Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 203 / 3

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