Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 203 / 1

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.