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

Set $\tilde{\Gamma }(x)=\Gamma (x)-\Gamma (b)$. Then $\tilde{\Gamma }(b)=0$, and subtraction of a constant leaves every edge difference unchanged:The linear maphas determinant $\pm 1$: its inverse is $\gamma (x)=\tilde{\gamma }(x)-\tilde{\gamma }(a)$, because $\gamma (a)=0$. The change of variables formula therefore shows that the new density is proportional toon functions vanishing at $b$. This is precisely the discrete Gaussian free field with Dirichlet boundary condition at $b$.

Only the $2d$ edges incident to $x_0$ depend on $\Gamma (x_0)$. Conditional on the remaining field, writeCompleting the square givesThe conditional density is therefore proportional to $e^{-(u-m{)}^2}$, and henceThis is the Gibbs-Markov property of the discrete Gaussian free field in the normalization used by the paper.

The covariance isIndeed, the killed-walk Green function satisfiesOn the other hand, the conditional mean from part 2 makes the covariance harmonic in $x\ne y,a$. At $x=y$, the conditional variance $1/2$ gives the same equation with source $\frac{1}{2}{1}_{\{x=y\}}$. Uniqueness of the Dirichlet problem therefore identifies the covariance with $G_{\{a\}}/2$. The factor $1/2$ comes from the coefficient $1/(2d)$ in this paper's energy density.

Reveal the field in the order $x_1,\dots ,x_K$ and evaluate its joint density at zero. By the Spatial Markov property of the Gaussian free field, after the values onhave been fixed to zero, the remaining field is the GFF killed on $R_i$. Part 3 therefore givesFactoring the joint density into conditional probability densities now yieldsThe left side is intrinsic and does not depend on the order used to factor the density. Thereforeis independent of the ordering.