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Eigenvalue gaps allow us to upper bound the rate of convergence of random walks on graphs.
In this note, we will look at the connection between the eigenvalue gap and a combinatorial graph parameter, called “expansion”"
Let \(G\) be a \(d\)-regular graph with \(n\) vertices. We identify \(G\) with its random-walk matrix (normalized adjacency matrix).
Definition: The eigenvalue gap of \(G\) is defined as \[
\gamma(G)=\min_{f\bot u} \frac{\langle f,L_G f\rangle }{\|f\|_2^2},
\] where \(L_G=I-G\) is the Laplacian of \(G\).
Exercise: Check that \(\gamma(G)=\lambda_1(G)-\lambda_2(G)\) is indeed the gap between the largest and second largest eigenvalue of \(G\).
We would like to characterize the eigenvalue gap of a graph in terms of a combinatorial property. The right combinatorial property turns out to be expansion, which measures how well \(G\) is connected.
Definition: The expansion of \(G\) is defined as \[
\phi(G)
= \min_{S\subseteq V,~|S|\le n/2}
\frac{\tfrac1d |E(S,\bar S)|}{|S|}.
\] (Here, \(E(S,\bar S)\) is the set of edges from \(S\) to \(\bar S\).) The quantity \(\phi(S)=\tfrac1d|E(S,\bar S)|/|S|\) is the expansion of \(S\).
Exercise: Check that \[
{\langle f,L_G f\rangle}
= \sum_i \E_{j\sim i}\tfrac12(f_i-f_j)^2.
\] (So the Laplacian exactly measures how close the values of \(f\) are for a typical edge of the graph.)
It follows that \({\langle 1_S,L_G 1_S\rangle}=\tfrac1d |E(S,\bar S)|\) and therefore \[\phi(G) = \min_{S\subseteq V,~|S|\le n/2} {\langle 1_S,L_G 1_S\rangle}/{\|1_S\|_2^2}.\]
Cheeger’s inequality asserts that \(\gamma(G)\) is the same as \(\phi(G)\) up to quadratic factor.
Theorem: (Cheeger’s inequality) \[
\gamma(G)/2 \le \phi(G) \le 2\sqrt{\gamma(G)}.
\]
The following direction is the easier one.
Lemma: \(\phi(G)\ge \gamma(G)/2\)
Proof: Let \(S\) be a subset of vertices with \(|S|\le n/2\). We like to lower bound its expansion. Let \(p=1_S/|S|\) be the uniform distribution over \(S\). Then \(p=u+f\), where \(u\) is the uniform distribution and \(f\) is orthogonal to \(u\). Then, \[
\phi(S) = \frac{\langle p,L_G p\rangle}{\|p\|_2^2}
%\frac{\langle u,L_G u\rangle+\langle f,L_G f\rangle}{\|u\|_2^2+\|f\|_2^2}
=\frac{\langle f,L_G f\rangle}{\|u\|_2^2+\|f\|_2^2}
\ge \frac{\gamma\|f\|_2^2}{\|u\|_2^2+\|f\|_2^2}.
\]
To finish the proof, it is enough to show that \(\|f\|_2^2 \ge \|u\|_2^2\). (This means that the distribution \(p\) is “far” from \(u\).) Recall that \(f=1_S/|S|-u\). It follows that the distance of \(1_S/|S|\) and \(u\) decreases as the size of \(S\) grows. Hence, the distance is smallest if \(|S|=n/2\). In this case, \(f\) takes values \(\pm 1/n\). Hence, \(\|f\|_2^2=n\cdot 1/n^2=1/n=\|u\|_2^2\).
Next, we will prove the other direction of Cheeger’s inequality.
The first step is the following lemma (which we have seen in class).
Lemma: Suppose \(h\) is a function with \(|\mathrm{supp}(h)|\le n/2\) and \(\langle h,L_G h\rangle\le \epsilon\|h\|_2^2\). Then there exists a set \(S\) with \(|S|\le n/2\) and \(\phi(S)\le \sqrt{2\epsilon}\).
W.l.o.g. we may assume \(0\le h\le 1\). The main idea is to construct a distribution over sets \(S\) by choosing a random threshold \(\tau\in[0,1]\) and setting \(S=\{i \mid h_i^2>\tau\}\). It is easy to verify that \(\E_S |S|=\|h\|_2^2\). Using Cauchy-Schwarz, one can also show \(\E_S \tfrac1d|E(S,\bar S)|\le \sqrt{2\epsilon}\|h\|_2^2\). Hence, there exists a set in the support of the distribution with expansion at most \(\sqrt{2\epsilon}\).
The next step is to construct such a function \(h\).
Lemma: There exists a function \(h\) with \(|\mathrm{supp}(h)|\le n/2\) and \(\langle h,L_G h\rangle\le 2\gamma(G)\|h\|_2^2\).
Proof: Let \(f\) be a function such that \(f\bot u\) and \(\langle f,L_G f\rangle \le \gamma(G)\|f\|_2^2\). Let \(m\) be the median of this function. (So that \(f\) has value larger than \(m\) for exactly half of the vertices.) Let \(h_1=(f-m)_+\) be the positive part of \(f-m\) and let \(h_2=(f-m)_{-}\) be the negative part of \(f-m\). Notice that both \(h_1\) and \(h_2\) have exactly half of the vertices in their support. We claim that one of the functions \(h_1,h_2\) satisfies also the other condition of the lemma.
Indeed, we can see that \(\langle h_1,L_G h_1\rangle \le \langle f,L_G f\rangle\) and \(\langle h_2,L_G h_2\rangle\le \langle f,L_G f\rangle\). On the other hand, \(\|f\|_2^2=\|h_1\|_2^2+\|h_2\|^2\). Hence, we can choose \(h\) to be the function \(h_1\) or \(h_2\) with larger norm.