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\subsubsection*{Vertex expansion}

\gls{ndlam} $G = (V, E)$.
Have for all $S, T \subseteq V$,
\[
  \left| e(S, T) - \frac{d}{n} |S| |T| \right|
  \le \frac{\lambda}{n} \sqrt{|S| |S^c| |T| |T^c|}
.\]
$\setexp_v(S) = \frac{|\partial S|}{|S|}$.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/9ef9e422ad90431f.png}
\end{center}
\begin{align*}
  |\partial S|
  &\ge \frac{e(S, S^c)}{d} \\
  \setexp_v(S)
  &\ge \frac{e(S, S^c)}{d|S|} \\
  &= \setexp_e(S)
\end{align*}
If $S \subseteq V$, $|S| \le n/2$, then 
\begin{align*}
  \setexp_V(S)
  &\ge \setexp_e(G) \\
  &\ge \frac{\lambda_2(\tilde{L}_G)}{2} \\
  \setexp_v(S)
  &\ge \frac{\left( 1 - \frac{\lambda}{d} \right)}{2} \\
  |S \cup \partial S|
  &\ge \left( 1 + \half - \frac{\lambda}{2d} \right)
\end{align*}
$n \to \infty$, $d$, $\lambda$ fixed.
$\lambda \le d - \delta$, $\delta > 0$.
$|S \cup \partial S| \ge (1 + \kappa)|S|$ for some $\kappa > 0$.
Hence
\[
  |B(x, r)|
  \ge (1 + \kappa)^r
\]
if $|B(x, r - 1)| \le n/2$.

$\diam G = O_{\frac{\lambda}{d}}(\log n)$ (by considering ball around start and
end).

\subsubsection*{Why aren't Cayley graphs of abelian groups expanders?}

Let $\Gamma$ be an abelian group and $S \subseteq \Gamma$ a set of generators of
size $d$.
Let $|\Gamma| = n \to \infty$.
Let $G = \Cay(\Gamma, S)$.
Then
\[
  |B(x, r)| \le (2r + 1)^d
.\]
This is not exponential in $r$, so the Cayley graph can't be an expander.

\begin{fcthm}[Alon-Boppana]
  Assuming:
  - $G = (V, E)$ an \gls{ndlam}
  Then:
  as $n \to \infty$
  \[
    \lambda
    \ge 2\sqrt{d - 1}
    - O(1)
  .\]
\end{fcthm}

\begin{proof}[Proof 1]
  Pick edge $st$, pick $r \in \Zbb_{\ge 0}$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/343e62b5f1b34fce.png}
  \end{center}
  $\varphi : V \to \Rbb$,
  \[
    \varphi(x)
    = \begin{cases}
      (d - 1)^{-i/2}
      & \text{if $x \in V_i$, $i \le r$} \\
      0
      & \text{if $x \in V_i$, $i > r$}
    \end{cases}
  \]
  \[
    \q_\lapm(\varphi)
    = \frac{\langle \varphi, \lapm \varphi \rangle}{\langle \varphi, \varphi
    \rangle}
    \qquad
    \langle \varphi, \varphi \rangle
    = \sum_{i = 0}^r \frac{|V_i|}{(d - 1)^i}
  .\]
  \begin{align*}
    \langle \varphi, \lapm \varphi \rangle
    &= \sum_{x \sim y} (\varphi(x) - \varphi(y))^2 \\
    &= \sum_{i = 0}^{r - 1} e(V_i, V_{i + 1}) \left( \frac{1}{(d - 1)^{i/2}} -
    \frac{1}{(d - 1)^{(i + 1)/2}} \right)^2
    + \frac{e(V_r, V_{r + 1})}{(d - 1)^r} \\
    &= \sum_{i = 0}^{r - 1} \frac{e(V_i, V_{i + 1})}{(d - 1)^i} \left( 1 -
    \frac{1}{\sqrt{d - 1}} \right)^2 + \frac{e(V_r, V_{r + 1})}{(d - 1)^r}
  \end{align*}
  $e(V_i, V_{i + 1}) \le (d - 1) |V_i|$.
  \begin{align*}
    \langle \varphi, \lapm \varphi \rangle
    &\le \sum_{i = 0}^{r - 1} \frac{|V_i|}{(d - 1)^i} (\sqrt{d - 1} - 1)^2 +
    \frac{|V_r|}{(d - 1)^r} (d - 1) \\
    (\sqrt{d - 1} - 1)^2
    &= (d - 1) - 2\sqrt{d - 1} + 1 = d - 2\sqrt{d - 1} \\
    \langle \varphi, \lapm \varphi \rangle
    &\le (d - 2\sqrt{d - 1}) \sum_{i = 0}^{r - 1} \frac{|V_i|}{(d - 1)^i} + (d -
    1) \frac{|V_r|}{(d - 1)^r} \\
    |V_r|
    &\le (d - 1) |V_{r - 1}|
    \le \cdots
    \le (d - 1)^{r - i} |V_i| \\
    \implies \frac{|V_r|}{(d - 1)^r}
    &\le \frac{1}{r} \sum_{i = 0}^{r} \frac{|V_i|}{(d - 1)^i} \\
    \langle \varphi, \lapm \varphi \rangle
    &\le (d - 2\sqrt{d - 1}) \sum_{i = 0}^{r}  \frac{|V_i|}{(d - 1)^i}
    + (2\sqrt{d - 1}) \frac{|V_r|}{(d - 1)^r} \\
    &\le \left(d - 2\sqrt{d - 1} + \frac{2\sqrt{d - 1} - 1}{r + 1}\right)
    \langle \varphi, \varphi \rangle \\
    \implies \lambda_2(\lapm)
    &= \min_{\dim W = 2} \max_{\substack{f \in W \\ f \neq 0}} \frac{\langle f,
    \lapm f \rangle}{\langle f, f \rangle}
  \end{align*}
  Suppose $G$ has 2 deges at distance $> 2r + 2$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/917ab3256183488c.png}
  \end{center}
  $f$, $f'$ as above.
  $\langle f, f' \rangle = 0$.
  \[
    \Q_\lapm(\alpha f + \beta f')
    = \Q_\lapm(\alpha f) + \Q_\lapm(\beta f')
  .\]
  Let $W = \Span \{f, f'\}$.
  \begin{align*}
    \lambda_2(\lapm)
    &\le \max_{(\alpha, \beta) \neq (0, 0)} \frac{\Q_\lapm(\alpha f + \beta
    f')}{\|\alpha f + \beta f'\|} \\
    &\le \frac{\alpha^2 \Q_\lapm(f) + \beta^2 \Q_\lapm(f')}{\alpha^2 \|f\| +
    \beta^2 \|f'\|} \\
    &\le d - 2\sqrt{d - 1} + \frac{2\sqrt{d - 1} - 1}{r + 1}
    \qedhere
  \end{align*}
\end{proof}

\begin{fccoro}[]
  For all $d$, \cloze{there are finitely many \glspl{ndlam} with $\lambda <
  2\sqrt{d - 1}$.}
\end{fccoro}

$r = c\log n$.
\[
  \lambda_{n - 1}(\adjm_G)
  \ge 2\sqrt{d - 1} - O_d \left( \frac{1}{\log n} \right) 
.\]

For Alon-Boppana:

\begin{proof}[Proof 2]
  $\tr A^{2k} = \sum_x A^{2k} (x, x)$.
  Note that
  \[
    \#\{\text{closed walks of length $2k$ in $G$ starting from $x$}\}
  .\]
  is at least
  \[
    \#\{\text{closed walks of length $2k$ in $\prod_d$ starting from $0$}\}
  \]
  ($\prod_d$ is an infinite $d$-regular tree).
  The latter is at least
  \[
    (d - 1)^k \frac{1}{k + 1} {2k \choose k}
    \approx (2 \sqrt{d - 1})^{2k + o(1)}
    2^{2k}
  .\]
  \[
    \le \sum_{i = 1}^n \lambda_i^{2k}
    \le d^{2k} + (n - 1) \lambda^{2k}
  .\]
  Exercise: finish details.
\end{proof}

$\min\{|\lambda_1(A)|\lambda_{n - 1}(A)\} \ge \cdots$
