Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 214 / 3 / a

Fix an edge $e=\{u,v\}$ and orient it from $u$ to $v$. For each spanning tree $T$, let ${\gamma }_T$ be its unique graph path from $u$ to $v$. Assign a signed unit graph path flow $j_T$ to oriented edges: value $+1$ in the direction used by ${\gamma }_T$, value $-1$ on the reversed orientation, and zero otherwise. Define its meanEach $j_T$ has net divergence one at $u$, minus one at $v$ and zero elsewhere, so $J$ is a unit flow. The mean spanning-tree path current also satisfies the Kirchhoff cycle law. Although its verification is permitted to be omitted, a short counting argument makes it explicit. Let ${\mathcal{F}}_{uv}$ be the two-component spanning forests separating $u$ and $v$, write $A_F$ for the component containing $u$, and let $N_T$ be the number of spanning trees. Deleting an edge used by the oriented tree graph path gives one such forest. Conversely adjoining an edge across its two components gives a unique tree with that edge on its $u$-to-$v$ graph path. ConsequentlyThe gradient form proves the Kirchhoff cycle law directly and is Ohm's law for unit edge resistance. The Kirchhoff node law then gives $Lh={\delta }_u-{\delta }_v$, and the electrical unit flow is unique: the difference between two such voltages is a harmonic function on a graph and hence constant on the connected finite graph.

If $e\in T$, its single-edge route is the unique tree graph path from $u$ to $v$, so $j_T(u,v)=1$. If $e\notin T$, that tree graph path does not use $e$, so $j_T(u,v)=0$. ThereforeBy Ohm's law, $J(u,v)=h(u)-h(v)$; for a unit terminal current, this voltage difference is exactly the effective resistance between the terminals. We have proved the requested edge-inclusion formula for a uniform spanning tree:Now use linearity of expectation and the fact that every spanning tree has exactly $n-1$ edges. This gives Foster's theorem:The sums here use unoriented edges exactly once. For unit conductance this is the displayed identity; with conductances $c_e$ the inclusion probability and summand become $c_e{R}_{\mathrm{eff}}(e)$.