Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 27 / 2

Use the Chordal Loewner equation with Loewner driving function ${\xi }_t=\sqrt{\kappa }{W}_t$, and set $Z_t=g_t(z)-{\xi }_t=X_t+i{Y}_t$. Before the Loewner swallowing time,The upper-half-plane branch of the logarithm has imaginary part $h_t$. The Itô formula givesand therefore the SLE angle process satisfiesAt $\kappa =4$ the drift vanishes. Since $0<h_t<\pi$, the stopped local martingale is a true bounded martingale, and the SLE4 angle martingale is global for each fixed point almost surely, using fixed-interior-point avoidance of SLE4 for the simple parameter-four Loewner trace.

Conversely, take $\kappa >0$. If this semimartingale were a local martingale, uniqueness of its finite-variation decomposition would force $(\kappa -4)X_t{Y}_t=0$ throughout every compact interval before swallowing. Because $Y_t>0$, for $\kappa \ne 4$ this would force $X_t$ to be identically zero on such an interval. Its Brownian component has quadratic variation $\kappa t$, which makes that impossible. Thus for positive $\kappa$, the martingale parameter is exactly four.

If the degenerate value $\kappa =0$ is admitted, there is one exception to a fixed-point reading of the assertion: for $z$ on the positive imaginary axis the deterministic flow stays on that axis until swallowing, and $h_t=\pi /2$ is constant. For all starting points simultaneously, the unique parameter giving the martingale property is still four.

For a simple Loewner trace, the domain mapped by $g_t$ is $D_t=\mathbb{H}\setminus \gamma [0,t]$. For a non-simple Loewner trace, use instead the unbounded component $D_t=\mathbb{H}\setminus K_t$; the map $g_t$ is not defined on swallowed bounded components. This is the necessary domain interpretation when $\kappa >4$.

The function $\mathrm{Im}\log (g_t(z)-{\xi }_t)$ is harmonic in $D_t$ because the logarithm is holomorphic on the complex upper half-plane. Let $S_t^{-}$ and $S_t^{+}$ be the intrinsic boundary sets mapped to $(-\infty ,{\xi }_t)$ and $({\xi }_t,\infty )$ respectively. The Dirichlet problem has boundary dataFor a simple Loewner trace these are the left bank of the Loewner trace together with the negative real boundary, and the right bank together with the positive real boundary. Left and right refer to the orientation from the starting point towards the tip. The tip and infinity correspond to discontinuities of the data; no unique limit is imposed there. They have zero harmonic measure.

More precisely, the bounded solution isby the Poisson kernel for the upper half-plane and conformal invariance of planar Brownian motion. This establishes the boundary values in the intrinsic sense and uniqueness among bounded solutions of the Dirichlet problem, without treating the two banks as one Euclidean boundary point.

For parameter four, the SLE4 angle martingale is bounded, so the Continuous-time martingale convergence theorem gives an almost sure limit. The imaginary part of the Chordal Loewner equation givesFor a point off the full simple Loewner trace the flow exists at every finite time. If the limiting angle lay strictly between $0$ and $\pi$, the last derivative would eventually be bounded above by a strictly negative constant. That would make $Y_t^2$ negative, a contradiction. Hence the terminal angle is either zero or $\pi$.

The simple transient chordal Loewner trace from $0$ to infinity splits the complex upper half-plane into a left component $D^{-}$, adjacent to the negative real boundary, and a right component $D^{+}$, adjacent to the positive real boundary. The Dirichlet boundary values of the SLE angle process identify which terminal value occurs. Indeed, run an independent planar Brownian motion from $z$ until it first hits the full Loewner trace or the real axis. This time is finite. The Brownian path up to that time is bounded, and Transience of chordal SLE ensures that any Loewner trace point it meets belongs to a finite initial segment. Thus its exit side for the truncated domains eventually agrees with its exit side for the full component. In $D^{-}$ every such exit is through the left bank or negative real boundary; in $D^{+}$ every exit is through the right bank or positive real boundary. Dominated convergence in the harmonic-measure representation therefore givesThis is a random side indicator, not the deterministic value $\arg z$. Uniform integrability also gives $\mathbb{E}[F(z)]=\arg z$, so the SLE4 left-passage probability is