= Solution
Work in the <prime end> <closure> of the marked domain, so that its two conformal <boundary> points are distinct even if their physical <boundary> representatives coincide. A <chordal filling> is a closed connected set joining those points, meeting the conformal <boundary> nowhere else, and full relative to the two <boundary> arcs between them: complementary pockets that do not reach either <boundary> arc have been added. In the normalized <complex upper half-plane>, its two complementary sides are <simply connected> and abut the negative and positive real axes. For a proper boundary-to-boundary curve, <chordal filling> means adjoining all such trapped pockets to its range. Independent curves are combined by taking their union and then applying this same <chordal filling> operation.
An admissible marked subdomain $V\subset U$ contains neighborhoods of the two marked <boundary> points and has a <simply connected> component joining them. The restriction property says that, conditional on the random <chordal filling> $K$ being contained in $V$, its law is the <chordal filling> law for $(V,z_0,z_1)$. Equivalently, mapping the conditional <chordal filling> conformally back to $(U,z_0,z_1)$ recovers the original law. The contained-set events are understood away from the shared marked endpoints.
By transporting the family under <conformal maps>, it suffices to work in $(\mathbb H,0,\infty)$. Let $A$ be an admissible <compact H-hull> whose <closure> avoids zero, and normalize
$$
\Phi_A:\mathbb H\setminus A\longrightarrow\mathbb H,
\qquad\Phi_A(0)=0,\quad\Phi_A'(\infty)=1.
$$
In terms of the single normalized measure $\mu$, restriction is precisely
$$
\boxed{\mathcal L(\Phi_A(K)\mid K\cap A=\varnothing)=\mu.}
$$
The <Schwarz reflection principle> extends $\Phi_A$ analytically near zero, with $\Phi_A'(0)>0$, so the stated conditioning events have positive probability for the two laws considered below. This is the <chordal restriction property> for filled sets. An unbounded marked subdomain can be approximated by such compact-hull removals; <conformal maps> transport the formulation to general domains.
We need a determination principle for filled sets, since a probability of one avoidance event alone would not establish equality of laws. Choose countably many thin polygonal <compact H-hulls> with rational data, attached to real <boundary> intervals away from zero. For each <chordal filling>, every point of its complement has a path within one of the two complementary sides to the appropriate real <boundary> interval. A sufficiently thin tube along such a path avoids the <chordal filling> and contains a neighborhood of that point. Hence the union of interiors of all avoided test tubes is exactly the complement.
Their joint avoidance probabilities are also determined by single-hull probabilities. For a finite union of test <compact H-hulls>, either the union separates zero from infinity, in which case a connected <chordal filling> cannot avoid it, or its unbounded complementary component contains both marked endpoints. In the latter case fill the union's bounded pockets to get an admissible <compact H-hull> $C$. A connected set joining zero to infinity and avoiding the union cannot enter one of these separated pockets, so it avoids $C$ as well. Thus joint avoidance is either impossible or is the avoidance event for $C$. The <inclusion-exclusion principle> gives every finite joint <probability distribution> of the tube-avoidance indicators, and these indicators determine the random complement and <chordal filling>. This proves <avoidance probabilities determine a chordal filling law>.
In the given avoidance formulas, the argument of the conformal map must be the retained open domain, rather than the random closed filling. We express that domain as $\mathbb H\setminus A$ to keep the two objects distinct. The formulas then imply for the two filled processes
$$
p_\gamma(A)=\Phi_A'(0)^{5/8},\qquad p_E(A)=\Phi_A'(0).
$$
The <chordal filling> operation does not change avoidance of an admissible <compact H-hull>: every piece of that <compact H-hull> is attached to the real <boundary> away from the endpoints, and so cannot be inside a trapped pocket while its generating curve avoids the <compact H-hull>. These laws are conformally invariant as unparameterized <chordal filling> laws. For <SLE> this follows from Question 1; for the <Brownian half-plane excursion>, <Brownian scaling> of its independent one- and three-dimensional coordinates shows that its unparameterized law is unchanged by positive dilations, and mapping the <Brownian half-plane excursion> to another marked domain gives the conformally natural law. <Conformal invariance of planar Brownian motion> describes the corresponding change of time, which does not alter the <chordal filling>.
Here is the conditional-law calculation for either exponent $\alpha\in\{5/8,1\}$. For another admissible <compact H-hull> $B$ in the image <complex upper half-plane>, set
$$
C=A\cup\Phi_A^{-1}(B),\qquad\Phi_C=\Phi_B\circ\Phi_A.
$$
The complement of $C$ is mapped to $\mathbb H$ by this composition, so $C$ is again admissible; both maps fix zero and have <derivative> one at infinity. The chain rule gives $\Phi_C'(0)=\Phi_B'(0)\Phi_A'(0)$. Therefore
$$
\begin{aligned}
\mathbb P(\Phi_A(K)\cap B=\varnothing\mid K\cap A=\varnothing)
&=\frac{\mathbb P(K\cap C=\varnothing)}{\mathbb P(K\cap A=\varnothing)}\\
&=\frac{\Phi_C'(0)^\alpha}{\Phi_A'(0)^\alpha}
=\Phi_B'(0)^\alpha.
\end{aligned}
$$
These are exactly the original <chordal filling>'s avoidance probabilities for every $B$. The determination principle proves equality of the conditional image law with the original law. Hence \b[both generated <chordal fillings> have restriction], with exponents $5/8$ and $1$ respectively. This is a complete implication from the two allowed avoidance formulas; the restriction conclusion was not assumed.
For independent <chordal fillings>, avoidance of $A$ by their filled union is equivalent to avoidance by every component. <Independence> multiplies the probabilities, and thus adds their <restriction exponents of a chordal filling>. Consequently
$$
\mathbb P(\overline\gamma^{\otimes8}\cap A=\varnothing)
=\bigl(\Phi_A'(0)^{5/8}\bigr)^8
=\Phi_A'(0)^5
=\bigl(\Phi_A'(0)\bigr)^5
=\mathbb P(\overline E^{\otimes5}\cap A=\varnothing).
$$
The determination principle applies once more, giving
$$
\boxed{\mathcal L(\overline\gamma^{\otimes8})=\mathcal L(\overline E^{\otimes5}).}
$$
The equality concerns the filled, unparameterized closed sets. The numerical reason is $8\cdot(5/8)=5\cdot1=5$.
Back to article page
