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A compact H-hull is a bounded relatively closed subset $K$ of $\mathbb{H}$ with simply connected complement, considered together with its compact closure in $\overline{\mathbb{H}}$. Its unique mapping-out function has hydrodynamic normalization at infinityThe nonnegative coefficient $a_K$ is its half-plane capacity. Thus the half-plane-capacity parameterization here gives $g_t(z)=z+2t/z+O(z^{-2})$. Strict increase means $K_s⊊K_t$ whenever $s<t$; the strictly increasing capacity already excludes equality.

The Loewner local growth property means that the new growth becomes small after removing the old compact H-hull. More precisely set $K_{t,t+h}=\overline{g_t(K_{t+h}\setminus K_t)}$, taking the closure from the complex upper half-plane. On each bounded time interval, the diameters of these image compact H-hulls tend uniformly to zero as $h\downarrow 0$. Equivalently the new growth can be separated from infinity by crosscuts of vanishing diameter in the mapped domain. It is this single-point local growth, together with capacity continuity, that gives a continuous real Loewner driver.

The Loewner transform ${\xi }_t$ is the limiting point at which this mapped growth occurs:The limit is independent of the chosen new-growth points. The Loewner correspondence theorem says that the resulting mapping-out functions satisfyConversely, given a continuous real $\xi$, solve this ordinary differential equation until the maximal lifetime $T_z$ of each $z\in \mathbb{H}$. The surviving set $H_t=\{z:T_z>t\}$ is mapped conformally onto $\mathbb{H}$ by $g_t$; its complement, with the appropriate boundary closure, is $K_t$. The Laurent coefficient is $2t$, and the composition law reconstructs the local growth. A continuous Loewner driver always determines the compact H-hulls; generation by a continuous Loewner trace is an additional property, not a conclusion for every continuous Loewner driver.

A chordal SLE Loewner trace in the complex upper half-plane with parameter $\kappa \ge 0$ is the continuous curve associated with the Loewner driver ${\xi }_t=\sqrt{\kappa }{B}_t$, where $B$ is standard real Brownian motion started at zero. It starts at zero, its compact H-hull at time $t$ is the curve's past with all pockets cut off from infinity filled, andThe capacity clock has $\mathrm{hcap}(K_t)=2t$. For $\kappa =0$ this gives the vertical slit $\gamma (t)=2i\sqrt{t}$.

To prove Scaling invariance of SLE, fix $r>0$ and putThe Laurent series of ${\hat{g}}_t$ has coefficient $2t$, and direct differentiation givesBrownian scaling says $(r^{-1}{B}_{r^{2}t}{)}_{t\ge 0}$ has the same law as $B$. The Loewner drivers therefore agree in law, and uniqueness of the Chordal Loewner equation gives equality in law of the maps, compact H-hulls, and continuous Loewner traces:This identifies the time change $t\mapsto r^{2}t$ as well as the spatial scaling.

For a simply connected Jordan domain $D$ with distinct marked boundary points, choose a conformal bijection $f:\mathbb{H}\to D$ taking zero to the initial point and infinity to the target, and use $f(\gamma )$ as an unparameterized curve. The Caratheodory boundary extension theorem supplies the boundary values of $f$. Any second such map is $f\circ (z\mapsto rz)$ for $r>0$, because the automorphisms of $\mathbb{H}$ fixing zero and infinity are positive dilations. The Scaling invariance of SLE just proved makes the two image laws identical, up to the corresponding deterministic change of capacity clock. Thus this defines SLE consistently in marked Jordan domains. The assumed $|\gamma (t)|\to \infty$ and the continuous boundary extension give $f(\gamma (t))\to f(\infty )$, the specified target. The law is conformally natural and does not impose a distinguished half-plane-capacity parameterization on an arbitrary domain.

Let ${\xi }_t=\sqrt{\kappa }{B}_t$ and $0<\kappa <4$. We use these precise Bessel process facts, which the question permits: for dimension $\delta >2$ a process started at $r>0$ exists for all time, never reaches zero, and has a strictly positive all-time minimum. More quantitatively,No assertion about SLE simplicity or transience is being assumed.

For a real point $x>0$ before its Loewner lifetime, set $Z_t^x=g_t(x)-{\xi }_t$. Its equation isThus $Z^x/\sqrt{\kappa }$ is a Bessel process of dimension $1+4/\kappa >2$. For each rational $x>0$ it stays positive on every finite interval, almost surely. These countably many events hold together. They cover every real $y>0$ as well: choose rational $0<x<y$ and observe that the noise cancels in the difference,Consequently $Z^y$ cannot reach zero while $Z^x$ remains positive. The negative side has the identical argument with the reflected Brownian motion.

This excludes Loewner trace contact with every real point other than zero, rather than just with each fixed rational point. On any finite interval a positive gap at a real point stays bounded away from zero. Continuous dependence of the Chordal Loewner equation then solves it on a complex neighborhood of that point, and the analytic flow extends the map conformally across a real interval there. Its real derivative is $\exp (-2{\int }_0^t(Z_s^x{)}^{-2}ds)>0$, and its image stays away from the driving point. The Loewner trace tip therefore cannot be there. We have proved $\gamma [0,\infty )\cap \mathbb{R}\subseteq \{0\}$.

Now use Brownian motion independent increments to restart at every deterministic rational time $s>0$. The centered future maps have Loewner driver ${\xi }_{s+t}-{\xi }_s$ and therefore the original SLE law, independently of the past. Its Loewner trace stays in $\mathbb{H}\cup \{0\}$ by the preceding argument. The inverse map $g_s^{-1}$ extends continuously to the real boundary for a continuous trace-generated compact H-hull: the finite Loewner trace gives a locally connected boundary, and the Caratheodory boundary extension theorem applies. Its value at ${\xi }_s$ is $\gamma (s)$. Thus the future in the original plane can meet $K_s\cup \mathbb{R}$ only at $\gamma (s)$. All these restart conclusions hold simultaneously for rational $s$.

Suppose $\gamma (u)=\gamma (v)$ with $u<v$. The Loewner trace cannot be constant throughout $(u,v)$, since its compact H-hull capacity increases strictly. By continuity there is a rational $s\in (u,v)$ with $\gamma (s)\ne \gamma (u)$. But $\gamma (v)=\gamma (u)$ lies in $K_s\cup \mathbb{R}$, contradicting the restart conclusion at $s$. This also excludes any return to the starting point. Hence the Loewner trace is simple and lies in the open complex upper half-plane at every positive time.

For transience, first strengthen the boundary conclusion to avoidance of the closure at each fixed real point. Fix $x>0$ and $0<r<x/4$. If the now-simple curve first enters the closed half-disc of radius $r$ about $x$ at time $\sigma$, join its tip to $x$ by the segment inside that disc. Before $\sigma$ the segment misses the Loewner trace. Together with the Loewner trace and the real interval from zero to $x$, it bounds a pocket whose boundary includes the positive side of the Loewner trace and $[0,x/2]$. A Brownian motion path from high on the imaginary axis reaching that boundary arc must enter the pocket through the segment. The maximum principle for harmonic functions bounds its harmonic measure by the chance of hitting the entire half-disc.

The boundary arc maps under $g_{\sigma }$ to $[{\xi }_{\sigma },g_{\sigma }(x/2)]$. Its harmonic measure from $iY$, as $Y\to \infty$, isThe half-disc maps out by $z\mapsto z+r^2/(z-x)$; its semicircle becomes an interval of length $4r$. Its harmonic measure has asymptotic $4r/(\pi Y)$. Comparing and letting $Y\to \infty$ yieldsBut $m_x={\inf }_{t\ge 0}(g_t(x/2)-{\xi }_t)>0$ by the all-time minimum of a transient Bessel process fact. The curve cannot enter any such disc with $4r<m_x$. Therefore it stays a strictly positive distance from $x$ for all time. Reflection proves the same for fixed $x<0$. This is the boundary approach forces a small SLE Bessel gap argument; the crosscut comparison is also described in the boundary-closure lemma in the primary SLE paper.

Map out the simple initial slit at time one. The starting point zero has two distinct real prime end images $b_{-},b_{+}$ under $g_1$, lying on opposite sides of ${\xi }_1$. Conditionally on the initial slit, the image future is a fresh independent SLE translated by ${\xi }_1$. The just-proved fixed-point closure avoidance applies to each of $b_{-}-{\xi }_1$ and $b_{+}-{\xi }_1$. Therefore neither $b_{-}$ nor $b_{+}$ lies in the closure of the mapped future. If a sequence of future Loewner trace points approached zero in the original domain, their images would approach one of these two prime ends. This is impossible. ThusFinally Scaling invariance of SLE gives $R_T:={\inf }_{t\ge T}|\gamma (t)|\stackrel{d}{=}\sqrt{T}{R}_1$. For every fixed $R>0$,The events of returning to the closed radius-$R$ disc after time $T$ decrease as $T$ increases. Continuity of probability on decreasing events implies that the probability of returns at arbitrarily late times is zero. Taking the countable intersection over positive integer $R$ provesThis proves actual eventual escape, not merely unboundedness of the growing compact H-hulls.

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.

The Scaling invariance of SLE says that $r^{-1}\gamma (r^{2}t)$ has the same law as $\gamma (t)$. Therefore its first time reaching height one is $r^{-2}{\sigma }_r$, andfor every $r>0$.

To identify $p$, it suffices to look at small heights. The trace must have a point of positive imaginary part: otherwise it lies entirely on the real axis and its hull in $\mathbb{H}$ is empty, contradicting $\mathrm{hcap}(A_t)=2t>0$. By continuity from $\gamma (0)=0$, any positive height attained by the trace forces it to pass through every smaller positive height.

Consequently the increasing union of events $\{{\sigma }_{1/n}<\infty \}$ has probability one. Each event has probability $p$, so continuity of probability gives $p=1$. This proves that SLE reaches every positive height:Equivalently, a positive random height supremum whose law is invariant under every dilation must be infinite almost surely. This argument uses neither Transience of chordal SLE nor any assertion about where the trace goes as time tends to infinity.

If $g_t$ is driven by $U_t=\sqrt{\kappa }{B}_t$, then the mapping-out functions ofareand their driver is ${\tilde{U}}_t=r{U}_{t/r^2}$. By Brownian scaling, $(r{B}_{t/r^2}{)}_{t\ge 0}$ is Brownian. Thus $\tilde{U}$ has the same law as $U$, and uniqueness of the Chordal Loewner equation provesThis is the Scaling invariance of SLE.

Let $D$ be simply connected domain with distinct marked boundary points $a,b$, and choose a conformal map $f:D\to \mathbb{H}$ with $f(a)=0$ and $f(b)=\infty$. Chordal ${\mathrm{SLE}}_{\kappa }$ from $a$ to $b$ is the unparameterized curve $f^{-1}(\gamma )$, where $\gamma$ is chordal SLE in $(\mathbb{H},0,\infty )$.

Any other such map is $\tilde{f}=rf$ for some $r>0$. The Scaling invariance of SLE says that $r^{-1}\gamma$ has the same unparameterized law as $\gamma$; only its capacity clock changes. Hence the pullback law is independent of $f$. This proves the Conformal invariance of SLE definition is well-defined.

For $0<\kappa \le 4$, the trace meets the real boundary only at its starting point, so the intersection has zero Lebesgue measure. Suppose $4<\kappa <8$. A boundary point swallowed by an interval need not itself lie on the trace. The stated fact that $\mathbb{P}({\tau }_x={\tau }_y)>0$ for $0<x<y$, combined with the Strong Markov property and Scaling invariance of SLE at successively nested swallowed intervals, implies that a fixed deterministic $x\ne 0$ is swallowed in an interval rather than hit by the trace with probability one. Equivalently,Applying the Tonelli theorem to the random indicator of the boundary trace givesThe nonnegative random measure is therefore zero almost surely.

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.

It is enough by Scaling invariance of SLE and reflection in the imaginary axis to treat $x=1$. Take $r=1+ϵ$. The drifttends to $(\kappa -8)/2<0$ as $z\to -\infty$. Choose $L<0$ and $a<0$ such that $b(z)\le a$ for $z\le L$. Before $\tilde{Z}$ reaches $L$,The infinite-horizon crossing probability for Brownian motion with negative drift shows that this process has a finite running maximum almost surely, so

Order preservation for the Loewner flow gives ${\tau }_1\le {\tau }_{1+ϵ}$. On $\{{\tau }_1<{\tau }_{1+ϵ}\}$, one has $V_t^1\to 0$ while $V_t^{1+ϵ}$ remains positive, so ${\tilde{Z}}_t\to +\infty$ and in particular $T_L<\infty$. HenceIf the trace itself hit the fixed boundary point $1$, then $1$ would be the right endpoint of the swallowed interval and ${\tau }_1<{\tau }_{1+ϵ}$ for every $ϵ>0$. Its probability is therefore zero. Scaling and reflection prove that SLE does not hit a fixed nonzero boundary point for every $x\in \mathbb{R}\setminus \{0\}$.

LetBy the stated fact, $R>0$ almost surely. Given $ϵ\in (0,1)$, choose a deterministic $r_0>0$ with $\mathbb{P}(R\ge r_0)\ge 1-ϵ$. The Scaling invariance of SLE givesand therefore, for every $t\ge 0$,

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.

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 every $x\in \mathbb{R}\setminus \{0\}$, the probability that a chordal ${\mathrm{SLE}}_{\kappa }$ trace in $(\mathbb{H},0,\infty )$ passes through $x$ is zero. The SLE boundary-point logarithmic separation diffusion shows that the chance for $x$ to be swallowed strictly before a nearby point tends to zero as that point approaches $x$; Scaling invariance of SLE and reflection then handle every nonzero $x$.

For a chordal SLE trace $\gamma$, the Scaling invariance of SLE makes $\mathbb{P}({\sup }_t\mathrm{Im}\gamma (t)>r)$ independent of $r>0$. The trace must enter the upper half-plane because its positive-time hull has positive half-plane capacity. Letting $r\downarrow 0$ therefore makes this probability one, and continuity implies that every positive height is reached. This argument does not assume Transience of chordal SLE.

The chordal SLE Loewner trace eventually leaves every bounded subset of the complex upper half-plane. For $0<\kappa <4$, its time-one compact H-hull is a simple slit and does not contain a half-disc. Instead, the boundary approach forces a small SLE Bessel gap estimate shows that every fixed nonzero real point is outside the closure of the full Loewner trace. After mapping out the initial simple slit, the starting point has two distinct real prime end images. A fresh independent SLE avoids approaching either image, so the original future stays a positive distance from its starting point. Scaling invariance of SLE gives ${\inf }_{t\ge T}|\gamma (t)|\stackrel{d}{=}\sqrt{T}{\inf }_{t\ge 1}|\gamma (t)|$, and decreasing-event continuity proves escape from every fixed disc. The separate argument based on a swallowed half-disc at positive time applies in the nonsimple regime $\kappa >4$; it must not be applied to a simple slit. The critical parameters require their own limiting or trace-existence arguments.