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 uth entry of C(\mathbf x) is just the average over all entries x_{(u,i)} in the uth “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.


|\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.

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