---
title: '18.156 notes 18: random walks on finite groups ii'
description: We start with some easy observations.
date: '2025-04-22'
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---
**Setup**:

- Let $G$ be a finite group.
- $\mu:G\to \mathbb R$ is a probability measure on $G$, i.e., $\mu(g)\ge 0,\sum_{g\in G}\mu(g)=1$.
- Starting with $g_0\in G$, let $h_t\in G,t=1,2,\dots$ be sampled according to $\mu$, and define the random walk on $G$ by $g_t=g_{t-1}\cdot h_t,t=1,2,\dots$.

**Question**: how evenly distributed is $g_K$ on $G$ for large $K$?

To state our question more precisely, we introduce some definitions. For two functions $f_1,f_2:G\to \mathbb C$, define

$$
f_1\ast f_2(g)=\sum_{g_1,\cdot g_2=g}f_1(g_1)f_2(g_2)\,,\quad\forall g\in G\,.
$$

Define the operator $T_\mu:\ell^2(G)\to \ell^2(G)$ by $T_\mu f=f\ast \mu$. It is straightforward to check that for any $K$, $T_\mu^K\delta_{g_0}$ is the distribution of $g_K$ defined as above. Our main question is to estimate

$$
\|T_\mu^K\delta_{g_0}-\tfrac{1}{|G|}\mathbf{1}\|_{\ell^2(G)}
$$

for large $K\in \mathbb N$.

We start with some easy observations.

> \[!lemma|latex\] Lemma 18.1.
>
> $T_\mu \mathbf 1=\mathbf 1$, and $\|T_\mu f\|_{\ell^2(G)}\le \|f\|_{\ell^2(G)},\forall f\in \ell^2(G)$.
>
> <ref slug="courses/18-156-projection-theory-spring-2025/resources/mit18_156_s25_lec18_pdf/#page=2">

^lemma-18-1

> \[!proof|latex\]+ Proof of lemma 18.1.
>
> The first claim can be checked straightforwardly. For the second claim, we define the right shift operator $R_g:\ell^2(G)\to \ell^2(G)$ by
>
> $$
> R_g f(h)=f(f\cdot g^{-1})\,,\quad\forall f\in \ell^2(G),g,h\in G\,.
> $$
>
> It is easy to check that $R_g:\ell^2(G)\to \ell^2(G)$ is an isometry, and it holds that
>
> $$
> T_\mu f=f\ast \mu=\sum_{g\in G}\mu(g)R_gf\,,\quad \forall f\in \ell^2(G)\,.
> $$
>
> Therefore, it follows from the triangle inequality that
>
> $$
> \|T_\mu f\|_{\ell^2(G)}\le \sum_{g\in G}\mu(g)\|R_gf\|_{\ell^2(G)}=\|f\|_{\ell^2(G)}\,.
> $$

^proof-18-1

Denote $\ell^2(G)_0$ as the orthogonal complement of the constant functions in $\ell^2(G)$. One can verify that $T_\mu$ maps $\ell^2(G)_0$ to itself. Denote by $\sigma_1(T_\mu)$ the largest singular value of the operator $T_\mu:\ell^2(G)_0\to \ell^2(G)_0$.

> \[!lemma|latex\] Lemma 18.2.
>
> For any $K\in \mathbb N$, it holds that
>
> $$
> \|T_\mu^K\delta_{g_0}-\tfrac{1}{|G|}\mathbf{1}\|_{\ell^2(G)}\le \sigma_1(T_\mu)^K\,.
> $$
>
> <ref slug="courses/18-156-projection-theory-spring-2025/resources/mit18_156_s25_lec18_pdf/#page=2">

^lemma-18-2

> \[!proof|latex\]+ Proof of lemma 18.2.
>
> Write $\delta_{g_0}=\tfrac{1}{|G|}\mathbf{1}+(\delta_{g_0})_h$, where $(\delta_{g_0})_h=\delta_{g_0}-\tfrac{1}{|G|}\mathbf{1}\in \ell^2(G)_0$. We have for any $K\in \mathbb N$, $T_\mu^K\delta_{g_0}=\tfrac{1}{|G|}\mathbf{1}+T_\mu^K(\delta_{g_0})_h$, and thus
>
> $$
> \|T_\mu^K\delta_{g_0}-\tfrac{1}{|G|}\mathbf{1}\|_{\ell^2(G)}\le \|T_\mu^K(\delta_{g_0})_h\|_{\ell^2(G)}\le \sigma_1(T_\mu)^K\|(\delta_{g_0})_h\|_{\ell^2(G)}\le \sigma_1(T_\mu)^K\,.
> $$

^proof-18-2

For a subset $A$ of $G$, we denote $\mu_A=\tfrac{1}{|A|}\mathbf{1}_A$ and abbreviate $T_{\mu_A}$ as $T_A$. Of particular interest of us is the following concrete example: let $G=\operatorname{SL}_2(\mathbb{F}_p)$ where $p$ is a large prime and

$$
A_{\operatorname{sel}}=\left\{\left(\begin{matrix}
    1&\pm 1\\
    0&1
\end{matrix}\right),\left(\begin{matrix}
    1&0\\
    \pm 1&1
\end{matrix}\right)\right\}\subset \operatorname{SL}_2(\mathbb{F}_p)\,.
$$

We will focus on the following theorem of Selberg.

> \[!theorem|latex\] Theorem 18.3.
>
> There exists a universal constant $c>0$ such that, for every $p$, $\sigma_1(T_{A_{\operatorname{sel}}})\le 1-c$.
>
> <ref slug="courses/18-156-projection-theory-spring-2025/resources/mit18_156_s25_lec18_pdf/#page=2">

^theorem-18-3

In general, for a pair $(G,A)$ where $A$ is a subset of the group $G$, we are interested in $\sigma_1(T_A)$. This is not only because it is related to the mixing of random walks on $G$ with steps in $A$ (see Lemma [[courses/18-156-projection-theory-spring-2025/notes/18#^lemma-18-2|18.2]]), but also because the spectral gap $1-\sigma_1(T_A)$ reflects a certain expansion property of the corresponding Cayley graph.

More precisely, for $A \subset G$ that is symmetric and generates $G$, we define $\operatorname{C}(G,A)$ as the graph with vertices corresponding to the elements of $G$, and with edges $(g_1, g_2) \in E$ if there exists $a \in A$ such that $g_2 = g_1 \cdot a$. Note that $A$ generates $G$, which implies that $\operatorname{C}(G,A)$ is connected. Moreover, for a subset $S \subset G$, we denote $E(S, S^c)$ as the set of edges $(g_1, g_2) \in E$ such that $g_1 \in S$ and $g_2 \in S^c$.

> \[!lemma|latex\] Lemma 18.4.
>
> For any $S\subset G$, it holds that
>
> $$
> |E(S,S^c)|\ge (1-\sigma_1(T_A))\frac{|A||S||S^c|}{|G|}\,.
> $$
>
> <ref slug="courses/18-156-projection-theory-spring-2025/resources/mit18_156_s25_lec18_pdf/#page=2">

^lemma-18-4

> \[!proof|latex\]+ Proof of lemma 18.4.
>
> It is straightforward to check that
>
> $$
> \begin{align*}
> |E(S,S^c)|=&\ |A|\langle T_A\mathbf{1}_S,\mathbf{1}_{S^c}\rangle \\
> =&\ |A|\langle\tfrac{|S|}{|G|}\mathbf{1},\tfrac{|S^c|}{|G|}\mathbf{1}\rangle+|A|\langle T_A(\mathbf{1}_S)_h,(\mathbf{1}_{S^c})_h\rangle \\
> \ge&\ \frac{|A||S||S^c|}{|G|}-\sigma_1(T_A)\|(\mathbf{1}_S)_h\|_{\ell^2(G)}\|(\mathbf{1}_{S^c})_h\|_{\ell^2(G)} \\
> \ge&\ (1-\sigma_1(T_A))\frac{|A||S||S^c|}{|G|}\,.
> \end{align*}
> $$

^proof-18-4

Combining Lemma [[courses/18-156-projection-theory-spring-2025/notes/18#^lemma-18-2|18.2]] with Theorem [[courses/18-156-projection-theory-spring-2025/notes/18#^theorem-18-3|18.3]], we obtain that for any $S\subset G=\operatorname{SL}_2(\mathbb{F}_p)$ with $|S|\le \tfrac{|G|}{2}$, in the graph $\operatorname{C}(\operatorname{SL}_2(\mathbb{F}_p),A_{\operatorname{sel}})$,

$$
E(S,S^c)\ge \frac{c|A||S|}{2}\,.
$$

For a large graph $H=(V,E)$, we say $H$ is an expander graph, if there exists a universal constant $c>0$ such that for any $S\subset V$ with $S\le \tfrac{|V|}{2}$, $E(S,S^c)\ge \frac{c|E||S|}{|V|}$. The above result indicates that $\operatorname{C}(\operatorname{SL}_2(\mathbb{F}_p),A_{\operatorname{sel}})$ is a sparse expander graph (here sparse means that the graph has average degree $O(1)$). While Selberg did not state his theorem exactly in the form above, his work is in a sense the first proof of the existence of sparse expander graphs. In what follows we fix a large prime number $p$ and let $G=\operatorname{SL}_2(\mathbb{F}_p)$. We now discuss the proof of Theorem [[courses/18-156-projection-theory-spring-2025/notes/18#^theorem-18-3|18.3]]. We will not give a complete proof, but we will discuss some of the ideas in the proof, following the approach developed by Sarnak-Xue in the early 1990s.

**$\ell^2$-bound**. We claim the following $\ell^2$-estimate of $\sigma_1(T_\mu)$.

> \[!theorem|latex\] Theorem 18.5.
>
> There exists a universal constant $C>0$ such that
>
> $$
> \sigma_1(T_\mu)^2\le Cp^2\|\mu\|^2_{\ell^2(G)}\,.
> $$
>
> <ref slug="courses/18-156-projection-theory-spring-2025/resources/mit18_156_s25_lec18_pdf/#page=4">

^theorem-18-5

We begin with a lemma on non-trivial representations of $G=\operatorname{SL}_2(\mathbb{F}_p)$.

> \[!lemma|latex\] Lemma 18.6.
>
> Let $\rho:G\to \operatorname{U}(d)$ be a non-trivial representation of $G$, then $d\ge \tfrac{p-1}{2}$.
>
> <ref slug="courses/18-156-projection-theory-spring-2025/resources/mit18_156_s25_lec18_pdf/#page=4">

^lemma-18-6

> \[!proof|latex\]+ Proof of lemma 18.6.
>
> Consider the following two elements in $G$:
>
> $$
> u=\left(\begin{matrix}
>     1&1\\
>     0&1
> \end{matrix}\right)\,,\quad
> v=\left(\begin{matrix}
>     1&0\\
>     1&1
> \end{matrix}\right)\,.
> $$
>
> It is easy to check that $u,v$ generates $G$. Since $\rho$ is non-trivial, without loss of generality we may assume that $\rho(u)\neq I_d$. Note that
>
> $$
> \left(\begin{matrix}
>     a&0\\
>     0&a^{-1}
> \end{matrix}\right)\left(\begin{matrix}
>     1&1\\
>     0&1
> \end{matrix}\right)\left(\begin{matrix}
>     a^{-1}&0\\
>     0&a
> \end{matrix}\right)=\left(\begin{matrix}
>     1&a^2\\
>     0&1
> \end{matrix}\right)\,,\quad\forall a\in \mathbb{F}_p^*\,.
> $$
>
> This implies that $u$ is conjugate to $u^{a^2}$ for any $a\in \mathbb{F}_p^*$. Let $\Lambda$ be the multi-set of eigenvalues of $\rho(u)$, we have $\Lambda=\Lambda^{a^2},\forall a\in \mathbb{F}_p^*$. On the other hand, since $u^p=1$, we have $\Lambda\subset \{z\in \mathbb C,z^p=1\}$. Moreover, one can check that $\Lambda\neq \{1,\dots,1\}$, as this would imply that $\rho(u)^p\neq I_d$ (unless $\rho(u)=I_d$). Consequently, we can pick $\lambda\in \Lambda$ such that $\lambda\neq 1$. Then, the $\tfrac{p-1}{2}$ distinct elements $\lambda^{a^2},a\in \mathbb{F}_p^*$ all lie in $\Lambda$. We conclude that $d\ge |\Lambda|\ge \frac{p-1}{2}$, as desired.

^proof-18-6

The above lemma says that any non-trivial representation of $G=\operatorname{SL}_2(\mathbb{F}_p)$ has dimension at least of order $p$. This lower bound is order tight: consider the subgroup $U$ of $G$:

$$
U=\left\{\left(\begin{matrix}
    a&t\\
    0&a^{-1}
\end{matrix}\right)\,,a\in \mathbb{F}_p^*,t\in \mathbb{F}_p\right\}\,,
$$

which has size of order $p^2$. We have $G$ acts on $G/U$ induces a non-trivial representation with dimension of order $p$.

Before proving Theorem [[courses/18-156-projection-theory-spring-2025/notes/18#^theorem-18-5|18.5]], we first introduce some notations. For $\mu:G\to \mathbb{R}$, define $\mu^*(g)=\mu(g^{-1}),\forall g\in G$. One can check that $T_\mu^*$, the adjoint of $T_\mu$, equals $T_{\mu^*}$. Moreover, $T_{\mu}T_\mu^*=T_{\mu}T_{\mu^*}=T_{\mu\ast \mu^*}$. Denote $\nu=\mu\ast \mu^*$.

> \[!proof|latex\]+ Proof of Theorem <ref slug="courses/18-156-projection-theory-spring-2025/notes/18#^theorem-18-5"> of lemma 18.6
>
> Let $V\subset \ell^2(G)_0$ be the eigenspace of $T_{\nu}=T_\mu T_\mu^*$ that corresponds to the eigenvalue $\lambda_1(T_\nu)=\sigma_1(T_\mu)^2$. Consider the left shift operator $L_g:\ell^2(G)\to \ell^2(G)$ defined by $L_gf(h)=f(g^{-1}h),\forall f\in \ell^2(G),g,h\in G$. It is straightforward to check that $L_g$ commutes with $T_\nu$, and thus $L_g$ maps $V$ to itself. Since $V$ does not contain any constant function, $L_g$ induces a non-trivial representation of $G$ on $V$. By Lemma [[courses/18-156-projection-theory-spring-2025/notes/18#^lemma-18-6|18.6]], we have $\operatorname{dim}(V)\ge \frac{p-1}{2}$, and thus $\lambda_1(T_\nu)$ has multiplicity at least $\frac{p-1}{2}$. Therefore, by the trace formula we have
>
> $$
> \begin{align*}
> \frac{p-1}{2}\sigma_1(T_\mu)^2&\ =\frac{p-1}{2}\lambda_1(T_\mu)\le \sum_i \lambda_i(T_\nu)=\operatorname{Tr}(T_\nu) \\
> &\ =\operatorname{Tr}(T_\mu T_\mu^*)=\sum_{g_1,g_2\in G}T_{\mu,g_1,g_2}^2=|G|\sum_{g\in G}\mu(g)^2=|G|\|\mu\|^2_{\ell^2(G)}\,.
> \end{align*}
> $$
>
> Since $|G|\sim p^3$, the desired result follows.

^proof-18-6-b

As a corollary, we see that for any set $A\subset G$ with $|A|\ge 2Cp^2$, it holds that $\sigma_1(T_A)^2\le Cp^2\|\mu_A\|_{\ell^2(G)}^2\le 1/2$. Note that for $U$ the subgroup of $G$ defined as above, we have $|U|\sim p^2$ and $\sigma_1(T_U)=1$. This example also shows that the result of Theorem [[courses/18-156-projection-theory-spring-2025/notes/18#^theorem-18-5|18.5]] is order-tight.

We say $\mu$ is symmetric if $\mu=\mu^*$ (i.e. $\mu(g)=\mu(g^{-1})$). When $\mu$ is symmetric, we have $T_\mu$ is self-adjoint and thus $\sigma_1(T_\mu)^K=\lambda_1(T_\mu)^K=\lambda_1(T_\mu^K)=\lambda_1(T_{\mu^{\ast K}})=\sigma_1(T_{\mu^{\ast K}})$. Applying Theorem [[courses/18-156-projection-theory-spring-2025/notes/18#^theorem-18-5|18.5]], we obtain that for any $K\in \mathbb{N}$, it holds that

$$
\sigma_1(T_{A_{\operatorname{sel}}})^K\le Cp^2\|\mu_{A_{\operatorname{sel}}}^{\ast K}\|^2_{\ell^2(G)}\,.
$$

Our plan is to pick $K=C_0\log p$ for some universal constant $C_0>0$, and show that $\|\mu_{A_{\operatorname{se;}}}^{\ast K}\|^2_{|ell^2(G)}\le p^{-2.1}$. This would imply that $\sigma_1(T_{A_{\operatorname{sel}}})\le 1-c$ for some universal $c=c(C,C_0)>0$.

**Lifting to $\operatorname{SL}_2(\mathbb{Z})$**. Consider the projection $\pi_p:\mathbb Z\to \mathbb{F}_p$, which induces a group homomorphism $\Pi_p:\operatorname{SL}_2(\mathbb{Z})\to \operatorname{SL}_2(\mathbb{F}_p)$. Let $M$ be a probability measure on $\operatorname{SL}_2(\mathbb Z)$ and we let $\mu=\Pi_p(M)$ be its push-forward onto $\operatorname{SL}_2(\mathbb{F}_p)$. It holds that $\Pi_p(M^{\ast K})=\mu^{\ast K}$ for any $K\in \mathbb N$. Therefore, to understand $\mu_{A_{\operatorname{sel}}}^{\ast K}$ for large $K\in \mathbb N$, we may try to first understand $M_{A_{\operatorname{sel}}}^{\ast K}$, where $M_{A_{\operatorname{sel}}}=\frac{1}{4}\mathbf{1}_{{A}_{\operatorname{sel}}}$, and then understand how it projects onto $\operatorname{SL}_2(\mathbb{F}_p)$.

**Some good features about $\operatorname{SL}_2(\mathbb{Z})$**:

- **(1)** $\operatorname{SL}_2(\mathbb{Z})$ is virtually free, meaning that it has a finite index free subgroup.
- **(2a)** $\operatorname{SL}_2(\mathbb{Z})\subset \operatorname{SL}_2(\mathbb{R})$ closely related to Lie groups.
- **(2b)** $\operatorname{SL}_2(\mathbb{Z})$ acts nicely on the $\mathbb{H}^2$ hyperbolic plane.

**Intuition about convolution on $\operatorname{SL}_2(\mathbb{Z})$**: As a warm-up, consider the convolution on $\mathbb Z$. Let $\mu=\frac{1}{2}(\delta_1+\delta_{-1})$. By the central limit theorem, $\mu^K$ can be approximated by Gaussian, which intimately relates to the heat kernel on $\mathbb R$. In light of this, we might hope that for a probability measure $M$ on $\operatorname{SL}_2(\mathbb{Z})$, there is some central limit theorem for matrices, and the convolution $M^{\ast K}$ would be related to the ”heat kernel” on $\operatorname{SL}_2(\mathbb{R})$.

Consider the ”ball” in $\operatorname{SL}_2(\mathbb{R})$ with radius $T$, defined as follows:

$$
B_T:=\left\{\left(\begin{matrix}
    a&b\\
    c&d
\end{matrix}\right)\in \operatorname{SL}_2(\mathbb{R}):a^2+b^2+c^2+d^2\le T^2\right\}\,.
$$

Moreover, we denote $B_T(\mathbb{Z}):=B_T\cap \operatorname{SL}_2(\mathbb{Z})$.

> \[!lemma|latex\] Lemma 18.7.
>
> For $T$ large, we have $|B_T(\mathbb{Z})|\approx T^2$.
>
> <ref slug="courses/18-156-projection-theory-spring-2025/resources/mit18_156_s25_lec18_pdf/#page=5">

^lemma-18-7

> \[!proof|latex\]+ Proof sketch of lemma 18.7.
>
> We need to count the solutions of $ad-bc=1,a,b,c,d\in \mathbb Z,a^2+b^2+c^2+d^2\le T^2$. For a typical pair $(a,d)\in [-T,T]^2$, the number of pairs $(b,c)\in [-T,T]^2$ such that $bc=ad-1$ is at least $1$, and at most $T^{o(1)}$. This suggests $|B_T(\mathbb{Z})|\approx T^2$.

^proof-18-7

Vague statement: for large $K$, $M^{\ast K}$ is roughly equally distributed on $B_T(\mathbb{Z})$, where $T\sim \exp(c(M)\cdot K)$.

Let us see how a statement of this form about random walks on $SL_2(\mathbb{Z})$ leads to a spectral gap in $SL_2(\mathbb{F}_p)$.

> \[!lemma|latex\] Lemma 18.8.
>
> If $\mu$ is symmetric, then $\|\mu^{\ast K}\|_{\ell^2(G)}^2=\mu^{\ast 2K}(I)$, where $I \in G$ is the identity element.
>
> <ref slug="courses/18-156-projection-theory-spring-2025/resources/mit18_156_s25_lec18_pdf/#page=5">

^lemma-18-8

> \[!proof|latex\]+ Proof of lemma 18.8.
>
> By definition we have
>
> $$
> \|\mu^{\ast K}\|_{\ell^2(G)}^2=\sum_{g\in G}\mu^{\ast K}(g)^2=\sum_{g\in G}\mu^{\ast K}(g)\mu^{\ast K}(g^{-1})=\mu^{\ast 2K}(I)\,.
> $$

^proof-18-8

This leads us to examine $\|\mu_{A_{\operatorname{sel}}}^{\ast K}\|_{\ell^2(G)}^2=\mu_{A_{\operatorname{sel}}}^{\ast 2K}(I)$, where $I \in SL_2(\mathbb{F}_p)$ is the identity. We can relate this to a measure on $SL_2(\mathbb{Z})$. To set this up, let $\Gamma_p\subset\operatorname{SL}_2(\mathbb{Z})$ be the pre-image of $I \in \operatorname{SL}_2(\mathbb{F}_p)$ under $\Pi_p$, i.e.,

$$
\Gamma_p:=\left\{a,b,c,d\in \mathbb{Z},ad-bc=1,\left(\begin{matrix}
    a&b\\
    c&d
\end{matrix}\right)\equiv \left(\begin{matrix}
    1&0\\
    0&1
\end{matrix}\right)\pmod p\right\}
$$

Now we have

$$
\|\mu_{A_{\operatorname{sel}}}^{\ast K}\|_{\ell^2(G)}^2=\mu_{A_{\operatorname{sel}}}^{\ast 2K}(I_2)=\Pi_p(M_{\operatorname{sel}}^{\ast 2K}(I)),
$$

and by the vague statement, we expect

$$
\Pi_p(M_{\operatorname{sel}}^{\ast 2K}(I_2))\approx \frac{|\Gamma_p\cap B_T(\mathbb{Z})|}{|B_T(\mathbb{Z})|}\,
$$

where $T \sim \log p$. Since $\operatorname{SL}_2(\mathbb{F}_p)$ has size of order $p^3$, it is natural to expect that for large $T$, $\Gamma_p\cap B_T(\mathbb{Z})$ occupies nearly a $p^{-3}$-fraction in $B_T(\mathbb{Z})$. The following lemma shows that this is indeed the case.

> \[!lemma|latex\] Lemma 18.9.
>
> For $T>p^2$, it holds that $|\Gamma_p\cap B_T(\mathbb{Z})|\lessapprox p^{-3}T^2$.
>
> <ref slug="courses/18-156-projection-theory-spring-2025/resources/mit18_156_s25_lec18_pdf/#page=6">

^lemma-18-9

> \[!proof|latex\]+ Proof of lemma 18.9.
>
> For any $\left(\begin{matrix}
>  a&b\\
>  c&d
>  \end{matrix}\right)\in \Gamma_p$, we have $p\mid b,p\mid c$, and thus $p^2\mid bc=ad-1$. Meanwhile, we have $p\mid a-1,p\mid d-1$, which implies $p^2\mid (a-1)(d-1)=ad-a-d+1$. Altogether we conclude that $p^2\mid a+d-2$. In light of this, we see that for $\left(\begin{matrix}
>  a&b\\
>  c&d
>  \end{matrix}\right)\in \Gamma_p\cap B_T(\mathbb{Z})$, $a\in [-T,T]$ has at most $O(p^{-1}T)$ choices, and given $a$, $d$ satisfies $d\equiv -2-a\pmod{p^2}$ has at most $O(p^{-2}T)$ choices (here we use the fact that $T>p^2$). Finally, given $a,d$, $b,c$ satisfies $bc=ad-1$ has at most $T^{o(1)}$ choices. Combining things together, we obtain the desired bound.

^proof-18-9

> \[!proof|latex\]+ Proof sketch for Theorem <ref slug="courses/18-156-projection-theory-spring-2025/notes/18#^theorem-18-3"> of lemma 18.9
>
> Assuming the vague statement about random walks on $SL_2(\mathbb{Z})$ we can now assemble our ingredients to give a proof sketch of Selberg’s theorem.
>
> We pick $K$ such that $T\sim\exp(c(M_{\operatorname{sel}})K)\sim p^{1.1}$, and thus $K\le C_0\log p$ for a universal constant $C_0>0$. By the vague statement and Lemmas [[courses/18-156-projection-theory-spring-2025/notes/18#^lemma-18-8|18.8]], [[courses/18-156-projection-theory-spring-2025/notes/18#^lemma-18-9|18.9]], we have
>
> $$
> \|\mu_{A_{\operatorname{sel}}}^{\ast K}\|_{\ell^2(G)}^2\lessapprox \frac{|\Gamma_p\cap B_T(\mathbb{Z})|}{|B_T(\mathbb{Z})|}\lessapprox \frac{p^{-3}T^2}{T^2}=p^{-3}\,.
> $$
>
> Applying Theorem [[courses/18-156-projection-theory-spring-2025/notes/18#^theorem-18-5|18.5]], we obtain that $\sigma_1(T_{A_{\operatorname{sel}}})^K\lessapprox p^{-1}$. This yields that $\sigma_1(T_{A_{\operatorname{sel}}})\le 1-c$ for some universal constant $c>0$.

^proof-18-9-b

**Connection to hyperbolic geometry**

Selberg’s theorem is closely connected to hyperbolic geometry. In fact, Selberg’s original theorem was about the eigenvalues of the Laplacian on certain hyperbolic manifolds. The hyperbolic manifold perspective also gives a nice approach to the vague statement in the proof sketch above. In this short section, we briefly introduce these ideas.

Recall that $\operatorname{SL}_2(\mathbb{Z})$ acts isometrically on $\mathbb{H}^2$. Let $X(p)=\mathbb{H}^2/\Gamma_p$. If $p$ is large, then the action is properly discontinuous, and so $X(p)$ is a hyperbolic surface. It is a complete surface with finite area and with some cusps. Note that $X(p)$ is a cover of $X(1)$, and the group of deck tranform of $X(p)$ is $\operatorname{SL}_2(\mathbb{F}_p)$. So the ”large scale geometry” of $X(p)$ is closely related to the geometry of the Cayley graph of $SL_2(\mathbb{F}_p)$ with generators $A$, where $A$ is the reduction mod $p$ of some set of generators of $SL_2(\mathbb{Z})$. For instance, we could take $A = A_{sel}$.

Consider the spectrum of the Laplacian of $X(p)$. We have $0$ lies in the spectrum, but above $0$ there is a gap. Denote $\lambda_1(X(p))$ the smallest positive eigenvalue of the Laplacian of $X(p)$. Selberg proved that $\lambda_1(X(p))\ge \frac 3{16}$ and conjectured that $\lambda_1(X(p))\ge \frac{1}{4}-o(1)$. Because of the close connection between the geometry of $X(p)$ and the geometry of the Cayley graph of $SL_2(\mathbb{F}_p)$, it is not too hard to show that a lower bound for $\lambda_1(X(p))$ is equivalent to an upper bound for $\sigma_1(T_A)$, with $A$ as above.

The proof we sketched above can be translated into hyperbolic geometry using the heat kernel. The heat kernel describes a diffusion process on a Riemannian manifold, and it is a continuous analogue of a random walk. The heat kernel on a Riemannian manifold is written as $H_t(x,y)$, where $t$ represents time, and $x,y$ live in the Riemannian manifold. The probabilistic interpretation is that $H_t(x,y) dvol_y$ is the probability distribution for the position of a particle that started at $x$ and then diffused for time $t$.

We write $H_{t, X(p)}$ for the heat kernel on $X(p)$. We think of $H_{t, X(p)}$ as analogous to $\mu^{*k}$ in the proof sketch above, with $t$ analogous to $k$.

First big step: Prove that $H_{t, X(p)}$ is roughly evenly distributed on $X(p)$. We will discuss the proof of this more below.

In particular, we prove that there is a constant $C_0$ so that if $t = C_0 \log p$ and $x \in X(p)$, and for $t = C_0 \log p$, then

$$
\| H_{t, X(p)}(x, y) \|_{L^2_y}^2 \le p^{- 2.1}.
$$

This is analogous to proving that $\| \mu^{*k} \|_{L^2(SL_2(\mathbb{F}_p)}^2 \le p^{-2.1}$. There is a close connection between the mixing properties of the heat kernel and the eigenvalues of the Laplacian on a Riemannian manifold. This connection is analogous to the trace formula that we used in the finite group setting. On a closed manifold, the formula has a simple form closely parallel to the formulas we used above. If we let $0 = \lambda_0 < \lambda_1 \le \lambda_2 \le ...$ be the spectrum of the Laplacian on a compact Riemannian manifold $M$, then we have

$$
\sum_{j} e^{- 2t \lambda_j} = \int_M H_{2t}(x,x) dvol = \int_{M \times M} H_t(x,y)^2 dx dy.
$$

Since $X(p)$ is not compact, its spectral theory is a little more complicated, but this is a technical detail. This part of the proof is less elementary in the hyperbolic setting than in the finite group setting, but it is basically analogous.

Since $SL_2(\mathbb{F}_p)$ acts isometrically on $X(p)$, each eigenspace is a representation of $SL_2(\mathbb{F}_p)$. The main case is when the representation on the $\lambda_1$ eigenspace is non-trivial. Then it has dimension at least $(p-1)/2$ and so we get

$$
\frac{p-1}{2} e^{-2t \lambda_1} \le  \int_{M \times M} H_t(x,y)^2 dx dy.
$$

Then the first big step gives us, with $t = C \log p$,

$$
\frac{p-1}{2} e^{-2t \lambda_1} \le  \int_{M \times M} H_t(x,y)^2 dx dy \lesssim p^3 p^{-2.1}
$$

and so $e^{- 2 t \lambda_1} \le p^{-.1}$, and so $\lambda_1 \ge c > 0$ uniformly in $p$.

Now we return to the first big step.

We write $H_{t, X(p)}$ for the heat kernel on $X(p)$ and $H_{t, \mathbb{H}}$ for the heat kernel on the hyperbolic plane. These two heat kernels are closely connected: $H_{t, X(p)}$ is the pushforward of $H_{t, \mathbb{H}}$ by the covering map $\Pi_p: \mathbb{H} \rightarrow X(p)$. In other words, if $\Pi_p(\tilde x) = x$ and $\Pi_p(\tilde y = y)$, then

$$
H_{t,  X(p)} (x,x) = \sum_{\gamma \in \Gamma_p} H_{t,  \mathbb{H}}(\gamma \tilde x, \tilde y).
$$

In particular, to do the first big step, we have to estimate

$$
H_{2t, X(p)}(x,x) = \sum_{\gamma \in \Gamma_p} H_{2t,  \mathbb{H}}(\gamma \tilde x, \tilde x).
$$

This is analogous to estimate $M^{*k}(\Gamma_p)$ in the proof sketch above. This was a key moment in the proof sketch above where we made a vague statement. This part of the proof is easier in the hyperbolic context because there is a simple explicit formula for $H_{t, \mathbb{H}}$. Using this explicit formula and Lemma [[courses/18-156-projection-theory-spring-2025/notes/18#^lemma-18-9|18.9]], it is fairly easy to prove the desired bounds for $H_{2t, X(p)}$. So this part of the proof is actually easier in the hyperbolic setting than in the finite group setting.