Math-1724 · mathematics · Probability

Math-1724

Percolation nonuniqueness on products with the line

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Statement

Definitions and mathematical statement

For an infinite graph $H$, define its Cheeger constant by $h(H)=\inf_{0<|A|<\infty}|\partial_H A|/|A|$, using external vertex boundary. Call a connected locally finite graph $G$ strongly amenable in this problem's sense if no infinite connected subgraph $H\subseteq G$ has $h(H)>0$. This is a graph-theoretic definition, not a condition on group actions.

The Cartesian product $G\times\mathbb Z$ has vertices $(v,n)$, with edges between $(v,n)$ and $(w,n)$ whenever $v\sim_Gw$, and between $(v,n)$ and $(v,n+1)$. Independently retain its edges with probability $p$.

Does there exist a strongly amenable $G$ and numbers $0<p_1<p_2<1$ such that, for each fixed $p\in[p_1,p_2]$, this percolation has infinitely many infinite open components almost surely? Can such an example be obtained with polynomial volume growth, meaning that for some vertex $o$ and constants $C,D<\infty$, \[ |B_G(o,r)|\le Cr^D\qquad(r\ge1)? \] The question is existential over graphs, rather than an assertion that every strongly amenable graph has such an interval. The polynomial-growth part asks for an example in that narrower class.

Requiring a nondegenerate interval distinguishes the question from exceptional behavior at one parameter. Since these products need not be transitive, uniqueness of their infinite cluster need not be monotone in $p$; no equivalence with a simple $p_c<p_u$ criterion is assumed. The two product directions and the edges within each layer use the same Bernoulli parameter.

Short English statement

Can a product of a strongly amenable graph with the line have infinitely many infinite clusters throughout an interval of percolation parameters?

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