# Spectral gaps on simplicial complexes (Wed, Feb 22)

## Spectral expansion in graphs

• Consider an undirected graph $G=(V,E)$ and a probability distribution $\mu$ on $E$. Let $\mu^{(0)}$ be the probability measure induced on $V$ as follows: We sample a random edge $uv \in E$ according to $\mu$, and then choose one of the two endpoints uniformly at random. Equivalently, $\mu^{(0)}(u) = \frac12 \sum_{v : uv \in E} \mu(uv)$.

• Random walk. We can define a random walk $\{X_t\}$ on $V$ as follows. For $v \in V$, let $\mu_v$ denote the distribution $\mu$ conditioned on the event $\{ v \in E \}$, i.e.,

$\mu_v(e) = \frac{\mu(e) \mathsf{1}_e(v)}{\sum_{e' \in E} \mu(e') \mathsf{1}_{e'}(v)}\,.$

Now given $X_t = v$, we sample an edge $e \in E$ according to $\mu_v$ and let $X_{t+1} = e \setminus v$.

Define the associated random walk transition matrix by

$P_{uv} \seteq \frac{\mu(uv)}{\sum_{w : uw \in E} \mu(uw)} = \frac12 \frac{\mu(uv)}{\mu^{(0)}(u)}\,.$

We will think of $P$ as a linear operator on the space of functions $f : V \to \R$, where $P f(u) = \sum_{v \in V} P_{uv} f(v)$.

• Note that the matrix $P$ is not necessarily symmetric, i.e., it is not self-adjoint with respect to the standard inner product on $\R^{V}$. For $f,g : V \to \R$, define the inner product

$\langle f,g \rangle_{\mu^{(0)}} \seteq \sum_{u \in V} \mu^{(0)}(u) f(u) g(u)\,.$

Let us use $\ell_2(\mu^{(0)})$ to denote the Hilbert space of functions equipped with $\langle \cdot,\cdot\rangle_{\mu^{(0)}}$. Now let us verify that

$\langle f, P g\rangle_{\mu^{(0)}} = \sum_{u \in V} \mu^{(0)}(u) f(u) \sum_{v \in V} P_{uv} g(v) = \frac12 \sum_{u,v \in V} \mu(uv) f(u) g(v) = \sum_{uv \in E} \mu(uv) f(u) g(v)\,.$

Evidently, this is also equal to $\langle Pf, g\rangle_{\mu^{(0)}}$ so that $P$ is indeed self-adjoint with respect to $\langle \cdot,\cdot\rangle_{\mu^{(0)}}$. In particular, this implies that $P$ has a complete orthonormal basis of eigenfunctions and associated real eigenvalues.

• If we denote by $\mathbf{1}$ the constant function, then evidently $P \mathbf{1} = \mathbf{1}$, hence $P$ has $1$ as an eigenvalue. Moreover, $P$ is an averaging operator (it has nonnegative entries and is row-stochastic), so by Jensen’s inequality,

$(P f(u))^2 = \left(\sum_{v \in V} P_{uv} f(v)\right)^2 \leq \sum_{v \in P_{uv}} P_{uv} f(v)^2\,,$

and therefore

$\|P f\|_{\mu^{(0)}}^2 = \sum_{u \in V} \mu^{(0)}(u) (P f(u))^2 \leq \sum_{u \in V} \mu^{(0)}(u) \sum_{v \in V} P_{uv} f(v)^2 = \frac12 \sum_{u,v \in V} \mu(uv) f(u)^2 = \sum_{u \in V} \mu^{(0)}(u) f(u)^2 = \|f\|_{\mu^{(0)}}^2\,.$

This is usually summarized by saying: “$P$ is a contraction on $\ell_2(\mu^{(0)})$.” It implies that the spectrum of $P$ lies in $[-1,1]$.

• The spectral gap. Recall that $1$ is an eigenvalue of $P$ with eigenvector $\mathbf{1}$. The spectral gap of $P$ is the largest value $\varepsilon$ such that all the other eigenvalues of $P$ lie in the interval $[-(1-\e),1-\e]$.

Equivalently, this is the largest value $\e$ such that $\|P f\|_{\mu^{(0)}} \leq (1-\e) \|f\|_{\mu^{(0)}}$ for all $\langle f, \mathbf{1}\rangle_{\mu^{(0)}} = 0$.

Let us denote this value, the “spectral gap of $\mu$” by $\e(\mu)$

• The Laplacian. It is sometimes more convenient to work with the corresponding Laplacian operator $L \seteq I - P$. Note that

\begin{align} \langle f, L g\rangle_{\mu^{(0)}} &= \sum_{u \in V} \mu^{(0)}(u) g(u) \left(f(u) - \sum_{v \in V} P_{uv} g(v)\right) \nonumber \\ &= \frac12 \sum_{u,v \in V} \mu(uv) \left(f(u) g(u) - f(u) g(v)\right) \nonumber\\ &= \frac14 \sum_{u,v \in V} \mu(uv) \left(f(u) g(u) + f(v) g(v) - f(u) g(v) - f(v) g(u)\right) \nonumber\\ &= \frac14 \sum_{u,v \in V} \mu(uv) (f(u)-f(v)) (g(u)-g(v)) \nonumber\\ &= \frac12 \sum_{uv \in E} \mu(uv) \left(f(u)-f(v)\right) \left(g(u)-g(v)\right)\,. \label{eq:laplacian} \end{align}

The eigenvalue $1$ in $P$ corresponding to the constant function $\mathbf{1}$ becomes an eigenvalue $0$ in $L$. We have an equivalent characterization

$\e(\mu) = \min \left\{ \frac{\langle f, L f\rangle_{\mu^{(0)}}}{\langle f,f\rangle_{\mu^{(0)}}} : \langle f, \mathbf{1}\rangle_{\mu^{(0)}} = 0 \right\}.$
• Poincare inequalities. In order to eliminate the algebraic constraint $\langle f,\mathbf{1}\rangle_{\mu^{(0)}}=0$, we can simply project $f$ onto the orthogonal complement of the span of $\mathbf{1}$, i.e., we can replace $f$ by $f - \E_{\mu^{(0)}}[f]$.

Define

$\mathsf{Var}_{\mu^{(0)}}(f) \seteq \left\|f - \E_{\mu^{(0)}}[f]\right\|_{\mu^{(0)}}^2 = \E_{\mu^{(0)}} \left(f - \E_{\mu^{(0)}}[f]\right)^2 = \E_{\mu^{(0)}}(f^2) - (\E_{\mu^{(0)}} f)^2\,.$

Then we can write

$\e(\mu) = \min_f \frac{\langle f-\E_{\mu^{(0)}}[f], L(f-\E_{\mu^{(0)}}[f])\rangle_{\mu^{(0)}}}{\mathsf{Var}_{\mu^{(0)}}(f)} = \min_f \frac{\langle f,L f\rangle_{\mu^{(0)}}}{\mathsf{Var}_{\mu^{(0)}}(f)}\,,$

where the second equality uses the fact that $L$ annhilates constant functions.

This is often written in the form of a Poincare inequality: For all $f : V \to \R$, it holds that

$\mathsf{Var}_{\mu^{(0)}}(f) \leq K(\mu) \langle f,L f\rangle_{\mu^{(0)}}\,,$

where $K(\mu) \seteq 1/\e(\mu)$.

• Let $X$ be a simplicial complex. The dimension of $X$ is $\dim(X) \seteq \max \{ \abs{\sigma} - 1 : \sigma \in X \}$. Say that a $d$-dimensional complex $X$ is pure if every face of $X$ is contained in some face of dimension $d$.

Note that a pure simplicial complex is specified by a $d$-uniform hypergraph.

Write $X \seteq X^{(-1)} \cup X^{(0)} \cup \cdots \cup X^{(d)}$ where $X^{(i)}$ contains the faces of $X$ of dimension $i$. Note that $X^{(-1)} = \{ \emptyset \}$.

• Induced measures. Suppose now that $\pi^{(d)}$ is a probability measure on $X^{(d)}$. In analogy with the graph case, we can define probability measures $\pi^{(0)}, \pi^{(1)},\ldots, \pi^{(d-1)}$ as follows: To sample from $\pi^{(i)}$, choose $\sigma \in X^{(d)}$ according to $\pi^{(d)}$ and then remove $d-i$ uniformly random vertices from $\sigma$.

Equivalently, we can define the measures inductively by

$\pi^{(i-1)}(\tau) \seteq \sum_{\sigma \in X^{(i)} : \tau \subseteq \sigma} \frac{\pi^{(i)}(\sigma)}{i+1}\,.$
• Links. Recall that for $\sigma \in X$, the link of $\sigma$ in $X$ is defined by

$X_{\sigma} \seteq \{ \tau \in X : \sigma \cap \tau = \emptyset \textrm{ and } \sigma \cup \tau \in X \}\,.$

Note that if $X$ is a graph, i.e., a $1$-dimensional simplicial complex, then for a vertex $v \in X^{(0)}$, $X_v$ is precisely the set of neighbors of $v$ in $X$.

• Induced measures on links. Given a face $\sigma \in X^{(i)}$, the link $X_{\sigma}$ is a pure $(d-i-1)$-dimensional complex. We can define the induced measure $\pi_{\sigma}^{(d-i-1)}$ in the straightforward way (as a conditional measure): For any top-dimensional face $\tau \in X_{\sigma}^{(d-i-1)}$,

$\pi_{\sigma}^{(d-i-1)}(\tau) \seteq \frac{\pi^{(d)}(\sigma \cup \tau)}{\sum_{\tau' \in X_{\sigma}^{(d-i-1)}} \pi^{(d)}(\sigma \cup \tau')} = \P_{\rho \sim \pi^{(d)}}\left[\rho = \sigma \cup \tau \mid \sigma \subseteq \rho\right].$

Repeating the earlier construction, we obtain measures $\pi_{\sigma}^{(0)}, \pi_{\sigma}^{(1)}, \ldots, \pi_{\sigma}^{(d-i-1)}$.

• Skeletons. The $k$-skeleton of a simplicial complex $X$ is precisely the complex $X^{(-1)} \cup X^{(0)} \cup \cdots \cup X^{(k)}$.

• Link expansion. Say that a pure $d$-dimensional complex $X$ with measure $\pi^{(d)}$ is an $\e$-spectral link expander if, for every $\sigma \in X^{(d-2)} \cup X^{(d-1)} \cup \cdots \cup X^{(-1)}$, it holds that the $1$-skeleton of $X_\sigma$ with the induced measure $\pi_{\sigma}^{(1)}$ satisfies $\e(\pi_{\sigma}^{(1)}) \geq \e$.

## Theorem

• Trickle step: Suppose that $X$ is a pure $2$-dimensional simpicial complex with a probability measure $\pi^{(2)}$ on $X^{(2)}$ and that

• (a) The $1$-skeleton $(X^{(0)}, X^{(1)})$ is connected. More precisely, $\e(\pi^{(1)}) > 0$.

• (b) For every vertex $v \in X^{(0)}$, the ($1$-dimensional) link $X_v$ satisfies $\e(\pi_v^{(1)}) \geq \e$.

Then $\e(\pi^{(1)}) \geq 2 - 1/\e$.

• Note that the theorem is only nontrivial for $\e \geq 1/2$, i.e., the links have to have a substantial spectral gap.

## Proof:

• Let $L$ denote the Laplace operator on the graph $(X^{(0)}, X^{(1)})$ equipped with the measure $\mu = \pi^{(1)}$.

Use \eqref{eq:laplacian} to write

\begin{align*} \langle f, L f \rangle_{\pi^{(0)}} &= \frac12 \E_{uw \sim \pi^{(1)}} (f(u)-f(w))^2 \\ &= \frac12 \E_{v \sim \pi^{(0)}} \E_{uw \sim \pi_v^{(1)}} (f(u)-f(w))^2 \\ &= \E_{v \sim \pi^{(0)}} \langle f, L_v f\rangle_{\pi_v^{(0)}}\,, \end{align*}

where $L_v$ is the Laplacian on $X_v$ associated to the edge measure $\mu = \pi_v^{(1)}$.

Now using assumption (b) that $\e(\pi_v^{(1)}) \geq \e$ gives

$\langle f, L f\rangle_{\pi^{(0)}} \geq \e\ \E_{v \sim \pi^{(0)}} \mathsf{Var}_{\pi_v^{(0)}}(f)\,.$
• The law of total variance: If $X$ and $Y$ are random variables, then

$\mathsf{Var}(Y) = \E[\mathsf{Var}(Y \mid X)] + \mathsf{Var}(\E[Y \mid X]).$

Note that $\E_{u \sim \pi_v^{(0)}} f(u) = P f(v)$. Thus by the law of total variance, we have

$\mathsf{Var}_{\pi^{(0)}}(f) = \E_{v \sim \pi^{(0)}} \mathsf{Var}_{\pi_v^{(0)}} (f) + \mathsf{Var}_{\pi^{(0)}} (P f)\,.$

We conclude that

$\begin{equation}\label{eq:penult} \langle f, L f\rangle_{\pi^{(0)}} \geq \e \left(\mathsf{Var}_{\pi^{(0)}}(f) - \mathsf{Var}_{\pi^{(0)}}(Pf) \right). \end{equation}$
• We are happy as long as $\mathsf{Var}_{\pi^{(0)}}(Pf) \ll \mathsf{Var}_{\pi^{(0)}}(f)$. But in some sense this is what we are trying to prove! Indeed, the spectral gap can be characterized in terms of the rate of decay of the variance of functions acted on by $P$. Convergence to the stationary measure $\pi^{(0)}$ is precisely the assertion that $\mathsf{Var}_{\pi^{(0)}}(P^t f) \to 0$ as $t \to \infty$.

The key reason we can still say something interesting is that $\mathsf{Var}_{\pi^{(0)}}(Pf)$ involves terms of the form $\langle f, P^2 f\rangle_{\pi^{(0)}}$. To check what happens, let us write everything in terms of the Laplacian: Observe that $I - P^2 = 2 (I-P) - (I-P)^2 = 2 L - L^2$, and therefore \eqref{eq:penult} can be equivalently written as

$\langle f, L f\rangle_{\pi^{(0)}} \geq \e \left(2 \langle f,L f\rangle_{\pi^{(0)}} - \langle f,L^2 f\rangle_{\pi^{(0)}}\right).$

We immediately conclude that any eigenvalue $\lambda$ of $L$ satisfies

$\lambda \geq \e (2-\lambda) \lambda\,.$

If $\lambda > 0$, this gives $\lambda \geq 2-1/\e$, implying that the connected components of $(X^{(0)},X^{(1)})$ all have spectral gap at least $2-1/\e$. Since we assume this graph to be connected, the desired result follows.