Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 203 (Schramm-Loewner Evolutions)

A compact H-hull is a bounded, relatively closed set $A\subset \mathbb{H}$ such that $\mathbb{H}\setminus A$ is simply connected. Its mapping-out function of a compact H-hull is the unique conformal bijectionwith hydrodynamic normalization

Uniqueness under hydrodynamic normalization at infinity givesandIndeed, each right-hand side maps the required complement conformally onto $\mathbb{H}$ and has expansion $z+O(1/z)$. This is the scaling and translation of half-plane capacity.

The assertion is false. Here $\mathbb{H}\setminus S=\{z:\mathrm{Im}z>1\}$. If a normalized map $g_S$ existed, thenwould be a conformal automorphism of $\mathbb{H}$. Hencewith real coefficients and $ad-bc=1$. Hydrodynamic normalization forces $h(w)\to \infty$ linearly, so $c=0$ and $h(w)=\alpha w+\beta$ with $\alpha >0$ and $\beta \in \mathbb{R}$. Butand $g_S(z)-z\to 0$ would require $\alpha =1$ and $\beta =i$, contradicting $\beta \in \mathbb{R}$.

Suppose a subsequence had $\mathrm{Im}{g}_A(z_n)\ge ϵ>0$. The images cannot tend to infinity, because the inverse mapping-out function satisfies $g_A^{-1}(w)=w+o(1)$ at infinity while $z_n$ remains bounded. A further subsequence therefore converges to some $w\in \mathbb{H}$. Continuity of $g_A^{-1}$ inside $\mathbb{H}$ would givecontradicting $z\in A$. Thus $\mathrm{Im}{g}_A(z_n)\to 0$.

Convergence of the real parts can fail. For the vertical slitthe two sides of the slit at $z=iy$, $0<y<1$, map to the two boundary values $\pm \sqrt{1-y^2}$. Alternating sequences approaching $iy$ from the two sides make $g_A(z_n)$ alternate between neighborhoods of these distinct limits.

Let $B$ be planar Brownian motion started at $z\in D$, let ${\tau }_D$ be its exit time, and let $\varphi :D\to D^{\prime }$ be conformal. DefineThe conformal invariance of planar Brownian motion states thatis Brownian motion started at $\varphi (z)$ and stopped when it exits $D^{\prime }$. Thus conformal maps preserve Brownian paths after this random time-change.

No. A conformal bijection is a homeomorphism and therefore induces an isomorphism of fundamental groups. The simply connected domain $D$ has trivial fundamental group, whereasThis contradiction rules out such a map.

By conformal invariance, $g_A(B)$ after its quadratic-variation time-change is Brownian motion in $\mathbb{H}$, started atSince $E\subset \mathbb{R}\setminus [-1,1]$ and $A\subset \overline{\mathbb{D}}$, the boundary correspondence is regular there, and the exit event maps to $g_A(E)$. The Poisson kernel of $\mathbb{H}$ therefore givesFor bounded subsets, multiplication by $y$ makes the integrand converge uniformly to $1/\pi$. Approximation by increasing bounded subsets and monotone convergence then giveswith both sides allowed to be infinite.

Let ${\tau }_{\mathbb{D}}$ be the first hit of the closed unit disc. The conformal map $z\mapsto 1/z$ sends its exterior to the punctured unit disc and sends $iy$ to $-i/y$. By conformal invariance, the hitting distribution on the unit circle is harmonic measure viewed from $-i/y$. As $y\to \infty$, this point tends to zero, where harmonic measure is normalized arc length by rotational invariance. Hence for every Borel $E\subset \partial \mathbb{D}$,

For a continuous real driving function $U$, solve the Chordal Loewner equationup to the first time $T_z$ at which $g_t(z)-U_t$ tends to zero. The generated hulls areThey form an increasing Loewner chain of compact H-hulls, satisfy $\mathrm{hcap}(K_t)=2t$, and $g_t$ is the mapping-out function of $K_t$.

The Conformal Markov property of SLE says that, conditionally on $(K_s{)}_{s\le t}$, the future hullshave the law of a fresh SLE in $(\mathbb{H},0,\infty )$ and are independent of the past, with the usual interpretation of mapping out the old hull.

Under the Loewner correspondence, mapping out the past replaces the driver byThe conformal Markov property therefore makes $U$ a continuous process with stationary independent increments. Every such process has the formScale invariance of SLE says that $r^{-1}{U}_{r^{2}t}$ has the same law as $U_t$. Comparing means forces $ar=a$ for every $r>0$, hence $a=0$; comparison of variances leaves the constant $\kappa$. Nondegeneracy gives $\kappa >0$, so

For $t<T_x$, differentiate the Loewner equation with respect to $x$:ThereforeThe exponent is nonpositive and decreases with $t$, so $0<g_t^{\prime }(x)\le 1$ and $g_t^{\prime }(x)$ is decreasing before $x$ is swallowed.

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