Set $\widetilde\Gamma(x)=\Gamma(x)-\Gamma(b)$. Then $\widetilde\Gamma(b)=0$, and subtraction of a constant leaves every edge difference unchanged:
$$
\mathcal E(\widetilde\gamma)=\mathcal E(\gamma).
$$
The linear map
$$
(\gamma(x))_{x\ne a}\longmapsto
(\widetilde\gamma(x))_{x\ne b}
$$
has determinant $\pm1$: its inverse is $\gamma(x)=\widetilde\gamma(x)-\widetilde\gamma(a)$, because $\gamma(a)=0$. The <change of variables formula> therefore shows that the new density is proportional to
$$
\exp\{-\mathcal E(\widetilde\gamma)/(2d)\}
$$
on functions vanishing at $b$. This is precisely the discrete <Gaussian free field with Dirichlet boundary condition> at $b$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 209 / 3 / 1 / Solution

{c}

A Gaussian free field with Dirichlet boundary condition is fixed to zero on a boundary set and has density proportional to the exponential of minus one half of its discrete Dirichlet energy on the remaining vertices.


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