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Every fixed $z\in \mathbb{H}$ has infinite interior-point swallowing time for a Loewner chain under Schramm–Loewner evolution at $\kappa =4$. The centered flow is bounded on any finite horizon, making $\log |Z_t|$ bounded above. The one-sided bound criterion for a martingale clock gives a finite logarithmic limit at any putative finite swallowing time. Thus $|Z_t|$ stays away from zero and the imaginary coordinate remains positive, so the differential equation extends, a contradiction. In particular the trace misses each fixed interior point almost surely.

A compact H-hull is a bounded set $K\subset \mathbb{H}$ that is closed relative to the complex upper half-plane and whose complement $D=\mathbb{H}\setminus K$ is a simply connected domain. Its Euclidean closure is compact in $\overline{\mathbb{H}}$; “compact” here does not require separation from the real axis. One may equivalently describe the closed hull in $\overline{\mathbb{H}}$, with its real-boundary convention understood. Its mapping-out function is the unique conformal map to $\mathbb{H}$ with hydrodynamic normalization at infinity.

Use the capacity convention $\mathrm{hcap}(K_t)=2t$. A chordal Schramm–Loewner evolution from $0$ to infinity is the hull family of the Chordal Loewner equationwhere $W$ is a real standard Brownian motion and $\kappa \ge 0$. The points with interior-point swallowing time for a Loewner chain at most $t$ constitute $K_t$. For each $t$, $g_t$ maps $\mathbb{H}\setminus K_t$ conformally onto $\mathbb{H}$ and has expansion $g_t(z)=z+2t/z+O(z^{-2})$. The case $\kappa =0$ is the deterministic vertical-slit chain.

First identify the relevant transformed Loewner driving functions directly. For $\lambda >0$, defineDifferentiation gives ${\partial }_t{\tilde{g}}_t=2/({\tilde{g}}_t-{\lambda }^{-1}{U}_{{\lambda }^{2}t})$. Its initial value is $z$, its domain is $\mathbb{H}\setminus {\lambda }^{-1}{K}_{{\lambda }^{2}t}$, and its expansion is $z+2t/z+O(z^{-2})$. Thus its driving function is ${\lambda }^{-1}{U}_{{\lambda }^{2}t}$, by uniqueness of the differential equation and of the hydrodynamically normalized map. Brownian scaling provesThis is capacity-parametrized scale invariance of a Loewner chain.

For a fixed time $t$, map the future remaining domains by $z\mapsto g_t(z)-U_t$. The corresponding hulls ${\hat{K}}_s$ are specified without any boundary-image ambiguity byTheir mapping-out functions areAt $s=0$ this is the identity. Differentiating at a point in its domain yieldsThe Laurent series is $z+2s/z+O(z^{-2})$. Hence the transformed Loewner driver is exactly $U_{t+s}-U_t$, with the original capacity clock unchanged. This proves the composition rule for chordal Loewner driving functions, rather than assuming an identification of transforms.

The Brownian motion increments after $t$ are independent of its past and have the original law. Since past hulls are measurable functions of the past Loewner driver, conditionally on the past, the centered mapped future is an independent copy of the original chain. This is the domain Markov property of a chordal Loewner chain. At an almost surely finite stopping time it follows likewise from the Strong Markov property.

For the converse, a precise class is essential: take capacity-parametrized Loewner chains generated by a continuous real Loewner driving function $U$, with $U_0=0$, and require the two displayed hull properties, including independence from the entire hull past in the domain Markov property. Within this class they characterize ${\mathrm{SLE}}_{\kappa }$, for some $\kappa \ge 0$.

To see why the hull assertions determine the Loewner driver assertions, the hull domains determine their normalized maps uniquely. For any surviving point,The left time derivative suffices for $t>0$, since $U$ is continuous. A sufficiently high point survives up to any given finite time, so points $in$, $n\in \mathbb{N}$, suffice to reconstruct the Loewner driver locally and measurably from the hull past. Thus the driving-function reconstruction for a chordal Loewner chain identifies the two past filtrations and makes the preceding scale and composition computations reversible.

The domain Markov property therefore gives stationary independent increments for $U$. Its continuity makes it a continuous Lévy process. Its classification gives $U_t=bt+\sqrt{\kappa }{W}_t$: in the Lévy–Khintchine formula continuity removes the jump measure, leaving characteristic function $\exp (t(ibu-\kappa u^2/2))$. Driver scaling would turn the drift $b$ into $\lambda b$, so invariance for every $\lambda$ forces $b=0$. The diffusion coefficient is unchanged. This proves the scale-and-domain-Markov characterization of SLE with its regularity and parametrization hypotheses stated explicitly.

Use $U_t=2W_t$ and write $Z_t=X_t+i{Y}_t$ up to its interior-point swallowing time for a Loewner chain $T_z$. The Chordal Loewner equation givesThe branch of the complex logarithm with argument in $(0,\pi )$ is well-defined while $t<T_z$. The Itô formula yields the crucial cancellationThe $2/Z_t^2$ drift is cancelled by the quadratic variation term, because $\kappa =4$. Thus both $L_t=\log |Z_t|$ and ${\theta }_t=\arg Z_t$ are continuous local martingales, withThis is the logarithmic martingale for SLE4.

We justify absence of a finite swallowing time rather than presuming that the logarithm survives forever. On a finite horizon $H$, $0<Y_t\le \mathrm{Im}z$. Also $|X_t|$ is bounded pathwise before $T_z\wedge H$. Indeed, while $|X_t|\ge 1$ its drift $2X_t/|Z_t{|}^2$ has absolute value at most $2$; on each excursion outside $[-1,1]$, integrate from its starting point and bound the Brownian oscillation on $[0,H]$. For example,Therefore $L_t$ is bounded above pathwise on this interval.

Suppose $T_z\le H$. By the Dambis-Dubins-Schwarz theorem, $L$ is a Brownian motion run at its own quadratic variation. If that clock diverged as $t\uparrow T_z$, Brownian oscillation would make $L$ unbounded above, contradicting the preceding bound. The one-sided bound criterion for a martingale clock therefore gives a finite clock limit and a finite real limit for $L_t$. Hence $|Z_t|$ is bounded away from zero near $T_z$.

Now $Y_t=Y_0\exp (-2{\int }_0^t|Z_s{|}^{-2}ds)$ has a strictly positive limit at $T_z$. The drift in $X$ is integrable there, so continuity of $W$ gives a finite limit for $X$ as well. The limiting point is in $\mathbb{H}$ and away from the Loewner driver singularity, and the differential equation extends past $T_z$, a contradiction. Thus $T_z=\infty$ almost surely. A point of the Loewner trace at a finite time belongs to that time's hull, soThis proves the fixed-interior-point avoidance of SLE4 without using simplicity as an input.

The angle remains in $(0,\pi )$ at all finite times. The bounded local martingale criterion upgrades its local martingale equation to a genuine martingale:In particular this is the SLE4 angle martingale, and it converges almost surely and in $L^1$ by bounded martingale convergence theorem.

It remains to identify the limiting angle using the assumed simple path tending to infinity. Orient that path from $0$ to infinity. Its left component is the one adjacent to the negative real half-axis. Under $g_t$, the left boundary of the slit domain maps to $(-\infty ,U_t)$ and the right boundary to $(U_t,\infty )$. The harmonic measure of the former as seen from $z$ isThis follows by conformal invariance of planar Brownian motion: in the upper half-plane, $\arg (w-U_t)/\pi$ is the bounded harmonic function with values $1$ on the left half-axis and $0$ on the right.

To justify the limiting boundary classification, condition on a simple proper realization of the path and use an independent planar Brownian motion from $z$. It exits the upper half-plane in finite time almost surely, so its path up to that time is compact. The curve tends to infinity, so its intersection with this compact set is contained in a finite initial curve segment. Once that segment has been drawn, the Brownian path exits the slit domain through its left boundary exactly when $z$ is in the final left component: a path from that component cannot reach the right boundary without crossing the curve, and the reverse assertion holds on the right. Endpoints have zero harmonic measure. Bounded convergence of these exit indicators provesTaking expectations in the bounded angle martingale gives the SLE4 left-passage probabilityFor the imaginary axis it is $1/2$; near the negative real axis it tends to $1$, fixing the orientation of “left”.