= Solution
Use the <Chordal Loewner equation> with <Loewner driving function> $\xi_t=\sqrt\kappa W_t$, and set $Z_t=g_t(z)-\xi_t=X_t+iY_t$. Before the <Loewner swallowing time>,
$$
dZ_t=\frac2{Z_t}\,dt-\sqrt\kappa\,dW_t.
$$
The upper-half-plane branch of the logarithm has imaginary part $h_t$. The <Itô formula> gives
$$
d\log Z_t=\frac{4-\kappa}{2Z_t^2}\,dt
-\frac{\sqrt\kappa}{Z_t}\,dW_t,
$$
and therefore the <SLE angle process> satisfies
$$
\boxed{dh_t=(\kappa-4)\frac{X_tY_t}{|Z_t|^4}\,dt
+\sqrt\kappa\,\frac{Y_t}{|Z_t|^2}\,dW_t.}
$$
At $\kappa=4$ 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 $(\kappa-4)X_tY_t=0$ throughout every compact interval before swallowing. Because $Y_t>0$, for $\kappa\ne4$ 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 \b[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.
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