Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 203 (Schramm-loewner Evolutions)

A compact H-hull is a bounded, relatively closed set $A\subset \mathbb{H}$ for which $\mathbb{H}\setminus A$ is simply connected domain. Its mapping-out function of a compact H-hull is the unique conformal map $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with hydrodynamic normalization at infinityThe half-plane capacity is

Let $g=g_A$ and define $u(z)=\mathrm{Im}(z-g(z))$. This is harmonic on $\mathbb{H}\setminus A$, has boundary values $\mathrm{Im}z$ on the hull boundary and zero on the real boundary, and tends to zero at infinity. The representation by harmonic measure and optional sampling theorem therefore giveAt $z=iy$, the hydrodynamic expansion givesConsequently the Brownian representation of half-plane capacity is

Map out $A$ first. The imagewith its bounded filling is a compact H-hull, and uniqueness of hydrodynamic normalization givesComparing the coefficients of $1/z$ at infinity yields the half-plane-capacity composition ruleThus half-plane capacity is monotone under inclusion.

The statement is true. In the notation of part (ii), equality of the capacities forces $\mathrm{hcap}(D)=0$. Every nonempty compact H-hull has strictly positive half-plane capacity: by the Brownian representation of half-plane capacity, Brownian motion started sufficiently high has positive harmonic measure of a boundary portion of positive height. Hence $D=\varnothing$, so $C\setminus A=\varnothing$ and

The family is a nondecreasing family of sets when

It has the half-plane-capacity parameterization under the standard chordal convention when

It has the Loewner local growth property when, after mapping out the old hull, each short new increment is small: for every $T,ϵ>0$ there is $\delta >0$ such thatEquivalent formulations use a crosscut of diameter below $ϵ$ separating the new increment from infinity.

Write $r_t=\sqrt{2t}$. The hulls $A_t=r_t(\mathbb{H}\cap \mathbb{D})$ are nested, so property (i) holds. Scaling the given map givesso $\mathrm{hcap}(A_t)=2t$ and property (ii) holds.

Property (iii) fails. For $s>0$, the image under $g_s$ of the outer semicircle of $A_t\setminus A_s$ isAs $t\downarrow s$, this converges to $2r_s\cos \theta$, which fills the real interval $[-2r_s,2r_s]$. Hence the diameter of the mapped new increment tends to $4r_s$, rather than zero. Therefore (i) and (ii) hold, while (iii) does not.

If $g_t$ is driven by $U_t=\sqrt{\kappa }{B}_t$, then the mapping-out functions ofareand their driver is ${\tilde{U}}_t=r{U}_{t/r^2}$. By Brownian scaling, $(r{B}_{t/r^2}{)}_{t\ge 0}$ is Brownian. Thus $\tilde{U}$ has the same law as $U$, and uniqueness of the Chordal Loewner equation provesThis is the Scaling invariance of SLE.

Let $D$ be simply connected domain with distinct marked boundary points $a,b$, and choose a conformal map $f:D\to \mathbb{H}$ with $f(a)=0$ and $f(b)=\infty$. Chordal ${\mathrm{SLE}}_{\kappa }$ from $a$ to $b$ is the unparameterized curve $f^{-1}(\gamma )$, where $\gamma$ is chordal SLE in $(\mathbb{H},0,\infty )$.

Any other such map is $\tilde{f}=rf$ for some $r>0$. The Scaling invariance of SLE says that $r^{-1}\gamma$ has the same unparameterized law as $\gamma$; only its capacity clock changes. Hence the pullback law is independent of $f$. This proves the Conformal invariance of SLE definition is well-defined.

Putso $M_t={{\rm Y}}_t^{-p}{S}_t^q$. Before ${\tau }_{ϵ}$, one has ${{\rm Y}}_t\ge ϵ$ and $0<S_t\le 1$. ThereforeThe supplied continuous local martingale is thus bounded after stopping, and a bounded local martingale is a true martingale. Hence

At time zero, ${{\rm Y}}_0(z)=\mathrm{Im}z$. Compactness of $K\subset \mathbb{H}$ givesAlso $M_0(z)$ is uniformly bounded above on $K$. Onone has $M_{{\tau }_{ϵ}}\ge {ϵ}^{-p}{2}^{-q}$. Optional stopping, Fatou lemma, and the assumed conditional angular estimate giveThusThe exponent $(8-\kappa )/\kappa$ requested in the question does not follow and is false as written. The SLE Green-function estimate gives probability comparable to ${ϵ}^{1-\kappa /8}$, confirming that the denominator in the requested exponent should be $8$.

For compact $K\subset \mathbb{H}$, letBy the Tonelli theorem and part (ii),Every point in the SLE range has conformal radius tending to zero and therefore belongs to every $E_{ϵ}$. Hence the range inside $K$ has zero expected Lebesgue measure, and so has zero measure almost surely. Exhausting $\mathbb{H}$ by countably many compact sets proves

The Phase classification of the SLE trace isAt $\kappa =0$ the trace is the deterministic vertical slit.

For a real boundary point $x\ne 0$, setAfter a deterministic rescaling of time, the Boundary-point Bessel flow for SLE says that $X$ is a Bessel process of dimensionWhen $0<\kappa \le 4$, one has $\delta \ge 2$, and the Hitting-zero classification for a Bessel process says that $X$ never reaches zero. Thus no nonzero real boundary point is swallowed. The standard Loewner trace criterion then implies that each new tip is attached only to the preceding tip and the trace never intersects its past, so it is simple. For $\kappa =0$, the equation is driven by zero and generates a vertical slit. Hence ${\mathrm{SLE}}_{\kappa }$ is simple for $0\le \kappa \le 4$.

Let $Z_t=g_t(z)-U_t$. For ${\mathrm{SLE}}_4$, $d{U}_t=2d{B}_t$, and the Chordal Loewner equation givesThe complex Itô formula yieldsTherefore

The imaginary partis a bounded local martingale and hence a martingale. As the simple transient trace passes $z$, this angle converges to $\pi$ if the trace passes to the right of $z$ and to $0$ if it passes to the left. Bounded convergence therefore givesso the SLE4 left-passage probability is

Fix $z$ and writeBy assumption, $L(z)$ is a continuous local martingale, so $Z_t(z)=e^{L_t(z)}$ is a semimartingale. The Chordal Loewner equation giveswhich has finite variation. Thereforeis a semimartingale. Thus the Loewner driver $U$ is a continuous semimartingale.

Write the semimartingale decomposition as $U=M+A$, where $M$ is a continuous local martingale and $A$ has finite variation. Applying Itô formula to $\log Z_t(z)$ shows that its finite-variation part isIt vanishes for every $z$. Multiplying by $Z_t(z{)}^2$ givesSubtract this identity for two points with distinct $Z_t$ to obtain $d{A}_t=0$; then $d⟨M{⟩}_t=4dt$. Since the curve starts at zero, $U_0=0$. The Lévy characterization of Brownian motion now gives $U_t=2B_t$. Hence the Loewner chain is

The law satisfies the chordal restriction property when, for every $A\in {\mathcal{Q}}_{\pm }$, conditional on $\gamma \cap A=\varnothing$, the mapped curve ${\psi }_A(\gamma )$ has the same unparameterized law as $\gamma$ in $(\mathbb{H},0,\infty )$.

Assume the avoidance formula. Given another admissible hull $C$, put $D=A\cup {\psi }_A^{-1}(C)$ with the bounded filling. Uniqueness of the normalized maps givesThereforeThese avoidance events determine the law of a simple closed random set. They agree with those of $\gamma$, so the conditional mapped law equals the original law. Hence the avoidance formula implies the chordal restriction property.

Let $J_t={\psi }_t^{\prime }(U_t)$, $q_t={\psi }_t^{\prime \prime }(U_t)/{\psi }_t^{\prime }(U_t)$, and $r_t=({\partial }_z^3{\psi }_t)(U_t)/{\psi }_t^{\prime }(U_t)$. Since $d{U}_t=\sqrt{8/3}d{B}_t$, the supplied identity and Itô formula giveFor $M_t=J_t^{\alpha }$, its drift coefficient isThus the nonzero choice isand $M_{t\wedge \tau }$ is a continuous local martingale. The boundary Schwarz lemma for mapping-out maps gives $0\le J_{t\wedge \tau }\le 1$, so $0\le M_{t\wedge \tau }\le 1$. A bounded local martingale is a true martingale. This is the SLE eight-thirds restriction martingale.

Let $A=\mathbb{H}\cap B(1,ϵ)$. Its normalized mapping-out map isso ${\psi }_A(0)=0$ andUsing the restriction exponent $5/8$ gives

For the vertical slit $A=(1,1+iϵ]$, choose the square-root branch asymptotic to $z-1$ at infinity. The normalized map isfor which ${\psi }_A(0)=0$ andTherefore