Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 203 / 2

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