= Solution
After the fixed time $s$, the <mapping-out function> of the uncentered mapped future is
$$
\bar g_t=g_{s+t}\circ g_s^{-1}.
$$
Its expansion is $\bar g_t(z)=z+2t/z+O(z^{-2})$, so the elapsed time $t$ is still the <half-plane-capacity parameterization>. Differentiating in $t$ yields
$$
\partial_t\bar g_t(z)=\frac2{\bar g_t(z)-\xi_{s+t}},\qquad\bar g_0(z)=z.
$$
Therefore \b[the <Loewner transform> of $\bar\gamma_t=g_s(\gamma_{s+t})$ is $\xi_{s+t}$], with starting point $\xi_s$.
Centering gives $\tilde g_t(w)=\bar g_t(w+\xi_s)-\xi_s$, whose driver is $\tilde\xi_t=\xi_{s+t}-\xi_s$. By independent stationary <Brownian increments>, $(B_{s+t}-B_s)_{t\geq0}$ is a standard <Brownian motion> independent of the past through time $s$. Hence
$$
\boxed{(\tilde\gamma_t)_{t\geq0}\text{ is a fresh chordal SLE}(\kappa)
\text{ from }0\text{ to }\infty,\text{ independent of the past}.}
$$
This proves the <domain Markov property of a chordal Loewner chain> from the <composition rule for chordal Loewner driving functions>. For possible contacts with the old hull, the mapped trace is understood through the continuous boundary extension or <prime ends>; on the open half-plane the displayed maps are ordinary <conformal maps>. The PDF asks first for the driver of the uncentered curve $\bar\gamma$; the centered driver is included as well.
Back to article page
