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For a continuous Loewner trace generating its hulls, the inverse mapping-out function of a compact H-hull $g_t^{-1}$ extends continuously to the closed complex upper half-plane. The finite trace is a continuous map image of an interval, and the boundary of its unbounded complementary component has local connectedness; the Caratheodory boundary extension theorem gives the extension. In particular $g_t^{-1}(U_t)=\gamma (t)$, while $g_t^{-1}$ maps the open complex upper half-plane into $D_t$. This is the deterministic boundary-extension input of Rohde and Schramm, Theorem 4.1. Under a Conformal Markov property of SLE restart, this relates boundary contacts of the mapped future to contacts with the old hull, without assuming the old trace is simple.

The boundary assertion is interpreted at positive finite times: a chordal SLE already starts on the domain boundary, and its marked target is another boundary point. The precise conclusion in the complex upper half-plane is $\gamma (0,\infty )\subseteq \mathbb{H}$ almost surely.

For $0<\kappa \le 4$ and a fixed real $x>0$, the centered Boundary-point Bessel flow for SLE $V_t=g_t(x)-U_t$ solvesThus $R_t=V_t/\sqrt{\kappa }$ is a Bessel process driven by $-B$ of dimensionThe Hitting-zero classification for a Bessel process says that, started positively, it never hits zero when $\delta \ge 2$. For $x<0$, reflect the equation and apply the same result. Therefore every fixed nonzero boundary point has infinite swallowing time. A countable intersection gives this simultaneously for all nonzero rational boundary points.

To extend this countable conclusion to every real $x>0$, choose rational $q$ with $0<q<x$. The difference of two centered boundary solutions satisfiesso it remains positive while both flows exist. On any finite time interval, $V_t(q)$ is bounded away from zero, hence so is $V_t(x)$. Moreover $V_t(x)=x-U_t+{\int }_0^{t}2/V_s(x)ds$ remains bounded there, so the differential equation continues for the whole interval. Its uniform separation from the Loewner driving function also allows each flow map to continue as a holomorphic function on a complex neighborhood of $x$. Thus $x$ is outside $K_t$ and cannot be visited by the Loewner trace. Reflection gives the negative real axis. This uses countably many Bessel processes followed by deterministic flow comparison, rather than an uncountable union of probability-zero events.

To exclude a return to the starting point, use the boundary extension of the inverse map for a continuous Loewner trace: $g_s^{-1}$ is continuous on the closed complex upper half-plane and maps $U_s$ to $\gamma (s)$. This follows from the continuous trace theorem and the Caratheodory boundary extension theorem; it does not assume simplicity. At each deterministic rational time $s$, the Conformal Markov property of SLE makes the mapped future a fresh SLE, so its trace lies in $\mathbb{H}\cup \{0\}$ by the preceding argument. Mapping back with $g_s^{-1}$ therefore showsIf a point $q=\gamma (r)$ were revisited at $t>r$, choose rational $s\in (r,t)$ with $\gamma (s)\ne q$. Such an $s$ exists because continuity and strictly increasing half-plane capacity forbid a constant trace on an interval. But $q\in K_s\cup \mathbb{R}$, contradicting the displayed inclusion. This excludes all self-contacts, including returns to zero, and proves the simple curve property needed when discussing later swallowing times.

For $\kappa =0$, the deterministic Loewner trace is $\gamma (t)=2i\sqrt{t}$, so the conclusion follows directly. Hence the boundary-intersection threshold for SLE givesMapping to another marked simply connected domain carries positive-time points into its interior. The literal statement including the starting point is false, since $\gamma (0)$ is prescribed to be on the domain boundary.

Before the Loewner swallowing time $T_z$, put $Z_t=g_t(z)-U_t=X_t+i{Y}_t$ and ${\theta }_t=\arg Z_t\in (0,\pi )$. The Itô formula applied to the holomorphic logarithm in the complex upper half-plane givesTaking imaginary parts yields the SLE angle processFor $\kappa =4$ the drift vanishes, sois a continuous local martingale. Since $0<{\theta }_t<\pi$, every localized stopped version is a bounded true martingale. After resolving the lifetime issue below, the dominated convergence theorem applied to conditional expectations removes localization and gives a true bounded martingale.

For completeness, there is no finite lifetime ambiguity for a fixed interior $z$ in this parameter range. The preceding Bessel process and Conformal Markov property of SLE argument gives a simple trace staying in the interior, so no point is swallowed in a disconnected pocket. If $T_z$ were finite, the trace would have to reach $z$, forcing the Loewner conformal radius to tend to zero by the Koebe quarter theorem. ButThe Dambis-Dubins-Schwarz theorem says that a bounded continuous local martingale cannot have infinite quadratic variation before a finite terminal time: that would require a Brownian motion to stay in a bounded interval for all clock times. Therefore ${{\rm Y}}_t$ cannot tend to zero at a finite $T_z$, a contradiction. Thus $T_z=\infty$ almost surely for each fixed $z$. This proves the SLE4 angle martingale assertion: the angle is a bounded continuous martingale. In particular $\mathbb{E}[{\theta }_t]=\arg z$. The upper-half-plane branch of the argument is used throughout.

Work with a locally growing Loewner chain started at $0$, with $A_0=\varnothing$ and the half-plane-capacity parameterization $\mathrm{hcap}(A_t)=2t$. The Conformal Markov property of SLE combines invariance under conformal changes of the marked domain with restarting after an initial hull has been removed.

In the half-plane this means, first, that $(r^{-1}{A}_{r^{2}t}{)}_{t\ge 0}$ has the same law as $(A_t{)}_{t\ge 0}$ for every $r>0$. Second, conditionally on ${\mathcal{F}}_t$, the future hulls mapped by $g_t-U_t$,with the usual hull closure, have the original law and are independent of ${\mathcal{F}}_t$. Their capacity is $2s$ by the half-plane-capacity composition rule. One may require the same restarting property at finite stopping times.

Equivalently, the driver has the scaling law and conditional increment lawThe conformal invariance clause matters: the restarting property alone would also allow a Brownian driver with deterministic drift.

The Loewner local growth property gives a continuous Loewner driving function, and we normalize $U_0=0$. Mapping out an initial hull transforms the future driver to $U_{t+s}-U_t$. The Conformal Markov property of SLE therefore gives stationary increments and independent increments.

Consequently $U$ is a continuous Lévy process. In the Lévy–Khintchine formula, continuity excludes its jump measure, so its only possible components are a deterministic linear drift and Brownian motion:Conformal invariance under dilations gives $U_{r^{2}t}/r\stackrel{d}{=}{U}_t$. Comparing the expectations of these Gaussian random variables yields $\mu rt=\mu t$ for every $r>0$, hence $\mu =0$. The Brownian term already has the required law by Brownian scaling. Therefore the characterization of the SLE driving function isThe case $\kappa =0$ is the constant zero driver. No reflection-symmetry assumption is needed to remove the drift; scaling suffices.

For $\kappa >0$ and $x>0$, before the boundary point is swallowed, put $X_t=(g_t(x)-U_t)/\sqrt{\kappa }$. The Chordal Loewner equation and $d{U}_t=\sqrt{\kappa }d{B}_t$ giveSince $-B$ is again a standard Brownian motion, this is the Boundary-point Bessel flow for SLE withFor $x<0$, it is $-X_t$ that is the nonnegative Bessel process, with the same dimension; the formula without this sign convention is a signed Bessel flow.

When $0<\kappa \le 4$, one has $\delta \ge 2$, and the Hitting-zero classification for a Bessel process says that $X_t$ never reaches zero. The borderline $\delta =2$ uses the scale function of a one-dimensional diffusion $\log x$: for $0<r<x<R$,Thus the critical case also cannot hit zero in finite time, though its all-time infimum is zero.

Apply the non-swallowing assertion simultaneously to all nonzero rational boundary points. The order-preserving real Loewner flow then keeps every compact real interval away from the origin in the surviving boundary. A boundary contact away from the starting point would cut off a nonempty real interval, swallowing a rational point, so the trace avoids $\mathbb{R}\setminus \{0\}$.

The same argument can be applied to the future after every rational time, using the Conformal Markov property of SLE. If the trace revisited an earlier point, choose a rational time strictly between the two visits. In the domain with that initial segment removed, the later visit would be a contact with an old boundary point away from the current growing tip. Under the mapping-out map, such a contact cuts off a real interval and contradicts the preceding non-swallowing result. This is the usual crosscut argument converting boundary non-swallowing into absence of self-intersections. Taking the countable intersection over rational restart times provesFor $\kappa =0$, the equation is deterministic and gives the simple slit $(0,2i\sqrt{t}]$. The division by $\sqrt{\kappa }$ is unnecessary in that case.

For $\kappa \le 4$, part i gives no nonzero boundary intersection. For $\kappa >4$, part i gives a boundary hit almost surely. After any hit, map out the past hull and recenter at the current tip. The Conformal Markov property of SLE says that the future is again an SLE in the remaining domain. Applying part i repeatedly and using Scaling invariance of SLE produces another boundary hit after every finite number of hits. Therefore there are almost surely infinitely many.

The Conformal Markov property of SLE states that, conditional on the hull $K_t$, the image under $g_t-U_t$ of the future hull has the same law as the original hull and is independent of the past.

For a Loewner chain with continuous driver $U$, this property says that $U_{t+s}-U_t$ is independent of the past and has the same distribution as $U_s$. Thus $U$ has stationary increments and independent increments. Every continuous process with those properties is a Brownian motion with drift, so $U_t=at+\sigma B_t$. Conformal scale invariance giveswhich forces $a=0$. Writing $\kappa ={\sigma }^2$ yields

The Conformal Markov property of SLE says that, conditionally on $(K_s{)}_{s\le t}$, the future hullshave the law of a fresh SLE in $(\mathbb{H},0,\infty )$ and are independent of the past, with the usual interpretation of mapping out the old hull.

Stop after a positive initial segment and map it out. By the Conformal Markov property of SLE, the future is a fresh SLE in the remaining domain. Part b says that it hits that domain's boundary away from its current and terminal points. Choosing a bounded crosscut neighborhood whose real-boundary part is shielded from the future forces such a hit to occur on the image of the earlier trace with positive probability. Scale invariance and repeated conditional trials upgrade this to probability one. Thus the trace has self-intersections for $\kappa >4$, in agreement with the Phase classification of the SLE trace.

The Conformal Markov property of SLE says that, conditionally on the curve through time $t$, mapping out the initial hull by $g_t-U_t$ turns the future into an independent Schramm–Loewner evolution in $(\mathbb{H},0,\infty )$. More precisely,has driving function ${\tilde{U}}_s=U_{t+s}-U_t$. Since $U_t=\sqrt{\kappa }{B}_t$, the stationary increments and independent increments of Brownian motion give the asserted independence and equality in law.

For a real boundary point $x\ne 0$, the Boundary-point Bessel flow for SLE says that $V_t^x/\sqrt{\kappa }$ is a Bessel process of dimensionWhen $\kappa \le 4$, one has $d\ge 2$, so part 2(b), including its logarithmic $d=2$ case, shows that no fixed boundary point is swallowed.

If the trace touched the real line away from its starting point, or if a later segment crossed an earlier segment, the resulting hull would disconnect from infinity a nonempty real interval, which contains a rational number. Applying the preceding argument after every rational time and using the Conformal Markov property of SLE rules this out on a countable probability-one event. Continuity of the trace then shows that no two distinct times have the same image. Thus ${\mathrm{SLE}}_{\kappa }$ is simple for $0<\kappa \le 4$.

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.