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For a Loewner chain with half-plane capacity $2t$, scale invariance means $({\lambda }^{-1}{K}_{{\lambda }^{2}t}{)}_{t\ge 0}$ has the original law for every $\lambda >0$. Its mapping-out maps are ${\lambda }^{-1}{g}_{{\lambda }^{2}t}(\lambda z)$ and its Loewner driving function is ${\lambda }^{-1}{U}_{{\lambda }^{2}t}$. This follows by differentiating the Chordal Loewner equation. Brownian scaling gives this property for Schramm–Loewner evolution.

After time $t$, map remaining domains by $g_t-U_t$. The new mapping-out function at elapsed time $s$ is ${\hat{g}}_s(z)=g_{t+s}(g_t^{-1}(z+U_t))-U_t$. Direct differentiation of the Chordal Loewner equation proves that its driver is $U_{t+s}-U_t$, and its half-plane capacity is $2s$. Defining transformed hulls as complements of transformed domains avoids ambiguity at the old hull boundary.

Before the Loewner swallowing time of $z$, set $Z_t=g_t(z)-U_t=X_t+i{Y}_t$ and ${{\rm Y}}_t=Y_t/|g_t^{\prime }(z)|$. Since the conformal radius of the complex upper half-plane at $Z_t$ is $2Y_t$, the transformation rule gives $2{{\rm Y}}_t={\mathrm{crad}}_{D_t}(z)$. The Koebe quarter theorem and the interior-disc comparison give $\mathrm{dist}(z,\partial D_t)/2\le {{\rm Y}}_t\le 2\mathrm{dist}(z,\partial D_t)$. Differentiating the Chordal Loewner equation gives $\boxed{{\partial }_t\log {{\rm Y}}_t=-4Y_t^2/|Z_t{|}^4}$. Thus this quantity decreases as the remaining domain shrinks.

If a compact H-hull lies in a disc of radius $r$ centred at $\xi \in \mathbb{R}$, the displayed estimate holds for $|z-\xi |>2r$, with an absolute constant $C$. It refines the Laurent expansion uniformly in the hull and supplies the first-order increment needed to derive the Chordal Loewner equation from the Loewner local growth property.

The domains of a capacity-parametrized Loewner chain determine their hydrodynamically normalized mapping-out functions of compact H-hulls. At a surviving point, the Chordal Loewner equation gives $U_t=g_t(z)-2/{\partial }_t{g}_t(z)$. For a continuous driver and $t>0$, the left derivative uses only the hull past. High points $in$, $n\in \mathbb{N}$, survive up to any fixed finite time for sufficiently large $n$, making this a countable local reconstruction of the driver and its past filtration.

A growing hull is parameterized by half-plane capacity when $\mathrm{hcap}(K_t)=2t$. The factor two makes its Chordal Loewner equation take the conventional form ${\partial }_t{g}_t=2/(g_t-U_t)$.

For a continuous Loewner driving function $U$, $T_z$ is the maximal lifetime of $g_t(z)$ in the complex upper half-plane under the Chordal Loewner equation. A finite lifetime has $\mathrm{Im}{g}_t(z)\to 0$ as $t\uparrow T_z$: otherwise the continuous driver and the ordinary differential equation continue the solution. A swallowed point need not have been visited by the Loewner trace; it may lie in a disconnected pocket of the filled hull.

A chordal Loewner chain is an increasing family of compact H-hulls, usually parameterized by half-plane capacity, whose mapping-out functions evolve according to the Chordal Loewner equation.

An increasing family of compact H-hulls of half-plane capacity $2t$ with the Loewner local growth property corresponds uniquely to a continuous real Loewner driving function. Its mapping-out functions solve the Chordal Loewner equation up to the maximal lifetime of each point. Conversely each continuous driver generates that hull family. A continuous Loewner trace is an additional property, not a consequence for every continuous driver. For SLE it is supplied by the trace-existence theorem.

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.

For chordal SLE with parameter $\kappa$, the Loewner trace $\gamma$ starts at zero and is parametrized by half-plane capacity $2t$. Let $D_t$ be the unbounded component of $\mathbb{H}\setminus \gamma [0,t]$, and let $K_t=\mathbb{H}\setminus D_t$ be its filled compact H-hull. When the trace is not simple, the hull includes the regions it disconnects from infinity. The associated mapping-out functions of compact H-hulls have expansionand satisfy the Chordal Loewner equationwhere $B$ is standard real Brownian motion. The equation is solved until $z$ is swallowed by the hull. The Loewner transform $\xi$ records the real boundary point at which the mapped hull grows; its relation to the trace isEquivalently the tip maps to ${\xi }_t$ in the appropriate boundary prime-end sense. Thus the flow encodes the surviving domains, the transform drives that flow, and the trace generates its hulls. For $\kappa =0$, the driver is identically zero, $g_t(z)=\sqrt{z^2+4t}$ with the branch asymptotic to $z$, and ${\gamma }_t=2i\sqrt{t}$.

Let $g_t$ map out the original filled hull and let $g_t^{\ast }$ map out the filled hull generated by $\Phi (\gamma [0,t])$. Both have hydrodynamic normalization at infinity. WriteAlthough the original $\Phi$ is specified only on $N$, this conjugacy is defined in the surviving neighborhood of the tip. Boundary preservation gives a real analytic extension there by the Schwarz reflection principle, with $a_t>0$. The identities below are first applied on compactly localized intervals before exit and then continued up to $T$.

Set $u(t)=\mathrm{hcap}(K_t^{\ast })/2$, the capacity clock of the image hull. The Chordal Loewner equation for each flow and the chain rule giveAnalyticity at the driving point cancels the coefficient of $(z-{\xi }_t{)}^{-1}$, forcing $\dot{u}(t)=a_t^2$. The constant term, computed as in Question 3, is then ${\partial }_t{h}_t({\xi }_t)=-3b_t$. For a general SLE driver $d{\xi }_t=\sqrt{\kappa }d{B}_t$, the Itô formula consequently gives the transformed SLE driving functionAt $\kappa =6$, this becomesThus $W$ is a continuous local martingale up to the exit time. If the source notation ${\xi }_t^{\ast }$ indexes the image driver by original time, it denotes $W_t$. If it uses the image's own half-plane-capacity time, it denotes $W_{t(u)}$, with $t(u)$ the inverse clock; that time-changed process is also a local martingale. This distinguishes the two clocks without changing the requested conclusion.

Since $\dot{u}(t)=a_t^2>0$, invert the image clock before exit and put ${\hat{\xi }}_u=W_{t(u)}$. Its quadratic variation is $6u$ and its initial value is zero, because $\Phi (0)=0$. By the Lévy characterization of Brownian motion, or the localized Dambis-Dubins-Schwarz theorem,for a standard Brownian motion up to the image exit time. If that lifetime is finite, the stopped Brownian motion can be extended beyond it; only the stopped law is used here.

In this clock the image mapping-out functions obey the ordinary Chordal Loewner equation with driver $\sqrt{6}{\hat{B}}_u$. Their generated trace therefore has the law of SLE with parameter six, stopped when it leaves $N^{\ast }$. Meanwhile the original image segment is exactly $\Phi (\gamma )$ up to exit from $N$. The clock change is increasing, so it disappears after passing to $[\gamma ]$. This proves the local stopped-curve equality required in part (a).

For completeness, the original law is scale invariant: the maps $r^{-1}{g}_{r^{2}t}(rz)$ have driver $r^{-1}{\xi }_{r^{2}t}$, whose law is again $\sqrt{6}{B}_t$ by Brownian scaling. Hence $(r^{-1}{\gamma }_{r^{2}t}{)}_{t\ge 0}$ has the original trace law. Together with the stopped conformal-change argument, this establishesThe conclusion concerns local unparameterized segments up to exit; the capacity parameter itself changes according to $du=h_t^{\prime }({\xi }_t{)}^{2}dt$.

Write $g_t=g_{K_t}$ and, for $s<t$, let $K_{s,t}$ be the filled image hull $g_s(K_t\setminus K_s)$. The half-plane-capacity composition rule givesThe Loewner local growth property says that these mapped increments shrink uniformly on bounded time intervals: for every $A<\infty$,Using diameter instead of radius is equivalent, since a nonempty compact H-hull has its closure meeting the real axis. This is a condition on the increments after mapping out the past, not merely on the Euclidean size of $K_t\setminus K_s$ in the original domain.

For fixed $t$, the nonempty compact sets $\overline{K_{t,t+h}}$, $h>0$, are nested as $h$ decreases. Their diameters tend to zero by Loewner local growth property. The nested-compact-set property gives a single intersection point, and its imaginary part is zero because $K_{t,t+h}$ lies in a real-centred disc of radius tending to zero. Define the Loewner transform byOne may restrict the intersection to sufficiently small $h$. Every point of $K_{t,t+h}$ is within $2{\omega }_A(h)$ of ${\xi }_t$ when $t+h\le A$.

Here is a direct proof of continuity, using the permitted displacement estimate. Fix $s<t$ and put $\delta =t-s$. Take $0<h\le \delta$ and a point $v\in K_{t,t+h}$. Set $z=g_{K_{s,t}}^{-1}(v)$. The hull composition rule puts $z\in K_{s,t+h}$, while the displacement estimate givesSince $z$ is within $2{\omega }_A(\delta +h)$ of ${\xi }_s$, let $h\downarrow 0$, taking $v\to {\xi }_t$, to obtainUse a slightly larger fixed time interval if $t$ is its endpoint. Both terms tend to zero with $\delta$. This proves continuity from both sides, and indeed uniform continuity on every compact time interval. The real boundary value is obtained as a limit; evaluating the displacement estimate directly on a possibly rough boundary is unnecessary.

The Loewner correspondence theorem states that a capacity-$2t$ increasing hull family with Loewner local growth property has the unique continuous real Loewner driving function just defined, and its mapping-out functions solve the Chordal Loewner equationFor each $z\in \mathbb{H}$, this solution is taken up to its maximal lifetime; $\mathbb{H}\setminus K_t$ is precisely the set of points whose lifetimes exceed $t$. Conversely a continuous real driver gives such a hull family through this ordinary differential equation. A continuous driver alone need not have a continuous Loewner trace; that additional assertion is part of the trace theory of SLE, not of the deterministic correspondence theorem.

A continuous process $\gamma :[0,\infty )\to \overline{\mathbb{H}}$ is chordal Schramm–Loewner evolution with parameter $\kappa$ from $0$ to infinity when ${\gamma }_0=0$, its filled past generates these capacity-$2t$ hulls, and their Loewner driving function iswith $B$ standard real Brownian motion. The filling includes bounded components cut off from infinity. Equivalently its Loewner trace is the continuous extension of ${\gamma }_t={\lim }_{y\downarrow 0}{g}_t^{-1}({\xi }_t+iy)$. This last boundary limit is the trace-existence input for SLE.

Fix $r>0$ and scale space and time together:The half-plane capacity of ${\hat{K}}_t$ is $2t$ by part (a). Differentiation givesBy Brownian scaling, $r^{-1}{B}_{r^{2}t}$ is again standard Brownian motion, so the new driver has the same law as the old one. Uniqueness of the Chordal Loewner equation transfers this equality to the hulls and their continuous Loewner traces. ThusThis is capacity-parametrized scale invariance of a Loewner chain. It includes $\kappa =0$, when the driver is zero and the trace is the deterministic vertical slit ${\gamma }_t=2i\sqrt{t}$.

The common Brownian motion cancels from the difference $D_t=Y_t-X_t$. Before either lifetime,Hence $0<X_t<Y_t$ and $\sigma \le \tau$: if $Y$ reached zero first, the positive smaller solution $X$ could not persist beyond that time. WriteUsing the Itô formula for $Y^{-1}$ givesThe difference has finite variation, so there is no cross-variation term in the Itô product rule for $D/Y$. Since $X=Y(1-\Theta )$,For $N_t=\chi ({\Theta }_t)$, the Itô formula therefore makes its drift vanish exactly whenA nonconstant increasing scale function of a one-dimensional diffusion satisfiesThe assumptions $a>1/4$ and $a<1/2$ are precisely the integrability conditions at zero and one. Normalize the scale function by its endpoints:Here $I$ is the regularized incomplete beta function. The normalized function is continuous and strictly increasing on $[0,1]$, with $\varphi (0)=0$ and $\varphi (1)=1$. We have obtained the local martingale

The terminal identification needs care; mere boundedness does not by itself say what happens when both lifetimes coincide. Stop first on compact subsets of $0<X<Y$, and then exhaust the interval $[0,\sigma )$. The bounded local martingale criterion makes $N$ a bounded martingale with a limit $N_{\sigma -}$, and optional sampling theorem gives $\mathbb{E}[N_{\sigma -}]=N_0$. For completeness, applying the Itô formula to each stopped $N^2$ bounds the expected quadratic variation by one. Monotone convergence over the localization sequence impliesSince ${\varphi }^{-1}$ is continuous on $[0,1]$, ${\Theta }_t$ itself has a limit ${\Theta }_{\sigma -}$.

On $\{\sigma <\tau \}$, $Y_{\sigma }>0$ while $X_t\to 0$, so ${\Theta }_{\sigma -}=1$ and $N_{\sigma -}=1$. On $\{\sigma =\tau \}$, the supplied divergent clock isIf ${\Theta }_{\sigma -}>0$, then eventually ${\Theta }_t{\varphi }^{\prime }({\Theta }_t)$ is bounded below by a positive constant. This remains true if its limit is one: ${\varphi }^{\prime }(\theta )$ diverges there, rather than tending to zero. The divergent clock would then force $[N{]}_{\sigma -}=\infty$, contradicting the finite quadratic variation. Thus ${\Theta }_{\sigma -}=0$ and $N_{\sigma -}=0$ on the simultaneous-lifetime event. We have provedTaking expectations yields the complete probability formulaThis is the beta integral for strict ordering of coupled Bessel lifetimes; the complementary probability is the probability of simultaneous swallowing, not of the larger initial point dying first.

To connect the calculation to SLE, let its driver be $W_t=\sqrt{\kappa }{\beta }_t$. For a positive boundary point $r$ before its boundary swallowing time, put $V_t^r=(g_t(r)-W_t)/\sqrt{\kappa }$. The Chordal Loewner equation givesThus two positive boundary points give exactly the coupled Bessel processes above, with $B=-\beta$ and $a=2/\kappa$. The interval $1/4<a<1/2$ corresponds to $4<\kappa <8$. Their boundary swallowing times obeyThe formula quantifies whether the growing hull separates one boundary point from infinity before the other, or swallows them together inside the same boundary interval. It does not identify swallowing with visiting each point. At $\kappa =6$, the two beta parameters are $1/3$ and this becomes the Cardy boundary crossing formula in the corresponding cross-ratio coordinate, explaining its role in the conformal invariance of planar percolation.

Let $t(u)$ be the inverse image-capacity clock. Part (a) and the quadratic variation of a stochastic integral giveThe Dambis-Dubins-Schwarz theorem, or the Lévy characterization of Brownian motion, now shows that ${\tilde{\xi }}_u/\sqrt{6}$ is standard Brownian motion up to the terminal image time. If that terminal clock is finite, it can be extended by a fresh Brownian continuation; this does not assert anything about the subsequent image trace. The initial value is zero because $\Phi (0)=0$.

Differentiating the image Loewner chain with respect to its own half-plane capacity time givesThus, until its first hit of $[1,\infty )$, the reparametrized image $\Phi (\gamma )$ has the ordinary chordal SLE law from $0$ to infinity. This follows from the uniqueness of the Chordal Loewner equation for a given continuous driver; the continuous Loewner trace is recovered from the inverse-map limit. Localization can be removed by exhausting times strictly before this boundary contact. The terminal event is defined from the trace itself, so equality of the stopped trace laws follows as well.

On the other hand, the definition of Conformal invariance of SLE says that the full unparameterized image $\Phi (\gamma )$ has the law of chordal ${\mathrm{SLE}}_6$ from $\Phi (0)=0$ to $\Phi (\infty )=1$, namely the law of ${\gamma }^{\prime }$. Its stopping event at $[1,\infty )$ is exactly the image of the original stopping event at $(-\infty ,-1]$. Therefore the same stopped image has both laws: ordinary ${\mathrm{SLE}}_6$ from $0$ to infinity stopped at $T$, and ${\mathrm{SLE}}_6$ from $0$ to $1$ stopped at $T^{\prime }$. We concludeThe statement does not identify $T$ with $T^{\prime }$ in their original clocks. It is target-change locality of SLE6, derived from the zero drift and capacity time change rather than assumed as a theorem.

Use the half-plane-capacity parameterization in which $g_t(z)=z+2t/z+O(z^{-2})$ at infinity. The Chordal Loewner equation isIt holds up to the Loewner swallowing time of $z$. Although $\beta$ is nowhere differentiable, it is continuous: the equation for $g$ is an ordinary integral equation with a continuous time-dependent coefficient away from its pole.

For $x\in \mathbb{R}\setminus \{0\}$, the same real-valued equation has a unique solution until $g_t(x)-W_t$ first reaches zero. Its solution agrees with the boundary value of the conformal map from the unswallowed side. One can also obtain it by Schwarz reflection across an unswallowed real interval. Thusis the boundary-point swallowing time for a Loewner chain; it can be infinite. The collision definition agrees with membership in the closed hull. The initial point has $T(0)=0$. Swallowing a real point is not the same as visiting it: a curve can cut off an entire real interval in one step.

For $\lambda >0$, define ${\tilde{g}}_t(z)={\lambda }^{-1/2}{g}_{\lambda t}(\sqrt{\lambda }z)$. Differentiating the Chordal Loewner equation givesBrownian scaling makes $\tilde{W}$ have the original driver law. The corresponding hull is ${\lambda }^{-1/2}{K}_{\lambda t}$, soThe time scaling applies to the whole coupled hull process. In particular, the joint law of swallowing times satisfies $(T(cx),T(cy))\stackrel{d}{=}{c}^2(T(x),T(y))$.

Thus $F(cx,cy)=F(x,y)$. A pair $0<x<y$ is determined up to dilation by $z=y/(y-x)>1$: dilating by $(y-x{)}^{-1}$ turns it into $(z-1,z)$. Define $f(z)=F(z-1,z)$. Then