Local Spectral Expansion Approach to High Dimensional Expanders Part II: Mixing and Geometrical Overlapping

Abstract

We further explore the local-to-global approach for expansion of simplicial complexes that we call local spectral expansion. Specifically, we prove that local expansion in the links implies the global expansion phenomena of mixing and geometric overlapping. Our mixing results also give tighter bounds on the error terms compared to previously known results.

Appendices

Appendix A: Spectral Descent for Random Walks

In [13], it was shown that given a simplicial complex with connected links, one can derive bounds on the spectra of the Laplacians in links of every dimension, based on bounds on the spectrum of the Laplacians in the 1-dimensional links. In the paper above, we replaced the Laplacians with random walk operators.

In this appendix, we state the spectral descent results of [13] in the language of random walks. These results are easy to derive, because the graph Laplacian $\varDelta _0^+$ is basically defined by $\varDelta _0^+ = I-(M')_0^+$, and therefore we do not include the proof. We also use this result to find a sufficient condition on the spectral gaps of 1-dimensional links, so that the conditions of our mixing theorems are fulfilled.

Theorem 10.1

(Spectral descent for random walks) Let X be a pure n-dimensional weighted simplicial complex, such that all the links of X of dimension $\ge 1$ (including X itself) are connected. For $0 \le k \le n-1$, let $0 \le \mu _k \le 1,-1 \le \nu _k \le 0$ be constants such that for every $\sigma \in X(k-1)$, ${\text {Spec}}((M_\sigma ')_0^{+}) \subseteq [\nu _k, \mu _k] \cup \lbrace 1 \rbrace$. Then for every $0 \le k \le n-2$,

$$\begin{aligned}&\mu _k \le \dfrac{\mu _{k+1}}{1-\mu _{k+1}}, \\&\nu _k \ge \dfrac{\nu _{k+1}}{1-\nu _{k+1}}. \end{aligned}$$

A simple induction leads to the following:

Corollary 10.2

Let X be a pure n-dimensional weighted simplicial complex, such that all the links of X of dimension $\ge 1$ (including X itself) are connected. Then for every $0 \le k \le n-2$,

$$\begin{aligned}&\mu _k \le \dfrac{\mu _{n-1}}{1-(n-1-k)\mu _{n-1}}, \\&\nu _k \ge \dfrac{\nu _{n-1}}{1-(n-1-k)\nu _{n-1}}. \end{aligned}$$

A corollary from the above corollary is

Corollary 10.3

Let X be a weighted pure n-dimensional simplicial complex, such that all the links of X of dimension $\ge 1$ (including X itself[) are connected, and let $0 < \lambda \le 1$ be some constant. If $\mu _{n-1} \le \frac{\lambda }{1+(n-1)\lambda }$, then for every $0 \le k \le n-1$, $\mu _k \le \lambda$, i.e., for every $\sigma \in \bigcup _{k=-1}^{n-2} X(k)$, ${\text {Spec}}((M_\sigma ')_0^{+}) \subseteq [-1, \lambda ] \cup \lbrace 1 \rbrace$.

Moreover, if $\mu _{n-1} \le \frac{\lambda }{1+(n-1)\lambda }$ and $\frac{-\lambda }{1+(n-1)\lambda } \le \nu _{n-1}$, then for every $0 \le k \le n-1$, $\mu _k \le \lambda$ and $-\lambda \le \nu _k$, i.e., for every $\sigma \in \bigcup _{k=-1}^{n-2} X(k)$, ${\text {Spec}}((M_\sigma ')_0^{+}) \subseteq [-\lambda , \lambda ] \cup \lbrace 1 \rbrace$.

Proof

By the above corollary, if $\mu _{n-1} \le \frac{\lambda }{1+(n-1)\lambda }$ then for every $0 \le k \le n-2$ we have that

$$\begin{aligned} \mu _k\le & {} \dfrac{\mu _{n-1}}{1-(n-1-k)\mu _{n-1}}\le \dfrac{\frac{\lambda }{1+(n-1)\lambda }}{1-(n-1-k)\frac{\lambda }{1+(n-1)\lambda }}\\\le & {} \dfrac{\frac{\lambda }{1+(n-1)\lambda }}{1-(n-1)\frac{\lambda }{1+(n-1)\lambda }} = \lambda , \end{aligned}$$

The proof of the second assertion is similar. $\square$

Appendix B: Alternative Proof of Theorem 7.1 by Passing to a Partite Cover

Let X be a pure n-dimensional weighted simplicial complex X with a weight function m. Below, we will prove a version of Theorem 7.1 for which we will need the following notation: for non-empty sets $U_0,\dots ,U_n \subseteq X(0)$, define

$$\begin{aligned} x(U_0,\dots ,U_n) = \bigl \lbrace (u_0,\dots ,u_n) \in U_0 \times \dots \times U_n : \lbrace u_0,\dots ,u_n \rbrace \in X(n) \bigr \rbrace , \end{aligned}$$

and

$$\begin{aligned} m( x(U_0,\dots ,U_n)) = \sum _{(u_0,\dots ,u_n) \in x(U_0,\dots ,U_n)} m(\lbrace u_0,\dots ,u_n \rbrace ). \end{aligned}$$

We note that $m( x(U_0,\dots ,U_n)))$ is defined so that it accounts for the intersections between the sets. For example, if the sets $U_0,\dots ,U_n \subseteq X(0)$ are pairwise disjoint, then $m( x(U_0,\dots ,U_n))) = m(X(U_0,\dots ,U_n))$, and, on the other hand, if $U_0 = \dots = U_n$, then $m( x(U_0,\dots ,U_n)) = (n+1)!\, m(X(U_0,\dots ,U_n))$. After these notations, we can state the version of Theorem 7.1 to be proven.

Theorem 11.1

Let X be as above and assume that there is a constant $0 \le \lambda < 1$ such that for every $0 \le k \le n-1$ and every $\tau \in X(k-1)$, ${\text {Spec}}((M')_{\tau ,0}^+) \subseteq [-(n-k) \lambda , \lambda ] \cup \lbrace 1 \rbrace$. Then for every non-empty sets $U_0,\dots ,U_n \subseteq X(0)$,

$$\begin{aligned} \biggl \vert m(x(U_0,\dots ,U_n)) - \dfrac{m(U_0) m (U_1) \dots m(U_n)}{m(X(0))^{n}} \biggr \vert \le C_n \lambda \min _{0 \le i < j \le n} \sqrt{m(U_i) m(U_j)}, \end{aligned}$$

where

$$\begin{aligned} C_n&= \sum _{k=0}^{n-1} \dfrac{n!}{(n-1-k)!} \frac{(n-k) (k+1)}{n+2}\\&\quad \times \bigl ( (n-k+1)^{n-k} (k+2)^{n-k} - (n-k)^{n-k} (k+1)^{n-k} \bigr ). \end{aligned}$$

Remark 11.2

When comparing Theorem 7.1 to Theorem 11.1, we note that Theorem 7.1 gives a better bound, i.e., the constant $C_n$ is smaller in Theorem 7.1. On the other hand, Theorem 11.1 is more general in the sense that it allows the sets $U_0,\dots ,U_n$ to have non-empty intersections and that the condition on the lower bound of the spectra of the random walks on the links is relaxed for links of dimension $>1$.

The proof of Theorem 11.1 uses the partite cover of X which we will now define. Given X as above, we construct a $(n+1)$-partite cover of X, denoted ${\widetilde{X}}$ as follows:

• The complex ${\widetilde{X}}$ has $n+1$-sides that are copies of X(0). Formally, ${\widetilde{X}} (0) = S_0 \cup \dots \cup S_n$, where the sets $S_0,\dots ,S_n$ are disjoint and for every $0 \le i \le n$,

  $$\begin{aligned} S_i = \lbrace v^i_u : u \in X(0) \rbrace . \end{aligned}$$

• The n-simplices of ${\widetilde{X}}$ are those with vertices corresponding to n-simplices of X that are all in different sides of ${\widetilde{X}}$ (and the simplices of all other dimensions are defined to be faces of n-simplices). Formally, for every $1 \le k \le n$, $\lbrace v^{i_0}_{u_0},\dots ,v^{i_k}_{u_k} \rbrace \in {\widetilde{X}} (k)$ if and only if $\lbrace u_0,\dots ,u_k \rbrace \in X(k)$, $0 \le i_0,\dots ,i_k \le n$ and $i_{j_1} \ne i_{j_2}$ for $0 \le j_1, j_2 \le k$.

• The weight function ${\widetilde{m}}$ of ${\widetilde{X}}$ is defined so that for every $\lbrace v^{0}_{u_0},\dots ,v^{n}_{u_n} \rbrace \in {\widetilde{X}} (n)$, ${\widetilde{m}} (\lbrace v^{0}_{u_0},\dots ,v^{n}_{u_n} \rbrace ) = m (\lbrace u_0,\dots ,u_n \rbrace )$. Explicitly, for every $-1 \le k \le n$ and every $\lbrace v^{i_0}_{u_0},\dots ,v^{i_k}_{u_k} \rbrace \in {\widetilde{X}} (k)$,

  $$\begin{aligned} {\widetilde{m}} (\lbrace v^{i_0}_{u_0},\dots ,v^{i_k}_{u_k} \rbrace ) = (n-k)!\, m(\lbrace u_0,\dots ,u_k \rbrace ). \end{aligned}$$

Observation 11.3

For every $\lbrace v^{i_0}_{u_0},\dots ,v^{i_k}_{u_k} \rbrace \in {\widetilde{X}} (k)$, an link ${\widetilde{X}}_{\lbrace v^{i_0}_{u_0},\dots ,v^{i_k}_{u_k} \rbrace }$ is the $(n-k)$-partite cover of the link $X_{\lbrace u_0,\dots ,u_k \rbrace }$.

The 1-skeleton of ${\widetilde{X}}$ is the tensor product of the 1-skeleton of X with the complete graph on $n+1$ vertices. We recall that it is known that given two graphs $G_1, G_2$, the spectrum of the random walk matrix on $G_1 \otimes G_2$ is

$$\begin{aligned} \bigl \lbrace \lambda _1 \lambda _2 : \lambda _i \text { is an eigenvalue of the random walk matrix on } G_i, i=1,2 \bigr \rbrace . \end{aligned}$$

Below, we prove a generalization of this fact in the weighted setting (this is probably a well-known fact, but we could not find a proof in the literature).

Proposition 11.4

For a finite weighted graph $G = (V,E)$, with a weight function m, denote $M_G'$ to be the weighted random walk operator on G, i.e., for $\phi \in \ell ^2 (V)$,

$$\begin{aligned} (M_G' \phi ) (v) = \frac{1}{m (v)} \sum _{\begin{array}{c} u \in V\\ \lbrace u,v \rbrace \in E \end{array}} m (\lbrace u,v \rbrace ) \phi (u). \end{aligned}$$

Let $G_1 = (V_1,E_1) ,G_2 = (V_2,E_2)$ be finite weighted graphs with weight functions $m_i :E_i \rightarrow {\mathbb {R}}^+$. Define $G_1 \otimes G_2$ to be the weighted graph defined as $V_{G_1 \otimes G_2} = V_1 \times V_2$,

$$\begin{aligned} E_{G_1 \otimes G_2} = \bigl \lbrace \lbrace (v_1,v_2), (u_1,u_2) \rbrace : \lbrace v_1, u_1 \rbrace \in E_1 \text { and } \lbrace v_2, u_2 \rbrace \in E_2 \bigr \rbrace , \end{aligned}$$

and

$$\begin{aligned} m_{G_1 \otimes G_2} (\lbrace (v_1,v_2), (u_1,u_2) \rbrace ) = m_1 (\lbrace v_1, u_1 \rbrace ) m_2 (\lbrace v_2, u_2 \rbrace ). \end{aligned}$$

Then

$$\begin{aligned} {\text {Spec}}(M_{G_1 \otimes G_2}') = \lbrace \lambda _1 \lambda _2 : \lambda _i \in {\text {Spec}}(M_{G_i}'), i=1,2 \rbrace . \end{aligned}$$

Proof

For $i=1,2$, let $\phi _i \in \ell ^2 (V_i)$ be an eigenfunction of $M_{G_i}'$ with eigenvalue $\lambda _i$. Define $\phi _1 \otimes \phi _2 \in \ell ^2 (V_1 \times V_2)$ by

$$\begin{aligned} \phi _1 \otimes \phi _2 ((v_1,v_2)) = \phi _1 (v_1) \phi _2 (v_2). \end{aligned}$$

We leave it to the reader to verify that for every $(v_1,v_2) \in V_1 \times V_2$,

$$\begin{aligned} (M_{G_1 \otimes G_2}' \phi _1 \otimes \phi _2 ) ((v_1,v_2))= & {} (M_{G_1}' \phi _1) (v_1) (M_{G_2}' \phi _2) (v_2)\\= & {} \lambda _1 \lambda _2 (\phi _1 \otimes \phi _2 ((v_1,v_2))). \end{aligned}$$

We also observe that if $\phi _1 \perp \phi _1'$ or $\phi _2 \perp \phi _2'$, then $(\phi _1 \otimes \phi _2) \perp (\phi _1' \otimes \phi _2')$. Therefore if for $i=1,2$, $\lbrace \phi _i^{j} : j=1,\dots ,\vert V_i \vert \rbrace$ is an orthogonal basis of eigenfunctions of $M_{G_i}'$, then $\lbrace \phi _1^{j_1} \otimes \phi _2^{j_2} : j_1=1,\dots ,\vert V_1 \vert , j_2=1,\dots ,\vert V_2 \vert \rbrace$ is an orthogonal set of eigenfunctions of $M_{G_1 \otimes G_2}'$ and since it is of size $\vert V_1 \times V_2|$, it is also a basis. Thus $\lbrace \phi _1^{j_1} \otimes \phi _2^{j_2} : j_1=1,\dots ,\vert V_1 \vert , j_2=1,\dots ,\vert V_2 \vert \rbrace$ is an orthogonal basis of eigenfunctions and all the eigenvalues are of the form $\lambda _1 \lambda _2$. $\square$

Corollary 11.5

Let X be a pure n-dimensional weighted simplicial complex and let ${\widetilde{X}}$ be the $(n+1)$-partite cover of X defined above. Denote $(M')_{X,0}^+$ to be the weighted random walk on the 1-skeleton of X and $(M')_{{\widetilde{X}},0}^+$ to be the weighted random walk on the 1-skeleton of ${\widetilde{X}}$. For $0 \le \lambda <1$, if ${\text {Spec}}\bigl ((M')_{X,0}^+\bigr ) \subseteq [-n\lambda , \lambda ] \cup \lbrace 1 \rbrace$, then ${\text {Spec}}\bigl ((M')_{{\widetilde{X}},0}^+\bigr ) \subseteq \bigl [-\frac{1}{n}, \lambda \bigr ] \cup \lbrace 1 \rbrace$.

Proof

We observe that the 1-skeleton of ${\widetilde{X}}$ is the tensor product of the 1-skeleton of X weighted by m and the complete graph on $n+1$ vertices with weight 1 on all the edges. The spectrum of the simple random walk on complete graph on $n+1$ vertices is $\bigl \lbrace - \frac{1}{n}, 1\bigr \rbrace$ and the corollary follows from Proposition 11.4. $\square$

Corollary 11.6

Let X be a pure n-dimensional weighted simplicial complex and let ${\widetilde{X}}$ be the $(n+1)$-partite cover of X defined above. Assume that there is a constant $0< \lambda < 1$ such that for every $0 \le k \le n-1$ and every $\tau \in X(k-1)$, ${\text {Spec}}\bigl ((M')_{X_\tau ,0}^+\bigr ) \subseteq [-(n-k)\lambda , \lambda ] \cup \lbrace 1 \rbrace$. Then for every $0 \le k \le n-1$ and every ${\widetilde{\tau }} \in {\widetilde{X}}(k-1)$, ${\text {Spec}}\bigl ((M')_{{\widetilde{X}}_{{\widetilde{\tau }}},0}^+\bigr ) \subseteq \bigl [- \frac{1}{n-k}, \lambda \bigr ] \cup \lbrace 1 \rbrace$.

Proof

The proof follows directly from the previous corollary and Observation 11.3. $\square$

After this corollary, we are ready to prove Theorem 11.1:

Proof

Let X be a complex as in the statement of Theorem 11.1. By Corollary 11.6, ${\widetilde{X}}$ fulfils the conditions of Theorem 7.3 (with the same $\lambda$). For non-empty sets $U_0,\dots ,U_n \subseteq X(0)$, we define sets ${\widetilde{U}}_i \subseteq S_i$ in ${\widetilde{X}} (0)$ as

$$\begin{aligned} {\widetilde{U}}_i = \lbrace \lbrace v^i_u \rbrace : u \in U_i \rbrace . \end{aligned}$$

By Theorem 7.3, it follows that

$$\begin{aligned} \biggl \vert \dfrac{{\widetilde{m}}({\widetilde{X}} ({\widetilde{U}}_0,\dots ,{\widetilde{U}}_n))}{{\widetilde{m}}({\widetilde{X}} (n))} - \dfrac{{\widetilde{m}}({\widetilde{U}}_0) \dots {\widetilde{m}}({\widetilde{U}}_n)}{{\widetilde{m}}(S_0) \dots {\widetilde{m}}(S_n)} \biggr \vert \le C_n \lambda \min _{0 \le i < j \le n} \sqrt{\dfrac{{\widetilde{m}}({\widetilde{U}}_i) {\widetilde{m}}({\widetilde{U}}_j)}{{\widetilde{m}}(S_i) {\widetilde{m}}(S_j)}}, \end{aligned}$$

where

$$\begin{aligned} C_n&= \sum _{k=0}^{n-1} \dfrac{n!}{(n-1-k)!} \frac{(n-k) (k+1)}{n+2}\\&\quad \times \bigl ( (n-k+1)^{n-k} (k+2)^{n-k} - (n-k)^{n-k} (k+1)^{n-k} \bigr ). \end{aligned}$$

We observe that

$$\begin{aligned}&{\widetilde{m}}({\widetilde{X}} ({\widetilde{U}}_0,\dots ,{\widetilde{U}}_n)) = m(x(U_0,\dots ,U_n)),\\&\quad {\widetilde{m}}({\widetilde{X}} (n)) = (n+1)!\, m (X(n)),\\&\quad \forall \, 0 \le i \le n,\;\; {\widetilde{m}}({\widetilde{U}}_i) = n!\, m (U_i) \;\text { and } \;{\widetilde{m}} (S_i) = n!\, m (X(0)). \end{aligned}$$

Thus,

$$\begin{aligned} \biggl \vert \dfrac{m(x(U_0,\dots ,U_n))}{(n+1)!\; m (X(n))} - \dfrac{m(U_0) m (U_1) \dots m(U_n)}{m(X(0))^{n+1}} \biggr \vert \le C_n \lambda \min _{0 \le i < j \le n} \sqrt{\dfrac{m (U_i) m (U_j)}{m (X(0))^2}}, \end{aligned}$$

where $C_n$ as above. Multiplying this inequality by $m(X(0)) = (n+1)!\, m(X(n))$ (see Proposition 2.3) finishes the proof. $\square$