Reveal the field in the order $x_1,\ldots,x_K$ and evaluate its joint density at zero. By the <Spatial Markov property of the Gaussian free field>, after the values on
$$
R_i=\{a,x_1,\ldots,x_{i-1}\}
$$
have been fixed to zero, the remaining field is the GFF killed on $R_i$. Part 3 therefore gives
$$
\operatorname{Var}(\Gamma(x_i)\mid\Gamma(x_1)=\cdots=\Gamma(x_{i-1})=0)
=\frac12G_{R_i}(x_i,x_i).
$$
Factoring the joint density into <conditional probability> densities now yields
$$
p_\Gamma(0,\ldots,0)
=\prod_{i=1}^K
\frac1{\sqrt{\pi G_{R_i}(x_i,x_i)}}.
$$
The left side is intrinsic and does not depend on the order used to factor the density. Therefore
$$
\prod_{i=1}^KG_{R_i}(x_i,x_i)
$$
is independent of the ordering.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 209 / 3 / 3 / 4 / Solution

{c}

Conditioning a Gaussian free field on its values along a boundary leaves an independent zero-boundary Gaussian free field plus the harmonic extension of those boundary values in each remaining component.


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 Incoming links: Spatial Markov property of the Gaussian free field