Transfer-current theorem

For oriented edges $e_1,\dots ,e_k$, the probability that their unoriented versions all belong to a uniform spanning tree is the determinant of the matrix of transfer currents between them. In particular, every finite forest that can be extended to a spanning tree has positive inclusion probability.

In a finite unweighted connected graph, take distinct graph vertices $a\ne z$ and orient the unique graph path from $a$ to $z$ in a uniform spanning tree and assign signed unit current to its edges. Its mean over trees is the electrical unit flow from $a$ to $z$. The Kirchhoff node law follows by averaging graph path divergences. For the graph cycle law, let ${\mathcal{F}}_{az}$ be the two-component spanning forests separating the terminals and $A_F$ the component containing $a$. Deleting a tree-path edge and conversely adjoining an edge across a forest cut are inverse operations. Hence the mean signed current on an edge $u\to v$ isThis is the gradient of $h(u)=|\mathcal{T}(G){|}^{-1}{\sum }_F{1}_{A_F}(u)$, proving the graph cycle law. In particular $R_{\mathrm{eff}}(a,z)=|{\mathcal{F}}_{az}|/|\mathcal{T}(G)|$. For an oriented existing edge $e=(a,z)$, that edge is used by the tree graph path exactly when it belongs to the tree, and then has positive orientation. Thus its mean current is $\mathbb{P}(e\in T)$, giving the edge-inclusion formula for a uniform spanning tree by Ohm's law.

For an edge $e=\{a,b\}$ of conductance $c_e$ in a finite electrical network, its inclusion probability in the conductance-weighted uniform spanning tree is $c_e{R}_{\mathrm{eff}}(a,b)$. The same formula holds for the wired limit using the limiting wired effective resistance.