Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 36 / 2

For chordal SLE with parameter $\kappa$, the Loewner trace $\gamma$ starts at zero and is parametrized by half-plane capacity $2t$. Let $D_t$ be the unbounded component of $\mathbb{H}\setminus \gamma [0,t]$, and let $K_t=\mathbb{H}\setminus D_t$ be its filled compact H-hull. When the trace is not simple, the hull includes the regions it disconnects from infinity. The associated mapping-out functions of compact H-hulls have expansionand satisfy the Chordal Loewner equationwhere $B$ is standard real Brownian motion. The equation is solved until $z$ is swallowed by the hull. The Loewner transform $\xi$ records the real boundary point at which the mapped hull grows; its relation to the trace isEquivalently the tip maps to ${\xi }_t$ in the appropriate boundary prime-end sense. Thus the flow encodes the surviving domains, the transform drives that flow, and the trace generates its hulls. For $\kappa =0$, the driver is identically zero, $g_t(z)=\sqrt{z^2+4t}$ with the branch asymptotic to $z$, and ${\gamma }_t=2i\sqrt{t}$.

After the fixed time $s$, the mapping-out function of the uncentered mapped future isIts expansion is ${\bar{g}}_t(z)=z+2t/z+O(z^{-2})$, so the elapsed time $t$ is still the half-plane-capacity parameterization. Differentiating in $t$ yieldsTherefore the Loewner transform of ${\bar{\gamma }}_t=g_s({\gamma }_{s+t})$ is ${\xi }_{s+t}$, with starting point ${\xi }_s$.

Centering gives ${\tilde{g}}_t(w)={\bar{g}}_t(w+{\xi }_s)-{\xi }_s$, whose driver is ${\tilde{\xi }}_t={\xi }_{s+t}-{\xi }_s$. By independent stationary Brownian increments, $(B_{s+t}-B_s{)}_{t\ge 0}$ is a standard Brownian motion independent of the past through time $s$. HenceThis proves the domain Markov property of a chordal Loewner chain from the composition rule for chordal Loewner driving functions. For possible contacts with the old hull, the mapped trace is understood through the continuous boundary extension or prime ends; on the open half-plane the displayed maps are ordinary conformal maps. The PDF asks first for the driver of the uncentered curve $\bar{\gamma }$; the centered driver is included as well.

For $0<\kappa \le 4$, use the allowed probability-one event that the whole SLE trace lies in $\mathbb{H}$ at every positive time. At each deterministic rational $s>0$, part (b) makes the centered mapped future a fresh SLE with the same property. Consequently, almost surely, for every $t>0$,This inverse relation holds through the trace's boundary-extension construction and puts positive-time future points in the open surviving domain. It therefore excludes every contact with the old filled hull, not just contacts with a fixed old point.

Intersect these probability-one events over all positive rational $s$. If ${\gamma }_u={\gamma }_v$ for $0<u<v$, choose rational $s$ with $u<s<v$. The first occurrence belongs to $K_s$, whereas the second lies in $D_s$, a contradiction. A return to ${\gamma }_0=0$ is excluded by the original positive-imaginary-part assertion. This simplicity of SLE from boundary-avoiding restarts proves