 Solution

For Schramm–Loewner evolution in $(\mathbb{H},0,\infty )$, the Scaling invariance of SLE states that, for every $a>0$,has the same law as $K_t$. The scaled Loewner driving function is ${\tilde{U}}_t=a^{-1}{U}_{a^{2}t}$. Since $U_t=\sqrt{\kappa }{B}_t$, the Brownian scaling identity $a^{-1}{B}_{a^{2}t}\stackrel{d}{=}{B}_t$ proves the claim.

The Conformal Markov property of SLE states that, conditionally on the hull through time $t$, the future hull mapped by $g_t-U_t$ is an independent ${\mathrm{SLE}}_{\kappa }$ in $(\mathbb{H},0,\infty )$. More precisely,has driving function ${\tilde{U}}_s=U_{t+s}-U_t$. The stationary increments and independent increments of Brownian motion show that $\tilde{U}$ is independent of ${\mathcal{F}}_t$ and has the same law as $\sqrt{\kappa }B$. The deterministic correspondence between continuous drivers and Loewner chains completes the proof.