 Transfer current

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]