Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 214 / 4 / c

Solution

Exhaust

Z^{2}

by finite boxes

G_{n}

and let

T_{n}

G_{n}

. Every finite tree satisfies the handshaking lemma, so

\frac{1}{∣ V ( G _{n} ) ∣} \sum_{v \in V (G_{n})} de g_{T_{n}} (v) = \frac{2 ( ∣ V ( G _{n} ) ∣ - 1 )}{∣ V ( G _{n} ) ∣} ⟶ 2.

Choose the root uniformly from

V (G_{n})

. The proportion of roots within any fixed distance of the boundary tends to zero, and the rooted trees converge locally to the uniform spanning tree

T

Z^{2}

. Since every degree is at most four, expectations also converge. Translation invariance therefore gives

E [de g_{T} (0)] = 2.

The four edges incident to

0

have equal inclusion probability by the rotations and reflections of the square lattice. If that common probability is

r

, then

4 r = E de g_{T} (0) = 2

. Consequently

P (e \in T) = \frac{1}{2}

for every edge

e

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