# High-dimensional random walks (Mon, Feb 27)

## Trickle Down Theorem

• Suppose that $X$ is a pure $d$-dimensional simpicial complex with a probability measure $\pi^{(d)}$ on $X^{(d)}$.

Consider $0 \leq k \leq d-2$ and assume furthermore that:

• (a) For every $\sigma \in X^{(k-1)}$, the $1$-skeleton of $X_{\sigma}$ is connected. More precisely, that $\e(\pi^{(1)}_{\sigma}) > 0$.

• (b) For every $\sigma \in X^{(k)}$, there is a uniform spectral gap $\e(\pi_{\sigma}^{(1)}) \geq \e$.

Then for every $\rho \in X^{(k-1)}$, we have $\e(\pi_{\rho}^{(1)}) \geq 2 - 1/\e$.

## Proof:

• Fix some $\rho \in X^{(k-1)}$. Let $L$ denote the Laplace operator on the $1$-skeleton of $X_{\rho}$ equipped with the measure $\mu = \pi^{(1)}_{\rho}$.

Write:

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

where $L_v$ is the Laplacian on the $1$-skeleton of $X_{\rho \cup v}$ associated to the edge measure $\mu = \pi_{\rho \cup v}^{(1)}$.

Now assumption (b) gives $\e(\pi_{\rho \cup v}^{(1)}) \geq \e$, and therefore

$\langle f, L f\rangle_{\pi_{\rho}^{(0)}} \geq \e\ \E_{v \sim \pi_{\rho}^{(0)}} \mathsf{Var}_{\pi_{\rho \cup v}^{(0)}}(f)\,.$
• Note that $\E_{u \sim \pi_{\rho \cup v}^{(0)}} f(u) = (I-L) f(v)$. Thus by the law of total variance, we have

$\mathsf{Var}_{\pi_{\rho}^{(0)}}(f) = \E_{v \sim \pi_{\rho}^{(0)}} \mathsf{Var}_{\pi_{\rho \cup v}^{(0)}} (f) + \mathsf{Var}_{\pi_{\rho}^{(0)}} ((I-L) f)\,.$

We conclude that

\begin{align*} \langle f, L f\rangle_{\pi_{\rho}^{(0)}} &\geq \e \left(\mathsf{Var}_{\pi_{\rho}^{(0)}}(f) - \mathsf{Var}_{\pi_{\rho}^{(0)}}((I-L)f) \right) \\ &= \e \left(2 \langle f,L f\rangle_{\pi_{\rho}^{(0)}} - \langle f,L^2 f\rangle_{\pi_{\rho}^{(0)}}\right). \end{align*}
• Therefore if $\lambda$ is an eignevalue of $L$, then

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

If $\lambda > 0$, this gives $\lambda \geq 2-1/\e$. Now assumption (a) asserts that only the trivial eigenvalue of $L$ is $0$, and the desired result follows.

## Corollary

• Suppose that $X$ is a pure $d$-dimensional simplicial complex equipped with a probability measure $\pi^{(d)}$ on $X^{(d)}$, and that

• (a) For every face $\sigma \in X$ with $\dim(\sigma) \leq d-2$, it holds that the $1$-skeleton of $X_{\sigma}$ is connected. More precisely, $\e(\pi_{\sigma}^{(1)}) > 0$.

• (b) For every face $\sigma \in X^{(d-2)}$, there is a uniform spectral gap $\e(\pi_{\sigma}^{(1)}) \geq \e$.

Then for every $-1 \leq k \leq d-2$ and $\sigma \in X^{(k)}$, it holds that

$\e(\pi_{\sigma}^{(1)}) \geq 1 - \left(d-(k+1)\right) (1-\e) \geq 1 - d(1-\e) \,.$
• Note that to derive a non-trivial bound for the $1$-skeleton of $X$, we need a very strong spectral gap for the $(d-2)$-links: $\e \geq 1-1/d$.

## Proof:

• The proof is by induction using the Trickle Down Theorem:

Consider $\sigma \in X^{(k)}$:

• $k=d-2$. By assumption, $\e(\pi_{\sigma}^{(1)}) \geq \e$.

• $k=d-3$. The theorem yields $\e(\pi_{\sigma}^{(1)}) \geq 2 - 1/\e \geq 1 - 2(1-\e)$, where the last inequality holds for $\e \geq 1/2$.

• $k=d-4$. The theorem yields $\e(\pi_{\sigma}^{(1)}) \geq 2 - \frac{1}{1-2(1-\e)} \geq 1 - 3(1-\e)$, where the last inequality holds for $\e \geq 1/3$.

• Etc. using the inequality

$2 - \frac{1}{1-(j-1)(1-\e)} \geq 1 - j (1-\e),\qquad \e \geq 1 - \frac{1}{j}\,.$

Finally, note that the constraint $\e \geq 1 - 1/j$ is unnecessary since the spectral gap is always at least $0$.

## Random walks

• Example: Sampling random spanning trees. Consider an undirected graph $G=(V,E)$ and let $\mathcal{T}$ denote the set of spanning trees in $G$ (where we think of a spanning tree as a subset $T \subseteq E$). The downward closure $X$ of $\mathcal{T}$ is an $(n-2)$-dimensional simplicial complex, where $n = |V|$.

The edge exchange walk on $\mathcal{T}$ is the random walk defined as follows: If $T$ is a spanning tree, we remove a uniformly random edge $e \in E(T)$ and then define $T' \seteq (T \setminus \{e\}) \cup \{e'\}$, where $e' \notin E(T)$ is a uniformly random edge such that $T' \in \mathcal{T}$.

Our goal now is to define high-dimensional random walks so that the associated spectral gaps allow us to analyze the mixing time of such a random process. The edge exchange walk will be a particular example of the “down-up” walk analyzed next.

• Let $X$ denote a pure $d$-dimensional simplicial complex equipped with a probability measure $\pi^{(d)}$ on $X^{(d)}$.

Let us define up and down walk operators: $P_{\nearrow k} : \ell_2(\pi^{(k-1)}) \to \ell_2(\pi^{(k)})$ and $P_{\searrow k} : \ell_2(\pi^{(k+1)}) \to \ell_2(\pi^{(k)})$ by

\begin{align*} P_{\nearrow k} f(\sigma) &\seteq \E_{v \sim \pi_{\sigma}^{(0)}} f(\sigma \cup v) \\ P_{\searrow k} f(\sigma) &\seteq \E_{v \in \sigma} f(\sigma \setminus v)\,. \end{align*}

Note that these are “co-“operators: They operate on functions. This means that the adjoint of the “up” operator $P_{\nearrow k}^{\top}$ acts in the opposite direction on distributions:

\begin{align*} \pi^{(k)} &= P_{\searrow k-1}^{\top} \pi^{(k-1)} \\ \pi^{(k-1)} &= P_{\nearrow k}^{\top} \pi^{(k)}\,. \end{align*}

Correspondingly, we obtain two different walk operators on $X^{(k)}$: The up-down walk and the down-up walk, respectively

\begin{align*} P_k^{\triangle} &\seteq P_{\searrow k} P_{\nearrow k+1} \\ P_k^{\triangledown} &\seteq P_{\nearrow k} P_{\searrow k-1}\,. \end{align*}

Let us also define the non-lazy up-down walk operator by

$P_k^{\wedge} \seteq P_k^{\triangle} + \frac{1}{k+1} (P_k^{\triangle} - I)\,.$
• It is not difficult to check that all three walks have $\pi^{(k)}$ as the stationary measure.

• Let us define the corresponding Laplacians

\begin{align*} L_k^{\triangle} &\seteq I - P_k^{\triangle} \\ L_k^{\triangledown} &\seteq I - P_k^{\triangledown} \\ L_k^{\wedge} &\seteq I - P_k^{\wedge} = \frac{k+2}{k+1} L_k^{\triangle}\,. \end{align*}

Define $\e_k \seteq \min_{\sigma \in X^{(k)}} \e(\pi^{(1)}_{\sigma})$ as the minimum spectral gap over all $1$-skeleta of links of $k$-dimensional faces.

Then for every $0 \leq k \leq d-1$, and $f \in \ell_2(\pi^{(k)})$, it holds that

$$$\label{eq:al-thm} \langle f, L_k^{\wedge} f\rangle_{\pi^{(k)}} \geq \e_{k-1} \langle f, L_k^{\triangledown} f\rangle_{\pi^{(k)}}\,.$$$

## Corollary

• Using the relationship between $L_k^{\wedge}$ and $L_k^{\triangle}$, this gives

$\frac{k+2}{k+1} \e(L_k^{\triangle}) \geq \e_{k-1} \e(L_k^{\triangledown})\,.$

But recall that $P_k^{\triangle} = P_{\searrow k} P_{\nearrow k+1}$ and $P_{k+1}^{\triangledown} = P_{\nearrow k+1} P_{\searrow k}$.

It is straightforward to check that for rectangular matrices $A,B$, the two matrices $AB$ and $BA$ have the same spectrum, and therefore $P_k^{\triangle}$ and $P_{k+1}^{\triangledown}$ have the same spectrum. In particular, $\e(L_k^{\triangle}) = \e(L_{k+1}^{\triangledown})$, and therefore

$$$\label{eq:rec-formula} \e(L_{k+1}^{\triangledown}) \geq \frac{k+1}{k+2} \e_{k-1} \e(L_k^{\triangledown})\,.$$$
• Applying \eqref{eq:rec-formula} iteratively gives, for $1 \leq k \leq d$,

$\e(L_{k}^{\triangledown}) \geq \frac{1}{k+1} \prod_{j=-1}^{k-2} \e_j\,.$

In particular, we have

$\e(L_d^{\triangledown}) \geq \frac{1}{d+1} \prod_{j=-1}^{d-2} \e_j\,,$

yielding a spectral gap for the down-up walk whenever there are non-trivial spectral gaps for the $1$-skeleta of all the links in $X$.

## Proof of \eqref{eq:al-thm}:

• For $j \leq k$, $\sigma \in X^{(j)}$, and $f : X^{(k)} \to \R$, let us define $f_{\sigma}^{(k-j-1)} : X_{\sigma}^{(k-j-1)} \to \R$ by $f^{(k-j-1)}_{\sigma}(\tau) \seteq f(\sigma \cup \tau)$.

• Let $L_{\sigma}$ denote the Laplacian on the $1$-skeleton of $X_{\sigma}$ with respect to the edge measure $\mu = \pi_{\sigma}^{(1)}$. Then we have

\begin{align} \langle f, L_k^{\wedge} f\rangle_{\pi^{(k)}} &= \E_{\sigma \in \pi^{(k-1)}} \langle f^{(0)}_{\sigma}, L_{\sigma} f^{(0)}_{\sigma} \rangle_{\pi_{\sigma}^{(0)}} \label{eq:down1} \\ \langle f, L_k^{\triangledown} f \rangle_{\pi^{(k)}} &= \E_{\sigma \in \pi^{(k-1)}} \mathsf{Var}_{\pi_{\sigma}^{(0)}}(f_{\sigma}^{(0)})\,, \label{eq:down2} \end{align}

and \eqref{eq:al-thm} follows immediately from the definition of the spectral gap.

• We are left to verify \eqref{eq:down1} and \eqref{eq:down2}.

Use \eqref{eq:laplacian} to write

\begin{align*} \langle f, L_k^{\wedge} f\rangle_{\pi^{(k)}} &= \frac14 \E_{\alpha \sim \pi^{(k)}} \E_{v \in \alpha} \E_{u \sim \pi^{(0)}_{\alpha}} (f(\alpha)-f(\alpha + u - v))^2 \\ &=\frac12 \E_{\sigma \sim \pi^{(k-1)}} \E_{uv \sim \pi^{(1)}_{\sigma}} (f(\sigma + u)-f(\sigma + v))^2 \\ &=\frac12 \E_{\sigma \sim \pi^{(k-1)}} \E_{uv \sim \pi^{(1)}_{\sigma}} (f_{\sigma}^{(0)}(u) - f_{\sigma}^{(0)}(v))^2 \\ &= \E_{\sigma \sim \pi^{(k-1)}} \langle f_{\sigma}^{(0)}, L_{\sigma} f_{\sigma}^{(0)}\rangle_{\pi_{\sigma}^{(0)}}\,. \end{align*}
• Next write

\begin{align*} \langle f, L_k^{\triangledown} f \rangle_{\pi^{(k)}} &= \frac14 \E_{\alpha \sim \pi^{(k)}} \E_{v \in \alpha} \E_{u \sim \pi^{(0)}_{\alpha \setminus v}} (f(\alpha) - f(\alpha + u - v))^2 \\ &= \frac12 \E_{\sigma \sim \pi^{(k-1)}} \E_{u,v \sim \pi^{(0)}_{\sigma}} (f(\sigma + u) - f(\sigma + v))^2 \\ &= \frac12 \E_{\sigma \sim \pi^{(k-1)}} \E_{u,v \sim \pi^{(0)}_{\sigma}} (f^{(0)}_{\sigma}(u) - f^{(0)}_{\sigma}(v))^2 \\ &= \E_{\sigma \sim \pi^{(k-1)}} \mathsf{Var}_{\pi^{(0)}_{\sigma}}(f_{\sigma}^{(0)})\,, \end{align*}

where the last line follows from the general fact

$\mathsf{Var}(X) = \frac12 \E |X-X'|^2\,,$

when $X$ and $X'$ are two independent copies of $X$.