Past exam of the mathematics course of the University of Cambridge / 2023 / 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 a simply connected domain. Its mapping-out function of a compact H-hull is the unique conformal map $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with hydrodynamic normalization at infinityThe nonnegative coefficient $a$ is the half-plane capacity $\mathrm{hcap}(A)$.

Put $F(z)=z+z^{-1}$. On the unit semicircle, $F(e^{i\theta })=2\cos \theta$, andThe conformal invariance of planar Brownian motion and the Poisson kernel for the upper half-plane therefore give the exit density with respect to $d\theta$:As $z\to \infty$ in $\mathbb{H}$,uniformly in $\theta \in [0,\pi ]$. Hence

Let $\sigma$ be the first time Brownian motion started outside the unit half-disc reaches its semicircular boundary. A path that reaches $A$ must first cross that semicircle. The Strong Markov property at $\sigma$ givesTake $z=iy$, multiply by $y$, and let $y\to \infty$. The Brownian representation of half-plane capacity identifies the left side with $\mathrm{hcap}(A)$, while part b givesSince the exit height is between zero and one, the dominated convergence theorem applies and yields

Write $A_r=[-r,r]\times (0,1]$ and take $r\ge 1$. By the Brownian representation of half-plane capacity,On hitting $A_r$, the exit height is at most one. Moreover, $A_r$ lies in the half-disc of radius $\sqrt{r^2+1}\le 2r$. The harmonic measure of that semicircle as viewed from $iy$ is $O(r/y)$: mapping its exterior to $\mathbb{H}$ by $z\mapsto z+(2r{)}^2/z$ reduces the estimate to the Poisson kernel for the upper half-plane on an interval of length $O(r)$. Consequentlyfor large $y$, and $\mathrm{hcap}(A_r)\le Cr$. This is the half-plane capacity of a low rectangle.

Let $R_r=[-r,r]\times (0,1]$ and setThe scaling and translation of half-plane capacity and part i giveOn the other hand,Thus very long, very shallow hulls can have vanishing half-plane capacity.

The Conformal Markov property of SLE says that, conditionally on the curve through time $t$, mapping out the initial hull by $g_t-U_t$ turns the future into an independent Schramm–Loewner evolution in $(\mathbb{H},0,\infty )$. More precisely,has driving function ${\tilde{U}}_s=U_{t+s}-U_t$. Since $U_t=\sqrt{\kappa }{B}_t$, the stationary increments and independent increments of Brownian motion give the asserted independence and equality in law.

For the Bessel processthe scale function of a one-dimensional diffusion is $s(x)=x^{2-d}$ when $d\ne 2$. If $0<ϵ<x<R$, the boundary hitting probability from a diffusion scale function gives

If $d<2$, then $s(0)=0$ and $s(R)\to \infty$. Letting $ϵ\downarrow 0$ and then $R\to \infty$ shows that the process cannot escape to infinity before reaching zero. The exit time from each bounded interval is finite almost surely, so $T_0<\infty$ almost surely.

If $d>2$, then $s(ϵ)\to \infty$ in absolute value as $ϵ\downarrow 0$. Equivalently,Thus the process does not hit zero. The borderline case $d=2$ has scale function $\log x$ and also does not hit zero. This is the Hitting-zero classification for a Bessel process.

For $x>0$, the Chordal Loewner equation and $U_t=\sqrt{\kappa }{B}_t$ givePut $D_t=V_t^r-V_t^1$. The common Brownian term cancels, soThe Itô formula applied to $Z_t=\log D_t-\log V_t^1$ givesSincethe clock $q(u)={\int }_0^u(V_s^1{)}^{-2}ds$ and its inverse $\sigma$ turn the local-martingale term into Brownian motion by the Dambis-Dubins-Schwarz theorem. ThereforeThis is the SLE boundary-point logarithmic separation diffusion.

It is enough by Scaling invariance of SLE and reflection in the imaginary axis to treat $x=1$. Take $r=1+ϵ$. The drifttends to $(\kappa -8)/2<0$ as $z\to -\infty$. Choose $L<0$ and $a<0$ such that $b(z)\le a$ for $z\le L$. Before $\tilde{Z}$ reaches $L$,The infinite-horizon crossing probability for Brownian motion with negative drift shows that this process has a finite running maximum almost surely, so

Order preservation for the Loewner flow gives ${\tau }_1\le {\tau }_{1+ϵ}$. On $\{{\tau }_1<{\tau }_{1+ϵ}\}$, one has $V_t^1\to 0$ while $V_t^{1+ϵ}$ remains positive, so ${\tilde{Z}}_t\to +\infty$ and in particular $T_L<\infty$. HenceIf the trace itself hit the fixed boundary point $1$, then $1$ would be the right endpoint of the swallowed interval and ${\tau }_1<{\tau }_{1+ϵ}$ for every $ϵ>0$. Its probability is therefore zero. Scaling and reflection prove that SLE does not hit a fixed nonzero boundary point for every $x\in \mathbb{R}\setminus \{0\}$.

The Phase classification of the SLE trace is:

For a real boundary point $x\ne 0$, the Boundary-point Bessel flow for SLE says that $V_t^x/\sqrt{\kappa }$ is a Bessel process of dimensionWhen $\kappa \le 4$, one has $d\ge 2$, so part 2(b), including its logarithmic $d=2$ case, shows that no fixed boundary point is swallowed.

If the trace touched the real line away from its starting point, or if a later segment crossed an earlier segment, the resulting hull would disconnect from infinity a nonempty real interval, which contains a rational number. Applying the preceding argument after every rational time and using the Conformal Markov property of SLE rules this out on a countable probability-one event. Continuity of the trace then shows that no two distinct times have the same image. Thus ${\mathrm{SLE}}_{\kappa }$ is simple for $0<\kappa \le 4$.

Let ${\hat{B}}_t=r{B}_{t/r^2}$. By Brownian scaling, $\hat{B}$ is standard Brownian motion. Substituting $Y_t=r{X}_{t/r^2}$ into the Bessel equation givesWeak uniqueness for the Bessel equation shows that $Y$ is a Bessel process of dimension $d$ started at $rx$. This is the Scaling invariance of a Bessel process.

LetBy the stated fact, $R>0$ almost surely. Given $ϵ\in (0,1)$, choose a deterministic $r_0>0$ with $\mathbb{P}(R\ge r_0)\ge 1-ϵ$. The Scaling invariance of SLE givesand therefore, for every $t\ge 0$,

Fix $R>0$. Part i, with $ϵ$ arbitrarily small and $t$ sufficiently large, shows thatThe hulls increase with time. Once this half-disc lies in $K_t$, the stated future-avoidance property givesIntersecting these probability-one events over positive integer $R$ provesalmost surely. This is Transience of chordal SLE.

The space of test functions $C_0^{\infty }(D)$ consists of infinitely differentiable real-valued functions whose support of a function is a compact subset of $D$.

Equip $C_0^{\infty }(D)$ with the Dirichlet inner productThe zero-boundary Sobolev space $H_0^1(D)$ is its Hilbert space completion in the norm $\Vert f{\Vert }_{\nabla }=(f,f{)}_{\nabla }^{1/2}$.

The zero-boundary Gaussian free field $h$ on $D$ is the isonormal Gaussian process over $H_0^1(D)$: for $f,g\in H_0^1(D)$, the variables $(h,f{)}_{\nabla }$ are centered jointly Gaussian and satisfyEquivalently, for an orthonormal basis $(e_n)$ of $H_0^1(D)$ and independent standard normal variables $({\xi }_n)$,as a random generalized function.

For an open subdomain $U\subset D$, identify $H_0^1(U)$ with the closed subspace of $H_0^1(D)$ obtained by zero extension. Its orthogonal complement isIf $f\in {\mathcal{H}}_U$, then testing against $C_0^{\infty }(U)$ and integrating by parts gives $\Delta f=0$ in $U$ in the distributional derivative sense. The Weyl lemma therefore gives a representative harmonic on $U$.

The orthogonal decomposition by a closed subspace givesProject the isonormal process defining $h$ onto these two orthogonal subspaces. The projections are jointly Gaussian and uncorrelated, hence independent. The first projection is a zero-boundary Gaussian free field $h_U$ on $U$; the second is a random distribution $h^{\mathrm{har}}$ that is harmonic on $U$. Thuswith independent summands. This proves the Domain Markov property of the Gaussian free field.

The zero-Dirichlet Green function of the Laplacian $G_D(x,y)$ is symmetric, vanishes at the boundary in the appropriate sense, is harmonic in $x$ away from $y$, and satisfiesas a distribution. Equivalently, for suitable $\varphi$,

For $\varphi \in C_0^{\infty }(D)$, setand define the distributional pairing byIt is centered Gaussian by the definition of the GFF. Integration by parts givesThis is the Test-function pairing with a Gaussian free field.

Choose an orthonormal basis $(e_n)$ of $H_0^1(D)$ and write the GFF formally as $h={\sum }_n{\xi }_n{e}_n$, where the ${\xi }_n$ are independent standard normal variables. The Green-kernel expansion in the Dirichlet space givesConsequently the seriesconverges in $L^2$. Its partial sums are centered Gaussian and their variances converge to the displayed Green energy. The $L^2$ limit is therefore Gaussian with mean zero and varianceThis constructs the Finite-Green-energy measure pairing with a Gaussian free field independently of the chosen orthonormal basis.