Spectral Convergence of the Dirac Operator on Typical Hyperbolic Surfaces of High Genus

Abstract

In this article, we study the Dirac spectrum of typical hyperbolic surfaces of finite area, equipped with a nontrivial spin structure (so that the Dirac spectrum is discrete). For random Weil–Petersson surfaces of large genus g with $o(\sqrt{g})$ cusps, we prove convergence of the spectral density to the spectral density of the hyperbolic plane, with quantitative error estimates. This result implies upper bounds on spectral counting functions and multiplicities, as well as a uniform Weyl law, true for typical hyperbolic surfaces equipped with any nontrivial spin structure.

1 Introduction

1.1 Setting and Motivation

The objective of this article is to provide information on the Dirac spectrum of typical hyperbolic surfaces of genus g with k cusps, where g, k are non-negative integers such that $2g-2+k>0$. In order to do so, we equip the moduli space of hyperbolic surfaces of signature (g, k) with the Weil–Petersson probability measure $\mathbb {P}_{g,k}$. This is a natural model to study typical hyperbolic surfaces, as illustrated by the rich literature that has developed in the last few years [1, 6,7,8, 10, 12, 14, 20, 21]. By typical, we mean that we wish to prove properties true with probability going to one in a certain asymptotic regime.

An example of interesting regime is the large-scale regime, i.e. the situation when g and/or k go to infinity. Indeed, by the Gauss–Bonnet formula, the area of any hyperbolic surface of signature (g, k) is $2\pi (2g-2+k)$. Since g and k are two independent parameters, we can a priori expect typical surfaces of large genus or large number of cusps to exhibit different geometric and spectral properties. This has been confirmed by the recent complementary works of Hide [7] and Shen–Wu [16], which prove very different behaviours for the spectral gap of the Laplacian of random surfaces depending on whether $k \ll \sqrt{g}$ or $k \gg \sqrt{g}$. In this article, we fix a sequence of non-negative integers $(k(g))_{g \ge 2}$ such that $k(g) = o(\sqrt{g})$, i.e. $k(g)/\sqrt{g} \rightarrow 0$ as $g \rightarrow + \infty$. In other words, we assume we are in the genus-dominated regime. Under this hypothesis, the first author [13] and Le Masson–Sahlsten [9] have proven that there exists a set $\mathcal {A}_{g,k(g)}$ of “good hyperbolic surfaces” of signature (g, k(g)) such that:

• the Weil–Petersson probability of $\mathcal {A}_{g,k(g)}$ goes to one as $g \rightarrow + \infty$;

• the systole of any $X \in \mathcal {A}_{g,k(g)}$ is bounded below by $g^{- 1/24} \sqrt{\log (g)}$;

• elements $X \in \mathcal {A}_{g,k(g)}$ are close to the hyperbolic plane in the sense of Benjamini–Schramm, and more precisely, the proportion of points on X of injectivity radius smaller than $\frac{1}{6} \log (g)+1$ is at most $g^{-1/3}$.

We prove upper bounds and asymptotics for the Dirac spectrum of hyperbolic surfaces in $\mathcal {A}_{g,k(g)}$, which we shall now present. Note that the third point, Benjamini–Schramm convergence to the hyperbolic plane, is crucial to our analysis; interestingly, we do not expect it to hold in the cusp-dominated regime, which means that the question might be quite different in this regime.

1.2 Spectral Convergence of the Dirac Operator

For a hyperbolic surface X of signature (g, k), a spin structure $\varepsilon$ on X, we denote as ${{\,\textrm{D}\,}}$ the Dirac operator on $(X, \varepsilon )$.

While the Laplacian spectrum of a hyperbolic surface with cusps always contains essential spectrum, equal to $[1/4, + \infty )$, for any X, one can pick $\varepsilon$ so that the spectrum of the Dirac operator ${{\,\textrm{D}\,}}$ is discrete [2]. We call such spin structures nontrivial. In that case, for $0 \le a \le b$, we denote as $N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )}(a,b)$ the number of eigenvalues of the absolute value ${{\,\mathrm{|D|}\,}}$ of the Dirac operator ${{\,\textrm{D}\,}}$, once all rigid multiplicities are removed (see Sect. 2.2.2).

Our main result is the spectral convergence of the Dirac operator on $(X,\varepsilon )$ to the Dirac operator on the hyperbolic plane ${\mathbb {H}}$, true for any typical hyperbolic surface X of high genus g with $o(\sqrt{g})$ cusps, and any nontrivial spin structure $\varepsilon$ on X.

Theorem 1

Let $(k(g))_{g \ge 2}$ be a sequence of non-negative integers such that $k(g) = o(\sqrt{g})$ as $g \rightarrow + \infty$. There exists a constant $C>0$ such that, for any $0\le a \le b$, any $g \ge 2$, any $X \in \mathcal {A}_{g,k(g)}$, and any nontrivial spin structure $\varepsilon$ on X, we have

$$\begin{aligned} \frac{N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )} (a,b)}{{{\,\textrm{Area}\,}}(X)}= \frac{1}{4\pi } \int _a^b \lambda \coth (\pi \lambda ) \, \textrm{d}\lambda + R(X,\varepsilon ,a,b), \end{aligned}$$

where the remainder satisfies:

$$\begin{aligned} - C \frac{b+1}{\sqrt{\log g}} \le R(X,\varepsilon ,a,b) \le C \frac{b + 1}{\sqrt{\log g}} \left( 1+ \sqrt{\log \left( 2 + (b - a) \sqrt{\log (g)}\right) } \right) . \end{aligned}$$

A similar statement holds for the Laplacian spectrum, by work of the first author [14] in the compact case and Le Masson–Sahlsten [9] when $k(g) \ll \sqrt{g}$. These results have further been extended to twisted Laplacians by Gong [6] very recently. The proof is similar to the proof in [14], replacing the Selberg trace formula by a Dirac version from the second author [17].

Note that the support of the limiting measure is $[0, + \infty )$, because the spectrum of ${{\,\mathrm{|D|}\,}}$ on ${\mathbb {H}}$ is $[0, + \infty )$. This contrasts with the (twisted) Laplacian setting, where the limiting spectral density is $\frac{1}{4 \pi } \tanh \left( \pi \sqrt{\lambda {-}\frac{1}{4}} \right) \mathbbm {1}_{[\frac{1}{4},{+}\infty )}(\lambda ) \, \textrm{d}\lambda$, supported on $[1/4, + \infty )$.

A remarkable aspect of Theorem 1 is that the limit we obtain is independent of the nontrivial spin structure $\varepsilon$. The reason for that is that the probabilistic assumption we make, and more precisely the Benjamini–Schramm hypothesis, makes the geometric term of trace formulae subdominant, i.e. the spectra of X converge to the spectra of ${\mathbb {H}}$, regardless of the precise geometry and spin structure on X.

1.3 Upper Bounds and Pathological Surfaces

In the process of proving Theorem 1, we prove the following upper bound on the Dirac spectrum of typical hyperbolic surfaces. Throughout this article, when we write $A = \mathcal {O}(B)$, we mean that there exists a constant $C>0$ such that, for any choice of parameters, $|A| \le C \, B$. We point out that the constant is allowed to depend on our choice of a fixed sequence $(k(g))_{g \ge 2}$. If the constant depends on a parameter p, e.g. the genus g, we rather write $A = \mathcal {O}_p(B)$.

Proposition 2

With the notations of Theorem 1,

$$\begin{aligned} \frac{N^{|D|}_{(X,\varepsilon )} (a,b)}{{{\,\textrm{Area}\,}}(X)} = \mathcal {O} \left( (b+1)\left( b-a + \frac{1}{\sqrt{\log g}} \right) \right) . \end{aligned}$$

Building on results of Bär on pinched surfaces [2], we prove that such a bound cannot be obtained for every spin hyperbolic surface, because there exist “pathological” examples for which the spectrum of the Dirac operator is discrete with arbitrarily many eigenvalues close to 0.

Proposition 3

Let (g, k) be integers such that $2g-2+k>0$ and $g \ge 1$. For any N, any $\eta >0$, there exists a hyperbolic surface X of signature (g, k) and a nontrivial spin structure $\varepsilon$ on X such that $N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )}(0,\eta ) \ge N$.

This is another interesting difference between Laplacian and Dirac spectra. Indeed, the Laplacian spectrum restricted to [0, 1/4] is discrete, and the number of eigenvalues under 1/4 is at most $2g-2+k$ by work of Otal–Rosas [15]. This is a topological bound, in the sense that it only depends on the topology of the hyperbolic surface. Proposition 3 proves that such a bound cannot exist in the Dirac setting for every surface, while Proposition 2 provides a bound for every typical hyperbolic surface.

1.4 Applications

We deduce from Theorem 1 a uniform version of the Weyl law for Dirac operators on typical hyperbolic surfaces (uniform in the sense that the rate of convergence is independent of the surface $X \in \mathcal {A}_{g,k(g)}$ and the nontrivial spin structure $\varepsilon$).

Corollary 4

For any $g \ge 2$, any $X \in \mathcal {A}_{g,k(g)}$, any nontrivial spin structure $\varepsilon$ on X,

$$\begin{aligned} \forall \lambda \ge 1, \quad \frac{N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )} (0,\lambda )}{{{\,\textrm{Area}\,}}(X)}= \frac{\lambda ^2}{8\pi } + \mathcal {O}_g \left( \lambda \sqrt{\log (\lambda )} \right) . \end{aligned}$$

Note that the implied constant above only depends on the genus g and the sequence k(g), in a way that can be made explicit using Theorem 1.

Taking a shrinking interval of size $1/\sqrt{\log (g)}$ above $\lambda$, we deduce from Proposition 2 the following topological bound on the multiplicity $\textrm{mult}_{(X,\varepsilon )}(\lambda )$ of any Dirac eigenvalue $\lambda$.

Corollary 5

For any $g \ge 2$, any $X \in \mathcal {A}_{g,k(g)}$, any $\lambda \ge 0$,

$$\begin{aligned} \frac{\textrm{mult}_{(X,\varepsilon )}(\lambda )}{{{\,\textrm{Area}\,}}(X)} = \mathcal {O} \left( \frac{\lambda +1}{\sqrt{\log (g)}} \right) . \end{aligned}$$

2 Preliminaries

2.1 Spin Structures and the Dirac Operator

In this subsection, we briefly describe spin structures and introduce the Dirac operator. For more details, we refer the reader to [4] and [5].

Let n be an even integer, and let us denote by ${\text {Cl}}_n$ the Clifford algebra associated with ${\mathbb {R}^{n}}$ with the standard scalar product. The subgroup ${\text {Spin}}(n) \subset {\text {Cl}}_n$ consists of all even products of unit vectors. It can be shown that ${\text {Spin}}(n)$ is a connected, two-sheeted covering of ${\text {SO}}(n)$, the special orthogonal group. Consider $\{e_1,...,e_{n} \}$ the standard basis in ${\mathbb {R}^{n}}$, and denote by J the standard almost complex structure. The representation:

$$\begin{aligned} cl (v) : = \frac{1}{\sqrt{2}}(v-iJv)\wedge (\cdot ) - \frac{1}{\sqrt{2}}(v+iJv)\lrcorner (\cdot ), \end{aligned}$$

where $\lrcorner$ is obtained from the ${\mathbb {C}}$-bilinear extension of the standard scalar product, acts on $\Sigma _n:= \wedge ^*W$, where $W \subset {\mathbb {C}}^n$ is generated by $\{\frac{1}{\sqrt{2}}(e_1-ie_2),...,\frac{1}{\sqrt{2}}(e_{n-1}-ie_{n}) \}$. One can check that this representation extends to the complexified Clifford algebra

$$\begin{aligned} cl:{\text {Cl}}_n\otimes _{{\mathbb {R}}}{\mathbb {C}}\longrightarrow {{\,\textrm{End}\,}}_{{\mathbb {C}}} (\Sigma _n). \end{aligned}$$

Consider X an n-dimensional oriented manifold with a Riemannian metric. A spin structure on X is a principal ${\text {Spin}}(n)$ bundle $P_{{\text {Spin}}(n)}X$ covering $P_{{\text {SO}}(n)}X$, the principal bundle of oriented orthonormal frames, with two sheets. Moreover, this covering must be compatible with the group covering ${\text {Spin}}(n) \longrightarrow {\text {SO}}(n)$. Once we have a fixed spin structure, we can define the spinor bundle as the associated vector bundle $S:=P_{{\text {Spin}}(n)} X\times _{cl}\Sigma _n$. The Dirac operator is defined as follows:

$$\begin{aligned} {{\,\textrm{D}\,}}:C^{\infty }(S) \longrightarrow C^{\infty } (S),{} & {} {{\,\textrm{D}\,}}:= cl \circ \nabla , \end{aligned}$$

where $\nabla$ is the connection on S induced by the Levi-Civita connection on X. One can see that ${{\,\textrm{D}\,}}$ is an elliptic, self-adjoint differential operator of order 1.

It is known that orientable, complete hyperbolic surfaces of finite area always admit spin structures. From now on, we restrict our attention to this type of surfaces. Let $X = {\Gamma }\setminus {\mathbb {H}}$, where ${\Gamma }$ is a subgroup of ${{\,\mathrm{PSL_2(\mathbb {R})}\,}}$ without elliptic elements for which the associated surface is of finite area. We denote as $\pi$ the standard projection $\pi : {\text {SL}}_2({\mathbb {R}})\longrightarrow {{\,\mathrm{PSL_2(\mathbb {R})}\,}}$. We can reinterpret a spin structure on X as a left splitting in the following short exact sequence:

$$\begin{aligned} 1 \longrightarrow \{ \pm 1 \} \longrightarrow \tilde{{\Gamma }} := \pi ^{-1}({\Gamma }) \longrightarrow {\Gamma }\longrightarrow 1, \end{aligned}$$

i.e. a morphism $\chi :\tilde{{\Gamma }}\longrightarrow \{ \pm 1\}$ for which $\chi \circ \iota = {\text {id}}_{\{ \pm 1 \}}$, where $\iota$ is the natural inclusion.

We define $\varepsilon :{\Gamma }\longrightarrow \{ \pm 1 \}$ by setting $\varepsilon ({\gamma }):= \chi (\tilde{{\gamma }})$, where $\tilde{{\gamma }} \in {\text {SL}}_2({\mathbb {R}})$ is the unique lift with positive trace of ${\gamma }\in {{\,\mathrm{PSL_2(\mathbb {R})}\,}}$. This function is a class function, i.e. constant along conjugacy classes. More details about the class function associated with a spin structure can be found in [17, Section 2]. Further details are also provided on the identifications between the frame bundle $P_{{\text {SO}}(2)}{\mathbb {H}}$ and the group of isometries ${{\,\mathrm{PSL_2(\mathbb {R})}\,}}$, and between the spin bundle $P_{{\text {Spin}}(2)}{\mathbb {H}}$ and the group ${\text {SL}}_2({\mathbb {R}})$. If we consider $\rho$, a right splitting in the above short exact sequence (which is uniquely determined by the left splitting $\chi$), we can define the action of a ${\gamma }\in {\Gamma }$ on $P_{{\text {Spin}}(2)}{\mathbb {H}}$ by left multiplication with $\rho ({\gamma })$. It can be easily seen that this action descends to the spinor bundle S.

2.2 The Spectrum of Dirac Operators

The aim of this article is to study the spectrum of the Dirac operator acting on a typical hyperbolic surface X of finite area.

2.2.1 Cusps and Nontrivial Spin Structures

The spectrum of the Dirac operator is always discrete when X is compact. Remarkably, when X admits some cusps, the spectrum can either be discrete or the real line ${\mathbb {R}}$, depending on the spin structure. More precisely, Bär showed in [2, Theorem 1] that the spectrum of the Dirac operator is discrete if and only if the spin structure is nontrivial along each cusp of X. It is shown in [17, Lemma 8] that this is equivalent to assuming that $\varepsilon ({\gamma }) = -1$ for any primitive parabolic element ${\gamma }\in {\Gamma }$.

Note that Bär further proved in [2, Corollary 2] that any finite area hyperbolic surface admits at least one spin structure such that the spectrum is discrete.

2.2.2 Multiplicities and Counting Functions

As explained in [3, Section 4], the spectrum of the Dirac operator admits two rigid sources of multiplicity: the chiral symmetry and the time-reversal symmetry. It follows that the spectrum of ${{\,\textrm{D}\,}}$ is symmetric about 0 (i.e. if $\lambda \in {\mathbb {R}}$ is an eigenvalue then $-\lambda$ is an eigenvalue), and every eigenvalue has even multiplicity. We conclude that the multiplicity of every eigenvalue of ${{\,\mathrm{|D|}\,}}$ is a multiple of 4.

In order to avoid counting every eigenvalue exactly four times, we shall study the reduced spectrum $(\lambda _j)_{j \ge 0}$, where we let $\lambda _j:= \Lambda _{4j}$ for $(\Lambda _j)_{j \ge 0}$ the ordered spectrum of ${{\,\mathrm{|D|}\,}}$ (with multiplicities). We then define, for $0 \le a \le b$, the counting function

$$\begin{aligned} N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )}(a,b) := \{j \ge 0 \, : \, a \le \lambda _j \le b\} = \frac{1}{4} \, \# \{ \text {eigenvalues of } {{\,\mathrm{|D|}\,}}\text { in } [a,b] \}. \end{aligned}$$

2.2.3 Pathological Examples of Dirac Spectra

Let us now prove Proposition 3, which claims that there is no topological bound on the number of eigenvalues in $[0, \eta ]$, provided $g >0$. The proof relies on the two following results, proven by Bär in [2].

Lemma 6

Let X be a finite area hyperbolic surface. For any simple non-separating closed geodesic $\gamma$ on X, there exists two nontrivial spin structures $\varepsilon _\pm$ on X such that $\varepsilon _\pm (\gamma ) = \pm 1$.

We precise that, in the previous statement, the geodesic $\gamma$ is simple if it has no self-intersection, and non-separating if the surface $X \setminus \gamma$ obtained by cutting X along $\gamma$ is connected. In other words, this lemma tells us that, if $\gamma$ is non-separating, then we can pick the value of a nontrivial spin structure at $\gamma$ freely.

Proof

This is a direct consequence of the discussion in [2, page 481]. $\square$

Lemma 7

Let X be a finite area hyperbolic surface equipped with a nontrivial spin structure $\varepsilon$. Let $\gamma$ be a simple non-separating geodesic on X such that $\varepsilon (\gamma ) = +1$. Let $(X_n)_{n \ge 1}$ be a sequence of finite area hyperbolic surfaces obtained from X by pinching the geodesic $\gamma$ so that its length goes to 0 as $n \rightarrow \infty$. Then, for any $\eta > 0$,

$$\begin{aligned} N^{{{\,\mathrm{|D|}\,}}}_{(X_n,\varepsilon )}(0,\eta ) = - \frac{\eta }{\pi } \log (\ell _{X_n}(\gamma )) + \mathcal {O}_{\eta }(1). \end{aligned}$$

The pinching procedure mentioned above is a classic way to construct pathological examples in hyperbolic geometry, and described in more detail in [2, Section 1]. Lemma 7 quantifies how the Dirac spectrum of $X_n$ converges to the Dirac spectrum of the limit $X_\infty$ of $(X_n)_n$ as $n \rightarrow \infty$. There is an accumulation process because, on top of the nontrivial cusps of X, $X_\infty$ has two new cusps with a trivial spin structure (coming from the pinched geodesic), and hence the Dirac spectrum of $(X_\infty ,\varepsilon )$ is ${\mathbb {R}}$.

Proof

This is a straightforward adaptation of [2, Theorem 2] when X is a surface of finite area equipped with a nontrivial spin structure, rather than a closed surface. No changes are required in the proof. $\square$

We are now ready to prove Proposition 3.

Proof of Proposition 3

Let X be an arbitrary hyperbolic surface of signature (g, k). The genus of X is nonzero, and hence there exists a simple non-separating geodesic $\gamma$ on X. By Lemma 6, since $\gamma$ is non-separating, there exists a nontrivial spin structure $\varepsilon$ on X such that $\varepsilon (\gamma ) = +1$. Then, we define a sequence of metrics $(X_n)_{n \ge 1}$ as in Lemma 7 by pinching the geodesic $\gamma$ so that its length goes to 0 as $n \rightarrow + \infty$. By Lemma 6, since $\varepsilon (\gamma )=+1$,

$$\begin{aligned} N^{{{\,\mathrm{|D|}\,}}}_{(X_n,\varepsilon )}(0,\eta ) = - \frac{\eta }{\pi } \log (\ell _{X_n}(\gamma )) + \mathcal {O}_{\eta }(1) \underset{n \rightarrow + \infty }{\longrightarrow }\ + \infty \end{aligned}$$

for any fixed $\eta >0$. In particular we can pick a n such that $N^{{{\,\mathrm{|D|}\,}}}_{(X_n,\varepsilon )}(0,\eta ) \ge N$. Then, $X_n$ satisfies our claim. $\square$

2.2.4 The Selberg Trace Formula for Dirac Operators

Our main tool to study the counting function $N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )}(a,b)$ is the Selberg trace formula for the Dirac operator on compact hyperbolic surfaces, developed by Bolte and Stiepan in [3] and generalised to hyperbolic surfaces of finite area by the second author [17, Theorem 13]. This formula relates the Dirac spectrum of a finite area hyperbolic surface X to its length spectrum, i.e. the list with multiplicities of the lengths of all closed geodesics on X, under the condition that the spin structure is nontrivial. There exists a one-to-one association between oriented closed geodesics and conjugacy classes of hyperbolic elements of ${\Gamma }$, namely each closed geodesic on X is associated with the conjugacy class of the (hyperbolic) isometries on X which preserve the said geodesic. In this article, following the line of [14], we will use the following pretrace formula, adapted from [17, formula (10)].

Theorem 8

Consider $X={\Gamma }\backslash {\mathbb {H}}$ a hyperbolic surface with k cusps, equipped with a nontrivial spin structure $\varepsilon$. Let $(\lambda _j)_{j\in {\mathbb {N}}}$ denote the reduced spectrum of ${{\,\mathrm{|D|}\,}}$. Then, for any admissible test function h,

$$\begin{aligned} \begin{aligned} \sum _{j=0}^{\infty } h(\lambda _j) =&\, \frac{{{\,\textrm{Area}\,}}(X)}{8\pi }\int _{{\mathbb {R}}} h(\lambda )\, \lambda \coth (\pi \lambda )\, \textrm{d}\lambda - \frac{ \log (2)}{2} \, k \, \check{h}(0) \\&+ \frac{1}{2} \sum _{{\gamma }\ne 1}\int _F\varepsilon ({\gamma }) \, K(d(z,{\gamma }z)) \, \tau _{z\mapsto {\gamma }^{-1}z} \, \textrm{d}z \end{aligned} \end{aligned}$$

(1)

where the sum is taken over all hyperbolic elements in ${\Gamma }$ and:

• the set $F$ is a fundamental domain of $X = {\Gamma }\backslash {\mathbb {H}}$;

• $\tau _{z\mapsto w} = - i \frac{z-\overline{w}}{|z-\overline{w}|}$ is the parallel transport of spinors from z to w with respect to $\nabla$;

• the kernel K can be expressed as:

  $$\begin{aligned} K(r) = -\frac{\cosh \left( \tfrac{r}{2} \right) }{\pi \sqrt{2}}\int _r^{\infty }\frac{\check{h}'(\rho )- \tfrac{1}{2} \check{h}(\rho ) \tanh \left( \tfrac{\rho }{2}\right) }{\cosh \left( \tfrac{\rho }{2}\right) \sqrt{\cosh (\rho ) - \cosh (r)}}\, \textrm{d}\rho \end{aligned}$$

  (2)

  where $\check{h}$ is the inverse Fourier transform of h, i.e.

  $$\begin{aligned} \check{h}(\rho ) = \frac{1}{2\pi }\int _{{\mathbb {R}}}h(\lambda )e^{i\lambda \rho }\, \textrm{d}\lambda . \end{aligned}$$

In the above statement, by admissible, we mean that there exists $\eta > 0$ such that h is an even holomorphic function defined on the strip $\{z = x+iy \,: \, |y| \le \tfrac{1}{2} + \eta \}$ which satisfies $|h(z)| \le C (|z|^2+1)^{-1-\eta }$ for a constant $C>0$, as in [3, 13].

Proof

In [17, formula (10)], the second author proved that, for any smooth compactly supported function $\phi$, if we set

$$\begin{aligned} w(x):=\frac{\phi (x)}{\sqrt{x+4}}, \quad \text {and} \quad \check{h}(x):=4\cosh \left( \tfrac{x}{2} \right) \int _{0}^{\infty } w\left( 4 \sinh ^2 \left( \tfrac{x}{2} \right) +y^2 \right) \, \textrm{d}y,\nonumber \\ \end{aligned}$$

(3)

then (1) holds with the kernel $K(r):= \phi (4 \sinh ^2 (r/2))$. Note that $\phi$ is of compact support if and only if $\check{h}$ is. Let us express the kernel K in terms of h. To do so, we shall consider the following operators ${{\,\textrm{A}\,}}$, ${{\,\textrm{B}\,}}$ acting on the set $\mathcal {S}([0,\infty ])$ of Schwartz functions on $[0,\infty ]$:

$$\begin{aligned} {{\,\textrm{A}\,}}\varphi (x) :=\int _0^{\infty }\varphi (x+y^2) \, \textrm{d}y,{} & {} {{\,\textrm{B}\,}}\psi (x) :=-\frac{4}{\pi }\int _0^{\infty }\psi '(x+y^2) \, \textrm{d}y. \end{aligned}$$

By direct computations, one can easily see that ${{\,\textrm{B}\,}}\circ {{\,\textrm{A}\,}}=1$. Indeed:

$$\begin{aligned} ({{\,\textrm{B}\,}}\circ {{\,\textrm{A}\,}})\psi (x)&=-\frac{4}{\pi } \int _0^{\infty }\int _0^{\infty }\psi '(x+y^2+z^2) \, \textrm{d}z \, \textrm{d}y\\&= \frac{-4}{\pi } \int _{[0,\tfrac{\pi }{2}]\times [0,\infty )}\psi '(x+r^2)r \, \textrm{d}\theta \, \textrm{d}r\\&=-2\int _0^{\infty }\frac{1}{2}\frac{\partial \psi (x+r^2)}{\partial r} \, \textrm{d}r = \psi (x). \end{aligned}$$

Writing the identity in such a way allows us to compute the kernel in terms of $\check{h}$. On the one hand, by the expression of $\check{h}$ in Eq. (3), we get that:

$$\begin{aligned} {{\,\textrm{A}\,}}w \circ \left( 4 \sinh ^2\left( \tfrac{\cdot }{2} \right) \right) = \frac{\check{h}}{4\cosh \left( \tfrac{\cdot }{2}\right) } \end{aligned}$$

thus, it follows that:

$$\begin{aligned} 4\sinh \left( \tfrac{\rho }{2}\right) \cosh \left( \tfrac{\rho }{2}\right) ({{\,\textrm{A}\,}}w)' \left( 4 \sinh ^2\left( \tfrac{\rho }{2} \right) \right) = \frac{\check{h}'(\rho ) \cosh \left( \frac{\rho }{2} \right) -\frac{1}{2}\check{h}(\rho ) \sinh \left( \frac{\rho }{2} \right) }{4\cosh ^2 \left( \frac{\rho }{2}\right) }. \end{aligned}$$

(4)

On the other hand, by definition of K and w, since $\sqrt{4 \sinh ^2(\tfrac{r}{2})+4}=2 \cosh (\tfrac{r}{2})$,

$$\begin{aligned} K(r)&= \phi \left( 4 \sinh ^2\left( \tfrac{r}{2} \right) \right) = 2 \cosh \left( \tfrac{r}{2} \right) w\left( 4 \sinh ^2\left( \tfrac{r}{2} \right) \right) \\&= - \frac{8}{\pi } \cosh \left( \tfrac{r}{2} \right) \int _0^{\infty }({{\,\textrm{A}\,}}w)' \left( 4 \sinh ^2\left( \tfrac{r}{2} \right) +y^2 \right) \, \textrm{d}y \end{aligned}$$

because ${{\,\textrm{B}\,}}\circ {{\,\textrm{A}\,}}= 1$. We then perform the change of variable $4 \sinh ^2 (\tfrac{\rho }{2}) = 4 \sinh ^2(\tfrac{r}{2})+y^2$ and obtain the claimed expression thanks to (4) and the fact that

$$\begin{aligned} y = \sqrt{4\sinh ^2 (\tfrac{\rho }{2}) - 4\sinh ^2 (\tfrac{r}{2})} = \sqrt{2 (\cosh (\rho ) - \cosh (r))}. \end{aligned}$$

Therefore, (1) holds as soon as $\check{h}$ is of compact support.

Let us now assume that h is admissible. For a cutoff function $\chi$ supported on $[-2,2]$ and identically equal to 1 on $[-1,1]$, we define for each integer n the function $h_n$ of inverse Fourier transform $\check{h}_n(r) = \chi (\frac{r}{n}) \, \check{h}(r)$. It is clear that $\check{h}_n$ has compact support, so that (1) holds for any n. We can straightforwardly check that it passes to the limit $n \rightarrow \infty$. Indeed, $h_n$ converges pointwise to h and a simple computation yields that $h_n(\lambda ) \le C(1+|\lambda |^{2})^{-1-\eta }$ for constants $\eta , C>0$ independent of n. From Weyl’s law, we obtain that the left hand side of (1) is bounded and therefore converges to its expected limit as $n\rightarrow \infty$. To tackle the terms on the right hand side of (1), we recall that the similar kernel in the Laplacian case is of exponential decay, with exponent strictly smaller than $-1$ (see e.g. [11, Proposition 3]). This, together with the fact that $|\check{h}(\rho )| = \mathcal {O}(e^{-|\rho |(1/2 + \eta )})$ ([11, formula (75)]), leads us to conclude that K also has exponential decay, of exponent strictly smaller than $-1$. Finally, because the number of hyperbolic elements in ${\Gamma }$ is locally bounded (Lemma 15), the sum in the right hand side is bounded and converges to its expected limit. $\square$

2.3 Random Hyperbolic Surfaces

In this article, we study the properties of random hyperbolic surfaces sampled with the Weil–Petersson probability measure. Let us provide the key elements that are necessary for the reading of this article – thorough presentations of this probabilistic model are provided in [14, 19].

Let g, k be integers such that $2g-2+k>0$. Our sample space is the moduli space

$$\begin{aligned} \mathcal {M}_{g,k} := \{ \text {hyperbolic surfaces of signature } (g,k) \} \diagup \text {isometries}. \end{aligned}$$

This space is an orbifold of dimension $6g-6+2k$. Weil introduced in [18] a natural symplectic structure on $\mathcal {M}_{g,k}$, called the Weil–Petersson form. It induces a volume form of finite volume, which can be renormalised to obtain a probability measure $\mathbb {P}_{g,k}$ on the moduli space $\mathcal {M}_{g,k}$.

Our objective is to describe “typical behaviour”, i.e. we will focus on proving properties true with probability going to one in a certain asymptotic regime. More precisely, for our fixed sequence $(k(g))_{g \ge 2}$, we will say a property is true with high probability in the large genus limit if

$$\begin{aligned} \lim _{g \rightarrow + \infty } \mathbb {P}_{g,k(g)}(X \in \mathcal {M}_{g,k(g)} \text { satisfies the property}) = 1. \end{aligned}$$

The following result states two key geometric properties true with high probability which we will use in this article.

Theorem 9

Let $(k(g))_{g \ge 2}$ be a sequence of non-negative integers such that $k(g) = o(\sqrt{g})$ as $g \rightarrow + \infty$. Then, for all $g\ge 2$, there exists a subset $\mathcal {A}_{g,k(g)}$ of the moduli space $\mathcal {M}_{g,k(g)}$ of probability $1 - \mathcal {O}(\log (g) g^{-1/12})$ such that any surface $X \in \mathcal {A}_{g, k(g)}$ satisfies the following.

• If $X^- (L)$ is the L-thin part of X, i.e. the set of points in X with injectivity radius smaller than L, then

  $$\begin{aligned} \frac{{{\,\textrm{Area}\,}}\left( X^- ( \frac{1}{6}\log g+1) \right) }{{{\,\textrm{Area}\,}}(X)} \le g^{-\frac{1}{3}}. \end{aligned}$$

• The systole of X (i.e. the length of its shortest closed geodesic) is larger than $g^{-\frac{1}{24}} \sqrt{\log g}$.

Proof

The first point was proven by the first author in [13, Corollary 4.4], and the second by Le Masson and Sahlsten in [9, Lemma A.1]. $\quad \square$

3 Plan of the Proof and First Estimates

In this section, we set up some notation in order to prove Theorem 1, following the lines of [14], and start with some easy estimates.

3.1 The Family of Test Functions

A key step of the proof is to construct a family of test functions such that the spectral side of the Selberg trace formula is a good approximation of the counting number $N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )} (a,b)$, for $0 \le a \le b$. Our choice of test function is a straightforward adaptation to the choice made in [14, Section 4].

For $t>0$, a parameter which will grow like $\sqrt{\log g}$, consider the family of test functions $h_t:{\mathbb {C}}\longrightarrow {\mathbb {C}}$ defined by the convolution

$$\begin{aligned} h_t(\lambda ) := (\mathbbm {1}_{[a,b]}\star v_t)(\lambda ) = \frac{t}{\sqrt{\pi }} \int _{a}^{b} \exp \left( -t^2(\lambda -\rho )^2\right) \, \textrm{d}\rho , \end{aligned}$$

where $\mathbbm {1}_{[a,b]}$ is the indicator of the segment [a, b] and $v_t(x):=\frac{t}{\sqrt{\pi }}\exp \left( -t^2x^2\right)$ is the Gaussian of mean 0 and variance $\frac{1}{t}$. One can easily see that $h_t$ is holomorphic. Since it is not even, we will rather apply Proposition 8 to the function $H_t(\lambda ):= h_t(\lambda )+h_t(-\lambda )$.

Let us present elementary properties of the functions $h_t$ and $g_t:= \check{H}_t$ proven in [14], which will be useful to the proof of Theorem 1.

Lemma 10

Let $0 \le a \le b$ and $t > 0$.

• (1) The function $H_t$ is admissible.

• (2) For any $t>0$, and for any $u>0$ we have:

  $$\begin{aligned} |g_t(u)|&\le \frac{2}{\pi u}\exp \left( -\frac{u^2}{4t^2} \right) \\ |g_t'(u)|&\le \left( \frac{1}{\pi t^2} + \frac{4b}{\pi u} \right) \exp \left( -\frac{u^2}{4t^2} \right) . \end{aligned}$$

• (3) As $t \rightarrow + \infty$, $h_t$ converges to the function $\lambda \mapsto \tilde{\mathbbm {1}}_{[a,b]}(\lambda )$ which coincides with $\mathbbm {1}_{[a,b]}$ except at $\lambda = a$ and b where it is equal to 1/2. More precisely, for $\lambda \in {\mathbb {R}}$,

  $$\begin{aligned} |h_t(\lambda )-\tilde{\mathbbm {1}}_{[a,b]}(\lambda )|\le {\left\{ \begin{array}{ll} s(t|\lambda -a|) &{} \text {if } \lambda \in (-\infty , a) \cup \{ b \} \\ s(t|\lambda -a|)+s(t|\lambda -b|) &{} \text {if } \lambda \in (a,b) \\ s(t|\lambda -b|) &{} \text {if } \lambda \in \{ a \} \cup ( b, \infty ) \end{array}\right. } \end{aligned}$$

  where $s:(0,\infty )\longrightarrow {\mathbb {R}}$ is the non-increasing function $s(\rho ):=\frac{\exp (-\rho ^2)}{2\sqrt{\pi }\rho }$.

Proof

These three points are respectively Lemma 10, 13 and 21 from [14]. $\quad \square$

3.2 Plan of the Proof

In order to prove our main result, we apply Theorem 8 to the family of functions $H_t$. We denote as $K_t$ the kernel associated with $H_t$, defined in (2). We prove that for any hyperbolic surface X of signature (g, k),

$$\begin{aligned} \frac{1}{{{\,\textrm{Area}\,}}(X)} \sum _{j=0}^{\infty } H_t(\lambda _j)&= \frac{1}{8\pi }\int _{{\mathbb {R}}}H_t(\lambda ) \lambda \coth (\pi \lambda )\, \textrm{d}\lambda - \frac{k \log (2)}{2{{\,\textrm{Area}\,}}(X)} \, g_t(0) \nonumber \\&\quad + \frac{1}{2{{\,\textrm{Area}\,}}(X)}\sum _{{\gamma }\ne 1}\int _F\varepsilon ({\gamma }) K_t(z,{\gamma }z) \tau _{z\mapsto {\gamma }^{-1}z}\, \textrm{d}z. \end{aligned}$$

(5)

The left hand side of this formula is an approximation of the ratio $N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )}(a,b)/{{\,\textrm{Area}\,}}(X)$, which we wish to estimate. Thus, we shall study the right hand side, term by term.

• In Sect. 3.3, we bound the difference between the integral term

  $$\begin{aligned} I(t,a,b) :=\frac{1}{8\pi }\int _{{\mathbb {R}}}H_t(\lambda ) \, \lambda \coth (\pi \lambda )\, \textrm{d}\lambda \end{aligned}$$

  and the integral that appears in Theorem 1.

• Then, in Sect. 3.4, we prove an easy bound on the cuspidal term

  $$\begin{aligned} C(X,t) := - \frac{k \log (2)}{2{{\,\textrm{Area}\,}}(X)} \, g_t(0). \end{aligned}$$

• Sect. 4 is dedicated to bounding the kernel term,

  $$\begin{aligned} R_K(X,\varepsilon ,t,a,b) :=\frac{1}{2{{\,\textrm{Area}\,}}(X)}\sum _{{\gamma }\ne 1}\int _F\varepsilon ({\gamma }) \, K_t(z,{\gamma }z) \tau _{z\mapsto {\gamma }^{-1}z}\, \textrm{d}z \end{aligned}$$

  which is the most difficult part of the analysis of the trace formula, where the probabilistic assumption on $X \in \mathcal {A}_{g,k(g)}$ is necessary.

We then conclude to the proof of Theorem 1 in Sect. 5, where we compare the left hand side of (5) with the rescaled number of ${{\,\mathrm{|D|}\,}}$-eigenvalues between a and b.

3.3 Asymptotic of the Integral Term

Let us prove the following result, which bounds the difference between the integral I(t, a, b) and integral appearing in our claim.

Proposition 11

For any $t>0$,

$$\begin{aligned} \left| I(t,a,b) - \frac{1}{4\pi }\int _a^b \lambda \coth (\pi \lambda )\, \textrm{d}\lambda \right| \le \frac{b + 1}{2t} + \frac{1}{2t^2}. \end{aligned}$$

Proof

We start by rewriting I(t, a, b) in a more convenient form. First we use the parity of $H_t$, then we write $H_t(\lambda ) = h_t(\lambda )+h_t(-\lambda )$, to obtain

$$\begin{aligned} \begin{aligned} I(t,a,b)&= \frac{1}{8\pi }\int _{{\mathbb {R}}} H_t(\lambda ) \, \lambda \coth (\pi \lambda )\, \textrm{d}\lambda = \frac{1}{4\pi }\int _{{\mathbb {R}}} h_t(\lambda ) \, \lambda \coth (\pi \lambda )\, \textrm{d}\lambda \\&= \frac{1}{4\pi }\int _{{\mathbb {R}}} \left( h_t(\lambda ) - \tilde{\mathbbm {1}}_{[a,b]}(\lambda ) \right) \lambda \coth (\pi \lambda )\, \textrm{d}\lambda +\frac{1}{4\pi }\int _{a}^{b}\lambda \coth (\pi \lambda )\, \textrm{d}\lambda . \end{aligned} \end{aligned}$$

We shall use Lemma 10.(3) to bound the difference $|h_t(\lambda ) - \tilde{\mathbbm {1}}_{[a,b]}(\lambda )|$ appearing in the equation above. Since the function s from this bound has a pole at 0, we shall use different estimates near $\lambda =a$ and $\lambda =b$. Thus we write the real line as a union of (up to) five intervals:

$$\begin{aligned} {\mathbb {R}}= \left( -\infty ,a-\frac{1}{t} \right] \cup \left[ a-\frac{1}{t},a+\frac{1}{t} \right] \cup \left[ a+\frac{1}{t},b-\frac{1}{t} \right] \cup \left[ b-\frac{1}{t},b+\frac{1}{t} \right] \cup \left[ b+\frac{1}{t},\infty \right) . \end{aligned}$$

If $a + \frac{1}{t} > b -\frac{1}{t}$ then the interval in the middle is omitted from the union. The integral splits accordingly into five parts, denoted $I_{j}$, for $1\le j \le 5$:

$$\begin{aligned} \begin{aligned}&\left| I(t,a,b) - \frac{1}{4\pi }\int _a^b \lambda \coth (\pi \lambda )\, \textrm{d}\lambda \right| \\&\quad \le \frac{1}{4\pi } \int _{{\mathbb {R}}}\left| h_t(\lambda ) -\tilde{\mathbbm {1}}_{[a,b]} (\lambda ) \right| \lambda \coth (\pi \lambda )\, \textrm{d}\lambda := \frac{1}{4\pi } \sum _{j=1}^5 I_j. \end{aligned} \end{aligned}$$

(6)

We start with $I_1, I_3$ and $I_5$. Throughout the computations, we will use the fact that:

$$\begin{aligned} \forall \lambda \in {\mathbb {R}}, \quad 0 \le \lambda \coth (\pi \lambda ) \le |\lambda |+\frac{1}{\pi }. \end{aligned}$$

(7)

Eq. 7 and Lemma 10.(3) for $\lambda < a$ together imply

$$\begin{aligned} I_1&\le \int _{-\infty }^{a-\frac{1}{t}} \frac{\exp \left( -t^2(\lambda -a)^2 \right) }{2\sqrt{\pi } \, t (a-\lambda )} \left( |\lambda |+\frac{1}{\pi } \right) \, \textrm{d}\lambda = \int _{1}^{\infty } \frac{e^{-x^2}}{2\sqrt{\pi }x} \left( \left| a - \frac{x}{t}\right| +\frac{1}{\pi } \right) \frac{\, \textrm{d}x}{t} \end{aligned}$$

by the change of variable $x = t (a - \lambda ) \in (1, + \infty )$. Then,

$$\begin{aligned} I_1 \le \frac{1}{2\sqrt{\pi }}\int _{1}^{\infty } e^{-x^2} \left( a +1 + \frac{x}{t} \right) \frac{\, \textrm{d}x}{t} \le \frac{a + 1 }{4t} + \frac{1}{4 t^2}. \end{aligned}$$

With similar approximations one can also prove that:

$$\begin{aligned} I_3 \le \frac{a+b +2}{4t} + \frac{1}{2t^2} \quad \text {and} \quad I_5 \le \frac{b + 1}{4t} + \frac{1}{4t^2}. \end{aligned}$$

For $I_2$ and $I_4$ we rather use the loose bound $\left| h_t(\lambda ) -\tilde{\mathbbm {1}}_{[a,b]} (\lambda ) \right| \le 1$, which yields

$$\begin{aligned} I_2&\le \int _{a -\frac{1}{t}}^{a +\frac{1}{t}} \left( |\lambda |+\frac{1}{\pi }\right) \, \textrm{d}\lambda \le \int _{a -\frac{1}{t}}^{a +\frac{1}{t}} \left( a +1 + \frac{1}{t} \right) \, \textrm{d}\lambda \le \frac{2a +2}{t} + \frac{2}{t^2}. \end{aligned}$$

In the same way we also get $I_4 \le \frac{2b +2}{t} + \frac{2}{t^2}$.

Combining those bounds and using $a \le b$, we obtain

$$\begin{aligned} I_1+I_3+I_5 \le \frac{b + 1}{t} + \frac{1}{t^2} \quad \text {and} \quad I_2+I_4 \le \frac{4b + 4}{t} + \frac{4}{t^2} \end{aligned}$$

which allows us to conclude using equation (6) and $5/(4\pi ) < 1/2$. $\square$

3.4 Bond of the Cusps Contribution

Let us now prove the following.

Proposition 12

For any X of signature (g, k) and any $t > 0$,

$$\begin{aligned} |C(X,t)| \le \frac{k}{2g-2+k} (b-a). \end{aligned}$$

Remark 13

We note that, in Theorem 1, we are placed in the regime $k = k(g) = o(\sqrt{g})$. Then, Proposition 12 implies

$$\begin{aligned} C(X,t) = o \left( \frac{b}{\sqrt{g}} \right) . \end{aligned}$$

(8)

Proof of Proposition 12

By definition, the cusp term is

$$\begin{aligned} C(X,t) = - \frac{k \log (2)}{2{{\,\textrm{Area}\,}}(X)} \, g_t(0) = - \frac{k \log (2)}{4\pi (2g-2+k)} \, g_t(0) \end{aligned}$$

by the Gauss–Bonnet theorem. Thus we simply have to estimate $g_t(0) = \frac{1}{2 \pi } \int _{\mathbb {R}}H_t(\lambda ) \, \textrm{d}\lambda$. We observe that

$$\begin{aligned} \int _{{\mathbb {R}}}h_t(\lambda )\, \textrm{d}\lambda= & {} \frac{t}{\sqrt{\pi }} \int _{{\mathbb {R}}}\int _{a}^{b} \exp \left( -t^2(\lambda -\rho )^2 \right) \, \textrm{d}\rho \, \textrm{d}\lambda \\{} & {} =\frac{t}{\sqrt{\pi }}\int _{a}^{b} \int _{{\mathbb {R}}} \exp \left( -t^2x^2 \right) \, \textrm{d}x \, \textrm{d}\rho =b - a. \end{aligned}$$

Similarly we have that $\int _{{\mathbb {R}}}h_t(-\lambda )\, \textrm{d}\lambda = b - a$. Hence, $g_t(0)= (b- a) / \pi$, which leads to the claim.$\square$

4 Bound of the Kernel Term

In what follows we show a bound on the kernel term of the trace formula,

$$\begin{aligned} R_K(X,\varepsilon ,t,a,b) :=\frac{1}{2{{\,\textrm{Area}\,}}(X)}\sum _{{\gamma }\ne 1} \int _F\varepsilon ({\gamma }) \, K_t(z,{\gamma }z) \, \tau _{z\mapsto {\gamma }^{-1}z}\, \textrm{d}z \end{aligned}$$

where the summation runs over hyperbolic elements in the group ${\Gamma }$ which are not the identity, and $F$ is a fundamental domain of $X = {\Gamma }\backslash {\mathbb {H}}$.

The steps of the kernel bound are as follows.

• First, in Sect. 4.1, we prove an upper bound on the values $K_t$ of the kernel appearing in $R_K$.

• We prove a classic counting bound on hyperbolic elements of ${\Gamma }$ in Sect. 4.2, in order to deal with the summation.

• We then cut the fundamental domain $F$ in a thick and thin part, $F^\pm (L)$, in Sect. 4.3. We bound the integrals over these two sets separately.

• Finally, in Sect. 4.4, we conclude to a quantitative probabilistic bound on $R_K$, using the probabilistic assumption from Theorem 9, and in particular the Benjamini–Schramm hypothesis.

4.1 Kernel Estimate

Let us prove the following bound on $K_t$.

Proposition 14

For any $\rho , t >0$ we have:

$$\begin{aligned} |K_t(\rho )| \le \left( \frac{1}{ t^2} + \frac{b + 1}{\rho } \right) \left( 1 + \frac{t}{\rho } \right) \exp \left( -\frac{\rho ^2}{4t^2} \right) . \end{aligned}$$

Proof

We start from the kernel formula, equation (2), and make use of the inequalities on $g_t$ and $g_t'$ obtained in Lemma 10.(2). More precisely,

$$\begin{aligned} |K_t(\rho )| =&\frac{\cosh \left( \tfrac{\rho }{2} \right) }{2\pi \sqrt{2}} \left| \int _\rho ^{\infty }\frac{2 g_t'(u)- g_t(u) \tanh \left( \tfrac{u}{2}\right) }{\cosh \left( \tfrac{u}{2}\right) \sqrt{\cosh (u) - \cosh (\rho )}}\, \textrm{d}u \right| \\&\le \frac{1}{2\pi \sqrt{2}} \int _\rho ^{\infty }\frac{2 |g_t'(u)| + |g_t(u)|}{\sqrt{\cosh (u) - \cosh (\rho )}}\, \textrm{d}u \end{aligned}$$

by the triangle inequality. By Lemma 10.(2),

$$\begin{aligned} 2 |g_t'(u)| + |g_t(u)| \le \frac{2}{\pi } \left( \frac{1}{t^2} +\frac{4b+1}{u} \right) \exp \left( - \frac{u^2}{4t^2} \right) \end{aligned}$$

and hence

$$\begin{aligned} |K_t(\rho )| \le \frac{1}{\pi ^2 \sqrt{2}} \left( \frac{1}{ t^2} + \frac{4b + 1}{\rho } \right) \int _{\rho }^{\infty } \frac{ \exp \left( -\tfrac{u^2}{4t^2} \right) }{\sqrt{\cosh u - \cosh \rho }} \, \textrm{d}u. \end{aligned}$$

We then proceed with the splitting of the integral at $u=2\rho$. If $u\in [\rho , 2\rho ]$, then $\cosh u - \cosh \rho \ge (u-\rho )\sinh \rho \ge (u-\rho )\rho$. Hence:

$$\begin{aligned} \int _{\rho }^{2\rho } \frac{ \exp \left( -\frac{u^2}{4t^2} \right) }{\sqrt{\cosh u - \cosh \rho }} \, \textrm{d}u \le \exp \left( -\frac{\rho ^2}{4t^2} \right) \int _{\rho }^{2\rho } \frac{\, \textrm{d}u}{\sqrt{(u-\rho )\rho }} = 2 \exp \left( -\frac{\rho ^2}{4t^2} \right) . \end{aligned}$$

In the other case, if $u\in [2\rho , \infty )$, we can deduce that $\cosh u - \cosh \rho \ge \frac{1}{2}(u^2-\rho ^2)\ge \frac{3}{2}\rho ^2$. It follows that:

$$\begin{aligned} \int _{2\rho }^{\infty } \frac{\exp \left( -\tfrac{u^2}{4t^2} \right) }{\sqrt{\cosh u - \cosh \rho }} \, \textrm{d}u&\le \frac{\sqrt{2}}{\sqrt{3} \, \rho } \int _{2\rho }^{\infty } \exp \left( -\frac{u^2}{4t^2} \right) \, \textrm{d}u \\&\le \frac{\sqrt{2}}{\sqrt{3} \, \rho } \int _{0}^{\infty } \exp \left( -\frac{u^2}{4t^2} - \frac{\rho ^2}{t^2} \right) \, \textrm{d}u\\&= \frac{\sqrt{2 \pi } \, t}{\sqrt{3}\rho } \exp \left( - \frac{\rho ^2}{t^2} \right) . \end{aligned}$$

Finally, putting everything together, we obtain:

$$\begin{aligned} |K_t(\rho )| \le \frac{1}{\pi ^2 \sqrt{2}} \left( \frac{1}{ t^2} + \frac{4b + 1}{\rho } \right) \left( 2 + \frac{\sqrt{2 \pi } \, t}{\sqrt{3}\rho } \right) \exp \left( -\frac{\rho ^2}{4t^2} \right) \end{aligned}$$

which implies our claim. $\square$

4.2 Bound on the Number of Hyperbolic Elements

We shall use the following classic bound in order to control the summations over hyperbolic elements of ${\Gamma }$.

Lemma 15

Let $r \le 2$ be a positive number and let $X = {\Gamma }\backslash {\mathbb {H}}$ be a hyperbolic surface whose systole is larger than r. Then, for any $j>0$, any $z \in {\mathbb {H}}$,

$$\begin{aligned} \# \{ {\gamma }\in {\Gamma }\setminus \{1\} : {\gamma }\text { hyperbolic, } d(z,{\gamma }z) \le j \} \le \frac{4e^{1+j} }{r^2}. \end{aligned}$$

Proof

Choose $z\in {\mathbb {H}}$ a point. The family of discs centred at ${\gamma }z$ and of radius r/2, for hyperbolic elements ${\gamma }\in {\Gamma }$, are disjoint. Since the area of the ball with radius R is $2\pi (\cosh R-1)$, by comparing areas, we get that the number of hyperbolic elements ${\gamma }$ for which $d(z,{\gamma }z)\le j$ must be smaller than:

$$\begin{aligned} \frac{\cosh \left( j+ \frac{r}{2} \right) -1}{\cosh \left( \frac{r}{2} \right) -1} \le \frac{e^{j+\frac{r}{2}}}{\frac{r^2}{4}} \le \frac{4e^{1+j}}{r^2}. \end{aligned}$$

$\square$

4.3 Thin-Thick Decomposition of the Fundamental Domain

For a positive real number L, we decompose the fundamental domain $F$ of $X = {\Gamma }\backslash {\mathbb {H}}$ as a disjoint union of two sets $F^-(L)$ and $F^+(L)$, the points of $F$ with injectivity radius smaller than L and larger than L respectively. Splitting the integral in the sum $R_K$ into two integrals, on those two sets, we can rewrite our sum as:

$$\begin{aligned} R_K(X,\varepsilon ,t,a,b)=R^-_K(X,\varepsilon ,t,a,b,L)+R^+_K(X,\varepsilon ,t,a,b,L). \end{aligned}$$

We shall start by bounding the contribution given by integration on $F^+(L)$, using the fact that all points on $F^+(L)$ have an injectivity radius larger than L.

Lemma 16

Let $t > 0$ and $0 < r \le 2$. Suppose that $X={\Gamma }\setminus {\mathbb {H}}$ is a hyperbolic surface whose systole is larger than r. If L is a real number such that $L \ge 8t^2+1$, then:

$$\begin{aligned} |R^+_K(X,\varepsilon ,t,a,b,L)| \le \frac{4e^2}{r^2} \left( \frac{1}{ t^2} + \frac{b + 1}{r} \right) \left( 1 + \frac{t}{r} \right) \exp \left( - L \right) . \end{aligned}$$

Proof

By definition of $F^+(L)$, for $z\in F^+(L)$, the sum defining $R^+_K$ contains no elements ${\gamma }$ such that $d(z,{\gamma }z) < L$. Thus, we can write:

$$\begin{aligned}&R^+_K(X,\varepsilon ,t,a,b,L)\\&\quad = \frac{1}{2{{\,\textrm{Area}\,}}(X)} \int _{F^+(L)}\sum _{j= \lfloor L \rfloor }^{\infty } \sum _{\begin{array}{c} {\gamma }\ne 1 \\ j\le d(z,{\gamma }z)<j+1 \end{array}} \varepsilon ({\gamma }) \, K_t(z,{\gamma }z) \, \tau _{z\mapsto {\gamma }^{-1}z}\, \textrm{d}z. \end{aligned}$$

When bounding the quantity above by the triangular inequality, we note that $|\varepsilon ({\gamma })|=|\tau _{z\mapsto {\gamma }^{-1}z}|=1$, which allows to safely ignore these terms. Moreover, since ${\gamma }$ is hyperbolic, the distance between z and ${\gamma }z$ is always larger than the length of the systole, and in particular larger than r. Thus, Proposition 14 together with Lemma 15 imply:

$$\begin{aligned}&|R^+_K(X,\varepsilon ,t,a,b,L)|\\&\quad \le \frac{2e^2}{r^2} \left( \frac{1}{ t^2} + \frac{b + 1}{r} \right) \left( 1 + \frac{t}{r} \right) \frac{{{\,\textrm{Area}\,}}(F^+(L))}{{{\,\textrm{Area}\,}}(X)} \sum _{j= \lfloor L \rfloor }^{\infty } \exp \left( j - \frac{j^2}{4t^2} \right) . \end{aligned}$$

First, we note that ${{\,\textrm{Area}\,}}(F^+(L)) \le {{\,\textrm{Area}\,}}(X)$. We then observe that, provided $\lfloor L \rfloor \ge 8t^2$, we have $j\le \frac{j^2}{8t^2}$ and hence by comparison of the sum with an integral,

$$\begin{aligned} \sum _{j= \lfloor L \rfloor }^{\infty } \exp \left( j - \frac{j^2}{4t^2} \right)&\le \sum _{j= \lfloor L \rfloor }^{\infty }\exp \left( -\frac{j^2}{8t^2} \right) \le \exp \left( - \frac{L^2}{8t^2} \right) + \int _L^{\infty }\exp \left( - \frac{x^2}{8t^2} \right) \, \textrm{d}x\\&\le \exp \left( - \frac{L^2}{8t^2} \right) + \frac{1}{L}\int _L^{\infty }x\exp \left( - \frac{x^2}{8t^2} \right) \, \textrm{d}x\\&= \left( 1+ \frac{4t^2}{L} \right) \exp \left( - \frac{L^2}{8t^2} \right) \end{aligned}$$

which is bounded by $2\exp (-L)$ as soon as $L \ge 8 t^2+1$, thus implying our claim. $\square$

We now bound the contribution of $F^-(L)$.

Lemma 17

With the notations of Lemma 16,

$$\begin{aligned} |R^-_K(X,\varepsilon ,t,a,b,L)| \le \frac{4 e^2}{r^2} \left( \frac{1}{ t^2} + \frac{b + 1}{r} \right) \left( 1 + \frac{t}{r} \right) \frac{{{\,\textrm{Area}\,}}(F^-(L))}{{{\,\textrm{Area}\,}}(X)} \left( 1+ L e^L \right) . \end{aligned}$$

We observe that, due to the fact that the injectivity radius on $F^-(L)$ is not bounded below by $L \gg 1$, we do not obtain an exponential decay like $e^{-L}$ in Lemma 16. However, the ratio ${{\,\textrm{Area}\,}}(F^-(L)) / {{\,\textrm{Area}\,}}(X)$ will decay under the Benjamini–Schramm hypothesis.

Proof

As before, we combine Proposition 14 together with Lemma 15 to obtain:

$$\begin{aligned}&|R^-_K(X,\varepsilon ,t,a,b,L)|\\&\quad \le \frac{2e^2}{r^2} \left( \frac{1}{ t^2} + \frac{b + 1}{r} \right) \left( 1 + \frac{t}{r} \right) \frac{{{\,\textrm{Area}\,}}(F^-(L))}{{{\,\textrm{Area}\,}}(X)} \sum _{j\ge 0} \exp \left( j - \frac{j^2}{4t^2} \right) . \end{aligned}$$

To deal with the last sum, we split it at $\lfloor 8t^2 \rfloor +1$. Proceeding as in Lemma 16, we deduce:

$$\begin{aligned} \sum _{j\ge \lfloor 8t^2 \rfloor +1} \exp \left( j - \frac{j^2}{4t^2} \right) \le \left( 1+ \frac{4t^2}{\lfloor 8t^2 \rfloor +1} \right) \exp \left( - \frac{(\lfloor 8t^2 \rfloor +1)^2}{8t^2} \right) \le 2. \end{aligned}$$

For remaining indices we just bound naively:

$$\begin{aligned} \sum _{j=0}^{\lfloor 8t^2 \rfloor } \exp \left( j - \frac{j^2}{4t^2} \right) \le \sum _{j=0}^{\lfloor 8t^2 \rfloor } \exp \left( j \right) \le 8t^2\exp (8t^2) \le L \exp (L) \end{aligned}$$

which leads to the claim. $\square$

4.4 Probabilistic Kernel Estimate

The last step of this section is to use our probabilistic hypotheses, presented in Sect. 2.3, to bound the kernel term. We prove the following.

Proposition 18

Let $0\le a \le b$, $g \ge 2$, and set $t:= \frac{1}{4\sqrt{3}} \sqrt{\log g}$. Then for any $X\in \mathcal {A}_{g,k(g)}$,

$$\begin{aligned} R_K(X,\varepsilon ,t,a,b) = \mathcal {O}\left( \frac{b+1}{\sqrt{ \log g }} \right) . \end{aligned}$$

Proof

Let us apply Lemmas 16 and 17 with the parameters

$$\begin{aligned} t := \frac{1}{4\sqrt{3}} \sqrt{\log g} \qquad L := 8 t^2 + 1= \frac{1}{6} \log (g) +1 \qquad r := \frac{\sqrt{\log g}}{ g^{\frac{1}{24}}} \end{aligned}$$

to a surface $X \in \mathcal {A}_{g,k(g)}$. By definition of the set $\mathcal {A}_{g,k(g)}$, the systole of X is bounded below by r. We observe that, for our choices of parameters,

$$\begin{aligned} \frac{1}{r^2} \left( \frac{1}{ t^2} + \frac{b + 1}{r} \right) \left( 1 + \frac{t}{r} \right) = \mathcal {O} \left( \frac{(b+1)t}{r^4} \right) = \mathcal {O} \left( \frac{b+1}{g^{- \frac{1}{6}} (\log g)^{\frac{3}{2}}} \right) . \end{aligned}$$

As a consequence, Lemma 16 implies

$$\begin{aligned} R^+_K(X,\varepsilon ,t,a,b,L) = \mathcal {O} \left( \frac{b+1}{g^{- \frac{1}{6}}(\log g)^{\frac{3}{2}}} \, e^{-L}\right) = \mathcal {O} \left( \frac{b+1}{(\log g)^{\frac{3}{2}}}\right) . \end{aligned}$$

Furthermore, by definition of $\mathcal {A}_{g,k(g)}$,

$$\begin{aligned} \frac{{{\,\textrm{Area}\,}}(F^-(L))}{{{\,\textrm{Area}\,}}(X)} \le g^{- \frac{1}{3}}. \end{aligned}$$

Lemma 17 then implies

$$\begin{aligned} R^-_K(X,\varepsilon ,t,a,b,L) = \mathcal {O} \left( \frac{b+1}{g^{- \frac{1}{6}} (\log g)^{\frac{3}{2}}} g^{- \frac{1}{3}} L e^L\right) = \mathcal {O} \left( \frac{b+1}{\sqrt{\log g}}\right) \end{aligned}$$

which allows us to conclude because $R_K = R_K^+ + R_K^-$. $\square$

5 Estimates for the Number of Eigenvalues

We are finally able to prove the main theorem. Throughout this section, we will take the parameter t to be equal to $\frac{\sqrt{\log g}}{4\sqrt{3}}$, for a large genus g. Combining results from the last three subsections we obtain the following:

Lemma 19

Let $g \ge 2$, $t = \frac{\sqrt{\log g}}{4\sqrt{3}}$, $0\le a \le b$. For any $X\in \mathcal {A}_{g,k(g)}$, any nontrivial spin structure $\varepsilon$ on X,

$$\begin{aligned} \frac{1}{{{\,\textrm{Area}\,}}(X)} \sum _{j=0}^{\infty } (h_t(\lambda _j)+ h_t(-\lambda _j))&=\frac{1}{4\pi }\int _a^b \lambda \coth (\pi \lambda )\, \textrm{d}\lambda + \mathcal {O}\left( \frac{b+1}{\sqrt{\log g}} \right) . \end{aligned}$$

Proof

Rewrite formula (5) using Proposition 11, 18 and Remark 13. $\square$

From this we easily deduce Proposition 2, the upper bound on the counting function $N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )} (a,b)$ defined in Sect. 2.2.2.

Proof of Proposition 2

First, we use the bound $0 \le \lambda \, \coth (\pi \lambda ) \le \lambda + 1/\pi$ for $\lambda >0$ to obtain

$$\begin{aligned} \int _a^b \lambda \coth (\pi \lambda )\, \textrm{d}\lambda \le \int _a^b \left( \lambda + \frac{1}{\pi }\right) \, \textrm{d}\lambda = \mathcal {O} \left( b^2-a^2 + b-a\right) = \mathcal {O} \left( (b+1) (b-a)\right) \end{aligned}$$

(9)

because $b^2-a^2 = (b-a)(b+a)$ and $a \le b$. Then, Lemma 19 implies

$$\begin{aligned} \frac{1}{{{\,\textrm{Area}\,}}(X)} \sum _{j=0}^{\infty } (h_t(\lambda _j)+ h_t(-\lambda _j))= \mathcal {O}\left( (b+1) \left( b-a + \frac{1}{\sqrt{\log g}} \right) \right) . \end{aligned}$$

(10)

We then observe that

$$\begin{aligned} N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )} (a,b) \inf _{\lambda \in [a,b]} h_t(\lambda ) \le \sum _{j=0}^{\infty } (h_t(\lambda _j)+ h_t(-\lambda _j)) \end{aligned}$$

(11)

by positivity of $h_t$, and because $N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )}$ counts the number of indices j such that $a \le \lambda _j \le b$ by definition. Moreover, the restriction of the function $h_t$ to [a, b] attains its infimum at both endpoints a and b. Then, we consider the two following regimes.

• If $t(b - a)\ge 1$, by Lemma 10.(3) for either one of these two values,

  $$\begin{aligned} \inf _{\lambda \in [a,b]} h_t(\lambda ) \ge \frac{1}{2}-\frac{\exp (-t^2(b-a)^2)}{2\sqrt{\pi } \, t(b-a)} \ge \frac{1}{2} - \frac{e^{-1}}{2\sqrt{\pi }}> \frac{1}{3}. \end{aligned}$$

  (12)

  Then equations (10), (11) and (12) together imply our claim.

• If $t(b-a)\le 1$, then we note that $b \le a + 1/t$ and hence

  $$\begin{aligned} N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )} (a,b) \le N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )} \left( a, a + \frac{1}{t} \right) . \end{aligned}$$

  We apply the first case to the parameters a and $b':= a+1/t$, and obtain

  $$\begin{aligned} \frac{N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )} (a,a + 1/t)}{{{\,\textrm{Area}\,}}(X)} = \mathcal {O}\left( \left( a + \frac{1}{t}+1 \right) \left( \frac{1}{t} + \frac{1}{\sqrt{\log g}} \right) \right) = \mathcal {O} \left( \frac{a+1}{\sqrt{\log g}} \right) , \end{aligned}$$

  which is enough to conclude because $a \le b$.

$\square$

We are now ready to conclude the proof of our main result, Theorem 1. We prove the upper and lower bounds separately, because they rely on a different method.

Proof of the upper bound of Theorem 1

We note that if $t(b - a)< \sqrt{2e}$, from (9):

$$\begin{aligned} \int _a^b \lambda \coth (\pi \lambda )\, \textrm{d}\lambda = \mathcal {O}((b+1)(b - a)) = \mathcal {O}\left( \frac{b +1 }{\sqrt{\log g}} \right) , \end{aligned}$$

and hence the upper bound is then a trivial consequence of Proposition 2. Thus, for the rest of the proof, we shall assume that $t(b - a) \ge \sqrt{2e}$. The control of the function $h_t$ given by Lemma 10.(3) is not optimal near a and b. Therefore we decompose the counting function as:

$$\begin{aligned} N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )} (a,b)= N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )} \left( a , a + \eta \right) + N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )} \left( a + \eta , b - \eta \right) + N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )} \left( b - \eta , b\right) \nonumber \\ \end{aligned}$$

(13)

for a number $\eta \in [\frac{1}{t}, \frac{b-a}{2}]$ that will be picked later (note that the hypothesis $t(b - a) \ge \sqrt{2e}$ implies that this interval is not empty).

On the one hand, we observe that the first and the third term on the right hand side of (13) can easily be bounded above using Proposition 2:

$$\begin{aligned} \frac{N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )} \left( a , a + \eta \right) + N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )} \left( b - \eta , b\right) }{{{\,\textrm{Area}\,}}(X)}&= \mathcal {O} \left( (b+1) \left( \eta + \frac{1}{\sqrt{\log g}} \right) \right) \nonumber \\&= \mathcal {O}\left( (b +1)\eta \right) . \end{aligned}$$

(14)

On the other hand, we can proceed as in the proof of Proposition 2 to bound the second term of the right hand side, except more finely this time. More precisely, we write again

$$\begin{aligned} \frac{N^{{{\,\mathrm{|D|}\,}}}_{(X, \varepsilon )}\left( a + \eta , b - \eta \right) }{{{\,\textrm{Area}\,}}(X)} \inf _{\lambda \in [a + \eta , b - \eta ]} h_t(\lambda ) \le \sum _{j=0}^{\infty }\left( h_t(\lambda _j)+h_t(-\lambda _j)\right) \end{aligned}$$

and then use Lemma 19 to obtain a constant $C>0$ such that

$$\begin{aligned} \frac{N^{{{\,\mathrm{|D|}\,}}}_{(X, \varepsilon )}\left( a + \eta , b - \eta \right) }{{{\,\textrm{Area}\,}}(X)} \inf _{\lambda \in [a + \eta , b - \eta ]} h_t(\lambda ) \le \frac{1}{4\pi }\int _a^b \lambda \coth (\pi \lambda )\, \textrm{d}\lambda + C \frac{b +1}{\sqrt{\log g}}. \end{aligned}$$

(15)

Let us estimate the infimum of $h_t$ on $[a+\eta , b-\eta ]$. This infimum is attained at both endpoints $a+\eta$ and $b-\eta$. Using Lemma 10.(3) we obtain:

$$\begin{aligned} \inf _{\lambda \in [a + \eta , b - \eta ]} h_t(\lambda ) \ge 1 - \frac{e^{-t^2\eta ^2}}{2\sqrt{\pi } t \eta } - \frac{e^{-t^2 (b - a -\eta )^2}}{2 \sqrt{\pi } t (b - a -\eta )} \ge 1 - \frac{e^{-t^2\eta ^2}}{\sqrt{\pi }t\eta } \end{aligned}$$

because $b - a - \eta \ge \eta$. We now observe that, for all $0 \le x \le 1/2$, $(1-x)^{-1} \le 1+2x$. We apply this inequality to $x = e^{-t^2\eta ^2}/(\sqrt{\pi }t\eta ) \le 1/2$ (thanks to the fact that $t\eta \ge 1$) and get:

$$\begin{aligned} \left( \inf _{\lambda \in [a + \eta , b - \eta ]} h_t(\lambda ) \right) ^{-1} \le 1 + 2 \frac{e^{-t^2\eta ^2}}{\sqrt{\pi }t\eta } \le 1 + 2 e^{-t^2\eta ^2}. \end{aligned}$$

(16)

We now use the bound (16) into equation (15), which yields

$$\begin{aligned} \frac{N^{{{\,\mathrm{|D|}\,}}}_{(X, \varepsilon )}\left( a + \eta , b - \eta \right) }{{{\,\textrm{Area}\,}}(X)}&\le (1 + 2 e^{-t^2\eta ^2}) \left( \frac{1}{4\pi }\int _a^b \lambda \coth (\pi \lambda )\, \textrm{d}\lambda + C \frac{b +1}{\sqrt{\log g}} \right) \\&\le \frac{1}{4\pi }\int _a^b \lambda \coth (\pi \lambda )\, \textrm{d}\lambda + e^{-t^2\eta ^2} \int _a^b \lambda \coth (\pi \lambda )\, \textrm{d}\lambda \\&\quad + 3 C \frac{b + 1}{\sqrt{\log g}}. \end{aligned}$$

By direct computations, one can check that the number

$$\begin{aligned} \eta := \frac{1}{t}\sqrt{\log \left( \sqrt{\frac{e}{2}} \, t(b - a) \right) } \end{aligned}$$

(17)

satisfies the hypothesis $\frac{1}{t} \le \eta \le \frac{b-a}{2}$ thanks to the assumption $t(b - a)\ge \sqrt{2e}$. Using (9) we obtain

$$\begin{aligned} e^{-t^2\eta ^2} \int _a^b \lambda \coth (\pi \lambda )\, \textrm{d}\lambda= & {} \mathcal {O} \left( e^{-t^2\eta ^2} (b+1)(b-a) \right) = \mathcal {O} \left( \frac{b+1}{t} \right) \\= & {} \mathcal {O} \left( \frac{b+1}{\sqrt{\log (g)}} \right) \end{aligned}$$

and therefore

$$\begin{aligned} \frac{N^{{{\,\mathrm{|D|}\,}}}_{(X, \varepsilon )}\left( a + \eta , b - \eta \right) }{{{\,\textrm{Area}\,}}(X)} \le \frac{1}{4\pi }\int _a^b \lambda \coth (\pi \lambda )\, \textrm{d}\lambda + C' \frac{b+1}{\sqrt{\log g}} \end{aligned}$$

(18)

for a constant $C'$. We can then conclude using the decomposition (13), the bounds (18) and (14) with our value of $\eta$ specified in (17). $\square$

Proof of the lower bound of Theorem 1

Since $0\le h_t\le 1$ everywhere one has:

$$\begin{aligned} N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )}(a,b) \ge \sum _{a \le \lambda _j \le b} h_t(\lambda _j) \end{aligned}$$

which we can rewrite as

$$\begin{aligned} N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )}(a,b)\ge & {} \sum _{j=0}^{\infty }\left( h_t(\lambda _j) +h_t(-\lambda _j)\right) \\{} & {} - \left( \sum _{\lambda _j < a} h_t(\lambda _j) + \sum _{\lambda _j> b} h_t(\lambda _j) + \sum _{j=0}^{\infty } h_t(-\lambda _j)\right) . \end{aligned}$$

By Lemma 19, there exists $C''>0$ such that:

$$\begin{aligned} \frac{1}{{{\,\textrm{Area}\,}}(X)}\sum _{j=0}^{\infty } \left( h_t(\lambda _j) +h_t(-\lambda _j)\right) \ge \frac{1}{4\pi }\int _a^b \lambda \coth (\pi \lambda )\, \textrm{d}\lambda - C'' \frac{b + 1}{\sqrt{\log g}}. \end{aligned}$$

It therefore suffices to prove that the three sums we subtract are $\mathcal {O}\left( \frac{b+1}{\sqrt{\log g}} \right)$ to conclude.

We shall only present the proof for the sum after eigenvalues larger than b, because one can treat the other cases similarly. For a non-negative integer k denote $b_k:= b + \frac{k}{t}$. Then,

$$\begin{aligned} \frac{1}{{{\,\textrm{Area}\,}}(X)}\sum _{\lambda _j> b} h_t(\lambda _j) =&\frac{1}{{{\,\textrm{Area}\,}}(X)} \sum _{k=0}^{\infty } \sum _{b_k \le \lambda _j < b_{k+1}} h_t(\lambda _j)\\&\le \sum _{k=0}^{\infty } \frac{N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )}(b_k,b_{k+1})}{{{\,\textrm{Area}\,}}(X)} \sup _{\lambda \in [b_k,b_{k+1}] }h_t(\lambda ). \end{aligned}$$

For $k \ge 0$, since $b_{k+1}-b_k = 1/t = \mathcal {O}(1/\sqrt{\log (g)})$, Proposition 2 implies that

$$\begin{aligned} \frac{N^{{{\,\mathrm{|D|}\,}}}_{(X,\varepsilon )}(b_k,b_{k+1})}{{{\,\textrm{Area}\,}}(X)} = \mathcal {O} \left( \frac{b_{k+1}+1}{\sqrt{\log (g)}}\right) = \mathcal {O} \left( \frac{b+1}{\sqrt{\log (g)}} (k+1)\right) . \end{aligned}$$

We then apply Lemma 10.(3) to bound the supremum of $h_t$ on $[b_k, b_{k+1}]$ for the terms $k \ge 1$, and obtain that

$$\begin{aligned} \sum _{k=0}^{\infty } (k+1) \sup _{\lambda \in [b_k,b_{k+1}] }h_t(\lambda ) \le 1&+ \sum _{k=1}^{\infty } (k+1) \, \frac{\exp \left( -t^2(b_k - b)^2\right) }{2\sqrt{\pi } t (b_k - b)} \\&= 1 + \sum _{k=1}^{\infty } (k+1) \, \frac{\exp \left( -k^2\right) }{2\sqrt{\pi } k} = \mathcal {O}(1) \end{aligned}$$

which is what we need to conclude.$\square$