Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 220 / 3 / d

Fix $x\in \mathbb{R}\setminus \{0\}$ and, before $x$ is swallowed, put $X_t=(g_t(x)-U_t)/\sqrt{\kappa }$. The Chordal Loewner equation and $U_t=\sqrt{\kappa }{B}_t$ show, after changing the sign of the Brownian motion, thatThus $X$ is a Bessel process of dimensionA Bessel process hits zero exactly when $\delta <2$, which here is equivalent to $\kappa >4$. Hitting zero is precisely the swallowing of a nonzero boundary point by the SLE hull. For $\kappa \le 4$ the trace is simple and swallows no such point; for $\kappa >4$ swallowing occurs through a boundary contact. Hence the trace intersects $\partial \mathbb{H}\setminus \{0\}$ exactly when $\kappa >4$.

For $\kappa \le 4$, part i gives no nonzero boundary intersection. For $\kappa >4$, part i gives a boundary hit almost surely. After any hit, map out the past hull and recenter at the current tip. The Conformal Markov property of SLE says that the future is again an SLE in the remaining domain. Applying part i repeatedly and using Scaling invariance of SLE produces another boundary hit after every finite number of hits. Therefore there are almost surely infinitely many.

For $0<\kappa \le 4$, the trace meets the real boundary only at its starting point, so the intersection has zero Lebesgue measure. Suppose $4<\kappa <8$. A boundary point swallowed by an interval need not itself lie on the trace. The stated fact that $\mathbb{P}({\tau }_x={\tau }_y)>0$ for $0<x<y$, combined with the Strong Markov property and Scaling invariance of SLE at successively nested swallowed intervals, implies that a fixed deterministic $x\ne 0$ is swallowed in an interval rather than hit by the trace with probability one. Equivalently,Applying the Tonelli theorem to the random indicator of the boundary trace givesThe nonnegative random measure is therefore zero almost surely.