Conformal invariance of planar Brownian motion

If $B$ is planar Brownian motion in a domain $D$ and $f:D\to D^{\prime }$ is a nonconstant holomorphic function, then $f(B)$, run with the clock ${\int }_0^t|f^{\prime }(B_s){|}^{2}ds$, is planar Brownian motion in $D^{\prime }$ until the relevant exit time. In particular, conformal maps transport harmonic measure between planar domains.

For planar Brownian motion started at radius $\epsilon$ and stopped on first hitting radius one, the net winding count divided by $\log \epsilon$ converges to a standard Cauchy random variable divided by $2\pi$. The complex exponential maps Brownian coordinates in the left half-plane to the punctured unit disk after the conformal Brownian clock. The unwrapped argument at exit is the vertical half-plane exit coordinate, with the Cauchy exit law from a Brownian half-plane. Counting integer turns introduces only a uniformly bounded error in winding units.

Planar Brownian motion started at $(-x,0)$ exits the left half-plane with ordinate distributed as $x$ times a standard Cauchy random variable. Conditional on the independent horizontal hitting time, the ordinate is normal with variance equal to that time. The inverse-square Gaussian law of Brownian first passage turns this into a ratio of independent standard normals. Its exit density is $x/(\pi (x^2+y^2))$.

For independent one-dimensional Brownian motions $\beta ,\theta$, the real and imaginary parts of $Z$ are continuous local martingales with equal bracket $A$ and zero cross bracket. The Divergence of a driftless Brownian exponential clock makes $A_{\infty }=\infty$, and the common-clock time change of orthogonal local martingales makes $Z_{A^{-1}(r)}$ planar Brownian motion started at $1$. The exponential never vanishes, proving that this Brownian motion never hits zero. Recurrent visits of $\beta$ to arbitrarily negative levels prove visits arbitrarily near zero at arbitrarily large Brownian times.

The map $z\mapsto z^{\pi /\alpha }$ takes a wedge of opening $\alpha$ to a half-plane and the circular boundary of radius $r$ to radius $r^{\pi /\alpha }$. Conformal invariance of planar Brownian motion preserves which boundary part is reached first, although it changes the clock. Use the branch defined by the wedge argument; the origin is a polar point for planar Brownian motion, so localization handles the vertex.

A planar Brownian motion starting at $1$ exits the disk of radius $r>1$ before leaving a wedge of opening $\alpha \in (0,2\pi ]$, centered on the positive axis, with probability $(4/\pi )\arctan (r^{-\pi /\alpha })$. The power-map reduction for Brownian exit from a wedge, the reflection identity for Brownian exit from a half-disc and the Möbius calculation of circular Brownian exit give the formula. Its large-radius decay exponent is $\pi /\alpha$.

For a conformal bijection $\varphi$ and planar Brownian motion $B$, this increasing clock measures the common quadratic variation of the two coordinates of $\varphi (B)$. The transformed Brownian path is $\varphi (B_{A^{-1}(t)})$. The inverse clock runs from image time to original time.

A planar Brownian motion exits a conformal image of a bounded planar domain in finite time almost surely. If the image lifetime were infinite, the inverse derivative is bounded away from zero on a small interior disc. Recurrence of planar Brownian motion and the Strong Markov property give infinite Brownian occupation time there, forcing the inverse conformal Brownian clock to diverge. It cannot exceed the finite original exit time. The expected image exit time can nevertheless be infinite.