= Transfer current
{title2=$K(e,f)$}
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 $\phi$ is the resulting vertex potential, <Ohm's law> gives
$$
K(e,f)=\phi(c)-\phi(d),\qquad L\phi=\delta_a-\delta_b,
$$
where $L$ 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.<Local Characteristics, Entropy and Limit Theorems for Spanning Trees and Domino Tilings via Transfer-Impedances by Robert Burton and Robin Pemantle>[Theorem1.1 and section4, transfer-impedance kernel and determinants]{ref}
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