Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 29 (Schramm--loewner Evolutions)

A compact H-hull is a bounded set $K\subset \mathbb{H}$ that is closed relative to the complex upper half-plane and whose complement $D=\mathbb{H}\setminus K$ is a simply connected domain. Its Euclidean closure is compact in $\overline{\mathbb{H}}$; “compact” here does not require separation from the real axis. One may equivalently describe the closed hull in $\overline{\mathbb{H}}$, with its real-boundary convention understood. Its mapping-out function is the unique conformal map to $\mathbb{H}$ with hydrodynamic normalization at infinity.

Use the capacity convention $\mathrm{hcap}(K_t)=2t$. A chordal Schramm–Loewner evolution from $0$ to infinity is the hull family of the Chordal Loewner equationwhere $W$ is a real standard Brownian motion and $\kappa \ge 0$. The points with interior-point swallowing time for a Loewner chain at most $t$ constitute $K_t$. For each $t$, $g_t$ maps $\mathbb{H}\setminus K_t$ conformally onto $\mathbb{H}$ and has expansion $g_t(z)=z+2t/z+O(z^{-2})$. The case $\kappa =0$ is the deterministic vertical-slit chain.

First identify the relevant transformed Loewner driving functions directly. For $\lambda >0$, defineDifferentiation gives ${\partial }_t{\tilde{g}}_t=2/({\tilde{g}}_t-{\lambda }^{-1}{U}_{{\lambda }^{2}t})$. Its initial value is $z$, its domain is $\mathbb{H}\setminus {\lambda }^{-1}{K}_{{\lambda }^{2}t}$, and its expansion is $z+2t/z+O(z^{-2})$. Thus its driving function is ${\lambda }^{-1}{U}_{{\lambda }^{2}t}$, by uniqueness of the differential equation and of the hydrodynamically normalized map. Brownian scaling provesThis is capacity-parametrized scale invariance of a Loewner chain.

For a fixed time $t$, map the future remaining domains by $z\mapsto g_t(z)-U_t$. The corresponding hulls ${\hat{K}}_s$ are specified without any boundary-image ambiguity byTheir mapping-out functions areAt $s=0$ this is the identity. Differentiating at a point in its domain yieldsThe Laurent series is $z+2s/z+O(z^{-2})$. Hence the transformed Loewner driver is exactly $U_{t+s}-U_t$, with the original capacity clock unchanged. This proves the composition rule for chordal Loewner driving functions, rather than assuming an identification of transforms.

The Brownian motion increments after $t$ are independent of its past and have the original law. Since past hulls are measurable functions of the past Loewner driver, conditionally on the past, the centered mapped future is an independent copy of the original chain. This is the domain Markov property of a chordal Loewner chain. At an almost surely finite stopping time it follows likewise from the Strong Markov property.

For the converse, a precise class is essential: take capacity-parametrized Loewner chains generated by a continuous real Loewner driving function $U$, with $U_0=0$, and require the two displayed hull properties, including independence from the entire hull past in the domain Markov property. Within this class they characterize ${\mathrm{SLE}}_{\kappa }$, for some $\kappa \ge 0$.

To see why the hull assertions determine the Loewner driver assertions, the hull domains determine their normalized maps uniquely. For any surviving point,The left time derivative suffices for $t>0$, since $U$ is continuous. A sufficiently high point survives up to any given finite time, so points $in$, $n\in \mathbb{N}$, suffice to reconstruct the Loewner driver locally and measurably from the hull past. Thus the driving-function reconstruction for a chordal Loewner chain identifies the two past filtrations and makes the preceding scale and composition computations reversible.

The domain Markov property therefore gives stationary independent increments for $U$. Its continuity makes it a continuous Lévy process. Its classification gives $U_t=bt+\sqrt{\kappa }{W}_t$: in the Lévy–Khintchine formula continuity removes the jump measure, leaving characteristic function $\exp (t(ibu-\kappa u^2/2))$. Driver scaling would turn the drift $b$ into $\lambda b$, so invariance for every $\lambda$ forces $b=0$. The diffusion coefficient is unchanged. This proves the scale-and-domain-Markov characterization of SLE with its regularity and parametrization hypotheses stated explicitly.

Set $\beta =\pi /\alpha$ and use the branch $\varphi (z)=z^{\beta }$ with argument in $(-\alpha /2,\alpha /2)$. This is a conformal map from the wedge to the right half-plane, fixes the starting point $1$, and sends the outer circle of radius $r$ to that of radiusThe two wedge sides map to the imaginary axis.

By conformal invariance of planar Brownian motion, the image of the stopped path is planar Brownian motion after the increasing conformal Brownian clock $A(t)={\int }_0^t|{\varphi }^{\prime }(B_s){|}^{2}ds$. This clock does not change which boundary portion is reached first. Localization away from the vertex justifies the map even when its derivative is unbounded there; the vertex is a polar point for planar Brownian motion and has zero hitting probability from $1$.

Consequentlywhere the probability on the right is for the right half-plane. The power-map reduction for Brownian exit from a wedge also works at $\alpha =2\pi$, when the wedge is the plane slit along the negative real axis.

Let $E_{+}$ and $E_{-}$ denote positive and negative real part at the circular exit. On $\{S<T(r)\}$, reflect the portion of the planar Brownian motion after $S$ by $z\mapsto -\overline{z}$. This reflection fixes the imaginary axis and preserves distances from the origin. The Strong Markov property and reflection symmetry show that the resulting path has the same law, its circular exit time is unchanged, and $E_{+}$ is exchanged with $E_{-}$. ThereforeAn exit with negative real part must first cross the imaginary axis. An exit before $S$ has positive real part. The two points $ir,-ir$ have zero circular exit probability, since circular harmonic measure has no atoms. HenceSubtracting provesThis is the reflection identity for Brownian exit from a half-disc.

Scale the disk to the unit disc, so that the starting point is $a=1/r\in (0,1)$. Use the Möbius transformationwhich maps the disk onto itself, sends $a$ to $0$, and sends $1$ to $1$. By conformal invariance of planar Brownian motion, the exit image is the circular exit of a Brownian motion starting at $0$. Rotational invariance of planar Brownian motion makes that exit uniform in angle.

The endpoints of the right semicircle satisfyIts image is the arc through $1$ between these points, of angular length $2\phi$. Thus the Möbius calculation of circular Brownian exit givesSince $E_{+}$ and $E_{-}$ partition the exit almost surely, part (b) yieldsThe last equality uses $2\arctan (1/r)\in (0,\pi /2)$, so the double-angle tangent identity uses the stated principal branch. As a check, the boundary angle derivative of ${\psi }_a$ is $(1-a^2)/|e^{i\theta }-a{|}^2$; integrating this circular exit density over the right semicircle gives exactly the integral supplied in the question.

Substitute $R=r^{\pi /\alpha }$ from part (a) into part (c). The Brownian wedge-exit probability isIt tends to $1$ as $r\downarrow 1$, and is asymptotic to $(4/\pi )r^{-\pi /\alpha }$ as $r\to \infty$. The exponent reflects how the conformal map stretches wedge angles; wider wedges give slower decay.

Write $D=\mathbb{H}\setminus K$. The mapping-out function is a conformal map $g_K:D\to \mathbb{H}$ with hydrodynamic normalization at infinityEquivalently, $g_K(z)-z\to 0$. The Riemann mapping theorem gives a map onto the half-plane. Since $K$ is bounded, infinity has an analytic real-boundary neighborhood. After sending its image to infinity, the Schwarz reflection principle gives a Laurent series $az+b+O(z^{-1})$ there, with $a>0$ and $b$ real. A real affine automorphism removes $a$ and $b$, giving the stated normalization. Reflection also makes $a_K$ real.

If $g_1,g_2$ both have this normalization, $g_2{g}_1^{-1}$ is a conformal automorphism of the upper half-plane. Such maps are real Möbius transformations. The expansion forces it to fix infinity, have leading coefficient $1$, and have zero translation. It is therefore the identity. Thus the normalized mapping-out function exists and is unique.

The half-plane capacity is the Laurent coefficientIt is real and nonnegative. For planar Brownian motion starting at $z\in D$, let ${\tau }_D$ be its first exit from $D$. The Brownian representation of half-plane capacity isOne may also state the underlying harmonic identity, valid throughout $D$,The process is killed at the hull or real boundary; real-boundary exits contribute zero. These are the Brownian characterizations, with the normalization that a Chordal Loewner equation at speed $2$ produces capacity $2t$.

Let $f=g_K^{-1}$. The reflection at infinity used in part (a) gives analytic expansions there for both inverse maps. In particular,uniformly for large $w\in \mathbb{H}$, including approach close to the real axis.

Fix $R<\infty$. If $g_K$ were unbounded on $D\cap \{|z|\le R\}$, there would be points $z_n$ there with $w_n=g_K(z_n)$ tending to infinity in modulus. The inverse expansion would imply $z_n=f(w_n)=w_n+O(1/w_n)$, contradicting boundedness of $z_n$. Thus $g_K$ is bounded on every bounded portion of its domain, including points arbitrarily near a rough hull boundary. This is the inverse-at-infinity criterion for local boundedness of a mapping-out function.

On the region $|z|>R$ for sufficiently large $R$, the expansion of $g_K$ gives $g_K(z)-z=O(1/z)$. On the remaining bounded region, the preceding bound for $g_K$ and the bound for $z$ give a finite bound for their difference. ThereforeThe argument uses reflection near infinity only. It does not assume that the real part of the map extends continuously at every point of an arbitrary hull; the sharp displacement bound for a compact H-hull is a further quantitative version of this boundedness.

Set $v(z)=\mathrm{Im}{g}_K(z)$, a positive harmonic function on $D$. First justify its boundary behavior. For $z_n\to p\in \partial D$ with $p$ finite, part (c) makes $g_K(z_n)$ bounded. Any subsequential limit $w$ lies in $\overline{\mathbb{H}}$. If $\mathrm{Im}w>0$, continuity of the inverse inside $\mathbb{H}$ would give $p=g_K^{-1}(w)\in D$, a contradiction. ThusThis boundary degeneration under a mapping-out function is valid without a smooth or locally connected hull boundary.

The harmonic function $u(z)=v(z)-\mathrm{Im}z$ consequently has nonpositive finite-boundary values. Its value tends uniformly to zero at infinity, by the Laurent series. On the bounded domain $D\cap \{|z|<R\}$, the maximum principle for harmonic functions bounds $u$ by $\max (0,{\epsilon }_R)$, where ${\epsilon }_R={\sup }_{D\cap \{|z|=R\}}|g_K(z)-z|\to 0$. Let $R\to \infty$ with $z$ fixed. We obtainThis is the height contraction of a hydrodynamically normalized mapping-out function. The exhaustion controls infinity explicitly, which is necessary when using a maximum principle on an unbounded domain.

Here the otherwise undefined printed domain $H$ must mean $D=\mathbb{H}\setminus K$. Let the Brownian motion start at $z\in D$, and let ${\tau }_R$ stop it on reaching height $R$ or leaving $D$. Take $R>\max (\mathrm{Im}z,{\sup }_K\mathrm{Im}w)$. This stopping time is finite almost surely, since it is no later than exit of its imaginary coordinate from $(0,R)$.

By part (d), $0\le v(w)\le R$ in the stopped domain. Its finite-boundary values vanish except on the top boundary, by the argument of part (d). Localization and the optional stopping theorem for this bounded harmonic martingale therefore giveWrite $p_R={\mathbb{P}}_z(\mathrm{Im}{B}_{{\tau }_R}=R)$ and use the uniform displacement bound $M$ from part (c). On the top boundary, $|v(w)-R|\le M$, henceThe stopped imaginary coordinate is itself a bounded martingale. Since its exit height is nonnegative,Consequently $|v(z)-R{p}_R|\le M\mathrm{Im}z/R\to 0$, proving the high-level escape representation of a mapping-out height:If $H$ instead meant the whole upper half-plane, the right side would always be $\mathrm{Im}z$. For example, the mapping-out function of a vertical slit $K=(0,ia]$ is $g_K(w)=\sqrt{w^2+a^2}$ with branch asymptotic to $w$; at $z=iy$, $y>a$, its height is $\sqrt{y^2-a^2}<y$. This demonstrates why killing on the hull is essential to the printed formula.

Use $U_t=2W_t$ and write $Z_t=X_t+i{Y}_t$ up to its interior-point swallowing time for a Loewner chain $T_z$. The Chordal Loewner equation givesThe branch of the complex logarithm with argument in $(0,\pi )$ is well-defined while $t<T_z$. The Itô formula yields the crucial cancellationThe $2/Z_t^2$ drift is cancelled by the quadratic variation term, because $\kappa =4$. Thus both $L_t=\log |Z_t|$ and ${\theta }_t=\arg Z_t$ are continuous local martingales, withThis is the logarithmic martingale for SLE4.

We justify absence of a finite swallowing time rather than presuming that the logarithm survives forever. On a finite horizon $H$, $0<Y_t\le \mathrm{Im}z$. Also $|X_t|$ is bounded pathwise before $T_z\wedge H$. Indeed, while $|X_t|\ge 1$ its drift $2X_t/|Z_t{|}^2$ has absolute value at most $2$; on each excursion outside $[-1,1]$, integrate from its starting point and bound the Brownian oscillation on $[0,H]$. For example,Therefore $L_t$ is bounded above pathwise on this interval.

Suppose $T_z\le H$. By the Dambis-Dubins-Schwarz theorem, $L$ is a Brownian motion run at its own quadratic variation. If that clock diverged as $t\uparrow T_z$, Brownian oscillation would make $L$ unbounded above, contradicting the preceding bound. The one-sided bound criterion for a martingale clock therefore gives a finite clock limit and a finite real limit for $L_t$. Hence $|Z_t|$ is bounded away from zero near $T_z$.

Now $Y_t=Y_0\exp (-2{\int }_0^t|Z_s{|}^{-2}ds)$ has a strictly positive limit at $T_z$. The drift in $X$ is integrable there, so continuity of $W$ gives a finite limit for $X$ as well. The limiting point is in $\mathbb{H}$ and away from the Loewner driver singularity, and the differential equation extends past $T_z$, a contradiction. Thus $T_z=\infty$ almost surely. A point of the Loewner trace at a finite time belongs to that time's hull, soThis proves the fixed-interior-point avoidance of SLE4 without using simplicity as an input.

The angle remains in $(0,\pi )$ at all finite times. The bounded local martingale criterion upgrades its local martingale equation to a genuine martingale:In particular this is the SLE4 angle martingale, and it converges almost surely and in $L^1$ by bounded Martingale convergence theorem.

It remains to identify the limiting angle using the assumed simple path tending to infinity. Orient that path from $0$ to infinity. Its left component is the one adjacent to the negative real half-axis. Under $g_t$, the left boundary of the slit domain maps to $(-\infty ,U_t)$ and the right boundary to $(U_t,\infty )$. The harmonic measure of the former as seen from $z$ isThis follows by conformal invariance of planar Brownian motion: in the upper half-plane, $\arg (w-U_t)/\pi$ is the bounded harmonic function with values $1$ on the left half-axis and $0$ on the right.

To justify the limiting boundary classification, condition on a simple proper realization of the path and use an independent planar Brownian motion from $z$. It exits the upper half-plane in finite time almost surely, so its path up to that time is compact. The curve tends to infinity, so its intersection with this compact set is contained in a finite initial curve segment. Once that segment has been drawn, the Brownian path exits the slit domain through its left boundary exactly when $z$ is in the final left component: a path from that component cannot reach the right boundary without crossing the curve, and the reverse assertion holds on the right. Endpoints have zero harmonic measure. Bounded convergence of these exit indicators provesTaking expectations in the bounded angle martingale gives the SLE4 left-passage probabilityFor the imaginary axis it is $1/2$; near the negative real axis it tends to $1$, fixing the orientation of “left”.