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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.

Assume $\kappa >0$ and $x\ne 0$, since the initial driving point does not admit the displayed ordinary boundary flow. The sign of the driver in this part is negative, soAfter division by $\sqrt{\kappa }$, the Boundary-point Bessel flow for SLE isFor $x>0$ this is a Bessel process of dimension $\delta =1+2a$. For $x<0$, $-X_t$ obeys the same equation driven by $-W_t$ until hitting zero. It suffices to treat a positive initial value.

The infinitesimal generator is $\mathcal{L}f=\frac{1}{2}{f}^{\prime \prime }+(a/u)f^{\prime }$. An increasing scale function of a one-dimensional diffusion isIt satisfies $\mathcal{L}s=0$. For $0<\epsilon <x<b$, optional stopping theorem gives the boundary hitting probability from a diffusion scale functionWhen $a<1/2$, put $\beta =1-2a>0$. The inner boundary is reached in finite time: the nonnegative functionvanishes at $0,b$ and satisfies $\mathcal{L}v=-1$. Applying the Itô formula before exiting $(\epsilon ,b)$ gives $\mathbb{E}({\tau }_{\epsilon }\wedge {\tau }_b)\le v(x)$. As $\epsilon \downarrow 0$, these times increase to a finite limiting exit time almost surely. Thus the scale limit is an actual hitting event, not just asymptotic approach to zero, andLet $b\uparrow \infty$ to obtain ${\mathbb{P}}_x({\tau }_0<\infty )=1$.

If $a>1/2$, then $s(\epsilon )\to -\infty$, and the same formula makes the probability of hitting zero before any fixed $b$ equal to zero. A finite-time hit would occur before reaching some integer upper level, because the stopped path is continuous and bounded on a finite interval. Taking the countable union over those levels proves that no hit occurs. At $a=1/2$, $s(u)=\log u$ gives the same conclusion. ThereforeThe source's $x=0$ case must be excluded: zero is then already the initial value, and the displayed singular flow is undefined.

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:

At $\kappa =4$, the Boundary-point Bessel flow for SLE has dimension two. Its scale function of a one-dimensional diffusion is $s(x)=\log x$, givingThe same bounded-path and countable-union argument as in part (c) excludes a finite zero hit. Hence the critical case belongs to the boundary-avoiding regime:The countable argument of part (d) gives no visits to any positive real point almost surely. By Reflection invariance of Brownian motion, reflected SLE has the same law, so no negative real point is visited either. Thus the set of real points visited is exactly $\{0\}$. Recurrence of a two-dimensional Bessel process near zero does not mean that it hits zero.

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.

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 a real boundary point $x\ne 0$, setAfter a deterministic rescaling of time, the Boundary-point Bessel flow for SLE says that $X$ is a Bessel process of dimensionWhen $0<\kappa \le 4$, one has $\delta \ge 2$, and the Hitting-zero classification for a Bessel process says that $X$ never reaches zero. Thus no nonzero real boundary point is swallowed. The standard Loewner trace criterion then implies that each new tip is attached only to the preceding tip and the trace never intersects its past, so it is simple. For $\kappa =0$, the equation is driven by zero and generates a vertical slit. Hence ${\mathrm{SLE}}_{\kappa }$ is simple for $0\le \kappa \le 4$.

For $x\in \mathbb{R}\setminus \{0\}$, the centered Loewner imageis the Boundary-point Bessel flow for SLE, a Bessel process of dimensionWhen $\kappa >4$, one has $d<2$, and the Hitting-zero classification for a Bessel process says that $Y$ hits zero almost surely. Thus every fixed nonzero boundary point is swallowed in finite time.

A hull generated by a continuous trace cannot swallow a real point while the trace remains strictly inside $\mathbb{H}$: before any contact with the real line, the trace is a crosscut-free interior curve and no boundary interval is disconnected from infinity. Consequently finite swallowing forces the trace to meet $\mathbb{R}$ away from its initial point. SLE therefore almost surely intersects the boundary for every $\kappa \in \Lambda$.

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 $0<\kappa \le 4$.

Set $\kappa =4/(d-1)>4$. For chordal Schramm–Loewner evolution, the centered image of a real boundary point, divided by $\sqrt{\kappa }$, follows the Boundary-point Bessel flow for SLE; changing $B$ to $-B$ matches the sign convention in the question. Thus ${\tau }_x$ and ${\tau }_y$ are the times at which the marked boundary points $x$ and $y$ are swallowed, or equivalently disconnected from infinity, by the Loewner chain.

ConsequentlyIt is the event that the negative marked point is swallowed before the positive marked point.

The trace of ${\mathrm{SLE}}_{\kappa }$ is simple for $0<\kappa \le 4$, has self-intersections without being space-filling for $4<\kappa <8$, and is space-filling for $\kappa \ge 8$. The thresholds arise from the Boundary-point Bessel flow for SLE and estimates for the Loewner chain near its driving point.

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.

For a continuous chordal SLE trace started at zero, hitting the real interval $[b,\infty )$ is equivalent to finite boundary-point swallowing time for a Loewner chain at $b>0$. A boundary crosscut may swallow all points between its endpoints without visiting them individually. The Boundary-point Bessel flow for SLE determines the probability of this event.