Suppose I have a connected graph $G$ and a fixed edge $e = \langle u, v \rangle \in G$, and I want to count the number of spanning trees that involve $e$. I really only want to estimate the fraction of spanning trees containing $e$ compared to the total number of spanning trees $G$, that is, I want to find a lower bound $c \leq \kappa(G \backslash e) / \kappa(G)$
This lower bound should be in terms of the degrees of vertices $u,v$. Let $c(d,d')$ be the smallest possible value of $\kappa(G \backslash e) / \kappa(G)$ when the vertices have degree $d, d'$. What can one say about $c(d,d')$?
For example, if $d = 1$ or $d' = 1$, then $c(d,d') = 1$. (The edge must be part of a spanning tree of $G$.).
If $d = 2$, then $c(d,d') \geq 1/2$, as every spanning tree must involve one of the two edges incident on $u$, and the spanning trees using only $e$ are in bijection with the spanning trees using the other edge.
If $d = d' = 2$, then $c(d,d') \geq 2/3$; and so on.
Is there are general formula for $c(d,d')$?
Best Answer
The probability that an edge $e=(u,v)$ is part of a uniform spanning tree is equal to the resistance between $u$ and $v$ when the graph is considered as an electric network (see the book by Lyons with Peres, section 4.2). The bounds you get (in term of the degrees $d_u,d_v$) are
$$ \frac{1}{\min(d_u,d_v)} \le R_{eff}(u \leftrightarrow v) \le 1$$
when you allow multiple edges, or
$$ \frac{(d_u-1)+(d_v-1)}{(d_u-1)+(d_v-1)+(d_u-1)(d_v-1)} < R_{eff}(u \leftrightarrow v) \le 1$$
when the graph is simple, and these bounds are sharp.