{c}

Let $\rho_t$ be uniform measure on the circle of radius $e^{-t}$ around the origin in the unit disc. The process $X_t=(h,\rho_t)$ is the circle-average process. The <Domain Markov property of the Gaussian free field> and the <mean value property for harmonic functions> imply that its increments are independent and stationary.

Circle-average process of the Gaussian free field

The <Zero-boundary Gaussian free field> on the unit disc is the centered <Gaussian process> indexed by finite <Borel measures> of finite <Green energy>, with <covariance>
$$
\mathbb E[(h,\rho)(h,\nu)]=\iint_{\mathbb D\times\mathbb D}G_{\mathbb D}(x,y)\rho(dx)\nu(dy).
$$
This covariance determines all its <finite-dimensional distributions>.

The <Domain Markov property of the Gaussian free field> says that for every suitable open $U\subset\mathbb D$ one can write
$$
h=h_U+h^{\mathbb D\setminus U},
$$
where $h_U$ is a zero-boundary Gaussian free field on $U$, independent of $h^{\mathbb D\setminus U}$, while $h^{\mathbb D\setminus U}$ is harmonic on $U$ and carries the information from the field outside $U$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 203 / 4 / a / Solution

Take $U=B(0,e^{-t})$ in the <Domain Markov property of the Gaussian free field>. The field $h^{\mathbb D\setminus U}$ is a <harmonic function> throughout $U$. If $s>t$, the circle $\partial B(0,e^{-s})$ lies inside $U$, so the <mean value property for harmonic functions> gives
$$
(h^{\mathbb D\setminus U},\rho_s)=h^{\mathbb D\setminus U}(0).
$$

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 203 / 4 / b / Solution

{c}

For an open $U\subset D$, the field decomposes as $h=h_U+h^{D\setminus U}$, where $h_U$ is an independent zero-boundary Gaussian free field on $U$ and $h^{D\setminus U}$ is harmonic on $U$ and agrees with the outside field in the distributional sense.


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