Simplicial branching random walks

Abstract

We study a model of branching random walks on simplicial complexes, which can be seen as a natural generalization of random walks on graphs. Exploiting the fact that each of its particles is distributed like the $(d-1)$-walk introduced in Parzanchevski and Rosenthal (Random Struct Algorithms 50(2):225–261, 2017), we show the model is connected to the spectral and topological properties of the complex. The branching model is then used in order to calculate the spectral measure of the upper Laplacian associated with high-dimensional analogues of regular trees, thus obtaining a Kesten–McKay type distribution in arbitrary dimensions. Finally, we use the branching model in order to construct solutions to the Dirichlet problem on simplicial complexes.

Appendices

Proof of Claim 2.3

For every $f\in \Omega _{L^{2}}^{k}$,

$$\begin{aligned} \begin{aligned} \Vert \delta _{k}f\Vert ^{2}&=\sum _{\tau \in X^{k}}w(\tau )|\delta _{k}f(\tau )|^{2} \le \sum _{\tau \in X^{k}}w(\tau )\cdot \left( {\begin{array}{c}k\\ 2\end{array}}\right) \sum _{i=0}^{k}|f(\tau \backslash \tau _{i})|^{2}\\&=\sum _{\sigma \in X^{k-1}}\left( {\begin{array}{c}k\\ 2\end{array}}\right) \cdot \Big (\sum _{\sigma \vartriangleleft \tau }w(\tau )\Big )|f(\sigma )|^{2} \le \left( {\begin{array}{c}k\\ 2\end{array}}\right) \cdot \Big (\sup _{\sigma \in X^{k-1}}\frac{1}{w(\sigma )}\sum _{\sigma \vartriangleleft \tau }w(\tau )\Big )\Vert f\Vert ^{2}, \end{aligned} \end{aligned}$$

which proves that (2.4) implies that $\delta _{k}$ is bounded. As for the other direction for any $\sigma \in X_{\pm }^{k-1}$ the function $\mathbb {1}_{\sigma }$ satisfies $\Vert w(\sigma )^{-1/2}\mathbb {1}_{\sigma }\Vert ^{2}=1$ and

$$\begin{aligned} \Vert \delta _{k}\mathbb {1}_{\sigma }\Vert ^{2} =\sum _{\tau \in X^{k}}w(\tau )|\delta _{k}(w(\sigma )^{-1/2} \mathbb {1}_{\sigma }(\tau ))|^{2}=\frac{1}{w(\sigma )}\sum _{\sigma \vartriangleleft \tau }w(\tau ). \end{aligned}$$

Thus, whenever (2.4) does not hold the operator $\delta _{k}$ is not bounded.

If $\deg (\sigma )<\infty$ for every $\sigma$, then $\partial _{k}g(\sigma )$ is a finite sum for every $\sigma \in X_{\pm }^{d-1}$ and is thus well defined. If (2.4) holds, then for every $g\in \Omega _{L^{2}}^{k}$

$$\begin{aligned} \begin{aligned} \Vert \partial _{k}g\Vert ^{2}&=\sum _{\sigma \in X^{k-1}}\frac{1}{w(\sigma )}\Big |\sum _{\sigma \vartriangleleft v\sigma }w(v\sigma )g(v\sigma )\Big |^{2}\\&\le \sum _{\sigma \in X^{k-1}}\frac{1}{w(\sigma )}\Big (\sum _{\sigma \vartriangleleft \tau }w(\tau )\Big )\Big (\sum _{\sigma \vartriangleleft v\sigma }w(v\sigma )|g(v\sigma )|^{2}\Big )\\&\le \sup _{\sigma \in X^{k-1}}\Big (\frac{1}{w(\sigma )}\sum _{\sigma \vartriangleleft \tau }w(\tau )\Big )\cdot \Big (\sum _{\sigma \in X^{k-1}}\sum _{\sigma \vartriangleleft \tau }w(\tau )|g(\tau )|^{2}\Big )\\&=k\cdot \sup _{\sigma \in X^{k-1}}\Big (\frac{1}{w(\sigma )}\sum _{\sigma \vartriangleleft \tau }w(\tau )\Big )\cdot \Vert g\Vert ^{2}. \end{aligned} \end{aligned}$$

thus proving that $\partial _k$ is a bounded operator whenever (2.4) holds. $\square$

Invertability of $\Delta ^+_{X{\setminus } A}$

The invertability of $\Delta ^+_{X{\setminus } A}$ plays a crucial role in the proof of Theorem 5.1. Therefore, we wish to study it further and provide several simpler conditions which are either necessary or sufficient for its validity. We start with some simple observations.

Claim B.1

(Invertibility of $\Delta _{X\backslash A}^{+}$) Let X be a finite d-complex and $\emptyset \ne A\subsetneq X^{d-1}$. Then, the following are equivalent:

• 1. $\Delta _{X\backslash A}^{+}$ is invertible.

• 2. $\ker \delta _{d}^{X\backslash A}=\ker \Delta _{X\backslash A}^{+}=\{0\}$.

• 3. For every $(d-1)$-form $f:(X^{d-1}\backslash A)_{\pm }\rightarrow \mathbb {R}$ which is not identically zero, the extension ${\tilde{f}}:X_{\pm }^{d-1}\rightarrow \mathbb {R}$ given by ${\tilde{f}}(\sigma )={\left\{ \begin{array}{ll} f(\sigma ) &{} \sigma \in (X^{d-1}\backslash A)_{\pm }\\ 0 &{} \sigma \in A_{\pm } \end{array}\right. }$ is not in $\ker \delta _{d}=\ker \Delta ^{+}=Z^{d-1}$.

• 4. The relative homology $H_{d}(X,A)$ (see Hatcher 2002, Section 2.1 for the definition) is trivial.

Proof

The equivalence of the first three conditions follows from:

$$\begin{aligned} \langle \Delta _{X\backslash A}^{+}f,f\rangle _{X\backslash A} =\langle \delta _{d}^{X\backslash A}f,\delta _{d}^{X\backslash A}f\rangle =\langle \delta _{d}{\tilde{f}},\delta _{d}{\tilde{f}}\rangle =\sum _{\tau \in X^{d}}\left( \sum _{i=0}^{d}f(\tau \backslash \tau _{i})\cdot {{\textbf {1}}}_{\tau \backslash \tau _{i}\notin A_{\pm }}\right) ^{2}, \end{aligned}$$

where $\langle \cdot ,\cdot \rangle _{X\backslash A}$ is the inner product $\langle \cdot ,\cdot \rangle$ restricted to $X^{d-1}\backslash A$. As for the last claim, the equivalence between $\ker \delta _{d}^{X\backslash A}=0$ and $H_{d}(X,A)=0$ follows directly from the definition of the relative homology, see (Hatcher 2002, Section 2.1). $\square$

Using Claim B.1, we can identify several cases in which it is simpler to check the invertability of $\Delta _{X\backslash A}^{+}$.

Definition B.2

Let X be a finite d-complex and $\emptyset \ne A\subsetneq X^{d-1}$. The set A is called exhaustive for the complex X if there exists a finite sequence $A=A_{0}\subsetneq A_{1}\subsetneq A_{2}\subsetneq \cdots \subsetneq A_{N}=X^{d-1}$ such that for every $n\ge 1$ and $\sigma \in A_{n},$ one can find $\sigma \vartriangleleft \tau$ so that all faces of $\tau$ besides $\sigma$ are in $A_{n-1}$.

If X is a d-complex and $-1\le k\le d$, we define the k-skeleton of X to be $X^{(k)} = \bigcup _{j=1}^k X^j$.

Definition B.3

(Duval et al. 2009, Definition 3.1) Let X be a d-complex and $k\le d$. A k-dimensional simplicial spanning tree (k-SST for short) of X is a k-dimensional subcomplex $Y\subset X$ such that $Y^{(k-1)}=X^{(k-1)}$, $H_{k}(Y;\mathbb {Z})=0$ and $|H_{k-1}(Y;\mathbb {Z})|<\infty$, where for $l\ge 0$, $H_{l}(Y;\mathbb {Z})$ is the l-homology groups of X with coefficients in $\mathbb {Z}$ (see Hatcher 2002, Section 2.1) for further details).

Lemma B.4

Let X be a finite d-complex.

• 1. Assume $d=1$. Then $\Delta _{X\backslash A}^{+}$ is invertible if and only if A contains a vertex in each of the 0-components of X.

• 2. If A is an exhaustive set of X, then $\Delta _{X\backslash A}^{+}$ is invertible.

• 3. If A is a deformation retract of X and $\Delta _{X\backslash A}^{+}$ is invertible then $H_{d}(X)=0$.

• 4. If there exists $\varrho \in X^{d-2}$ all of its cofaces lie in $X^{d-1}\backslash A$, then $\Delta _{X\backslash A}^{+}$ is not invertible.

• 5. Following Duval et al. (2009). If $|A|=|X^{d}|$ then $\Delta _{X\backslash A}^{+}$ is invertible if and only if X is a d-SST of itself and $X^{(d-2)}\cup (X^{d-1}\backslash A)$ is a $\left( d-1\right)$-SST of X.

Proof

Let $f\in \Omega ^{d-1}(X)$.

• 1. Since $d=1$.

  $$\begin{aligned} \delta _{1}^{X\backslash A}f([x,y])=f(y){{\textbf {1}}}_{y\notin A}-f(x){{\textbf {1}}}_{x\notin A} \end{aligned}$$

  Thus $\delta _{1}^{X\backslash A}f=0$ implies that f is constant on every connected component and is zero on every component containing a vertex in A.

• 2. Assume that A is exhaustive with an exhausting sequence $\left( A_{n}\right) _{0\le n\le N}$ and that $f\in \ker \Delta _{X\backslash A}^{+}=\ker \delta _{d}^{X\backslash A}$. For every $\sigma \in A_{1}\backslash A_{0}$ one can find vertex v such that $\sigma \vartriangleleft v\sigma$ and all $(d-1)$-cells of $v\sigma$ except for $\sigma$ itself are in $A_{0}$. Therefore

  $$\begin{aligned} 0=\delta _{d}^{X\backslash A}f(v\sigma )=\sum _{i=0}^{d}f((v\sigma )\backslash (v\sigma )_{i}) \cdot {{\textbf {1}}}_{(v\sigma )\backslash (v\sigma )_{i}\notin A_{\pm }}=f(\sigma ). \end{aligned}$$

  Consequently $f|_{(A_{1}\backslash A_{0})_{\pm }}\equiv 0$. Proceeding by induction, gives $f_{(A_{n}\backslash A_{0})_{\pm }}\equiv 0$ for every $1\le n\le N$. Since $A_N{\setminus } A_0 = X^{d-1}{\setminus } A$, the kernel of $\delta _{d}^{X\backslash A}$ is trivial and the result follows.

• 3. Since $X^{d-2}\cup A$ is a $(d-1)$-complex, $H_{d}(X^{d-2}\cup A)=0$. In addition by Claim B.1 and the assumption that $\Delta _{X\backslash A}^{+}$ is invertible we have $H_{d}(X,A)=0$. The result now follows since whenever A is a deformation retract of X the sequence

  $$\begin{aligned} 0\rightarrow H_{d}(X^{d-2}\cup A)\rightarrow H_{d}(X)\rightarrow H_{d}(X,A)\rightarrow H_{d-1}(X^{d-2}\cup A)\rightarrow \cdots \end{aligned}$$

  is exact (see Hatcher 2002, Theorem 2.13).

• 4. Let $\varrho \in X^{d-2}$ be a $(d-2)$-cell all of its cofaces are contained in $X^{d-1}\backslash A$. With a slight abuse of notation, denote by $\varrho$ an orientation of the $(d-2)$-cell and define the $(d-1)$-forms ${\tilde{f}}=\delta _{d-1}\mathbb {1}_{\varrho }\ne 0$ and $f={\tilde{f}}|_{(X^{d-1}{\setminus } A)_\pm }$. Since $\textrm{Supp}({\tilde{f}})\subset (X^{d-1}{\setminus } A)_\pm$, we can think of ${\tilde{f}}$ as an extension of f in the sense of Claim B.1(3). Furthermore, by the chain structure, $\delta _{d}{\tilde{f}}=\delta _{d}\delta _{d-1}\mathbb {1}_{\varrho }=0$, that is ${\tilde{f}}\in \ker \delta _d$. Due to Claim B.1(3), this is equivalent to $\Delta _{X\backslash A}^{+}$ not being invertible.

• 5. This is the content of Duval et al. (2009, Proposition 4.1), see also Kalai (1983) for a discussion on the case case of d-simplexes.

$\square$