Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 36 (Stochastic Loewner Evolutions)

Write $D=\mathbb{H}\setminus K$. By definition of a compact H-hull, $D$ is a proper simply connected domain, and it agrees with the complex upper half-plane outside a sufficiently large disc. The Riemann mapping theorem supplies a conformal isomorphism $f:D\to \mathbb{H}$. Choose its boundary normalization so that the prime end at infinity maps to infinity. This prime end is unambiguous because the domain has an ordinary straight real boundary near infinity.

The Schwarz reflection principle, applied there in the coordinate $1/z$, extends $f$ across the real boundary near infinity. It has a simple pole there, with Laurent seriesThe pole is simple because the reflected map is locally conformal at infinity; its leading coefficient is positive because it maps the upper side to the upper side. Postcomposing with the half-plane automorphism $w\mapsto (w-b)/a$ givesThis is hydrodynamic normalization at infinity.

If $g_1,g_2$ both satisfy it, $g_2\circ g_1^{-1}$ is a conformal automorphism of the upper half-plane. The asymptotics force it to fix infinity, so it has the form $w\mapsto \alpha w+\beta$ with $\alpha >0$ and $\beta \in \mathbb{R}$. The same asymptotics force $\alpha =1$, $\beta =0$. Hence the normalized mapping-out function exists and is unique. The argument requires reflection only outside a large disc, not regularity of the hull's entire boundary.

For the Joukowski map $g(z)=z+z^{-1}$,It maps the exterior upper half-disc $D_0=\{z\in \mathbb{H}:|z|>1\}$ into $\mathbb{H}$. It is injective there: $g(z)=g(w)$ gives $(z-w)(1-1/(zw))=0$, and the second factor cannot vanish when both moduli exceed one. It is onto as well. For $v\in \mathbb{H}$, the two roots of $z^2-vz+1=0$ have product one. Neither lies on the unit circle, since $z+z^{-1}$ is real there. Exactly one root has modulus greater than one, and the displayed imaginary-part formula makes its imaginary part positive. That root belongs to $D_0$ and maps to $v$.

Since $g_{K_0}$ takes values in $\mathbb{H}$, its domain must be contained in $D_0$. The established injectivity and surjectivity of $g:D_0\to \mathbb{H}$ mean that no proper subdomain of $D_0$ can map onto all of $\mathbb{H}$ using this same function. ThereforeThe half-plane capacity is the coefficient of $1/z$ in the normalized expansion. If hulls are regarded as closed subsets of $\overline{\mathbb{H}}$, take the closure of this half-disc, including its diameter; its relative half-plane hull and capacity are unchanged. This is the half-plane capacity of a half-disc.

At a real point $x$ outside the hull's closure, the Schwarz reflection principle makes the mapping-out function analytic across the boundary and $g_K^{\prime }(x)>0$. Use the course's Brownian excursion restriction identity: for an excursion from $x$ to infinity in the complex upper half-plane,One way to obtain this identity is to start its Doob h-transform at $x+i\epsilon$: its avoidance probability is $\mathrm{Im}{g}_K(x+i\epsilon )/\epsilon$, which tends to $g_K^{\prime }(x)$ by reflection. Thus the boundary derivative is an excursion avoidance probability, and in particular it is at most one.

Let $D$ be the filled unit half-disc. Since $K\subseteq D$, avoiding $D$ implies avoiding $K$. The monotonicity of boundary derivatives of mapping-out functions and part (b) give, for $|x|>1$,The upper bound is attained by the empty hull and the lower by the filled half-disc, so the comparison is sharp on either real tail.

For chordal SLE with parameter $\kappa$, the Loewner trace $\gamma$ starts at zero and is parametrized by half-plane capacity $2t$. Let $D_t$ be the unbounded component of $\mathbb{H}\setminus \gamma [0,t]$, and let $K_t=\mathbb{H}\setminus D_t$ be its filled compact H-hull. When the trace is not simple, the hull includes the regions it disconnects from infinity. The associated mapping-out functions of compact H-hulls have expansionand satisfy the Chordal Loewner equationwhere $B$ is standard real Brownian motion. The equation is solved until $z$ is swallowed by the hull. The Loewner transform $\xi$ records the real boundary point at which the mapped hull grows; its relation to the trace isEquivalently the tip maps to ${\xi }_t$ in the appropriate boundary prime-end sense. Thus the flow encodes the surviving domains, the transform drives that flow, and the trace generates its hulls. For $\kappa =0$, the driver is identically zero, $g_t(z)=\sqrt{z^2+4t}$ with the branch asymptotic to $z$, and ${\gamma }_t=2i\sqrt{t}$.

After the fixed time $s$, the mapping-out function of the uncentered mapped future isIts expansion is ${\bar{g}}_t(z)=z+2t/z+O(z^{-2})$, so the elapsed time $t$ is still the half-plane-capacity parameterization. Differentiating in $t$ yieldsTherefore the Loewner transform of ${\bar{\gamma }}_t=g_s({\gamma }_{s+t})$ is ${\xi }_{s+t}$, with starting point ${\xi }_s$.

Centering gives ${\tilde{g}}_t(w)={\bar{g}}_t(w+{\xi }_s)-{\xi }_s$, whose driver is ${\tilde{\xi }}_t={\xi }_{s+t}-{\xi }_s$. By independent stationary Brownian increments, $(B_{s+t}-B_s{)}_{t\ge 0}$ is a standard Brownian motion independent of the past through time $s$. HenceThis proves the domain Markov property of a chordal Loewner chain from the composition rule for chordal Loewner driving functions. For possible contacts with the old hull, the mapped trace is understood through the continuous boundary extension or prime ends; on the open half-plane the displayed maps are ordinary conformal maps. The PDF asks first for the driver of the uncentered curve $\bar{\gamma }$; the centered driver is included as well.

For $0<\kappa \le 4$, use the allowed probability-one event that the whole SLE trace lies in $\mathbb{H}$ at every positive time. At each deterministic rational $s>0$, part (b) makes the centered mapped future a fresh SLE with the same property. Consequently, almost surely, for every $t>0$,This inverse relation holds through the trace's boundary-extension construction and puts positive-time future points in the open surviving domain. It therefore excludes every contact with the old filled hull, not just contacts with a fixed old point.

Intersect these probability-one events over all positive rational $s$. If ${\gamma }_u={\gamma }_v$ for $0<u<v$, choose rational $s$ with $u<s<v$. The first occurrence belongs to $K_s$, whereas the second lies in $D_s$, a contradiction. A return to ${\gamma }_0=0$ is excluded by the original positive-imaginary-part assertion. This simplicity of SLE from boundary-avoiding restarts proves

Use the PDF's starred image hull $K_t^{\ast }$ and write $h_t={\Phi }_t=g_{K_t^{\ast }}\circ \Phi \circ g_{K_t}^{-1}$. For $t<T$, this map is analytic across a real neighborhood of the driver. SetLet $u(t)=\mathrm{hcap}(K_t^{\ast })/2$. The classical conformal change of half-plane capacity gives $\dot{u}(t)=a_t^2$, and the image driver, indexed by original time, is $h_t({\xi }_t)$. Differentiating the conjugacy relation givesThese are precisely the standard classical identities allowed in this question. The singularities cancel at $z={\xi }_t$. To calculate the remaining derivative, put $\delta =z-{\xi }_t$ and expand at fixed $t$:Thereforewhere each ${\partial }_t$ keeps the spatial variable fixed before evaluation. Applying the Itô formula to $h_t^{\prime }({\xi }_t)$ gives the boundary derivative diffusion under conformal Loewner conjugacyAt $\kappa =8/3$, the third-derivative drift cancels. A second application of the Itô formula to $a_t^p$ givesChoosing $p=1-1/\kappa =5/8$ cancels the remaining drift. Hence the SLE eight-thirds restriction martingale isLocalizing away from $T$ and from vanishing derivatives makes the displayed stochastic integral a martingale on each localized interval, so it defines a continuous local martingale up to $T$.

Put $A=\mathbb{H}\setminus U$. The neighborhood assumptions at zero and infinity mean that $A$ is a bounded hull away from the starting boundary point. For $t<T$, the conjugacy map $h_t$ is the hydrodynamically normalized mapping-out function of the remaining forbidden hull after removing $K_t$. Its closure avoids the driver, so the boundary derivative is an excursion avoidance probability yields $0<{\Sigma }_t\le 1$. Thus $0<M_t\le 1$, and the bounded local martingale criterion makes the localized martingales uniformly bounded and uniformly integrable.

Choose localization times ${\tau }_n\uparrow T$, also bounded by $n$. Optional stopping gives $\mathbb{E}{M}_{{\tau }_n}=M_0$. The supplied terminal limit and continuity of the power function giveSince these variables are bounded by one, dominated convergence yieldsAs usual for a chord starting at zero, the time-zero point is understood as the boundary prime end of $U$; the avoidance assertion concerns the subsequent curve. The boundedness step is what upgrades the local martingale calculation to an equality of probabilities.

Regard a chord as an oriented curve modulo increasing reparameterization, and write $[\gamma ]$ for that class. Scale invariance of $\mu$ means that $[r\gamma ]$ has the same law for each $r>0$. The locality of a chord law under conformal changes of neighborhoods is the following stopped-curve property: if $N,N^{\ast }$ are neighborhoods of the starting boundary point and $\Phi :N\to N^{\ast }$ is a boundary-preserving conformal isomorphism with $\Phi (0)=0$, thenThe two stopped curves are sampled from the same original law $\mu$; equality is for their unparameterized initial segments. A normalization of ${\Phi }^{\prime }(0)$ can be imposed in an equivalent formulation, since scale invariance removes the positive dilation. In a hull-removal formulation, the Locality property of SLE says that a modification away from the starting point leaves the initial law unchanged until the curve reaches the modified part of the domain.

Let $g_t$ map out the original filled hull and let $g_t^{\ast }$ map out the filled hull generated by $\Phi (\gamma [0,t])$. Both have hydrodynamic normalization at infinity. WriteAlthough the original $\Phi$ is specified only on $N$, this conjugacy is defined in the surviving neighborhood of the tip. Boundary preservation gives a real analytic extension there by the Schwarz reflection principle, with $a_t>0$. The identities below are first applied on compactly localized intervals before exit and then continued up to $T$.

Set $u(t)=\mathrm{hcap}(K_t^{\ast })/2$, the capacity clock of the image hull. The Chordal Loewner equation for each flow and the chain rule giveAnalyticity at the driving point cancels the coefficient of $(z-{\xi }_t{)}^{-1}$, forcing $\dot{u}(t)=a_t^2$. The constant term, computed as in Question 3, is then ${\partial }_t{h}_t({\xi }_t)=-3b_t$. For a general SLE driver $d{\xi }_t=\sqrt{\kappa }d{B}_t$, the Itô formula consequently gives the transformed SLE driving functionAt $\kappa =6$, this becomesThus $W$ is a continuous local martingale up to the exit time. If the source notation ${\xi }_t^{\ast }$ indexes the image driver by original time, it denotes $W_t$. If it uses the image's own half-plane-capacity time, it denotes $W_{t(u)}$, with $t(u)$ the inverse clock; that time-changed process is also a local martingale. This distinguishes the two clocks without changing the requested conclusion.

Since $\dot{u}(t)=a_t^2>0$, invert the image clock before exit and put ${\hat{\xi }}_u=W_{t(u)}$. Its quadratic variation is $6u$ and its initial value is zero, because $\Phi (0)=0$. By the Lévy characterization of Brownian motion, or the localized Dambis-Dubins-Schwarz theorem,for a standard Brownian motion up to the image exit time. If that lifetime is finite, the stopped Brownian motion can be extended beyond it; only the stopped law is used here.

In this clock the image mapping-out functions obey the ordinary Chordal Loewner equation with driver $\sqrt{6}{\hat{B}}_u$. Their generated trace therefore has the law of SLE with parameter six, stopped when it leaves $N^{\ast }$. Meanwhile the original image segment is exactly $\Phi (\gamma )$ up to exit from $N$. The clock change is increasing, so it disappears after passing to $[\gamma ]$. This proves the local stopped-curve equality required in part (a).

For completeness, the original law is scale invariant: the maps $r^{-1}{g}_{r^{2}t}(rz)$ have driver $r^{-1}{\xi }_{r^{2}t}$, whose law is again $\sqrt{6}{B}_t$ by Brownian scaling. Hence $(r^{-1}{\gamma }_{r^{2}t}{)}_{t\ge 0}$ has the original trace law. Together with the stopped conformal-change argument, this establishesThe conclusion concerns local unparameterized segments up to exit; the capacity parameter itself changes according to $du=h_t^{\prime }({\xi }_t{)}^{2}dt$.