 Solution

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.