Expander mixing lemma

The expander mixing lemma intuitively states that the edges of certain $$d$$-regular graphs are evenly distributed throughout the graph. In particular, the number of edges between two vertex subsets $$S$$ and $$T$$ is always close to the expected number of edges between them in a random $$d$$-regular graph, namely $$\frac dn|S||T|$$.

d-Regular Expander Graphs
Define an $$(n, d, \lambda)$$-graph to be a $$d$$-regular graph $$G$$ on $$n$$ vertices such that all of the eigenvalues of its adjacency matrix $$A_G$$ except one have absolute value at most $$\lambda.$$ The $$d$$-regularity of the graph guarantees that its largest absolute value of an eigenvalue is $$d.$$ In fact, the all-1's vector $$\mathbf1$$ is an eigenvector of $$A_G$$ with eigenvalue $$d$$, and the eigenvalues of the adjacency matrix will never exceed the maximum degree of $$G$$ in absolute value.

If we fix $$d$$ and $$\lambda$$ then $$(n, d, \lambda)$$-graphs form a family of expander graphs with a constant spectral gap.

Statement
Let $$G = (V, E)$$ be an $$(n, d, \lambda)$$-graph. For any two subsets $$S, T \subseteq V$$, let $$e(S, T) = |\{(x,y) \in S \times T : xy \in E(G)\}|$$ be the number of edges between S and T (counting edges contained in the intersection of S and T twice). Then


 * $$\left|e(S, T) - \frac{d |S| |T|}{n}\right| \leq \lambda \sqrt{|S| |T| }\,.$$

Tighter Bound
We can in fact show that


 * $$\left|e(S, T) - \frac{d |S| |T|}{n}\right| \leq \lambda \sqrt{|S| |T| (1 - |S|/n)(1 - |T|/n)}\,$$

using similar techniques.

Biregular Graphs
For biregular graphs, we have the following variation, where we take $$\lambda $$ to be the second largest eigenvalue.

Let $$G = (L, R, E)$$ be a bipartite graph such that every vertex in $$L$$ is adjacent to $$d_L$$ vertices of $$R$$ and every vertex in $$R$$ is adjacent to $$d_R$$ vertices of $$L$$. Let $$S \subseteq L, T \subseteq R$$ with $$|S| = \alpha|L|$$ and $$|T| = \beta |R|$$. Let $$e(G) = |E(G)|$$. Then
 * $$\left|\frac{e(S, T)}{e(G)} - \alpha \beta\right| \leq \frac{\lambda}{\sqrt{d_Ld_R}} \sqrt{\alpha \beta (1 - \alpha) (1 - \beta)} \leq \frac{\lambda}{\sqrt{d_Ld_R}} \sqrt{\alpha \beta}\,.$$

Note that $$\sqrt{d_L d_R}$$ is the largest eigenvalue of $$G$$.

Proof of First Statement
Let $$A_G$$ be the adjacency matrix of $$G$$ and let $$\lambda_1\geq\cdots\geq\lambda_n$$ be the eigenvalues of $$A_G$$ (these eigenvalues are real because $$A_G$$ is symmetric). We know that $$\lambda_1=d$$ with corresponding eigenvector $$v_1=\frac 1{\sqrt n}\mathbf{1}$$, the normalization of the all-1's vector. Define $$\lambda=\sqrt{\max\{\lambda_2^2,\dots,\lambda_n^2\}}$$ and note that $$\max\{\lambda_2^2,\dots,\lambda_n^2\}\leq\lambda^2\leq\lambda_1^2=d^2$$. Because $$A_G$$ is symmetric, we can pick eigenvectors $$v_2,\ldots,v_n$$ of $$A_G$$ corresponding to eigenvalues $$\lambda_2,\ldots,\lambda_n$$ so that $$\{v_1,\ldots,v_n\}$$ forms an orthonormal basis of $$\mathbf R^n$$.

Let $$J$$ be the $$n\times n$$ matrix of all 1's. Note that $$v_1$$ is an eigenvector of $$J$$ with eigenvalue $$n$$ and each other $$v_i$$, being perpendicular to $$v_1=\mathbf{1}$$, is an eigenvector of $$J$$ with eigenvalue 0. For a vertex subset $$U \subseteq V$$, let $$1_U$$ be the column vector with $$v^\text{th}$$ coordinate equal to 1 if $$v\in U$$ and 0 otherwise. Then,


 * $$\left|e(S,T)-\frac dn|S||T|\right|=\left|1_S^\operatorname{T}\left(A_G-\frac dnJ\right)1_T\right|$$.

Let $$M=A_G-\frac dnJ$$. Because $$A_G$$ and $$J$$ share eigenvectors, the eigenvalues of $$M$$ are $$0,\lambda_2,\ldots,\lambda_n$$. By the Cauchy-Schwarz inequality, we have that $$|1_S^\operatorname{T}M1_T|=|1_S\cdot M1_T|\leq\|1_S\|\|M1_T\|$$. Furthermore, because $$M$$ is self-adjoint, we can write


 * $$\|M1_T\|^2=\langle M1_T,M1_T\rangle=\langle 1_T,M^21_T\rangle=\left\langle 1_T,\sum_{i=1}^nM^2\langle 1_T,v_i\rangle v_i\right\rangle=\sum_{i=2}^n\lambda_i^2\langle 1_T,v_i\rangle^2\leq\lambda^2\|1_T\|^2$$.

This implies that $$\|M1_T\|\leq\lambda\|1_T\|$$ and $$\left|e(S,T)-\frac dn|S||T|\right|\leq\lambda\|1_S\|\|1_T\|=\lambda\sqrt{|S||T|}$$.

Proof Sketch of Tighter Bound
To show the tighter bound above, we instead consider the vectors $$1_S-\frac{|S|}n\mathbf 1$$ and $$1_T-\frac{|T|}n\mathbf 1$$, which are both perpendicular to $$v_1$$. We can expand


 * $$1_S^\operatorname{T}A_G1_T=\left(\frac{|S|}n\mathbf 1\right)^\operatorname{T}A_G\left(\frac{|T|}n\mathbf 1\right)+\left(1_S-\frac{|S|}n\mathbf 1\right)^\operatorname{T}A_G\left(1_T-\frac{|T|}n\mathbf 1\right)$$

because the other two terms of the expansion are zero. The first term is equal to $$\frac{|S||T|}{n^2}\mathbf 1^\operatorname{T}A_G\mathbf 1=\frac dn|S||T|$$, so we find that


 * $$\left|e(S,T)-\frac dn|S||T|\right|

\leq\left|\left(1_S-\frac{|S|}n\mathbf 1\right)^\operatorname{T}A_G\left(1_T-\frac{|T|}n\mathbf 1\right)\right|$$

We can bound the right hand side by $$\lambda\left\|1_S-\frac{|S|}{|n|}\mathbf 1\right\|\left\|1_T-\frac{|T|}{|n|}\mathbf 1\right\| =\lambda\sqrt{|S||T|\left(1-\frac{|S|}n\right)\left(1-\frac{|T|}n\right)}$$ using the same methods as in the earlier proof.

Applications
The expander mixing lemma can be used to upper bound the size of an independent set within a graph. In particular, the size of an independent set in an $$(n, d, \lambda)$$-graph is at most $$\lambda n/d.$$ This is proved by letting $$T=S$$ in the statement above and using the fact that $$e(S,S)=0.$$

An additional consequence is that, if $$G$$ is an $$(n, d, \lambda)$$-graph, then its chromatic number $$\chi(G)$$ is at least $$d/\lambda.$$ This is because, in a valid graph coloring, the set of vertices of a given color is an independent set. By the above fact, each independent set has size at most $$\lambda n/d,$$ so at least $$d/\lambda$$ such sets are needed to cover all of the vertices.

A second application of the expander mixing lemma is to provide an upper bound on the maximum possible size of an independent set within a polarity graph. Given a finite projective plane $$\pi$$ with a polarity $$\perp,$$ the polarity graph is a graph where the vertices are the points a of $$\pi$$, and vertices $$x$$ and $$y$$ are connected if and only if $$x\in y^{\perp}.$$ In particular, if $$\pi$$ has order $$q,$$ then the expander mixing lemma can show that an independent set in the polarity graph can have size at most $$q^{3/2} - q + 2q^{1/2} - 1,$$ a bound proved by Hobart and Williford.

Converse
Bilu and Linial showed that a converse holds as well: if a $$d$$-regular graph $$G = (V, E)$$ satisfies that for any two subsets $$S, T \subseteq V$$ with $$S \cap T = \empty $$ we have


 * $$\left|e(S, T) - \frac{d |S| |T|}{n}\right| \leq \lambda \sqrt{|S| |T|},$$

then its second-largest (in absolute value) eigenvalue is bounded by $$O(\lambda (1+\log(d/\lambda)))$$.

Generalization to hypergraphs
Friedman and Widgerson proved the following generalization of the mixing lemma to hypergraphs.

Let $$H$$ be a $$k$$-uniform hypergraph, i.e. a hypergraph in which every "edge" is a tuple of $$k$$ vertices. For any choice of subsets $$V_1, ..., V_k$$ of vertices,


 * $$\left| |e(V_1,...,V_k)| - \frac{k!|E(H)|}{n^k}|V_1|...|V_k| \right| \le \lambda_2(H)\sqrt{|V_1|...|V_k|}.$$