Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 203 / 2

Let $B$ be planar Brownian motion started at $z\in D$, let ${\tau }_D$ be its exit time, and let $\varphi :D\to D^{\prime }$ be conformal. DefineThe conformal invariance of planar Brownian motion states thatis Brownian motion started at $\varphi (z)$ and stopped when it exits $D^{\prime }$. Thus conformal maps preserve Brownian paths after this random time-change.

No. A conformal bijection is a homeomorphism and therefore induces an isomorphism of fundamental groups. The simply connected domain $D$ has trivial fundamental group, whereasThis contradiction rules out such a map.

By conformal invariance, $g_A(B)$ after its quadratic-variation time-change is Brownian motion in $\mathbb{H}$, started atSince $E\subset \mathbb{R}\setminus [-1,1]$ and $A\subset \overline{\mathbb{D}}$, the boundary correspondence is regular there, and the exit event maps to $g_A(E)$. The Poisson kernel of $\mathbb{H}$ therefore givesFor bounded subsets, multiplication by $y$ makes the integrand converge uniformly to $1/\pi$. Approximation by increasing bounded subsets and monotone convergence then giveswith both sides allowed to be infinite.

Let ${\tau }_{\mathbb{D}}$ be the first hit of the closed unit disc. The conformal map $z\mapsto 1/z$ sends its exterior to the punctured unit disc and sends $iy$ to $-i/y$. By conformal invariance, the hitting distribution on the unit circle is harmonic measure viewed from $-i/y$. As $y\to \infty$, this point tends to zero, where harmonic measure is normalized arc length by rotational invariance. Hence for every Borel $E\subset \partial \mathbb{D}$,