one precise pair of pants $\mathcal{O}$. And it starts from a
boundary curve in this pair of pants and then comes back to this boundary
curve again. Since $\gamma$ is a closed geodesic, this arc $\tau^{\prime}$ can
not be homotopic to certain arc in the boundary curve. And this arc
$\tau^{\prime}$ has two intersection points with one boundary curve of
$\mathcal{O}$. So the length $\ell(\tau^{\prime})$ of this arc must be bigger
than or equal to the length of the arc $\tau$ as shown in figure 1. By (4) we
know that

$\ell(\tau^{\prime})\geq\ell(\tau)>\frac{\varepsilon}{2}.$

Since $\gamma$ is a closed geodesic and $\Gamma\subset G_{g}$ is trivial, the
complement $\gamma\setminus\tau^{\prime}$ of $\tau^{\prime}$ also contains an
arc $\tau^{\prime\prime}$ which has two intersection points with one boundary
curve of another pair of pants. Similar as above we also have

$\ell(\tau^{\prime\prime})\geq\ell(\tau)>\frac{\varepsilon}{2}.$

Thus, we have

$\displaystyle\ell(\gamma)$ $\displaystyle\geq$
$\displaystyle\ell(\tau^{\prime})+\ell(\tau^{\prime\prime})$ $\displaystyle>$
$\displaystyle\varepsilon.$

This again contradicts our assumption that $\gamma$ is a shortest non-trivial
loop.

Thus, only all the curves in $\partial P$ realize the systole of
$\mathcal{X}_{g}$. So we have

$\mathop{\rm sys}(\mathcal{X}_{g})=\varepsilon.$

In particular, part (a) also holds. We finish the proof. ∎

##  5\. Upper Bound

In this section we will prove the upper bound in Theorem 1. Let
$\mathcal{X}_{g}$ be a closed surface of genus $g$ in Proposition 3 as
constructed in the last section. Recall that $\mathop{\rm
sys}(\mathcal{X}_{g})=\varepsilon$. We will show that

######  Proposition 15.

For all integer $k\geq 1$ there exists a positive constant
$\beta(\varepsilon)$, only depending on $\varepsilon$, such that for large
enough $g$,

$\lambda_{k}(\mathcal{X}_{g})\leq\frac{\beta(\varepsilon)\cdot k^{2}}{g^{2}}.$

By construction in last section, in figure 5 (or figure 6 for small
$\varepsilon$), we let $\\{\gamma_{i}\\}_{1\leq i\leq g_{0}}$ be these $g_{0}$
separating (thick) closed geodesics, which are listed in order from left to
right. The union $\cup_{1\leq i\leq g_{0}}\gamma_{i}$ separates the surface
$\mathcal{X}_{g}$ into $(g_{0}+1)$ components. We denote them by
$M_{0},M_{1},...,M_{g_{0}-1},M_{g_{0}}$ in order from left to right. For each
$1\leq i\leq g_{0}-1$, each component $M_{i}$ is a hyperbolic surface of genus
$\frac{2+V_{0}}{2}$ with $2$ closed geodesic boundaries. So the Euler
characteristic

$\chi(M_{i})=2-2\cdot\frac{2+V_{0}}{2}-2=-2-V_{0}.$

By the Gauss-Bonnet formula we know that for each $1\leq i\leq g_{0}-1$,

(16) $\displaystyle\mathop{\rm Area}(M_{i})=2\pi(V_{0}+2).$

For $1\leq i\leq g_{0}$, we let $K_{i}$ be the collar of $\gamma_{i}$ (as
defined in the Collar Lemma: see Lemma 6). For each collar $K_{i}$, we let
$w(\varepsilon)$ to be its half width. That is,

$w(\varepsilon)=\mathop{\rm arcsinh}\frac{1}{\sinh\frac{1}{2}\varepsilon}.$

Recall by Lemma 6 the hyperbolic metric on $K_{i}$ is

$ds^{2}=d\rho^{2}+\varepsilon^{2}\cosh^{2}{\rho}dt^{2}.$

Thus, the area of each collar is

$\displaystyle\mathop{\rm Area}(K_{i})$ $\displaystyle=$
$\displaystyle\int_{K_{i}}dV$ $\displaystyle=$
$\displaystyle\int_{0}^{1}\int_{-w}^{w}\varepsilon\cosh\rho d\rho dt$
$\displaystyle=$ $\displaystyle 2\varepsilon\sinh w$ $\displaystyle=$
$\displaystyle\frac{2\varepsilon}{\sinh\frac{1}{2}\varepsilon}.$

It is clear that $\mathop{\rm Area}(K_{i})<4$.

For each $1\leq i\leq g_{0}-1$, we set

$M_{i}^{0}=M_{i}-\cup_{j=1}^{g_{0}}K_{j}$

to be the complement. Thus, for each $1\leq i\leq g_{0}-1$ the area satisfies
that

$\displaystyle\mathop{\rm Area}(M_{i}^{0})$ $\displaystyle=$ $\displaystyle
2\pi(V_{0}+2)-\frac{1}{2}\mathop{\rm Area}(K_{i})-\frac{1}{2}\mathop{\rm
Area}(K_{i+1})$ $\displaystyle=$ $\displaystyle
2\pi(V_{0}+2)-\frac{2\varepsilon}{\sinh\frac{1}{2}\varepsilon}$

It is clear that $\mathop{\rm Area}(M_{i}^{0})>2\pi(V_{0}+2)-4>0$.

Now we are ready to prove Proposition 15.

###### Proof of Proposition 15.

For any fixed integer $k>0$, let $g_{1}>0$ be an integer with

(19) $\displaystyle g_{0}-1=g_{1}(k+1)+r$

where $0\leq r\leq k$.

For each $0\leq i\leq k$ and $1\leq j\leq g_{1}$, we set

$M^{0}_{i,j}=M^{0}_{ig_{1}+j}$

and

$A_{i}=\cup_{1\leq j\leq g_{1}}M^{0}_{i,j}.$

In this way, we divide the union $\cup_{1\leq l\leq g_{0}-1}M^{0}_{l}$ into
$(k+1)$ piles $A_{0},A_{1},...,A_{k}$ and $r$ remains. For each pile $A_{i}$,
it has $g_{1}(\asymp_{g}g)$ components.

For $i=0,1,...,k$, we define a function $\varphi_{i}$ on $\mathcal{X}_{g}$ as
follows.

Set $\varphi_{i}=0$ on $M_{j}$ for those $M_{j}\cap A_{i}=\emptyset$.

On $M_{i,j}^{0}\subset A_{i}$, set

(20) $\varphi_{i}=\left\\{\begin{array}[]{ll}0&\text{on $M_{i,1}^{0}\cup
M_{i,g_{1}}^{0}$},\\\ 1&\text{on $M_{i,2}^{0}\cup M_{i,g_{1}-1}^{0}$},\\\
2&\text{on $M_{i,3}^{0}\cup M_{i,g_{1}-2}^{0}$},\\\ 3&\text{on
$M_{i,4}^{0}\cup M_{i,g_{1}-3}^{0}$},\\\ \cdots\\\ \frac{g_{1}}{2}-1&\text{on
$M_{i,g_{1}/2}^{0}\cup M_{i,(g_{1}/2+1)}^{0}$ if $g_{1}$ is even},\\\
\frac{g_{1}-1}{2}&\text{on $M_{i,(g_{1}+1)/2}^{0}$ if $g_{1}$ is odd}.\\\
\end{array}\right..$

Now we have defined $\varphi_{i}$ on each $M_{j}^{0}$. It remains to define
$\varphi_{i}$ on the collars of these $g_{0}$ separating (thick) closed
geodesics (as shown in figure 5 or 6). We define $\varphi_{i}$ as shown in the
following picture.

$M_{l}^{0}$$M_{l+1}^{0}$$\rho=$$-w$$0$$w$$\varphi_{i}=$$a$$\frac{a+b}{2}$$b$$\rho$$t$
Figure 7.

For each collar, we use polar coordinate
$(\rho,t)\in[-w,w]\times\mathbb{S}^{1}$ where $w=w(\varepsilon)=\mathop{\rm
arcsinh}\frac{1}{\sinh\frac{1}{2}\varepsilon}$ is the half width (see Lemma
6). As figure 7 shows, the two boundary cycles are parts of boundaries of two
consecutive components $M_{l}^{0}$ and $M_{l+1}^{0}$ for some $0\leq l\leq
g_{0}$. So $\varphi_{i}$ takes constant values $a$ and $b$ on each boundary
loop respectively. We linearly extend $\varphi_{i}$ onto the collar in the
following sense

$\varphi_{i}(\rho,t)=\varphi_{i}(\rho)=\frac{a+b}{2}+\frac{b-a}{2\arctan(\tanh\frac{w}{2})}\arctan(\tanh\frac{\rho}{2})$

for every $(\rho,t)\in[-w,w]\times\mathbb{S}^{1}$.

In this way, for each $0\leq i\leq k$ we have defined a $H^{1}$ function
$\varphi_{i}$ on $\mathcal{X}_{g}$. Now we make estimations of $\varphi_{i}$
on $\mathcal{X}_{g}$.

First by definition, for each $0\leq i\leq k$ we have

$\displaystyle\int_{\mathcal{X}_{g}}\varphi_{i}^{2}$ $\displaystyle>$
$\displaystyle\int_{\bigcup_{1\leq
j\leq[\frac{g_{1}+1}{2}]}M_{i,j}^{0}}\varphi_{i}^{2}$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{[\frac{g_{1}+1}{2}]}(j-1)^{2}\cdot\mathop{\rm
Area}(M_{i,j}^{0})$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{[\frac{g_{1}+1}{2}]}(j-1)^{2}\cdot(2\pi(V_{0}+2)-\frac{2\varepsilon}{\sinh\frac{1}{2}\varepsilon})$

where we apply (5) in the last equation.

For all integer $n\geq 1$ the following elementary equation is well-known.

$\sum_{j=1}^{n}j^{2}=\frac{n(n+1)(2n+1)}{6}\geq\frac{n^{3}}{3}.$

Thus, for each $0\leq i\leq k$ we have

(22)
$\displaystyle\int_{\mathcal{X}_{g}}\varphi_{i}^{2}>(2\pi(V_{0}+2)-\frac{2\varepsilon}{\sinh\frac{1}{2}\varepsilon})\cdot\frac{(g_{1}-2)^{3}}{24}.$

On each collar where $\varphi_{i}$ is not a constant map, the function
$\varphi_{i}$ takes values of difference equal to $1$ on the two boundaries
curves of the collar. By the definition of $\varphi$ we have at most
$(g_{1}-1)$ such collars (if $g_{1}$ is even, we only have $(g_{1}-2)$ such
collars). We denote them by $K_{l}^{\prime}$. So on each $K_{l}^{\prime}$,

$\varphi_{i}=c_{l}\pm\frac{1}{2\arctan(\tanh\frac{w}{2})}\arctan(\tanh\frac{\rho}{2})$

for some constant $c_{l}$.

Recall the hyperbolic metric $ds^{2}=d\rho^{2}+\varepsilon^{2}\cosh^{2}\rho
dt^{2}$ on each collar (see Lemma 6). For a smooth function $f(\rho,t)$ on
each collar, the gradient $\nabla f$ of $f$ is

$\nabla f=\frac{\partial
f}{\partial\rho}\frac{\partial}{\partial\rho}+\frac{1}{l^{2}\cosh^{2}\rho}\frac{\partial
f}{\partial t}\frac{\partial}{\partial t}.$

So on each collar the energy density of $\varphi_{i}$ is

$|\nabla\varphi_{i}|^{2}(\rho,t)=|\frac{1}{2\arctan(\tanh\frac{w}{2})}\frac{1}{2\cosh\rho}|^{2}.$

Recall that $\varphi_{i}$ is constant on
$\mathcal{X}_{g}\setminus\cup_{l=1}^{g_{1}-1}K^{\prime}_{l}$. In particular,
$\nabla\varphi_{i}\equiv 0$ on
$\mathcal{X}_{g}\setminus\cup_{l=1}^{g_{1}-1}K^{\prime}_{l}$. Thus, for each
$0\leq i\leq k$ we have,

$\displaystyle\int_{\mathcal{X}_{g}}|\nabla\varphi_{i}|^{2}$
$\displaystyle\leq$
$\displaystyle\sum_{l=1}^{g_{1}-1}\int_{K_{l}^{\prime}}|\nabla(\frac{1}{2\arctan(\tanh\frac{w}{2})}\arctan(\tanh\frac{\rho}{2}))|^{2}$
$\displaystyle=$
$\displaystyle(g_{1}-1)\int_{0}^{1}\int_{-w}^{w}|\frac{1}{2\arctan(\tanh\frac{w}{2})}\frac{1}{2\cosh\rho}|^{2}\varepsilon\cosh\rho\
d\rho dt$ $\displaystyle=$
$\displaystyle(g_{1}-1)\frac{1}{(2\arctan(\tanh\frac{w}{2}))^{2}}\varepsilon\arctan(\tanh\frac{w}{2})$
$\displaystyle\leq$ $\displaystyle
g_{1}\cdot\frac{\varepsilon}{4\arctan(\tanh\frac{w}{2})}$

By (22) and (5), for each $0\leq i\leq k$ we have

(24)
$\displaystyle\frac{\int_{\mathcal{X}_{g}}\lvert\nabla\varphi_{i}\rvert^{2}}{\int_{\mathcal{X}_{g}}\varphi_{i}^{2}}\leq\frac{24\varepsilon}{4\arctan(\tanh\frac{w}{2})\cdot(2\pi(V_{0}+2)-\frac{2\varepsilon}{\sinh\frac{1}{2}\varepsilon})}\cdot\frac{g_{1}}{(g_{1}-2)^{3}}.$

By (11) and (19) we know that

$\lim_{g\to\infty}\frac{g_{1}}{g}=\frac{2}{(k+1)(2+V_{0})}$

where $V_{0}$ only depends on $\varepsilon$. Actually the following inequality
always holds:

$\frac{g_{1}}{g}\geq\frac{1}{24V_{0}k}.$

Thus, for each $0\leq i\leq k$ we have that for large enough $g$ ,

(25)
$\displaystyle\frac{\int_{\mathcal{X}_{g}}\lvert\nabla\varphi_{i}\rvert^{2}}{\int_{\mathcal{X}_{g}}\varphi_{i}^{2}}\leq\beta(\varepsilon)\cdot\frac{k^{2}}{g^{2}}$

where

(26) $\displaystyle\beta(\varepsilon)=\frac{10^{10}\varepsilon
V_{0}^{2}}{\arctan(\tanh\frac{w}{2})\cdot(\pi(V_{0}+2)-\frac{\varepsilon}{\sinh\frac{1}{2}\varepsilon})}.$

Recall that our construction of $\varphi_{i}$ ensures that the supports
satisfy

(27) $\displaystyle supp(\varphi_{i})\cap
supp(\varphi_{j})=\emptyset,\quad\forall 0\leq i\neq j\leq k.$

In particular, these functions $\varphi_{0},\varphi_{1},...,\varphi_{k}$ are
linearly independent. So

$\dim(span\\{\varphi_{i}\\}_{0\leq i\leq k})=k+1.$

One may choose a non-constant function

$\varphi=a_{0}\varphi_{0}+...+a_{k}\varphi_{k}$

for certain constants $a_{i}\ (0\leq i\leq k)$ such that

$\int_{\mathcal{X}_{g}}\varphi\cdot f_{l}=0,\quad\forall l\in[0,k-1]$

where $f_{l}$ is the normalized $l$-th eigenfunction of $\mathcal{X}_{g}$.

Recall equation (6) says that the $k$-th eigenvalue satisfies that

$\lambda_{k}(\mathcal{X}_{g})=\inf\\{\frac{\int_{\mathcal{X}_{g}}|\nabla
f|^{2}}{\int_{\mathcal{X}_{g}}f^{2}};\ f\in H^{1}\ \text{and}\
\int_{\mathcal{X}_{g}}f\cdot f_{l}=0\ \forall l\in[0,k-1]\\}.$

Therefore, we have that for large enough $g$,

$\displaystyle\lambda_{k}(\mathcal{X}_{g})$ $\displaystyle\leq$
$\displaystyle\frac{\int_{\mathcal{X}_{g}}\lvert\nabla\varphi\rvert^{2}}{\int_{\mathcal{X}_{g}}\varphi^{2}}$
$\displaystyle=$
$\displaystyle\frac{\sum_{i=0}^{k}a_{i}^{2}\cdot\int_{\mathcal{X}_{g}}\lvert\nabla\varphi_{i}\rvert^{2}}{\sum_{i=0}^{k}a_{i}^{2}\cdot{\int_{\mathcal{X}_{g}}\varphi_{i}^{2}}}\quad\quad\text{(by
\eqref{empty})}$ $\displaystyle\leq$
$\displaystyle\beta(\varepsilon)\cdot\frac{k^{2}}{g^{2}}.\quad\quad\quad\quad\quad\quad\quad(\text{by
\eqref{upp-i}})$

The proof is complete. ∎

###### Proof of Theorem 1.

It clearly follows by Proposition 9 and 15. ∎

We finish this section by the following several remarks.

######  Remark 16.

For any constant $\varepsilon$ with

$0<\varepsilon<2\mathop{\rm arcsinh}1=2\ln(1+\sqrt{2}),$

the Riemann surface $\mathcal{X}_{g}$ may be shown as in figure 6. For this
case, the constant $d=d(\varepsilon)>\varepsilon$. As in Remark 13 one may
take $V_{0}=1$. So the constant $\beta(\varepsilon)$ (see (26)) in Proposition
15 can be taken to be

$\beta(\varepsilon)=c\cdot\varepsilon$

where $c>0$ is a universal constant. More precisely, the upper bound in
Theorem 1 can be stated as follows.

######  Theorem 17.

For small $\varepsilon$ with $0<\varepsilon<2\mathop{\rm
arcsinh}1=2\ln(1+\sqrt{2})$ and any positive integer $k\geq 1$, then there
exists a universal constant $c>0$ such that for all $g$ large enough,

$\inf_{X\in\mathcal{M}_{g}^{\geq\varepsilon}}\lambda_{k}(X)\leq
c\cdot\varepsilon\cdot\frac{k^{2}}{g^{2}}.$

######  Remark 18.

By adding one more edge between the left subgraph $G(V_{0})$ and the right
subgraph $G(V_{1})$ in $G_{g}$ as shown in Figure 3, we get a new trivalent
graph denoted by $G^{\prime}_{g+1}$. Since we have two more vertices, the new
corresponding Riemann surface $\mathcal{X}^{\prime}_{g+1}$ has genus of
$(g+1)$. The proof of Proposition 3 can also give that

$\mathcal{X}^{\prime}_{g+1}\in\mathcal{M}_{g+1}^{\geq\varepsilon}.$

Moreover, the proof of Theorem 1 also yields that for any integer $k>0$,

$\lambda_{k}(\mathcal{X}^{\prime}_{g})\asymp_{g}\frac{1}{g^{2}}.$

$\Sigma(G(V_{1}))$$\Sigma(G(V_{0}))$$\Sigma(G(V_{0}))$$\Sigma(G(V_{0}))$
Figure 8. surface $\mathcal{X}^{\prime}_{g+1}$

In this case, we remark that the surface $\mathcal{X}^{\prime}_{g}$ does not
satisfy that the number of separating systolic curves on
$\mathcal{X}^{\prime}_{g}$ is uniformly comparable to the genus $g$ of
$\mathcal{X}^{\prime}_{g}$. Actually in the proof of Proposition 15, the
essential point is existence of a collection of systolic curves, which has
number uniformly comparable to the genus $g$ and can be ordered from left to
right in some sense.

######  Remark 19.

The reason why we choose the function $\arctan(\tanh\frac{\rho}{2})$ to extend
$\varphi_{i}$ on collars is that this makes $\int_{\text{one
collar}}|\nabla\varphi_{i}|^{2}$ to be minimal (by fixing constant boundary
values). For the proof of Proposition 15, this choice is not essential. One
may also choose some other function $\varphi_{i}$ to make
$\int_{\text{collar}}|\nabla\varphi_{i}|^{2}$ to be comparable to $1$. It will
also yield the uniform upper bound in Theorem 1.

The following lemma may tell that why the function
$\arctan(\tanh\frac{\rho}{2})$ makes the energy $\int_{\text{one
collar}}|\nabla\varphi_{i}|^{2}$ to be minimal.

######  Lemma 20.

For a collar $K$ with center geodesic length $\ell$ and half width $w$ and
using the coordinate $(\rho,t)$ defined in Lemma 6, we have for two given
constants $a$ and $b$,

(29) $\min\int_{K}|\nabla
f|^{2}=(a-b)^{2}\frac{l}{4\arctan(\tanh\frac{w}{2})}$

where the minimum is taken over all functions $f(\rho,t)$ with

$f|_{\\{\rho=-w\\}}=a\ \text{and}\ f|_{\\{\rho=w\\}}=b.$

The equality in (29) holds if and only if

(30)
$f(\rho,t)=\frac{a+b}{2}+\frac{b-a}{2\arctan(\tanh\frac{w}{2})}\arctan(\tanh\frac{\rho}{2}).$

###### Proof.

We use variation method.

Recall that the hyperbolic metric on the collar is

$ds^{2}=d\rho^{2}+l^{2}\cosh^{2}\rho dt^{2}.$

So for any function $f$ we have

$\nabla f=\frac{\partial
f}{\partial\rho}\frac{\partial}{\partial\rho}+\frac{1}{l^{2}\cosh^{2}\rho}\frac{\partial
f}{\partial t}\frac{\partial f}{\partial t},$ $\lvert\nabla
f\rvert^{2}=(\frac{\partial
f}{\partial\rho})^{2}+\frac{1}{l^{2}\cosh^{2}\rho}(\frac{\partial f}{\partial
t})^{2},$ $\int_{K}\lvert\nabla
f\rvert^{2}=\int_{0}^{1}\int_{-w}^{w}l\cosh\rho\ (\frac{\partial
f}{\partial\rho})^{2}+\frac{1}{l\cosh\rho}(\frac{\partial f}{\partial
t})^{2}d\rho dt.$

For fixed $t$, let

$A_{t}(f):=\int_{-w}^{w}\cosh\rho\ (\frac{\partial
f}{\partial\rho})^{2}d\rho.$

Let $h(\rho)$ be an arbitrary smooth function on $[-w,w]$ such that
$h(-w)=h(w)=0$. Then the first derivative is

$\displaystyle\frac{d}{ds}|_{s=0}(\int_{-w}^{w}\cosh\rho\
(\frac{\partial(f+sh)}{\partial\rho})^{2}d\rho)$ $\displaystyle=$
$\displaystyle\int_{-w}^{w}\cosh\rho\ 2\frac{\partial
f}{\partial\rho}\frac{\partial h}{\partial\rho}d\rho$ $\displaystyle=$
$\displaystyle-2\int_{-w}^{w}\frac{\partial}{\partial\rho}(\cosh\rho\
\frac{\partial f}{\partial\rho})hd\rho$

Since $h$ is arbitrary, $A_{t}(f)$ takes the minimum only if

$\frac{\partial}{\partial\rho}(\cosh\rho\ \frac{\partial f}{\partial\rho})=0.$

Since $f|_{\\{\rho=-w\\}}=a$ and $f|_{\\{\rho=w\\}}=b$, by elementary ODE
theory we know that

$f(\rho)=\frac{a+b}{2}+\frac{b-a}{2\arctan(\tanh\frac{w}{2})}\arctan(\tanh\frac{\rho}{2}).$

Thus, for any $f$ we have

$\displaystyle A_{t}(f)$ $\displaystyle=$
$\displaystyle\int_{-w}^{w}\cosh\rho\ (\frac{\partial
f}{\partial\rho})^{2}d\rho$ $\displaystyle\geq$
$\displaystyle\int_{-w}^{w}\cosh\rho\
(\frac{\partial(\frac{a+b}{2}+\frac{b-a}{2\arctan(\tanh\frac{w}{2})}\arctan(\tanh\frac{\rho}{2}))}{\partial\rho})^{2}d\rho$
$\displaystyle=$
$\displaystyle\int_{-w}^{w}(\frac{b-a}{4\arctan(\tanh\frac{w}{2})})^{2}\frac{1}{\cosh\rho}d\rho$
$\displaystyle=$ $\displaystyle\frac{(b-a)^{2}}{4\arctan(\tanh\frac{w}{2})}.$

Therefore, for any $f$ we have

$\displaystyle\int_{K}\lvert\nabla f\rvert^{2}$ $\displaystyle=$
$\displaystyle\int_{0}^{1}lA_{t}(f)dt+\int_{0}^{1}\int_{-w}^{w}\frac{1}{l\cosh\rho}(\frac{\partial
f}{\partial t})^{2}d\rho dt$ $\displaystyle\geq$
$\displaystyle\int_{0}^{1}lA_{t}(f)dt$ $\displaystyle\geq$
$\displaystyle\frac{(b-a)^{2}l}{4\arctan(\tanh\frac{w}{2})}.$

It is clear that the equality holds if and only if

$f(\rho,t)=\frac{a+b}{2}+\frac{b-a}{2\arctan(\tanh\frac{w}{2})}\arctan(\tanh\frac{\rho}{2}).$

Which completes the proof. ∎

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