Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 27 (Schramm–Loewner Evolutions)

A compact H-hull is a bounded relatively closed set $K\subset \mathbb{H}$ whose complement $D=\mathbb{H}\setminus K$ is a simply connected domain. Equivalently one can retain its compact closure in $\overline{\mathbb{H}}$; real boundary points do not affect the mapping-out function. The Riemann mapping theorem and Schwarz reflection principle near infinity give a unique conformal map $g_K:D\to \mathbb{H}$ with hydrodynamic normalization at infinity. Its Laurent series isThe half-plane capacity is $\mathrm{hcap}(K)=a_K$. The missing constant term is part of the normalization, not a choice to be made again when comparing different hulls.

Here are the conformal map facts we use to justify positivity and comparison. The inverse $f_K:\mathbb{H}\to D\subset \mathbb{H}$ has expansion $f_K(w)=w-a_K/w+O(w^{-2})$. The upper-half-plane Schwarz-Pick theorem, applied to $w$ and $iY$ and then letting $Y\to \infty$, implies $\mathrm{Im}{f}_K(w)\ge \mathrm{Im}w$. Indeed the inequality between pseudohyperbolic distances givesand multiplying by $Y$ gives the height comparison in the limit. At $w=iY$, the inverse expansion reads $\mathrm{Im}{f}_K(iY)=Y+a_K/Y+O(Y^{-2})$, so $a_K\ge 0$.

For $r>0$, uniqueness of the mapping-out function givesConsequentlyTranslation by a real number $b$ similarly gives $g_{K+b}(z)=b+g_K(z-b)$ and leaves half-plane capacity unchanged.

If $K\subset A$ are compact H-hulls, their mapping-out functions compose as $g_A=g_L\circ g_K$, where $L$ is the filled image of $A\setminus K$. Comparing Laurent series yieldsThis proves monotonicity of half-plane capacity, rather than assuming it from inclusion alone. The filled half-disc $A=\{z\in \mathbb{H}:|z-b|\le R\}$ has mapping-out functionwhich takes its semicircular boundary to the real line and is a conformal bijection onto $\mathbb{H}$. Thus its half-plane capacity is $R^2$. Enclose $K$ in such a half-disc, compare capacities, and minimize over real centres and radii. This proves the sharp boundIf one defines the radius by an infimum, take $R$ strictly larger than that infimum and then let $R$ decrease; no attainment assumption is needed.

We use the following standard planar Brownian motion facts: its exit probability distribution is harmonic measure; conformal maps preserve that probability distribution after an increasing time change; and the Poisson kernel for the upper half-plane gives its real-boundary exit density. For rough hulls the boundary is interpreted through prime ends, with boundary correspondence at harmonic-measure-almost every exit point. This avoids imposing a smoothness or local-connectedness assumption on $K$.

Under $g_K$, the exit event through $K$ corresponds to a measurable set $E_K\subset \mathbb{R}$ of boundary images. It is bounded: the mapping-out function extends by Schwarz reflection along the real line outside a sufficiently large interval enclosing $K$, and its boundary near infinity corresponds to the original real boundary near infinity. All prime ends representing $K$ therefore have images in a bounded interval. The boundary correspondence assertion can equivalently be formulated using the almost-everywhere limits of $g_K^{-1}$; changing null sets does not change $E_K$'s measure.

Write $g_K(iy)=u_y+i{v}_y$. The hydrodynamic normalization at infinity gives $u_y=O(y^{-2})$ and $v_y=y-a_K/y+O(y^{-2})$. By conformal invariance of planar Brownian motion and the Poisson kernel for the upper half-plane,On the bounded set $E_K$, the integrand multiplied by $y$ tends uniformly to one. Therefore the limit exists and is finite:This is the harmonic capacity from infinity in the upper half-plane with the normalization $c(K)=\mathrm{cap}(K)/\pi$; it is different from half-plane capacity. If real boundary points are retained in the definition of $K$, include their boundary images in $E_K$ as well; the argument is unchanged.

For the scaling statement, Brownian scaling says that $r^{-1}{B}_{r^{2}t}$ started from $iy/r$ is again standard planar Brownian motion. The exit event scales with the hull, soSet $v=y/r$, multiply by $y=rv$ and take the limit. It follows thatThe same conclusion follows geometrically from $E_{rK}=r{E}_K$ and the formula above.

First derive the Brownian representation of half-plane capacity with the chosen normalization. SetThe height comparison from part (a), applied to $f_K=g_K^{-1}$, gives $h\ge 0$. On the boundary, $\mathrm{Im}{g}_K$ tends to zero; on $K$ the limiting value of $h$ is therefore $\mathrm{Im}z$, and on the real boundary it is zero. To see the asserted boundary-height limit without assuming a smooth hull, suppose a bounded sequence approaching the boundary had mapped images with imaginary part bounded away from zero. The inverse's expansion near infinity prevents these images from being unbounded, and an interior subsequential limit would map back to an interior point of $D$, a contradiction.

At infinity $h$ tends to zero by the Laurent series. The maximum principle for harmonic functions now bounds $h$ between zero and $R=\mathrm{rad}(K)$, since every point of $K$ has imaginary part at most $R$. Apply the optional sampling theorem to the bounded harmonic function $h(B_{t\wedge T})$. The exit time $T$ is almost surely finite because it is no larger than the first time the imaginary coordinate of planar Brownian motion reaches zero. Taking $t\to \infty$ givesThis uses the bounded harmonic difference, not optional sampling of the unbounded imaginary coordinate at an unbounded stopping time.

At $z=iy$, the expansion gives $yh(iy)\to a_K$. Sincewe obtainThus the quadratic half-plane capacity is controlled by a length times the linearly scaling harmonic capacity from infinity in the upper half-plane. The factor $1/\pi$ is essential if the alternative normalization $\mathrm{cap}(K)=\pi c(K)$ is used.

Write $g_t=g_{K_t}$ and, for $s<t$, let $K_{s,t}$ be the filled image hull $g_s(K_t\setminus K_s)$. The half-plane-capacity composition rule givesThe Loewner local growth property says that these mapped increments shrink uniformly on bounded time intervals: for every $A<\infty$,Using diameter instead of radius is equivalent, since a nonempty compact H-hull has its closure meeting the real axis. This is a condition on the increments after mapping out the past, not merely on the Euclidean size of $K_t\setminus K_s$ in the original domain.

For fixed $t$, the nonempty compact sets $\overline{K_{t,t+h}}$, $h>0$, are nested as $h$ decreases. Their diameters tend to zero by Loewner local growth property. The nested-compact-set property gives a single intersection point, and its imaginary part is zero because $K_{t,t+h}$ lies in a real-centred disc of radius tending to zero. Define the Loewner transform byOne may restrict the intersection to sufficiently small $h$. Every point of $K_{t,t+h}$ is within $2{\omega }_A(h)$ of ${\xi }_t$ when $t+h\le A$.

Here is a direct proof of continuity, using the permitted displacement estimate. Fix $s<t$ and put $\delta =t-s$. Take $0<h\le \delta$ and a point $v\in K_{t,t+h}$. Set $z=g_{K_{s,t}}^{-1}(v)$. The hull composition rule puts $z\in K_{s,t+h}$, while the displacement estimate givesSince $z$ is within $2{\omega }_A(\delta +h)$ of ${\xi }_s$, let $h\downarrow 0$, taking $v\to {\xi }_t$, to obtainUse a slightly larger fixed time interval if $t$ is its endpoint. Both terms tend to zero with $\delta$. This proves continuity from both sides, and indeed uniform continuity on every compact time interval. The real boundary value is obtained as a limit; evaluating the displacement estimate directly on a possibly rough boundary is unnecessary.

The Loewner correspondence theorem states that a capacity-$2t$ increasing hull family with Loewner local growth property has the unique continuous real Loewner driving function just defined, and its mapping-out functions solve the Chordal Loewner equationFor each $z\in \mathbb{H}$, this solution is taken up to its maximal lifetime; $\mathbb{H}\setminus K_t$ is precisely the set of points whose lifetimes exceed $t$. Conversely a continuous real driver gives such a hull family through this ordinary differential equation. A continuous driver alone need not have a continuous Loewner trace; that additional assertion is part of the trace theory of SLE, not of the deterministic correspondence theorem.

A continuous process $\gamma :[0,\infty )\to \overline{\mathbb{H}}$ is chordal Schramm–Loewner evolution with parameter $\kappa$ from $0$ to infinity when ${\gamma }_0=0$, its filled past generates these capacity-$2t$ hulls, and their Loewner driving function iswith $B$ standard real Brownian motion. The filling includes bounded components cut off from infinity. Equivalently its Loewner trace is the continuous extension of ${\gamma }_t={\lim }_{y\downarrow 0}{g}_t^{-1}({\xi }_t+iy)$. This last boundary limit is the trace-existence input for SLE.

Fix $r>0$ and scale space and time together:The half-plane capacity of ${\hat{K}}_t$ is $2t$ by part (a). Differentiation givesBy Brownian scaling, $r^{-1}{B}_{r^{2}t}$ is again standard Brownian motion, so the new driver has the same law as the old one. Uniqueness of the Chordal Loewner equation transfers this equality to the hulls and their continuous Loewner traces. ThusThis is capacity-parametrized scale invariance of a Loewner chain. It includes $\kappa =0$, when the driver is zero and the trace is the deterministic vertical slit ${\gamma }_t=2i\sqrt{t}$.

The common Brownian motion cancels from the difference $D_t=Y_t-X_t$. Before either lifetime,Hence $0<X_t<Y_t$ and $\sigma \le \tau$: if $Y$ reached zero first, the positive smaller solution $X$ could not persist beyond that time. WriteUsing the Itô formula for $Y^{-1}$ givesThe difference has finite variation, so there is no cross-variation term in the Itô product rule for $D/Y$. Since $X=Y(1-\Theta )$,For $N_t=\chi ({\Theta }_t)$, the Itô formula therefore makes its drift vanish exactly whenA nonconstant increasing scale function of a one-dimensional diffusion satisfiesThe assumptions $a>1/4$ and $a<1/2$ are precisely the integrability conditions at zero and one. Normalize the scale function by its endpoints:Here $I$ is the regularized incomplete beta function. The normalized function is continuous and strictly increasing on $[0,1]$, with $\varphi (0)=0$ and $\varphi (1)=1$. We have obtained the local martingale

The terminal identification needs care; mere boundedness does not by itself say what happens when both lifetimes coincide. Stop first on compact subsets of $0<X<Y$, and then exhaust the interval $[0,\sigma )$. The bounded local martingale criterion makes $N$ a bounded martingale with a limit $N_{\sigma -}$, and optional sampling theorem gives $\mathbb{E}[N_{\sigma -}]=N_0$. For completeness, applying the Itô formula to each stopped $N^2$ bounds the expected quadratic variation by one. Monotone convergence over the localization sequence impliesSince ${\varphi }^{-1}$ is continuous on $[0,1]$, ${\Theta }_t$ itself has a limit ${\Theta }_{\sigma -}$.

On $\{\sigma <\tau \}$, $Y_{\sigma }>0$ while $X_t\to 0$, so ${\Theta }_{\sigma -}=1$ and $N_{\sigma -}=1$. On $\{\sigma =\tau \}$, the supplied divergent clock isIf ${\Theta }_{\sigma -}>0$, then eventually ${\Theta }_t{\varphi }^{\prime }({\Theta }_t)$ is bounded below by a positive constant. This remains true if its limit is one: ${\varphi }^{\prime }(\theta )$ diverges there, rather than tending to zero. The divergent clock would then force $[N{]}_{\sigma -}=\infty$, contradicting the finite quadratic variation. Thus ${\Theta }_{\sigma -}=0$ and $N_{\sigma -}=0$ on the simultaneous-lifetime event. We have provedTaking expectations yields the complete probability formulaThis is the beta integral for strict ordering of coupled Bessel lifetimes; the complementary probability is the probability of simultaneous swallowing, not of the larger initial point dying first.

To connect the calculation to SLE, let its driver be $W_t=\sqrt{\kappa }{\beta }_t$. For a positive boundary point $r$ before its boundary swallowing time, put $V_t^r=(g_t(r)-W_t)/\sqrt{\kappa }$. The Chordal Loewner equation givesThus two positive boundary points give exactly the coupled Bessel processes above, with $B=-\beta$ and $a=2/\kappa$. The interval $1/4<a<1/2$ corresponds to $4<\kappa <8$. Their boundary swallowing times obeyThe formula quantifies whether the growing hull separates one boundary point from infinity before the other, or swallows them together inside the same boundary interval. It does not identify swallowing with visiting each point. At $\kappa =6$, the two beta parameters are $1/3$ and this becomes the Cardy boundary crossing formula in the corresponding cross-ratio coordinate, explaining its role in the conformal invariance of planar percolation.

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.