Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 203 / 4

The required set isThis is the boundary-intersection threshold in the Phase classification of the SLE trace.

For $x\in \mathbb{R}\setminus \{0\}$, the centered Loewner imageis the Boundary-point Bessel flow for SLE, a Bessel process of dimensionWhen $\kappa >4$, one has $d<2$, and the Hitting-zero classification for a Bessel process says that $Y$ hits zero almost surely. Thus every fixed nonzero boundary point is swallowed in finite time.

A hull generated by a continuous trace cannot swallow a real point while the trace remains strictly inside $\mathbb{H}$: before any contact with the real line, the trace is a crosscut-free interior curve and no boundary interval is disconnected from infinity. Consequently finite swallowing forces the trace to meet $\mathbb{R}$ away from its initial point. SLE therefore almost surely intersects the boundary for every $\kappa \in \Lambda$.

Stop after a positive initial segment and map it out. By the Conformal Markov property of SLE, the future is a fresh SLE in the remaining domain. Part b says that it hits that domain's boundary away from its current and terminal points. Choosing a bounded crosscut neighborhood whose real-boundary part is shielded from the future forces such a hit to occur on the image of the earlier trace with positive probability. Scale invariance and repeated conditional trials upgrade this to probability one. Thus the trace has self-intersections for $\kappa >4$, in agreement with the Phase classification of the SLE trace.

The statement is false; the displayed probability is zero. Let $\tau$ be the first exit from the fixed domain $\mathbb{D}$ and choose $x_n>0$ tending to zero. The Bessel scaling in part b givesHence $T_{x_n}\to 0$ in probability, while continuity of the trace gives $\tau >0$ almost surely. ThereforeSwallowing $x_n$ before leaving $\mathbb{D}$ forces a boundary contact away from zero before $\tau$. Taking the limit shows that such a contact occurs almost surely, so