Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 203 (Schramm–Loewner Evolutions)

A compact H-hull is a bounded relatively closed set $A\subset \mathbb{H}$ for which $\mathbb{H}\setminus A$ is a simply connected domain. The Riemann mapping theorem and hydrodynamic normalization at infinity give a unique mapping-out function of a compact H-hull $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ withThe coefficient $a\ge 0$ is the half-plane capacity $\mathrm{hcap}(A)$.

For $\lambda >0$ and $x\in \mathbb{R}$, uniqueness of the normalized map givesSubstituting the expansion of $g_A$ yieldsTherefore the scaling and translation of half-plane capacity is

Let $d=\mathrm{diam}(A)$. Unless $A$ is empty, its closure meets the real axis; otherwise a loop in $\mathbb{H}\setminus A$ surrounding $A$ could not contract, contrary to $\mathbb{H}\setminus A$ being a simply connected domain. Choose $x\in \overline{A}\cap \mathbb{R}$. Then $A\subseteq \{z\in \mathbb{H}:|z-x|\le d\}$.

The unit half-disc is a compact H-hull with mapping-out function $z+z^{-1}$, so its half-plane capacity is one. The monotonicity of half-plane capacity and its scaling rule now giveThus the assertion holds with the universal constant $c=1$ under this normalization.

For $r\ge 1$, let $R_r=[-r,r]\times (0,1]$. The Brownian representation of half-plane capacity givesOn hitting $R_r$ the imaginary part is at most one, while the harmonic measure estimate supplied in the question shows that the probability of reaching a disc of radius $O(r)$ containing $R_r$ is $O(r/y)$. Hence $\mathrm{hcap}(R_r)\le Cr$, the half-plane capacity of a low rectangle bound.

SetThe scaling and translation of half-plane capacity giveswhereas $\mathrm{diam}(A_n)=\sqrt{4+n^{-2}}\to 2$. This supplies the required sequence.

Let $S_r=\{s{e}^{i\theta }:0<s\le r\}$. Since $S_r=r{S}_1$, the scaling and translation of half-plane capacity givesThe half-plane-capacity parameterization requires $ar(t{)}^2=2t$, so

The parameterization in part (c) gives the exact self-similarityfor every $\lambda >0$. The Loewner local growth property supplies a continuous Loewner driving function $U$. Under this scaling of the hulls, the deterministic scaling rule for the Chordal Loewner equation givesTaking $t=1$ and $\lambda =\sqrt{t}$ shows thatfor the real constant $c=U_1$.

For Schramm–Loewner evolution in $(\mathbb{H},0,\infty )$, the Scaling invariance of SLE states that, for every $a>0$,has the same law as $K_t$. The scaled Loewner driving function is ${\tilde{U}}_t=a^{-1}{U}_{a^{2}t}$. Since $U_t=\sqrt{\kappa }{B}_t$, the Brownian scaling identity $a^{-1}{B}_{a^{2}t}\stackrel{d}{=}{B}_t$ proves the claim.

The Conformal Markov property of SLE states that, conditionally on the hull through time $t$, the future hull mapped by $g_t-U_t$ is an independent ${\mathrm{SLE}}_{\kappa }$ in $(\mathbb{H},0,\infty )$. More precisely,has driving function ${\tilde{U}}_s=U_{t+s}-U_t$. The stationary increments and independent increments of Brownian motion show that $\tilde{U}$ is independent of ${\mathcal{F}}_t$ and has the same law as $\sqrt{\kappa }B$. The deterministic correspondence between continuous drivers and Loewner chains completes the proof.

Put $w=g_T^{-1}(U_T+z)$ and defineThen $f_0(z)=z$. Differentiating with the Chordal Loewner equation givesUniqueness for this ordinary differential equation shows that $f_s=h_s$. At $s=T$,which is the endpoint identity for the Reverse Loewner flow.

For $0<s<T$, the pathwise identity $h_s(z)=g_s^{-1}(U_s+z)-U_s$ is generally false. The left side is built from the reversed final driver segment $(U_{T-r}-U_T{)}_{0\le r\le s}$, whereas the right side is built from the initial segment $(U_r{)}_{0\le r\le s}$. By time reversal and symmetry of Brownian motion they have the same probability distribution, but they are not equal for the given Brownian path.

For $z=x+iy$ with $|x|\le 1$ and $0<y\le 1$, the given Reverse SLE derivative martingale starts fromIt is a nonnegative local martingale and therefore a supermartingale. Since its second factor is at least one,

Choosewhich is possible exactly because $\kappa >8$. At height $y_n=2^{-n}$, take a horizontal grid of spacing comparable to $y_n$ in $[-1,1]$. The Markov inequality gives, at each grid point,There are $O(y_n^{-1})$ grid points, so the probability that the bound fails anywhere on level $n$ is at most $C{y}_n^{1-2\alpha -8/\kappa }$. These probabilities are summable. The Borel-Cantelli lemmas therefore give an almost surely finite random constant controlling every sufficiently fine grid, and enlarging it handles the finitely many remaining levels.

Every point of the half-rectangle lies within a fixed hyperbolic distance of one of these grid points at comparable height. The Koebe distortion theorem compares the two derivatives by a universal factor. Hence an almost surely finite random $C$ satisfiesfor all $x\in [-1,1]$ and $0<y\le 1$. This proves the Reverse SLE derivative bound above the space-filling threshold.

Apply the derivative criterion for Hölder continuity up to a boundary to the estimate from part (c). For two points at distance $r$, move each vertically to height at least $r$, join them horizontally there, and move back. The two vertical integrals are bounded byand the horizontal integral is at most $Cr{r}^{-1+\alpha }=C{r}^{\alpha }$. Thus $h_T$ extends continuously to the bottom edge and satisfieson the half-rectangle. In particular, it is almost surely a Hölder continuous function there.

Set $\kappa =4/(d-1)>4$. For chordal Schramm–Loewner evolution, the centered image of a real boundary point, divided by $\sqrt{\kappa }$, follows the Boundary-point Bessel flow for SLE; changing $B$ to $-B$ matches the sign convention in the question. Thus ${\tau }_x$ and ${\tau }_y$ are the times at which the marked boundary points $x$ and $y$ are swallowed, or equivalently disconnected from infinity, by the Loewner chain.

ConsequentlyIt is the event that the negative marked point is swallowed before the positive marked point.

For every $a>0$, defineThe Brownian scaling theorem makes $\hat{B}$ a standard Brownian motion, and substitution shows that $(\hat{X},\hat{Y})$ satisfies the same coupled Bessel process equations from initial values $(x/a,y/a)$. Both hitting times are divided by $a^2$, so their order is unchanged. Taking $a=x$ or comparing any two pairs with the same ratio proves that $\mathbb{P}({\tau }_x>{\tau }_y)$ depends only on $x/y$.

The Strong Markov property and part (b) show that, before $\tau ={\tau }_x\wedge {\tau }_y$,Thus $F(V_{t\wedge \tau })$ is a bounded martingale. From the given stochastic differential equation,The Itô formula says that the drift of $F(V_t)$ isIt must vanish. Dividing by $(1-v{)}^2/2$ and using the algebraic identity supplied in the question gives

Because $\tilde{F}$ solves the differential equation from part (c), the Itô formula makes $\tilde{F}(V_t)$ a local martingale before $\tau$. The defining improper integral converges at both endpoints: its integrand is asymptotic to $(-u{)}^{1-d}$ near zero and to $|u{|}^{d-3}$ at minus infinity. Since $1<d<2$, both exponents are integrable. Henceso the stopped local martingale is a bounded martingale.

If ${\tau }_x<{\tau }_y$, then $V_t\to 0$ and $\tilde{F}(V_t)\to 0$. If ${\tau }_y<{\tau }_x$, then $V_t\to -\infty$ and $\tilde{F}(V_t)\to I$. The hitting times cannot coincide because $X_t-Y_t$ stays positive. The optional sampling theorem for a supermartingale and bounded convergence theorem therefore giveConsequentlywhich is the Two-sided SLE boundary swallowing probability.

The Zero-boundary Gaussian free field on the unit disc is the centered Gaussian process indexed by finite Borel measures of finite Green energy, with covarianceThis covariance determines all its finite-dimensional distributions.

The Domain Markov property of the Gaussian free field says that for every suitable open $U\subset \mathbb{D}$ one can writewhere $h_U$ is a zero-boundary Gaussian free field on $U$, independent of $h^{\mathbb{D}\setminus U}$, while $h^{\mathbb{D}\setminus U}$ is harmonic on $U$ and carries the information from the field outside $U$.

Take $U=B(0,e^{-t})$ in the Domain Markov property of the Gaussian free field. The field $h^{\mathbb{D}\setminus U}$ is a harmonic function throughout $U$. If $s>t$, the circle $\partial B(0,e^{-s})$ lies inside $U$, so the mean value property for harmonic functions gives

Write $h=h_U+h^{\mathbb{D}\setminus U}$ with $U=B(0,e^{-t})$. By part (b), the harmonic part contributes the same value $h^{\mathbb{D}\setminus U}(0)$ to every inner circle average. At radius $e^{-t}$ the zero-boundary part contributes zero; equivalently, take the limit from inner circles and use the assumed continuity. HenceThe field $h_U$ is independent of $h^{\mathbb{D}\setminus U}$, so the increment has the required independence.

The dilation $z\mapsto e^{t}z$ maps $U$ onto the unit disc and maps the circle of radius $e^{-s}$ onto the circle of radius $e^{-(s-t)}$. The Conformal invariance of the two-dimensional Gaussian free field therefore gives

Part (c), iterated over disjoint nested annuli, gives independent increments, and the law $X_s-X_t\stackrel{d}{=}{X}_{s-t}$ gives stationary increments. Every finite vector is jointly Gaussian by the definition of the Zero-boundary Gaussian free field, and a continuous version was assumed. Moreover $X_0=0$, because the field has zero boundary values.

Thus $X$ is a continuous centered Gaussian process with stationary increments and independent increments. Its variance is a continuous additive function on the nonnegative real numbers, so $\mathrm{Var}(X_t)={\sigma }^{2}t$ for some ${\sigma }^2\ge 0$. The Gaussian-process characterization of Brownian motion now givesfor standard Brownian motion $B$. This is the Circle-average process of the Gaussian free field is Brownian motion.