In which we define and analyze the zig-zag graph product.

1. Replacement Product and Zig-Zag Product

In the previous lecture, we claimed it is possible to “combine” a ${d}$-regular graph on ${D}$ vertices and a ${D}$-regular graph on ${N}$ vertices to obtain a ${d^2}$-regular graph on ${ND}$ vertices which is a good expander if the two starting graphs are. Let the two starting graphs be denoted by ${H}$ and ${G}$ respectively. Then, the resulting graph, called the zig-zag product of the two graphs is denoted by ${G {\textcircled z} H}$.

Using ${\lambda(G)}$ to denote the eigenvalue with the second-largest absolute value for a graph ${G}$, we claimed that if ${\lambda(H) \leq b}$ and ${\lambda(G) \leq a}$, then ${\lambda(G {\textcircled z} H) \leq a+ 2b + b^2}$. In this lecture we shall describe the construction for the zig-zag product and prove this claim.

2. Replacement product of two graphs

We first describe a simpler product for a “small” ${d}$-regular graph on ${D}$ vertices (denoted by ${H}$) and a “large” ${D}$-regular graph on ${N}$ vertices (denoted by ${G}$). Assume that for each vertex of ${G}$, there is some ordering on its ${D}$ neighbors. Then we construct the replacement product (see figure) ${G {\textcircled r} H}$ as follows:

• Replace each vertex of ${G}$ with a copy of ${H}$ (henceforth called a cloud). For ${v \in V(G), i \in V(H)}$, let ${(v,i)}$ denote the ${i^{th}}$ vertex in the ${v^{th}}$ cloud.
• Let ${(u, v) \in E(G)}$ be such that ${v}$ is the ${i}$-th neighbor of ${u}$ and ${u}$ is the ${j}$-th neighbor of ${v}$. Then ${((u,i),(v,j)) \in E(G {\textcircled r} H)}$. Also, if ${(i,j) \in E(H)}$, then ${\forall u \in V(G)~ ((u,i),(u,j)) \in E(G {\textcircled r} H)}$.

Note that the replacement product constructed as above has ${ND}$ vertices and is ${(d+1)}$-regular.

3. Zig-zag product of two graphs

Given two graphs ${G}$ and ${H}$ as above, the zig-zag product ${G{\textcircled z} H}$ is constructed as follows (see figure):

• The vertex set ${V(G {\textcircled z} H)}$ is the same as in the case of the replacement product.
• ${((u,i),(v,j)) \in E(G {\textcircled z} H)}$ if there exist ${\ell}$ and ${k}$ such that ${((u,i)(u,\ell)}$, ${((u,\ell),(v,k))}$ and ${((v,k),(v,j))}$ are in ${E(G {\textcircled r} H)}$ i.e. ${(v,j)}$ can be reached from ${(u,i)}$ by taking a step in the first cloud, then a step between the clouds and then a step in the second cloud (hence the name!).

It is easy to see that the zig-zag product is a ${d^2}$-regular graph on ${ND}$ vertices.

Let ${M \in {\mathbb R}^{([N] \times [D]) \times ([N] \times [D])}}$ be the normalized adjacency matrix of ${G {\textcircled z} H}$. Using the fact that each edge in ${G {\textcircled r} H}$ is made up of three steps in ${G {\textcircled r} H}$, we can write ${M}$ as ${BAB}$, where

$\displaystyle \begin{array}{rcl} B[(u,i),(v,j)] = \left\{ \begin{array}{ll} 0 & if~u\neq v \\ M_H[i,j] & if~u=v \\ \end{array} \right. \end{array}$

And ${A[(u,i),(v,j)]=1}$ if ${u}$ is the ${j}$-th neighbor of ${v}$ and ${v}$ is the ${i}$-th neighbor of ${u}$, and ${A[(u,i),(v,j)]=0}$ otherwise.

Note that ${A}$ is the adjacency matrix for a matching and is hence a permutation matrix.

4. Preliminaries on Matrix Norms

Recall that, instead of bounding ${\lambda_2}$, we will bound the following parameter (thus proving a stronger result).

Definition 1 Let ${M}$ be the normalized adjacency matrix of a graph ${G=(V,E)}$, and ${\lambda_1 \geq \ldots \geq \lambda_n}$ be its eigenvalues with multiplicities. Then we use the notation

$\displaystyle \lambda(M) := \max_{i=2,\ldots,n} \{ |\lambda_i| \} = \max \{ \lambda_2,-\lambda_n \}$

The parameter ${\lambda}$ has the following equivalent characterizations.

Fact 2

$\displaystyle \lambda(M) = \max_{{\bf x} \in {\mathbb R}^V - \{ {\bf {0}} \} , {\bf x} \perp {\bf 1}} \frac {|| M x||}{||x||} = \max_{{\bf x} \in {\mathbb R}^v ,{\bf x} \perp {\bf 1}, ||{\bf x}||=1} || M x ||$

Another equivalent characterization, which will be useful in several contexts, can be given using the following matrix norm.

Definition 3 (Spectral Norm) The spectral norm of a matrix ${M\in {\mathbb R}^{n \times n}}$ is defined as

$\displaystyle ||M|| = \max_{{\bf x}\in {\mathbb R}^V, || {\bf x} ||=1} || M x ||$

If ${M}$ is symmetric with eigenvalues ${\lambda_1,\ldots,\lambda_n}$, then the spectral norm is ${\max_i |\lambda_i|}$. Note that ${M}$ is indeed a norm, that is, for every two square real matrices ${A,B}$ we have ${||A+B||\leq ||A || + ||B||}$ and for every matrix ${A}$ and scalar ${\alpha}$ we have ${||\alpha A|| = \alpha ||A||}$. In addition, it has the following useful property:

Fact 4 For every two matrices ${A,B \in {\mathbb R}^{n\times n}}$ we have

$\displaystyle || A B || \leq ||A|| \cdot ||B||$

Proof: For every vector ${x}$ we have

$\displaystyle ||AB {\bf x}|| \leq ||A|| \cdot ||B {\bf x}|| \leq ||A|| \cdot ||B|| \cdot ||{\bf x}||$

where the first inequality is due to the fact that ${||Az|| \leq ||A|| \cdot ||{\bf z}||}$ for every vector ${{\bf z}}$, and the second inequality is due to the fact that ${||B{\bf x}|| \leq ||B|| \cdot ||{\bf x}||}$. So we have

$\displaystyle \min_{{\bf x}\in {\mathbb R}^n , {\bf x}\neq {\bf {0}}}$

$\Box$

We can use the spectral norm to provide another characterization of the parameter ${\lambda(M)}$ of the normalized adjacency matrix of a graph.

Lemma 5 Let ${G}$ be a regular graph and ${M\in {\mathbb R}^{n \times n}}$ be its normalized adjacency matrix. Then

$\displaystyle \lambda(M) = || M - \frac 1n J||$

where ${J}$ is the matrix with a 1 in each entry.

Proof: Let ${\lambda_1=1 \geq \lambda_2 \geq \cdots \lambda_n}$ be the eigenvalues of ${M}$ and ${{\bf v}_1 = \frac 1 {\sqrt n} {\bf 1}}$, ${{\bf v}_2}$, ${\ldots}$, ${{\bf v}_n}$ a corresponding system of orthonormal eigenvector. Then we can write

$\displaystyle M = \lambda_1 {\bf v}_1 {\bf v}_1^T + \cdots + \lambda_n {\bf v}_n {\bf v}_n^T$

Noting that ${{\bf v}_1{\bf v}_1^T = \frac 1n J}$, we have

$\displaystyle M -\frac 1n J = 0 \cdot {\bf v}_1{\bf v}_2^T + \sum_{i=2}^n \lambda_i {\bf v}_i {\bf v}_i^T$

and so ${{\bf v}_1,\ldots,{\bf v}_n}$ is also a system of eigenvectors for ${M-\frac 1n J}$, with corresponding eigenvalues ${0,\lambda_2,\ldots,\lambda_n}$, meaning that

$\displaystyle || M - \frac 1n J|| = \max \{ 0,\lambda_2,\ldots,\lambda_n \} = \lambda(M)$

$\Box$

The above lemma has several applications. It states that, according to a certain definition of distance, when a graph is a good expander then it is close to a clique. (The matrix ${\frac 1n J}$ is the normalized adjacency matrix of a clique with self-loops.) The proof of several results about expanders is based on noticing that the result is trivial for cliques, and then on “approximating” the given expander by a clique using the above lemma.

We need one more definition before we can continue with the analysis of the zig-zag graph product.

Definition 6 (Tensor Product) Let ${A \in {\mathbb R}^{N\times N}}$ and ${B\in {\mathbb R}^{D \times D}}$ be two matrices. Then ${A \otimes B \in {\mathbb R}^{ND \times ND}}$ is a matrix whose rows and columns are indexed by pairs ${(u,i)\in [N] \times [D]}$ such that

$\displaystyle (A\otimes B)_{(u,i),(v,j)} = A_{u,v} \cdot B_{i,j}$

For example ${I \otimes M}$ is a block-diagonal matrix in which every block is a copy of ${M}$.

5. Analysis of the Zig-Zag Product

Suppose that ${G}$ and ${H}$ are identical cliques with self-loops, that is, are both ${n}$-regular graphs with self-loops. Then the zig-zag product of ${G}$ and ${H}$ is well-defined, because the degree of ${G}$ is equal to the number of vertices of ${H}$. The resulting graph ${G{\textcircled z} H}$ is a ${n^2}$-regular graph with ${n^2}$ vertices, and an inspection of the definitions reveals that ${G{\textcircled z} H}$ is indeed a clique (with self-loops) with ${n^2}$ vertices.

The intuition for our analysis is that we want to show that the zig-zag graph product “preserves” distances measured in the matrix norm, and so if ${G}$ is close (in matrix norm) to a clique and ${H}$ is close to a clique, then ${G{\textcircled z} H}$ is close to the zig-zag product of two cliques, that is, to a clique. (Strictly speaking, what we just said does not make sense, because we cannot take the zig-zag product of the clique that ${G}$ is close to and of the clique that ${H}$ is close to, because they do not have the right degree and number of vertices. The proof, however, follows quite closely this intuition.)

Theorem 7 If ${\lambda(M_G) = a}$ and ${\lambda(M_H) = b}$, then

$\displaystyle \lambda( G {\textcircled z} H ) \leq a + 2b + b^2$

Proof: Let ${M}$ be the normalized adjacency matrix of ${G{\textcircled z} H}$, and let ${{\bf x}}$ be a unit vector such that ${{\bf x} \perp {\bf 1}}$ and

$\displaystyle \lambda(M) = || M {\bf x} ||$

Recall that we defined a decomposition

$\displaystyle M = BAB$

where ${A}$ is a permutation matrix, and ${B = I \otimes M_H}$. Let us write ${E:= M_H - \frac 1D J}$, then ${B = I \otimes \frac 1D J + I \otimes E}$. Let us call ${\bar J : = I \otimes \frac 1D J}$ and ${\bar E:= I \otimes E}$.

First, we argue that the matrix norm of ${\bar E}$ is small. Take any vector ${{\bf z} \in {\mathbb R}^{ND}}$ and write is as ${{\bf z} = ({\bf z}_1,\ldots,{\bf z}_N)}$, where, for each ${u\in [N]}$, ${{\bf z}_u}$ is the ${D}$-dimensional restriction of ${{\bf z}}$ to the coordinates in the cloud of ${u}$. Then

$\displaystyle ||( I \otimes E ){\bf z} ||^2 = \sum_u || E {\bf z}_u||^2 \leq \sum_u ||E||^2 \cdot ||{\bf z}_u ||^2 = ||E||^2 \cdot ||{\bf z}||^2$

and so we have

$\displaystyle || I \otimes E|| \leq ||E|| \leq b$

Then we have

$\displaystyle BAB = ( \bar J + \bar E) A (\bar J + \bar E )$

$\displaystyle = \bar J A \bar J + \bar J A \bar E + \bar E A \bar J + \bar E A \bar A$

and so, using the triangle inequality and the property of the matrix norm, we have

$\displaystyle ||BAB {\bf x}|| \leq ||\bar J A \bar J {\bf x} || + || \bar E A \bar J || + || \bar J A \bar E|| + || \bar E A \bar E||$

where

$\displaystyle || \bar E A \bar J || \leq || \bar E|| \cdot || A || \cdot || \bar J|| \leq ||\bar E|| \leq b$

$\displaystyle || \bar J A \bar E || \leq || \bar J|| \cdot || A || \cdot || \bar E|| \leq ||\bar E|| \leq b$

$\displaystyle || \bar E A \bar E || \leq || \bar E|| \cdot || A || \cdot || \bar E|| \leq ||\bar E||^2 \leq b^2$

It remains to prove that ${||\bar J A \bar J {\bf x} || \leq a}$. If we let ${A_G = D M_G}$ be the adjacency matrix of ${G}$, then we can see that

$\displaystyle ( \bar J A \bar J ) _{(u,i),(v,j)} = \frac {1}{D^2} (A_G)_{u,v} = \frac 1D (M_G)_{u,v} = ( M_G \otimes \frac 1D J)_{(u,i),(v,j) }$

That is,

$\displaystyle \bar J A \bar J = M_G \otimes \frac 1D J$

Finally, we write ${{\bf x} = ({\bf x}_1,\ldots,{\bf x}_N)}$, where ${{\bf x}_u}$ is the ${D}$-dimensional vector of entries corresponding to the cloud of ${u}$, we call ${y_u := \sum_i {\bf x}_u(i)/D}$, and we note that, by Cauchy-Schwarz:

$\displaystyle || {\bf y}||^2 = \sum_u \left( \sum_i \frac 1D {\bf x}_{u,i} \right)^2 \leq \sum_u \left( \sum_i \frac 1D^2 \right) \cdot \left ( \sum_i {\bf x}_{u,i}^2 \right) = \frac 1D || {\bf x}||^2$

The final calculation is:

$\displaystyle || \bar J A \bar J {\bf x} ||^2 = \left|| \left ( M_G \otimes \frac 1D J \right) {\bf x} \right||^2$

$\displaystyle = \sum_{u,i} \left( \sum_{v,j} \frac 1D (M_G)_{u,v} {\bf x}_{u,i} \right )^2$

$\displaystyle = \sum_{u,i} \left( \sum_{v} (M_G)_{u,v} y_u \right)^2$

$\displaystyle = D \cdot \sum_u \left( \sum_{v} (M_G)_{u,v} y_u \right)^2$

$\displaystyle = D \cdot || M_G {\bf y} ||^2$

$\displaystyle \leq D \cdot a^2 \cdot || {\bf y}||^2$

$\displaystyle \leq a^2 \cdot ||{\bf x}^2 ||^2$

$\Box$