Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 27 / 4

The conformal automorphism of the upper half-plane with the indicated three boundary values isIts Möbius transformation matrix has determinant one, so it preserves $\mathbb{H}$. The inverse is ${\Phi }^{-1}(w)=w/(1-w)$. In particularon the extended real boundary. Therefore the image trace's first contact with the latter ray corresponds to the original time $S$. The map sends the original target at infinity to the new target $1$; it does not preserve the target of the chordal curve.

We use the image half-plane capacity as its time parameter:Since ${\Phi }_t$ is a conformal automorphism of the upper half-plane analytic at ${\xi }_t$, its real derivative there is strictly positive. Hence $u$ is continuous and strictly increasing before $S$. The original time $t$ and image time $u$ must be kept distinct throughout the comparison.

By the tip correspondence in the conjugacy,The time-dependent Itô formula is valid after localization away from the pole and the terminal time $S$. The map's time variation is of finite variation, so there is no extra stochastic cross-variation term. Using the supplied conjugacy derivative,ConsequentlyThe cancellation is special to $\kappa =6$: for a general SLE parameter the drift would be $(\kappa /2-3){\Phi }_t^{\prime \prime }({\xi }_t)dt$. This is the transformed SLE driving function, and explains the special Locality property of SLE at six.

Let $t(u)$ be the inverse image-capacity clock. Part (a) and the quadratic variation of a stochastic integral giveThe Dambis-Dubins-Schwarz theorem, or the Lévy characterization of Brownian motion, now shows that ${\tilde{\xi }}_u/\sqrt{6}$ is standard Brownian motion up to the terminal image time. If that terminal clock is finite, it can be extended by a fresh Brownian continuation; this does not assert anything about the subsequent image trace. The initial value is zero because $\Phi (0)=0$.

Differentiating the image Loewner chain with respect to its own half-plane capacity time givesThus, until its first hit of $[1,\infty )$, the reparametrized image $\Phi (\gamma )$ has the ordinary chordal SLE law from $0$ to infinity. This follows from the uniqueness of the Chordal Loewner equation for a given continuous driver; the continuous Loewner trace is recovered from the inverse-map limit. Localization can be removed by exhausting times strictly before this boundary contact. The terminal event is defined from the trace itself, so equality of the stopped trace laws follows as well.

On the other hand, the definition of Conformal invariance of SLE says that the full unparameterized image $\Phi (\gamma )$ has the law of chordal ${\mathrm{SLE}}_6$ from $\Phi (0)=0$ to $\Phi (\infty )=1$, namely the law of ${\gamma }^{\prime }$. Its stopping event at $[1,\infty )$ is exactly the image of the original stopping event at $(-\infty ,-1]$. Therefore the same stopped image has both laws: ordinary ${\mathrm{SLE}}_6$ from $0$ to infinity stopped at $T$, and ${\mathrm{SLE}}_6$ from $0$ to $1$ stopped at $T^{\prime }$. We concludeThe statement does not identify $T$ with $T^{\prime }$ in their original clocks. It is target-change locality of SLE6, derived from the zero drift and capacity time change rather than assumed as a theorem.

The two curves have different destinations, so the equality of their initial segments is not an equality of their complete laws. At the first contact with $[1,\infty )$, the initial hull separates the target $1$ from infinity. Except for the null event of hitting the fixed point $1$ itself, this happens by a contact at a real point strictly larger than $1$. The fact that a fixed nonzero real point is almost surely not visited is the SLE does not hit a fixed nonzero boundary point property; it is consistent with the strictly positive probability of swallowing an entire interval at once in Question 3.

The original curve, whose target is infinity, continues in the closure of the unbounded surviving component. The curve targeted at $1$ continues in the surviving component whose boundary contains its target, which is now separated from infinity. Its domain normalization must follow that component. Ultimately one curve tends to infinity and the other to $1$. This is target-dependent continuation of SLE6.

The precise failure of the argument is the loss of the conformal conjugacy between the same surviving components. In the transformed construction, at time $S$ the pole $-1$ is swallowed; its image is infinity. The composition $g_t^{\ast }\circ \Phi \circ g_t^{-1}$ was a conformal automorphism of the upper half-plane only while both maps described corresponding surviving domains. After the boundary contact, the image of the original target-containing domain is the component containing $1$, whereas $g_t^{\ast }$ in the ordinary half-plane-capacity convention maps out the hull relative to the component containing infinity. These are no longer the same component. Equivalently, the swallowed pole collides with the growing hull, and the analytic-neighbourhood assumption at the Loewner driving function cannot be continued by the preceding argument.

Hence neither the given analytic identities nor the Brownian motion description obtained from their capacity clock licenses a continuation of the comparison beyond the stopping time. The zero drift before contact does not remove this geometric obstruction. The restriction to pre-contact segments is essential, even though the curves can continue in their respective target components.