Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 220 / 4

The Conformal Markov property of SLE states that, conditional on the hull $K_t$, the image under $g_t-U_t$ of the future hull has the same law as the original hull and is independent of the past.

For a Loewner chain with continuous driver $U$, this property says that $U_{t+s}-U_t$ is independent of the past and has the same distribution as $U_s$. Thus $U$ has stationary increments and independent increments. Every continuous process with those properties is a Brownian motion with drift, so $U_t=at+\sigma B_t$. Conformal scale invariance giveswhich forces $a=0$. Writing $\kappa ={\sigma }^2$ yields

The Locality property of SLE says the following. Let $D\subset \mathbb{H}$ be simply connected and agree with $\mathbb{H}$ in a neighborhood of $0$, and let $\psi :D\to \mathbb{H}$ be conformal with $\psi (0)=0$. The image under $\psi$ of an ${\mathrm{SLE}}_6$ in $\mathbb{H}$, stopped when it first leaves $D$, has the law of an ${\mathrm{SLE}}_6$ in $\mathbb{H}$, up to the corresponding stopping time and a change of half-plane-capacity parameterization.

With the notation supplied in the question, ${\tilde{U}}_t={\psi }_t(U_t)$ and $d{U}_t=\sqrt{\kappa }d{B}_t$. The Itô formula and ${\partial }_t{\psi }_t(U_t)=-3{\psi }_t^{\prime \prime }(U_t)$ giveFor $\kappa =6$, the drift vanishes. The resulting continuous local martingale has quadratic variationIf $s=\tilde{a}(t)/2$ is the usual half-plane-capacity time, then $⟨\tilde{U}⟩=6s$. The Dambis-Dubins-Schwarz theorem therefore givesfor a standard Brownian motion $\tilde{B}$. The mapped hulls are consequently an ${\mathrm{SLE}}_6$, which proves locality.

Explore the two stopped curves in the opposite order. Begin with an ${\mathrm{SLE}}_6$ from $1$ to $\infty$, stopped on leaving $B(1,1)$, and then, in its unbounded complementary component, draw an ${\mathrm{SLE}}_6$ from $-1$ to $\infty$, stopped on leaving $B(-1,1)$. Before their respective stopping times, each curve is separated from the neighborhood in which the other hull changes the domain. The Locality property of SLE therefore says that mapping out the other stopped hull does not change either stopped marginal law.

The two exploration orders consequently define the same joint law for the pair of stopped hulls. Disintegrating this joint law with respect to the second curve shows that, conditional on ${\gamma }_2{|}_{[0,{\tau }_2]}$, the first curve is an ${\mathrm{SLE}}_6$ in the unbounded component offrom $-1$ to $\infty$, stopped when it leaves $B(-1,1)$, as required.

Choose a Möbius transformation of $\mathbb{H}$ that fixes $0$ and exchanges $x$ with $\infty$. By Conformal invariance of SLE, it transforms an ${\mathrm{SLE}}_6$ from $0$ to $x$ into one from $0$ to $\infty$. Until $x$ is disconnected from infinity, the discrepancy between the two target domains lies beyond the component visible from the growing tip. The Locality property of SLE therefore makes the two initial curve laws identical up to that disconnection time. Hence an ${\mathrm{SLE}}_6$ from $0$ to $x$, stopped at ${\tau }_x$, has the law of an ${\mathrm{SLE}}_6$ from $0$ to $\infty$ stopped when it disconnects $x$ from infinity.