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 \(\e(\pi_{\rho}^{(1)}) \geq 2 - 1/\e\).
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.
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: \(\e \geq 1-1/d\).
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 \(0\).
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
\[\begin{equation}\label{eq:al-thm} \langle f, L_k^{\wedge} f\rangle_{\pi^{(k)}} \geq \e_{k-1} \langle f, L_k^{\triangledown} f\rangle_{\pi^{(k)}}\,. \end{equation}\]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
\[\begin{equation}\label{eq:rec-formula} \e(L_{k+1}^{\triangledown}) \geq \frac{k+1}{k+2} \e_{k-1} \e(L_k^{\triangledown})\,. \end{equation}\]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 \(X\).
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 \(X\).