Math-1725 · mathematics · Probability

Math-1725

Infinite-cluster intersections with vertical fibers

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Statement

Definitions and mathematical statement

Let $G$ be an infinite connected locally finite graph. For independent bond percolation on $G$, let \[ p_c(G)=\inf\{q\in[0,1]:\mathbb P_q(o\text{ lies in an infinite open component})>0\}, \] where $o$ is any vertex. Suppose $p_c(G)=1$, so no constant parameter below one produces an infinite component.

Form the Cartesian product $G\times\mathbb Z$. It has vertices $(v,n)$ and edges $(v,n)\sim(w,n)$ when $v\sim_Gw$, together with $(v,n)\sim(v,n+1)$. For a fixed $p\in[0,1]$, retain all product edges independently with probability $p$. The vertical fiber above $v$ is $F_v=\{v\}\times\mathbb Z$.

Is it almost surely true that every infinite open cluster $C$ and every vertex $v\in V(G)$ satisfy \[ C\cap F_v\ne\varnothing\quad\Longrightarrow\quad |C\cap F_v|=\infty? \] The quantification is simultaneous over all clusters and all fibers in one realization, for each fixed parameter. If no infinite cluster exists, the assertion is vacuous; at $p=1$, the entire connected product is one cluster.

A cluster meeting a fiber infinitely often need not contain any infinite vertical run of open edges in that fiber. Its repeated returns may be linked through arbitrarily distant horizontal excursions. The hypothesis concerns the critical probability of $G$ itself, whereas the random cluster lives in its product with the line, which can have a smaller threshold. No transitivity of $G$ is imposed.

Short English statement

When the base graph cannot percolate below parameter one, must every infinite cluster in its product with the line return infinitely often to each fiber it visits?

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