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 a nonconstant entire function choose a point with nonzero derivative and a small closed disk around it where $|f^{\prime }{|}^2$ has a positive lower bound. recurrence of planar Brownian motion gives infinitely many successive crossings from a concentric inner circle to the disk boundary. Their durations are positive iid variables by strong Markov and rotational invariance of Brownian motion. The strong law applied to their bounded positive truncations makes their sum infinite. Hence the conformal clock is infinite and the analytic Brownian image, under time change of a continuous process, has full infinite Brownian lifetime. Centering the disk at a noncritical point is essential.

For planar Brownian motion started at $p+iq$, $p>0$ and $0<q<a$, in $\{x+iy:x>0,0<y<a\}$, the displayed value is its probability of exiting through $y=a$. The half-strip conformal map by hyperbolic cosine sends the top edge to $(-\infty ,-1)$ and the starting point to $A+iB$, where $A=\cosh (\pi p/a)\cos (\pi q/a)$ and $B=\sinh (\pi p/a)\sin (\pi q/a)$. Integrating the Cauchy exit law from a Brownian half-plane over that ray proves the formula. As $p\to \infty$ it tends to $q/a$.

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.

For Brownian motion started at $i$ in the upper half-plane, the real-axis hit $T$ is finite and its Cauchy distribution is nonzero almost surely. The continuous path on $[0,T]$ therefore stays in a compact annulus avoiding zero. The derivative of $z^{\alpha }$ is bounded on that annulus, proving finite changed exit time from its wedge image. The exponent zero is excluded because the map is constant and its image is not an open domain.

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.

For a nonconstant entire function, its derivative is nonzero somewhere and therefore bounded away from zero on a small disc. Recurrence of planar Brownian motion and the Strong Markov property imply infinite total occupation time in that disc: repeat entrances into a smaller concentric disc after exits from the larger one. On each entrance there is a uniformly positive chance of remaining inside the larger disc for a fixed positive duration. Iterating the conditional failure bound gives infinitely many such sojourns almost surely.

The displayed integral is bounded below by a positive constant times this occupation time, so it diverges. The same conclusion follows by comparing with the divergence of a positive planar Brownian occupation integral. At every finite time the clock is finite, because the Brownian path has compact range and $f^{\prime }$ is continuous. Thus the conformal Brownian clock covers every nonnegative image time; conformal invariance of planar Brownian motion then yields a Brownian image with no missing finite or infinite lifetime.

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.