Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 220 (Random Planar Geometry)

A compact real tree is a compact metric space $(T,d)$ such that any $x,y\in T$ are joined by a unique arc, and that arc is isometric to $[0,d(x,y)]$. The multiplicity of a point in a real tree $a\in T$ is the number of connected components of $T\setminus \{a\}$.

For $s,t\in [0,1]$, defineThe function $d_g$ is a pseudometric. Declare $s\sim t$ when $d_g(s,t)=0$, and give the quotient set $T_g=[0,1]/\sim$ the induced metric, again denoted $d_g$. This is the real tree encoded by an excursion $g$.

The assertion is false. Join, at one common endpoint $o$, a line segment of length $1/n$ for every positive integer $n$, and use the intrinsic path metric. This is a real tree. It is totally bounded, because outside the first finitely many arms every point lies arbitrarily close to $o$, and it is complete; hence it is compact. Removing $o$ leaves one connected component for every arm, so $o$ has countably infinite multiplicity of a point in a real tree.

The assertion is false because excursion coding does not remember the speed of traversal. Let $f$ be any nonzero coding function and let $\varphi :[0,1]\to [0,1]$ be a nonidentity increasing homeomorphism. Set $g=f\circ \varphi$. ThenThus $[s]\mapsto [\varphi (s)]$ induces an isometry $T_g\to T_f$, although generally $g\ne f$.

For nonempty compact subsets $A,B$ of a metric space $(Z,d)$, the Hausdorff distance isFor compact metric spaces $X,Y$, the Gromov-Hausdorff distance iswhere $\phi$ and $\psi$ range over isometric embeddings into a common metric space $Z$.

The collection of compact real trees is not compact in the Gromov-Hausdorff topology. Indeed, the intervals $T_n=[0,n]$ are compact real trees andTheir diameters are unbounded, so $(T_n)$ has no convergent subsequence in the Gromov-Hausdorff topology.

Choose a root $o\in T$ and finite sets $F_n\subset T$ whose union is dense, arranging that $F_n$ is a $2^{-n}$-net and $F_n\subset F_{n+1}$. Let $T_n$ be the finite subtree spanned by $o$ and $F_n$. A depth-first contour traversal of $T_n$, recording distance from $o$, gives a continuous excursion $g_n$ whose real tree encoded by an excursion is $T_n$.

The traversals may be chosen compatibly: when passing from $T_n$ to $T_{n+1}$, insert the new branch traversals into small time intervals at their attachment points. Since every new component has height at most $2^{-n+1}$, choose the time changes so thatAfter harmlessly taking a faster sequence of nets, these errors are summable. Hence $(g_n)$ is uniformly Cauchy and converges uniformly to a continuous $g:[0,1]\to {\mathbb{R}}_{+}$ with $g(0)=g(1)=0$.

The net property gives $d_{GH}(T_n,T)\to 0$. By the stated continuity of excursion coding, $d_{GH}(T_{g_n},T_g)\to 0$. Since $T_{g_n}$ is isometric to $T_n$, uniqueness of limits in the Gromov-Hausdorff distance implies that $T_g$ is isometric to $T$. This proves the excursion coding theorem for compact real trees.

For $u\ge 0$, let ${\mathcal{C}}_u$ be the collection of interval components of the superlevel set $\{r:g(r)\ge u\}$. Two times $s,t$ lie in the same member of ${\mathcal{C}}_u$ exactly when $u\le m_g(s,t)$. ThereforeFor arbitrary real ${\lambda }_1,\dots ,{\lambda }_n$, the Tonelli theorem now givesThus $(m_g(s_i,s_j){)}_{i,j}$ is a positive semidefinite matrix.

The head of the Brownian snake driven by $g$ is the centered Gaussian process $(Z_t{)}_{0\le t\le 1}$ with covariance function $\mathbb{E}[Z_s{Z}_t]=m_g(s,t)$. Part i shows that these finite-dimensional distributions exist consistently. Moreover,If $g$ has Hölder constant $L$, then $d_g(s,t)\le 2L|t-s{|}^{\alpha }$. The absolute moment formula for a centered normal distribution consequently gives, for every $p\ge 2$,The Kolmogorov continuity theorem, with $p$ arbitrarily large, produces a modification that is $\gamma$-Hölder continuous for every $\gamma <\alpha /2$. Taking $\gamma =\alpha /2-\epsilon$ proves the claim whenever $0<\epsilon <\alpha /2$; for larger $\epsilon$ the assertion is vacuous.

The set ${\mathcal{Q}}_n$ consists of rooted planar maps with $n$ faces, every face having degree four. The set ${\mathcal{Q}}_n^{\bullet }$ consists of these quadrangulations with an additional distinguished vertex.

For the trivial bijection between planar maps and quadrangulations, start from a rooted planar map with $n$ edges. Put a new vertex in every face and join it to the original vertex at every incident corner. Delete the original edges. The two endpoints of each deleted edge and the new vertices in its two adjacent faces bound a quadrangular face, so the result lies in ${\mathcal{Q}}_n$. Its bipartition distinguishes old from new vertices and reconstructs the original map. With the standard root convention this is a bijection, and hence

Consider an occurrence counted by $G_m$. Before the next counted occurrence, the exploration must first take a downward step of the simple symmetric random walk, which has probability $1/2$, and then choose the decrement $-1$ among the three equally likely values of $\xi$, which has probability $1/3$. Thus it terminates the visits to the current record value with probabilityThe Strong Markov property at successive counted occurrences makes these trials independent. Therefore $G_m$ has the geometric distribution on $\{1,2,\dots \}$ with parameter $1/6$, andfor every $j\ge 1$.

In the standard labelled encoding of a pointed planar quadrangulation, incidences at the distinguished vertex are represented by visits counted by the record variables $G_m$. Consequently its degree of a vertex is bounded by the largest such count encountered before the coding walk first reaches $-1$.

Before time $2n+1$ there are at most $2n+1$ possible record levels. The union bound and part i therefore implyConditioning on $\sigma =2n+1$ and using the supplied lower bound givesSet $r=C\log n$. If $C>5/(2\log (6/5))$, the right-hand side tends to zero. Hence for some constant $C>0$,

A compact H-hull is a bounded relatively closed set $A\subset \mathbb{H}$ for which $\mathbb{H}\setminus A$ is a simply connected domain. Its mapping-out function of a compact H-hull is the unique conformal map $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with hydrodynamic normalization at infinityIts half-plane capacity is $\mathrm{hcap}(A)=a$.

Set $h_A(z)=\mathrm{Im}(z-g_A(z))$. This is a nonnegative harmonic function on $\mathbb{H}\setminus A$, has boundary value $\mathrm{Im}z$ on the hull boundary, and tends to zero on the real boundary. If $\tau$ is the first exit time of planar Brownian motion from $\mathbb{H}\setminus A$, the optional stopping theorem givesThe hydrodynamic expansion yields $h_A(iy)=\mathrm{hcap}(A)/y+O(y^{-2})$, and therefore the Brownian representation of half-plane capacity

If $A\subseteq C$, Brownian motion exits $\mathbb{H}\setminus C$ no later than it exits $\mathbb{H}\setminus A$. The Strong Markov property and the nonnegative harmonic function $h_A$ show that the expected exit height for $C$ is at least that for $A$. Taking the limits in the Brownian representation of half-plane capacity proves the monotonicity of half-plane capacity

On $\mathbb{H}\setminus (A\cup C)$, the nonnegative harmonic function $h_A+h_C$ dominates the boundary data $\mathrm{Im}z$ on $A\cup C$: at a point of $A$, for example, $h_A(z)=\mathrm{Im}z$ and $h_C(z)\ge 0$. The maximum principle for harmonic functions, or equivalently Brownian motion stopped on the union, givesComparing the coefficients at infinity proves

The assertion is false. LetAfter filling bounded complementary components if necessary, this is a compact H-hull with diameter asymptotic to $2n$. The half-plane capacity of a low rectangle estimate, together with scaling and translation of half-plane capacity, givesThus unbounded diameter need not force unbounded capacity.

The assertion is true. The standard estimate half-plane capacity is bounded by squared diameter givesConsequently $\mathrm{hcap}(A_n)\to \infty$ forces $\mathrm{diam}(A_n)\to \infty$.

Fix $x\in \mathbb{R}\setminus \{0\}$ and, before $x$ is swallowed, put $X_t=(g_t(x)-U_t)/\sqrt{\kappa }$. The Chordal Loewner equation and $U_t=\sqrt{\kappa }{B}_t$ show, after changing the sign of the Brownian motion, thatThus $X$ is a Bessel process of dimensionA Bessel process hits zero exactly when $\delta <2$, which here is equivalent to $\kappa >4$. Hitting zero is precisely the swallowing of a nonzero boundary point by the SLE hull. For $\kappa \le 4$ the trace is simple and swallows no such point; for $\kappa >4$ swallowing occurs through a boundary contact. Hence the trace intersects $\partial \mathbb{H}\setminus \{0\}$ exactly when $\kappa >4$.

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.

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.

The Conformal Markov property of SLE states that, conditional on the hull $K_t$, the image under $g_t-U_t$ of the future hull has the same law as the original hull and is independent of the past.

For a Loewner chain with continuous driver $U$, this property says that $U_{t+s}-U_t$ is independent of the past and has the same distribution as $U_s$. Thus $U$ has stationary increments and independent increments. Every continuous process with those properties is a Brownian motion with drift, so $U_t=at+\sigma B_t$. Conformal scale invariance giveswhich forces $a=0$. Writing $\kappa ={\sigma }^2$ yields

The Locality property of SLE says the following. Let $D\subset \mathbb{H}$ be simply connected and agree with $\mathbb{H}$ in a neighborhood of $0$, and let $\psi :D\to \mathbb{H}$ be conformal with $\psi (0)=0$. The image under $\psi$ of an ${\mathrm{SLE}}_6$ in $\mathbb{H}$, stopped when it first leaves $D$, has the law of an ${\mathrm{SLE}}_6$ in $\mathbb{H}$, up to the corresponding stopping time and a change of half-plane-capacity parameterization.

With the notation supplied in the question, ${\tilde{U}}_t={\psi }_t(U_t)$ and $d{U}_t=\sqrt{\kappa }d{B}_t$. The Itô formula and ${\partial }_t{\psi }_t(U_t)=-3{\psi }_t^{\prime \prime }(U_t)$ giveFor $\kappa =6$, the drift vanishes. The resulting continuous local martingale has quadratic variationIf $s=\tilde{a}(t)/2$ is the usual half-plane-capacity time, then $⟨\tilde{U}⟩=6s$. The Dambis-Dubins-Schwarz theorem therefore givesfor a standard Brownian motion $\tilde{B}$. The mapped hulls are consequently an ${\mathrm{SLE}}_6$, which proves locality.

Explore the two stopped curves in the opposite order. Begin with an ${\mathrm{SLE}}_6$ from $1$ to $\infty$, stopped on leaving $B(1,1)$, and then, in its unbounded complementary component, draw an ${\mathrm{SLE}}_6$ from $-1$ to $\infty$, stopped on leaving $B(-1,1)$. Before their respective stopping times, each curve is separated from the neighborhood in which the other hull changes the domain. The Locality property of SLE therefore says that mapping out the other stopped hull does not change either stopped marginal law.

The two exploration orders consequently define the same joint law for the pair of stopped hulls. Disintegrating this joint law with respect to the second curve shows that, conditional on ${\gamma }_2{|}_{[0,{\tau }_2]}$, the first curve is an ${\mathrm{SLE}}_6$ in the unbounded component offrom $-1$ to $\infty$, stopped when it leaves $B(-1,1)$, as required.

Choose a Möbius transformation of $\mathbb{H}$ that fixes $0$ and exchanges $x$ with $\infty$. By Conformal invariance of SLE, it transforms an ${\mathrm{SLE}}_6$ from $0$ to $x$ into one from $0$ to $\infty$. Until $x$ is disconnected from infinity, the discrepancy between the two target domains lies beyond the component visible from the growing tip. The Locality property of SLE therefore makes the two initial curve laws identical up to that disconnection time. Hence an ${\mathrm{SLE}}_6$ from $0$ to $x$, stopped at ${\tau }_x$, has the law of an ${\mathrm{SLE}}_6$ from $0$ to $\infty$ stopped when it disconnects $x$ from infinity.