Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 203 (Schramm-loewner Evolutions)

A compact H-hull is a bounded subset $A\subset \mathbb{H}$, closed relative to the complex upper half-plane, such that $\mathbb{H}\setminus A$ is a simply connected domain. The word compact refers to its bounded closure in $\overline{\mathbb{H}}$; a nonempty hull need not be a compact subset of the open half-plane.

The Riemann mapping theorem and normalization at infinity give a unique mapping-out function of a compact H-hull $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with $g_A(z)-z\to 0$ as $z\to \infty$. This hydrodynamic normalization at infinity yields a Laurent seriesThe coefficient is the half-plane capacity:

There is a small closure issue in the printed set: the unit disc is open, so its half-disc is not relatively closed in $\mathbb{H}$. As written, the set is not a compact H-hull. We compute the intended half-plane capacity after taking its relative closure in $\mathbb{H}$.

First remove the closed unit half-disc $K$. Its mapping-out function of a compact H-hull is the Joukowski mapFor $z=iy$ on the remaining portion of the vertical slit, $1<y\le 2$,Thus the image of the remaining slit is $(0,3i/2]$. By the half-plane capacity of a vertical slit, its capacity is $(3/2{)}^2/2=9/8$. The half-plane-capacity composition rule givesAs a direct check, composing with the slit map gives $g_A(z)=\sqrt{(z+z^{-1}{)}^2+9/4}$, with the branch asymptotic to $z$. Its $z^{-1}$ coefficient is $17/8$.

Work with a locally growing Loewner chain started at $0$, with $A_0=\varnothing$ and the half-plane-capacity parameterization $\mathrm{hcap}(A_t)=2t$. The Conformal Markov property of SLE combines invariance under conformal changes of the marked domain with restarting after an initial hull has been removed.

In the half-plane this means, first, that $(r^{-1}{A}_{r^{2}t}{)}_{t\ge 0}$ has the same law as $(A_t{)}_{t\ge 0}$ for every $r>0$. Second, conditionally on ${\mathcal{F}}_t$, the future hulls mapped by $g_t-U_t$,with the usual hull closure, have the original law and are independent of ${\mathcal{F}}_t$. Their capacity is $2s$ by the half-plane-capacity composition rule. One may require the same restarting property at finite stopping times.

Equivalently, the driver has the scaling law and conditional increment lawThe conformal invariance clause matters: the restarting property alone would also allow a Brownian driver with deterministic drift.

The Loewner local growth property gives a continuous Loewner driving function, and we normalize $U_0=0$. Mapping out an initial hull transforms the future driver to $U_{t+s}-U_t$. The Conformal Markov property of SLE therefore gives stationary increments and independent increments.

Consequently $U$ is a continuous Lévy process. In the Lévy–Khintchine formula, continuity excludes its jump measure, so its only possible components are a deterministic linear drift and Brownian motion:Conformal invariance under dilations gives $U_{r^{2}t}/r\stackrel{d}{=}{U}_t$. Comparing the expectations of these Gaussian random variables yields $\mu rt=\mu t$ for every $r>0$, hence $\mu =0$. The Brownian term already has the required law by Brownian scaling. Therefore the characterization of the SLE driving function isThe case $\kappa =0$ is the constant zero driver. No reflection-symmetry assumption is needed to remove the drift; scaling suffices.

The scaling rule for the Chordal Loewner equation shows that $A_t=\sqrt{t}{A}_1$, because $r^{-1}{U}_{r^{2}t}=a\sqrt{t}$. To identify the shape, rather than infer it from scaling alone, construct the inverse map explicitly. PutThen $r_{-}<0<r_{+}$, $r_{-}{r}_{+}=-4$, and $\alpha r_{+}+(1-\alpha )r_{-}=0$. Defineusing logarithms whose arguments lie in $(0,\pi )$ on $\mathbb{H}$. Its Laurent series is $f_t(w)=w-2t/w+O(w^{-2})$, so it has the inverse hydrodynamic normalization at infinity.

On the real interval $(r_{-}\sqrt{t},r_{+}\sqrt{t})$, the first factor has argument $\pi \alpha$ and the second has argument zero. The image therefore lies on one straight ray. Its modulus increases from zero to a maximum at $w=a\sqrt{t}$, and then decreases to zero. The complementary real intervals map to the negative and positive real axes. The argument principle, or the usual conformal slit-map construction, shows that $f_t$ maps $\mathbb{H}$ conformally onto the half-plane minus that segment; the interval traverses its two sides.

The boundary walk has winding number one around every point of the half-plane off the segment: the two traversals of the slit cancel, leaving the real boundary and a large semicircle. The argument principle therefore gives exactly one preimage of each such point.

It remains to check that this is the correct Loewner chain. For $f=f_1$,Together with $f_t(w)=\sqrt{t}f(w/\sqrt{t})$, this givesthe inverse form of the Chordal Loewner equation. Uniqueness therefore identifies its slit with $A_t$. Its endpoint is $\sqrt{t}{f}_1(a)$, where $f_1(a)\ne 0$ lies in $\mathbb{H}$. This proves that the square-root Loewner driving function generates a straight slit and thata straight ray from the boundary point $0$ to infinity.

By the Phase classification of the SLE trace, chordal SLE is a simple curve forIf the convention restricts the notation to $\kappa >0$, the range is $0<\kappa \le 4$. The degenerate case $\kappa =0$ is a deterministic vertical slit and is simple as well.

By the Phase classification of the SLE trace, the trace has self-intersections but is not a space-filling curve forThe endpoint $\kappa =4$ remains simple, while $\kappa =8$ belongs to the space-filling regime.

The Phase classification of the SLE trace gives a space-filling curve precisely forIn particular, the critical value $\kappa =8$ is included.

Start the Bessel process at $x>0$ and let ${\tau }_0$ be its first hit of zero. Away from zero its stochastic differential equation isApply the Itô formula to $f(x)=x^{2-\delta }$, localizing first to a compact interval in $(0,\infty )$. Its drift cancels:Thus the Bessel power local martingale satisfiesFor $\delta =2$, this power is simply the constant one. For $\delta >2$, it is a continuous local martingale for all time, as zero is inaccessible.

Here is a proof of that inaccessibility and of the stronger minimum assertion. For $0<r<x<R$, stop the power at ${\tau }_r\wedge {\tau }_R$. The stopped process is bounded, so the optional stopping theorem givesExit from the bounded interval occurs almost surely. Taking $r\downarrow 0$ with $R$ fixed shows that zero cannot be hit before $R$, hence cannot be hit in finite time. Taking $R\to \infty$ instead givesIf the all-time infimum were zero, every level $r>0$ below $x$ would be hit, by continuity. Letting $r\downarrow 0$ proves the all-time minimum of a transient Bessel process assertion:The first claim in the printed question needs a qualification when $\delta <2$: it holds before the first hit of zero, or for the process stopped there, rather than for the usual reflecting continuation. For example, at $\delta =1$ the proposed power is $X$ itself, a Reflected Brownian motion; the Tanaka formula contains a nonzero local time of a semimartingale term, so it is not a local martingale after reflection. A positive starting point is also needed for the negative power when $\delta >2$; an entrance process started at zero would give an infinite value at time zero.

For $\kappa >0$ and $x>0$, before the boundary point is swallowed, put $X_t=(g_t(x)-U_t)/\sqrt{\kappa }$. The Chordal Loewner equation and $d{U}_t=\sqrt{\kappa }d{B}_t$ giveSince $-B$ is again a standard Brownian motion, this is the Boundary-point Bessel flow for SLE withFor $x<0$, it is $-X_t$ that is the nonnegative Bessel process, with the same dimension; the formula without this sign convention is a signed Bessel flow.

When $0<\kappa \le 4$, one has $\delta \ge 2$, and the Hitting-zero classification for a Bessel process says that $X_t$ never reaches zero. The borderline $\delta =2$ uses the scale function of a one-dimensional diffusion $\log x$: for $0<r<x<R$,Thus the critical case also cannot hit zero in finite time, though its all-time infimum is zero.

Apply the non-swallowing assertion simultaneously to all nonzero rational boundary points. The order-preserving real Loewner flow then keeps every compact real interval away from the origin in the surviving boundary. A boundary contact away from the starting point would cut off a nonempty real interval, swallowing a rational point, so the trace avoids $\mathbb{R}\setminus \{0\}$.

The same argument can be applied to the future after every rational time, using the Conformal Markov property of SLE. If the trace revisited an earlier point, choose a rational time strictly between the two visits. In the domain with that initial segment removed, the later visit would be a contact with an old boundary point away from the current growing tip. Under the mapping-out map, such a contact cuts off a real interval and contradicts the preceding non-swallowing result. This is the usual crosscut argument converting boundary non-swallowing into absence of self-intersections. Taking the countable intersection over rational restart times provesFor $\kappa =0$, the equation is deterministic and gives the simple slit $(0,2i\sqrt{t}]$. The division by $\sqrt{\kappa }$ is unnecessary in that case.

The Scaling invariance of SLE says that $r^{-1}\gamma (r^{2}t)$ has the same law as $\gamma (t)$. Therefore its first time reaching height one is $r^{-2}{\sigma }_r$, andfor every $r>0$.

To identify $p$, it suffices to look at small heights. The trace must have a point of positive imaginary part: otherwise it lies entirely on the real axis and its hull in $\mathbb{H}$ is empty, contradicting $\mathrm{hcap}(A_t)=2t>0$. By continuity from $\gamma (0)=0$, any positive height attained by the trace forces it to pass through every smaller positive height.

Consequently the increasing union of events $\{{\sigma }_{1/n}<\infty \}$ has probability one. Each event has probability $p$, so continuity of probability gives $p=1$. This proves that SLE reaches every positive height:Equivalently, a positive random height supremum whose law is invariant under every dilation must be infinite almost surely. This argument uses neither Transience of chordal SLE nor any assertion about where the trace goes as time tends to infinity.

For real test functions $f,g\in C_c^{\infty }(D)$, the Dirichlet inner product in the normalization compatible with the logarithmic Green kernel isIts norm measures gradient energy rather than the values of the functions. A change of variables shows that this energy is invariant under planar conformal maps: the squared derivative in the transformation of the gradient cancels the area Jacobian determinant.

The convention used for planar Gaussian free field theory is the Dirichlet energy space:It is a real Hilbert space with the extended Dirichlet inner product. A conformal map from the unit disc onto $D$ identifies the two energy completions, so local representatives can be understood through the disc's zero-boundary space.

For an arbitrary unbounded domain this is the homogeneous energy completion; it should be distinguished from completion in the inhomogeneous norm $\Vert f{\Vert }_{L^2}+\Vert \nabla f{\Vert }_{L^2}$ used in another standard definition of the zero-boundary Sobolev space. On bounded domains with a Poincaré inequality, the two norms are equivalent.

Choose an orthonormal basis $(e_n)$ of the Dirichlet energy space and independent variables $({\xi }_n)$ with the standard normal distribution. The zero-boundary Gaussian free field is the formal seriesIts precise energy-indexed interpretation is the isonormal Gaussian processFor every fixed $f$, this sum converges in $L^2$ of the probability space by Parseval identity, and it is a centered Gaussian random variable. For $f,g\in H_0^1(D)$,This covariance characterizes its law independently of the chosen basis. The series does not define an $H_0^1$-valued random element: ${\sum }_n{\xi }_n^2=\infty$ almost surely. It is realized as a random distribution, and its pairing with ordinary test functions is constructed in part (b).

The expression is a Test-function pairing with a Gaussian free field, rather than the pointwise integral of an ordinary random function. For a real test function $\varphi$, take the Dirichlet solutionusing the Green-kernel identity allowed in the original PDF. In particular, $-\Delta f_{\varphi }=2\pi \varphi$. Integration by parts gives, initially for smooth compactly supported $u$ and then by continuity in the Dirichlet energy space,The integral functional is continuous in the energy norm: pull back to the unit disc, where the transformed test function has compact support, and use the Poincaré inequality. Hence $f_{\varphi }$ is its energy-space representer. DefineBy the isonormal Gaussian process definition, it has mean zero and varianceThe logarithmic singularity is locally integrable, so this variance is finite for smooth compactly supported test functions. ThusThe local TeX corrupted the double integral and the bracketed Green-kernel identity; the PDF supplies the identity above and does not assert an infinite variance.

Set $F(w)=(\varphi (w)-\varphi (0))/w$ for $w\ne 0$. The singularity is removable, with $F(0)={\varphi }^{\prime }(0)$. Injectivity of the conformal map makes $F(w)\ne 0$ away from zero, and conformality makes $F(0)\ne 0$ as well. Since the disc is simply connected, $F$ has a holomorphic logarithm. Therefore $\psi =\log |F|$ is a harmonic function, including at zero, and $\psi (0)=\log |{\varphi }^{\prime }(0)|$.

The mean value property for harmonic functions gives, for every $0<r<1$,For general $D$, boundary values mean radial limits, rather than a continuous extension of $\varphi$ to every point of the circle. The boundary logarithmic mean of a univalent function justifies taking $r\uparrow 1$: the Koebe distortion theorem bounds $|F|$ below by $|{\varphi }^{\prime }(0)|/4$, while the standard integral-mean bound for a univalent function, ${\sup }_{r<1}\int |\varphi (r{e}^{i\theta }){|}^{p}d\theta <\infty$ for $0<p<1/2$, gives uniform integrability of its positive logarithm. Radial limits exist almost everywhere, and passage to the integral follows. Thus, writing $dw=d\theta$ for arc length on the unit circle,The expression on the left is the logarithm of the conformal radius of $D$ at $z=\varphi (0)$.

The canonical Dirichlet Green function on the unit disc with its pole at zero is $-\log |w|$. Conformal invariance of the planar Green kernel therefore givesComparing with $G_D(z,y)=-\log |z-y|-G_z(y)$ yieldsThis also explains the boundary data: when $|w|=1$, the correction is $-\log |\varphi (w)-z|$. The composition with ${\varphi }^{-1}$ is harmonic by conformal invariance of harmonicity. Its apparent singularity at $z$ is removable because the quotient defining $\psi$ extends to ${\varphi }^{\prime }(0)$.

Evaluating there proves the regular part of the planar Dirichlet Green function formulaOn an unbounded domain, boundary values alone need not specify a unique harmonic function. Here $G_z$ is the correction belonging to the canonical Dirichlet Green kernel, which fixes that ambiguity.

For the Locality property of SLE, take a compact H-hull $A$ away from $0$, and map $\mathbb{H}\setminus A$ conformally to $\mathbb{H}$ by ${\psi }_A$, fixing $0$ and infinity and with derivative one at infinity. Up to the first contact with $A$, the image under ${\psi }_A$ of an ${\mathrm{SLE}}_6$ grown in $\mathbb{H}$ has the same law as ${\mathrm{SLE}}_6$ grown in $\mathbb{H}$, after the change to its own half-plane-capacity parameterization. Equivalently, ${\mathrm{SLE}}_6$ in the smaller domain has the same initial law as the curve in the larger domain until the latter encounters the removed hull. There is no conditioning on eventual avoidance.

The chordal restriction property for ${\mathrm{SLE}}_{8/3}$ concerns the whole curve. Conditional on avoiding $A$, applying ${\psi }_A$ gives the original ${\mathrm{SLE}}_{8/3}$ law, modulo increasing reparameterization:Its associated avoidance probability, from the SLE eight-thirds restriction martingale, isThus locality identifies a stopped, unconditioned law, whereas restriction identifies a conditioned law of the entire curve.

Interpret the printed $A_0=0$ as the empty initial hull. Write $a(t)=\mathrm{hcap}({\tilde{A}}_t)$ and $F_t={\psi }_t$. Conformal maps preserve inclusion and simple connectedness of the complementary domains, so the image sets form an increasing hull family. Boundedness of $\psi$ on bounded sets ensures that these image hulls are bounded. The Loewner local growth property is preserved under conformal transport: after mapping out the hull at time $t$, the small new hull is transported by $F_t$ near its single boundary growth point. The Schwarz reflection principle extends $F_t$ analytically across that point, with real positive derivative. The image diameters therefore tend to zero as the original ones do. These statements use the usual hull closures and prime ends; the initial image hull is empty.

For completeness, the infinitesimal capacity rule underlying this argument is that a shrinking hull $J$ attached near $u$, transported by a map $F$ analytic there, hasOne obtains this by rescaling at $u$: the transported map tends to its linear part, and half-plane capacity scales by the square of the dilation. Uniform analytic distortion near the growth point controls the error. Applying it to the mapped-out increments gives local absolute continuity of $a$ and $a^{\prime }(t)=2F_t^{\prime }(U_t{)}^2$.

The coefficient can also be read directly from the Chordal Loewner equation. The image driver is ${\tilde{U}}_t=F_t(U_t)$. Differentiate $F_t={\tilde{g}}_t\circ \psi \circ g_t^{-1}$, at a fixed point $w$, to obtainThe left side is regular at $w=U_t$. On the right, the coefficient of $(w-U_t{)}^{-1}$ is $a^{\prime }(t)/F_t^{\prime }(U_t)-2F_t^{\prime }(U_t)$, so it must vanish. This proves the conformal change of half-plane capacity rule and its integrated version:If the derivative at the initial boundary point is singular, the formula is interpreted by integrating from a positive time and taking the lower limit to zero; finite image capacity and kernel continuity give this limit. The duplicated incomplete normalization sentence in the TeX is absent from the PDF.

A Brownian excursion in the upper half-plane from $0$ to infinity is killed planar Brownian motion conditioned to stay in $\mathbb{H}$, with its entrance law at $0$. Precisely, it is the Doob h-transform for $h(z)=\mathrm{Im}z$, obtained by starting at $i\epsilon$ and letting $\epsilon \downarrow 0$. Equivalently,where $W$ is standard Brownian motion and the independent process $R$ is a dimension-three Bessel process started at zero. The transformed generator is $\frac{1}{2}\Delta +y^{-1}{\partial }_y$, which explains this representation.

First compute avoidance from an interior point $z$. Let $\tau$ be the exit time of ordinary planar Brownian motion from $\mathbb{H}\setminus A$. The harmonic correctionhas boundary values $\mathrm{Im}z$ on $A$ and zero on the real axis. The hydrodynamic normalization at infinity gives $g_A(z)=z+\mathrm{hcap}(A)/z+O(|z{|}^{-2})$. Together with the maximum principle for harmonic functions this gives the useful bound $0\le v\le {\sup }_{a\in A}\mathrm{Im}a$. Since ordinary Brownian motion hits the real axis almost surely, the optional stopping theorem applied to this bounded harmonic function givesThe Doob h-transform weights a stopped path by its final height divided by its initial height. Applying this at the first hit of $A$, with bounded-time localization and then a limit, therefore yieldsFor the entrance law the same result follows by conditioning at a small positive time and letting that time decrease to zero: the avoidance function has the boundary limit computed next, and the excursion is initially in a neighbourhood disjoint from $A$.

Because the closure of the hull avoids $0$, the Schwarz reflection principle makes $g_A$ analytic in a neighbourhood of zero, with real $g_A(0)$ and positive real $g_A^{\prime }(0)$. ThusIn fact, the reflected Taylor expansion gives $\mathrm{Im}{g}_A(z)/\mathrm{Im}z=g_A^{\prime }(0)+O(|z|)$ uniformly as $z\to 0$ inside the half-plane, so the same limit applies to the entrance process. The map in the question is ${\psi }_A=g_A-g_A(0)$, so its derivative agrees with $g_A^{\prime }$. We obtain the restriction probability of a Brownian half-plane excursion: