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Robert Burton coauthored the transfer-impedance analysis of uniform spanning trees with Robin Pemantle. Their original paper verifies this byline.^[ref]

The method computes probabilities for small collections of edges from determinants, even when the graph is large. The relevant matrix entries depend on global electrical information, so a small determinant does not mean the problem is entirely local. This distinction is essential in extending the calculations to infinite periodic graphs.^[ref]

Robin Pemantle studies probability theory and random combinatorial structures. With Robert Burton, he developed methods for computing local probabilities in uniform spanning trees using electrical transfer impedances.^[ref]

The probability that several specified edges occur is a determinant of a small transfer matrix. This avoids enumerating all trees and relates joint edge events to Green's functions and network currents. Their work also applies these methods to domino tilings and entropy.^[ref]

In a finite connected graph with unit edge electrical resistances, orient two edges $e=(a,b)$ and $f=(c,d)$. The transfer current $K(e,f)$ is the signed current through $f$ when one unit of electric current is injected at $a$ and removed at $b$. If $\varphi$ is the resulting vertex potential, Ohm's law giveswhere $L$ is the graph Laplacian. Potentials are determined up to an additive constant, which cancels in the difference.

On a unit-resistance triangle, injecting across one edge sends $2/3$ of the current along that edge and $1/3$ along the two-edge alternative path. Thus $K(e,e)=2/3$. The transfer-current theorem identifies this diagonal value with the probability that the edge belongs to a uniform spanning tree; exactly two of the triangle's three spanning trees contain it.

Electrical reciprocity makes the unit-resistance kernel symmetric. Reversing one edge orientation changes the appropriate sign, while the determinants giving joint tree-edge probabilities remain unchanged.^[ref]

In a finite connected graph with unit electrical resistance on each edge, choose distinct undirected edges $e_1,\dots ,e_k$ and orient each once. Define the transfer current $K(e,f)$ as the signed electric current through $f$ when one unit of electric current enters the tail of $e$ and leaves its head. For a uniform spanning tree $T$,Reversing one orientation negates its row and column, leaving the determinant unchanged. Distinctness is necessary: repeating an edge repeats a row and gives zero, although its inclusion can have positive probability.

Electrical reciprocity makes $K$ symmetric. The two-edge case therefore givesThus edge inclusions are negatively correlated.^[ref] The positive-weight version uses the corresponding conductance-normalized transfer current. Every extendable finite forest has positive inclusion probability, since at least one positively weighted spanning tree contains it.