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The Phase classification of the SLE trace has its simple rangewith $\kappa =0$ the deterministic vertical slit. For positive $\kappa$, the dividing parameter is precisely $\delta =1+4/\kappa =2$ in the Boundary-point Bessel flow for SLE.

Here is the reason this diffusion threshold controls simplicity. For $\kappa \le 4$, no nonzero real boundary point is swallowed. It suffices to check rational boundary points: a first meeting with either nonzero real half-axis would close a boundary crosscut and swallow a nonempty real interval, including a rational point. Thus the Loewner trace stays in the complex upper half-plane apart from its starting point. By the domain Markov property of a chordal Loewner chain, after any fixed rational time $s$ the future mapped and centred Loewner trace has the same boundary-avoidance property.

Suppose two Loewner trace times $r<r^{\prime }$ had the same image. Positive half-plane capacity growth rules out constancy on a nonempty time interval, so continuity supplies a rational $s\in (r,r^{\prime })$ with ${\gamma }_s\ne {\gamma }_r$. In the mapped future, the point at $r^{\prime }$ is either in the open upper half-plane or at the starting boundary point $0$; it cannot lie at another real point. The first case puts ${\gamma }_{r^{\prime }}$ inside the surviving domain at $s$, whereas ${\gamma }_r$ lies on the past Loewner trace. The second gives ${\gamma }_{r^{\prime }}={\gamma }_s$, also contradicting the choice of $s$. Boundary continuity of the inverse mapping-out function makes these identifications valid. This proves that the Loewner trace has no repeated points.

For $\kappa >4$, part (ii) makes a fixed positive real point have finite swallowing time, so the SLE boundary swallowing criterion ensures that the Loewner trace hits the positive real axis. In fact positive boundary-interval hitting probability for SLE above parameter four holds: any interval $J\subset (0,\infty )$ has positive hitting probability. To prove this, cover the positive axis by countably many dilates of $J$. If $J$ had zero hitting probability, Scaling invariance of SLE would give zero probability for every dilate, contradicting the almost sure hit of the positive axis. Reflection gives the same conclusion for negative intervals.

At a fixed positive time, if the past Loewner trace already repeats a point there is nothing to prove. Otherwise, boundary continuity of the inverse mapping-out function supplies a nonempty real interval, away from the current driving point, mapped back into the earlier Loewner trace in the open upper half-plane. Such an interval exists because positive capacity growth creates a genuine Loewner trace boundary in the interior; choose an accessible boundary point away from the tip and then a small interval around its preimage. Conditional on the past, the domain Markov property of a chordal Loewner chain gives a future centred SLE. With positive conditional probability its Loewner trace hits this interval, by the preceding boundary-interval argument. Mapping back then gives a visit to the earlier Loewner trace. Thus a repeated point occurs by some finite time with positive probability.

Finally let $E_t$ be the event of a repeated point by time $t$. Scaling invariance of SLE makes $\mathbb{P}(E_t)$ the same for every $t>0$. The positive finite-time probability just proved makes this common value positive. Hence ${\bigcap }_n{E}_{1/n}$ also has positive probability. This event belongs to the Brownian germ sigma-field; the Blumenthal zero-one law forces its probability to be one. Consequently the Loewner trace is not simple almost surely for every $\kappa >4$. At $\kappa =4$, the logarithmic scale function gives non-hitting of zero, so equality belongs to the simple range.

Use the following deterministic boundary fact from the Chordal Loewner equation. For a continuous trace of a Loewner chain started at zero and generating its hulls, the real flow at $b>0$ has a boundary swallowing time $T_b$. Until that time it is the reflected boundary value of $g_t$, andA finite $T_b$ is its first collision with the driver, $X_t\to 0$ as $t\uparrow T_b$. Topologically, the boundary point $b$ is then visited or separated from infinity precisely when the trace has reached some point in $[b,\infty )$. A boundary crosscut can swallow an interval, so this is not a claim that the trace visits the particular point $b$.

Consequently the SLE boundary swallowing criterion isFor the SLE driver this real flow obeysup to its first hit of zero. This is the Boundary-point Bessel flow for SLE.

For $\kappa >0$, divide the Boundary-point Bessel flow for SLE by $\sqrt{\kappa }$ and replace $W$ by $-W$. The resulting Bessel process has dimensionsince its drift is $\frac{\delta -1}{2R_t}=2/(\kappa R_t)$. The Hitting-zero classification for a Bessel process suggests the threshold $\delta =2$, or $\kappa =4$. Here is a direct verification including accessibility in finite time.

The infinitesimal generator of $X$ is $\mathcal{L}=\frac{\kappa }{2}{\partial }_x^2+(2/x){\partial }_x$. An increasing scale function of a one-dimensional diffusion isIt satisfies $\mathcal{L}s=0$. For $0<\epsilon <b<R$, the boundary hitting probability from a diffusion scale function, or optional stopping theorem applied to $s(X)$, givesFor $0<\kappa <4$, $s(\epsilon )\to -\infty$, so this probability tends to zero as $\epsilon \downarrow 0$. A finite zero hit has a bounded path before the hit and hence precedes $T_R$ for some integer $R$; taking a countable union proves that zero is never hit.

For $\kappa >4$, write $a=1-4/\kappa >0$. Taking the inner boundary to zero gives $1-(b/R{)}^a$. This limit really concerns a finite zero hit: the Itô formula givesStopped on $[\epsilon ,R]$, it impliesuniformly in $\epsilon$. The increasing limit of these exit times is finite almost surely; the stopped squared process has a continuous extension, so its limiting lower endpoint is zero. ThusLet $R\to \infty$ to obtain ${\mathbb{P}}_b(T_0<\infty )=1$. When $\kappa =0$, the explicit solution is $X_t=\sqrt{b^2+4t}$, which stays positive.

Combining this with the SLE boundary swallowing criterion gives the critical parameter and the two regimes:

For $\kappa <4$, each event $\{\gamma \text{ hits }[1/n,\infty )\}$ has probability zero by part (c). Every positive real point lies in one of these intervals. The countable union therefore has probability zero, proving simultaneous avoidance of the entire positive real axis:For $\kappa >4$, every event $\{\gamma \text{ hits }[n,\infty )\}$ has probability one. Their countable intersection still has probability one. On that event the set of positive real points visited contains a point at least $n$ for every integer $n$, so the visited positive boundary set is unbounded:This uses countably many SLE boundary swallowing criteria; no intersection over an uncountable family of probability-one events is needed.

For $\kappa >4$, the Boundary-point Bessel flow for SLE and the SLE boundary swallowing criterion imply that the full Loewner trace hits the positive real axis almost surely. Every fixed open interval in that axis has positive hitting probability: its countably many dilates cover the axis, while Scaling invariance of SLE makes all their hitting probabilities equal. Reflection gives the corresponding negative-axis statement. Together with the domain Markov property of a chordal Loewner chain, this produces visits to the old Loewner trace boundary in a mapped future domain.