Ramanujan graph

If ${\displaystyle G}$ is a ${\displaystyle d}$-regular graph with diameter ${\displaystyle m}$, a theorem due to Noga Alon[11] states

${\displaystyle \lambda _{2}(G)\geq 2{\sqrt {d-1}}-{\frac {2{\sqrt {d-1}}-1}{\lfloor m/2\rfloor }}.}$

Whenever ${\displaystyle G}$ is ${\displaystyle d}$-regular and connected on at least three vertices, ${\displaystyle |\lambda _{2}|<d}$, and therefore ${\displaystyle \lambda (G)\geq \lambda _{2}}$. Let ${\displaystyle {\mathcal {G}}_{n}^{d}}$ be the set of all connected ${\displaystyle d}$-regular graphs ${\displaystyle G}$ with at least ${\displaystyle n}$ vertices. Because the minimum diameter of graphs in ${\displaystyle {\mathcal {G}}_{n}^{d}}$ approaches infinity for fixed ${\displaystyle d}$ and increasing ${\displaystyle n}$, this theorem implies an earlier theorem of Alon and Boppana[12] which states

${\displaystyle \lim _{n\to \infty }\inf _{G\in {\mathcal {G}}_{n}^{d}}\lambda (G)\geq 2{\sqrt {d-1}}.}$

A slightly stronger bound is

${\displaystyle \lambda _{2}(G)\geq 2{\sqrt {d-1}}\cdot \left(1-{\frac {c}{m^{2}}}\right),}$

where ${\displaystyle c\approx 2\pi ^{2}}$. The outline of the proof is the following. Take ${\displaystyle k=\left\lfloor {\frac {m}{2}}\right\rfloor -1}$. Let ${\displaystyle T_{d,k}}$ be the complete ${\displaystyle (d-1)}$-ary tree of height ${\displaystyle k}$ (each internal vertex has ${\displaystyle d-1}$ children), and let ${\displaystyle A_{d,k}}$ be its adjacency matrix. We want to prove that ${\displaystyle \lambda _{2}(G)\geq \lambda _{1}(T_{d,k})=2{\sqrt {d-1}}\cos \theta }$, where ${\displaystyle \theta ={\frac {\pi }{k+2}}}$. Define a function ${\displaystyle g:V(T_{d,k})\rightarrow \mathbb {R} }$ by ${\displaystyle g(x)=(d-1)^{-\delta /2}\cdot \sin((k+1-\delta )\theta )}$, where ${\displaystyle \delta }$ is the distance from ${\displaystyle x}$ to the root of ${\displaystyle T_{d,k}}$. One can verify that ${\displaystyle A_{t,k}g=\lambda _{1}(T_{d,k})g}$ and that ${\displaystyle \lambda _{1}(T_{d,k})}$ is indeed the largest eigenvalue of ${\displaystyle A_{d,k}}$. Now let ${\displaystyle s}$ and ${\displaystyle t}$ be a pair of vertices at distance ${\displaystyle m}$ in ${\displaystyle G}$ and define

${\displaystyle f(v)={\begin{cases}c_{1}g(v_{s})&{\text{for }}v{\text{ at distance }}\leq k{\text{ from }}s,\\-c_{2}g(v_{t})&{\text{for }}v{\text{ at distance }}\leq k{\text{ from }}t,\\0&{\text{otherwise,}}\end{cases}}}$

where ${\displaystyle v_{s}}$ is a vertex in ${\displaystyle T_{d,k}}$ which distance to the root is equal to the distance from ${\displaystyle v}$ to ${\displaystyle s}$ and the symmetric for ${\displaystyle v_{t}}$. (One can think of this as "embedding" two disjoint copies of ${\displaystyle T_{d,k}}$, with some vertices collapsed into one.) By choosing the value of positive reals ${\displaystyle c_{1},c_{2}}$ properly we get ${\displaystyle f\perp 1}$, ${\displaystyle (Af)_{v}\geq \lambda _{1}(T_{d,k})f_{v}}$ for ${\displaystyle v}$ close to ${\displaystyle s}$ and ${\displaystyle (Af)_{v}\leq \lambda _{1}(T_{d,k})f_{v}}$ for ${\displaystyle v}$ close to ${\displaystyle t}$. Then by the min-max theorem we get

$${\displaystyle \lambda _{2}(G)\geq fAf^{T}/\|f\|^{2}\geq \lambda _{1}(T_{d,k})\approx 2{\sqrt {d-1}}\left(1-{\frac {1}{2}}\left({\frac {2\pi }{m}}\right)^{2}\right),}$$

as desired.