# Expanders, Property Testing and the PCP theorem

## Lecture 9: Zig-Zag Product

Posted in lectures by HQN on October 9, 2008

We talked about the zig-zag product today. Let $G, H$ be an $(n,m,\alpha)$-spectral expander and an $(m,d,\beta)$-spectral expander, respectively. Let’s use $\star$ to denote the zig-zag product. The graph $G\star H$ is an $(mn, d^2, f(\alpha,\beta))$-spectral expander where $f(\alpha,\beta)<1$ if $\alpha<1$ and $\beta<1$. Moreover, $f(\alpha,\beta) \leq \alpha+\beta+\beta^2$. For better upper-bounds on $f(\alpha,\beta)$, see the original zig-zag product paper. For a recent new construction of expanders, see this paper. Perhaps one of you can present this paper.

For each vertex $u \in V(G)$, label all edges incident to $u$ arbitrarily by $e_u^1, \dots, e_u^m$.  The product is defined as follows:

• The graph $G\star H$ has vertex set $V = \{(u,i) \ | \ u\in V(G), i \in [m]=V(H)\}$.
• There is an edge from $(u,i)$ to $(v,j)$ iff there are $k, l \in [m]$ such that $ik \in E(H)$, $e_u^k = e_v^l$, and $lj \in E{H}$. It is not difficult to see that the product has $mn$ vertices and degree $d^2$.

Let $\mathbf{\hat A}, \mathbf{\hat B}$ be the normalized adjacency matrices of $G, H$, respectively. Define an $mn \times mn$ matrix $\mathbf P$ where $p_{(u,k),(v,l)} = 1$ iff $e_u^k = e_v^l$. Then, $\mathbf P$ is a symmetric permutation matrix. Now, let $\mathbf{\tilde B} = \mathbf I_n \otimes \mathbf{\hat B}$ (where $\otimes$ denote the tensor product of two matrices). Then, it follows that $\mathbf M = \mathbf{\tilde B P \tilde B}$ is the normalized adjacency matrix of the product graph $G\star H$. So,

$f(\alpha,\beta) = \max_{\mathbf{x \perp u}} \frac{|\langle \mathbf{\tilde B P \tilde B x, x} \rangle|}{\langle x, x \rangle}$

Fix a vector $\mathbf{x \perp u} \in \mathbb R^{mn}$. Define a “collection vector” $C(\mathbf x) \in \mathbb R^n$ as follows: $C(\mathbf x)_u = \frac 1 m \sum_{i\in [m]} x_{(u,i)}$. So the $u$th entry of $C(\mathbf x)$ is just the average over all entries $x_{(u,i)}$ in the $u$th “cloud” of the vector $\mathbf x$. Also define

$\mathbf x^{\|} := C(\mathbf x) \otimes \mathbf 1_m$ where $\mathbf 1_m \in \mathbb R^m$ is the all-1 vector

$\mathbf x^\perp = \mathbf x - \mathbf x^\|$

Fix a vector $\mathbf x \perp \mathbf u$, we want to derive something like $|\langle \mathbf{\tilde B P \tilde B x, x} \rangle| \leq f(\alpha, \beta) \langle \mathbf{x,x} \rangle$. So, let’s start with the left hand side. Note that, since $\mathbf x^\|$ is uniform on each “cloud”, we have $\mathbf{\tilde Bx^\| = x^\|}$.

$|\langle \mathbf{\tilde B P \tilde B (x^\|+x^\perp), (x^\|+x^\perp)} \rangle| \leq |\langle \mathbf{Px^\|, x^\|} \rangle| + 2|\langle \mathbf{P\tilde B x^\perp, \tilde B x^\|} \rangle| + \|\mathbf{\tilde Bx^\perp} \|^2$

Consider the last term. Since $\mathbf x^\perp$ is anti-uniform on each “cloud” (i.e. the sum of $\mathbf x^\perp$‘s coordinates on each cloud is zero), implying that $\|\mathbf{\tilde Bx^\perp}\| \leq \beta \|\mathbf x^\perp\|$.

We now bound the second term. Since $\mathbf P$ is a permutation matrix, it does not changes a vector’s norm. We have

$2|\langle \mathbf{P\tilde B x^\perp, \tilde B x^\|} \rangle| \leq 2\| \mathbf{P\tilde Bx^\perp}\| \|\mathbf{\tilde B x^\|}\| \leq 2\beta \|x^\perp\| \|x^\|\|$

Finally, we bound the first term. Relatively straightforward computation leads to

$\langle \mathbf{Px^\|, x^\|} \rangle = m \langle \mathbf{\hat A} C(\mathbf x), C(\mathbf x)\rangle \leq m\alpha \|C(\mathbf x)\|^2 = \alpha \|\mathbf x^\|\|^2$.

Consequently,

$|\langle \mathbf{\tilde B P \tilde B x, x} \rangle| \leq \alpha \|\mathbf x^\|\|^2 + 2\beta \|x^\perp\| \|x^\|\| + \beta^2 \|\mathbf x^\perp\|^2$

Noting that $\|\mathbf x\|^2 = \|\mathbf x^\|\|^2 + \|\mathbf x^\perp\|^2$. It is easy to show that

$|\langle \mathbf{\tilde B P \tilde B x, x} \rangle| \leq \max(\alpha+\beta, \beta+\beta^2)\|\mathbf x\|^2 \leq (\alpha+\beta+\beta^2)\|\mathbf x\|^2$.

Now, suppose we start from a $(d^4,d,1/5)$-spectral expander $H$. Let $G_1=H^2$ and $G_n = G_{n-1}^2 \star H$. It is not difficult to prove by induction that $G_n$ is a $(d^{4n}, d^2, 2/5)$-spectral expander.

The zig-zag construction can also be made strongly explicit.