For <Schramm–Loewner evolution> in $(\mathbb H,0,\infty)$, the <Scaling invariance of SLE> states that, for every $a>0$,
$$
\widetilde K_t=a^{-1}K_{a^2t}
$$
has the same law as $K_t$. The scaled <Loewner driving function> is $\widetilde U_t=a^{-1}U_{a^2t}$. Since $U_t=\sqrt\kappa B_t$, the <Brownian scaling> identity $a^{-1}B_{a^2t}\overset d=B_t$ 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 $\operatorname{SLE}_\kappa$ in $(\mathbb H,0,\infty)$. More precisely,
$$
\widetilde K_s=(g_t(K_{t+s}\setminus K_t)-U_t)^{\mathrm{fill}}
$$
has driving function $\widetilde U_s=U_{t+s}-U_t$. The <stationary increments> and <independent increments> of <Brownian motion> show that $\widetilde U$ 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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 203 / 2 / a / Solution

{c}

For every $a>0$, if $(K_t)$ is $\operatorname{SLE}_\kappa$, then $(a^{-1}K_{a^2t})$ has the same law. Its driving function is $a^{-1}U_{a^2t}$, which has the same law as $U_t$ by <Brownian scaling>.

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 Incoming links: Scaling invariance of SLE