Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 204 / 1

A cylinder event in $\{0,1{\}}^{E(G)}$ is an event determined by the states of finitely many edges. It is increasing when $\omega \in A$ and ${\omega }^{\prime }\ge \omega$ coordinatewise imply ${\omega }^{\prime }\in A$.

Let ${\mu }_n$ be the uniform spanning tree measure on the finite connected induced graph $G_n$. Uniform spanning-tree edge indicators are negatively associated: increasing events depending on disjoint edge sets have nonpositive covariance. Together with the spatial Markov property, this implies the free-boundary monotonicity under graph enlargement: for every increasing cylinder event $A$, ${\mu }_n(A)$ is eventually nonincreasing once $G_n$ contains all edges on which $A$ depends.

Define the free uniform spanning forest measure byfor increasing cylinder events. These limits determine a unique probability measure, independent of the exhaustion; equivalently, ${\mu }_n$ converges weakly to ${\mu }_F$ on the product space.

Fix a finite nonempty vertex set $K$. Once $G_n$ contains $K$ and every edge incident to it, every spanning tree of $G_n$ contains an edge of the finite cutbecause otherwise $K$ is disconnected from the rest of $G_n$. Henceand the same holds in the weak limit. If the free spanning forest had a finite component, its vertex set would be some finite connected $K$ and all edges of ${\partial }_{E}K$ would be absent. Taking the countable union over finite $K$ proves that every component is infinite almost surely.

Because $T$ itself is a tree, every finite induced connected exhaustion has the unique spanning tree consisting of all its edges. Thus its free spanning forest is deterministically $T$.

By the stated transience criterion, choose an edge $e=uv$ whose two complementary subtrees are transient. Run Wilson algorithm rooted at infinity first from $u$. With positive probability its loop-erased walk remains forever in the $u$-side. Starting next from $v$, there is likewise positive conditional probability that its walk remains forever in the $v$-side. On this event the two rays never use $e$, so $e$ is absent from the wired uniform spanning forest. The wired law is therefore not the deterministic free law.