# Generating uniform trees

A long time ago, I wrote quite a few a things about uniform trees. That is, a uniform choice from the $n^{n-2}$ unrooted trees with vertex set [n]. This enumeration, normally called Cayley’s formula, has several elegant arguments, including the classical Prufer bijection. But making a uniform choice from a large set is awkward, and so we seek more probabilistic methods to sample such a tree, which might also give insight into the structure of a ‘typical’ uniform tree.

In another historic post, I talked about the Aldous-Broder algorithm. Here’s a quick summary. We run a random walk on the complete graph $K_n$ started from a uniformly-chosen vertex. Every time we arrive at a vertex we haven’t visited before, we record the edge just traversed. Eventually we have visited all n vertices, so have recorded n-1 edges. It’s easy enough to convince yourself that these n-1 edges form a tree (how could there be a cycle?) and a bit more complicated to decide that the distribution of this tree is uniform.

It’s worth noting that this algorithm works to construct a uniform spanning tree on any connected base graph.

This post is about a few alternative constructions and interpretations of the uniform random tree. The first construction uses a Galton-Watson process. We take a Galton-Watson process where the offspring distribution is Poisson(1), and condition that the total population size is n. The resulting random tree has a root but no labels, however if we assign labels in [n] uniformly at random, the resulting rooted tree has the uniform distribution among rooted trees on [n].

Proof

This is all about moving from ordered trees to non-ordered trees. That is, when setting up a Galton-Watson tree, we distinguish between the following two trees, drawn extremely roughly in Paint:

That is, it matters which of the first-generation vertices have three children. Anyway, for such a (rooted) ordered tree T with n vertices, the probability that the Galton-Watson process ends up equal to T is

$\mathbb{P}(GW = T) = \prod_{v\in T} \frac{e^{-1}}{C(v)!} = e^{-n} \prod_{v\in T}\frac{1}{C(v)!},$

where $C(v)$ is the number of children of a vertex $v\in T$. Then, since $\mathbb{P}( |GW|=n )$ is a function of n, we find

$\mathbb{P}(GW=T \,\big|\, |GW|=n) = f(n)\prod_{v\in T} \frac{1}{C(v)!},$

where f(n) is a function of n alone (ie depends on T only through its size n).

But given an unordered rooted tree t, labelled by [n], there are $\prod_{v \in t} C(v)!$ ordered trees associated to t in the natural way. Furthermore, if we take the Poisson Galton-Watson tree conditioned to have total population size n, and label uniformly at random with [n], we obtain any one of these ordered trees with probability $\frac{f(n)}{n!} \prod_{v\in t} \frac{1}{C(v)!}$. So the probability that we have t after we forget about the ordering is $\frac{f(n)}{n!}$, which is a function of n alone, and so the distribution is uniform among the set of rooted unordered trees labelled by [n], exactly as required.

Heuristic for Poisson offspring distribution

In this proof, the fact that $\mathbb{P}(C(v)=k)\propto \frac{1}{k!}$ exactly balances the number of orderings of the k children explains why Poisson(1) works out. Indeed, you can see in the proof that Poisson(c) works equally well, though when $c\ne 1$, the event we are conditioning on (namely that the total population size is n) has probability decaying exponentially in n, whereas for c=1, the branching process is critical, and the probability decays polynomially.

We can provide independent motivation though, from the Aldous-Broder construction. Both the conditioned Galton-Watson construction and the A-B algorithm supply the tree with a root, so we’ll keep that, and look at the distribution of the degree of the root as constructed by A-B. Let $\rho=v_1,v_2,v_3,\ldots$ be the vertices [n], ordered by their discovery during the construction. Then $\rho$ is definitely connected by an edge to $v_2$, but thereafter it follows by an elementary check that the probability $\rho$ is connected to $v_m$ is $\frac{1}{n-1}$, independently across all m. In other words, the distribution of the degree of $\rho$ in the tree as constructed by A-B is

$1+ \mathrm{Bin}\left(n-2,\frac{1}{n-1}\right) \approx 1+\mathrm{Poisson}(1).$

Now, in the Galton-Watson process, conditioning the tree to have fixed, large size changes the offspring distribution of the root. Conveniently though, in a limiting sense it’s the same change as conditioning the tree to have size at least n. Since these events are monotone in n, it’s possible to take a limit of the conditioning events, and interpret the result as the Galton-Watson tree conditioned to survive. It’s a beautiful result that this interpretation can be formalised as a local limit. The limiting spine decomposition consists of an infinite spine, where the offspring distribution is a size-biased version of the original offspring distribution (and so in particular, always has at least one child) and where non-spine vertices have the original distribution.

In particular, the number of the offspring of the root is size-biased, and it is well-known and not hard to check that size-biasing Poisson(c) gives 1+Poisson(c) ! So in fact we have, in an appropriate limiting sense in both objects, a match between the degree distribution of the root in the uniform tree, and in the conditioned Galton-Watson tree.

This isn’t supposed to justify why a conditioned Galton-Watson tree is relevant a priori (especially the unconditional independence of degrees), but it does explain why Poisson offspring distributions are relevant.

Construction via G(N,p) and the random cluster model

The main reason uniform trees were important to my thesis was their appearance in the Erdos-Renyi random graph G(N,p). The probability that vertices {1, …, n} form a tree component in G(N,p) with some particular structure is

$p^{n-1} (1-p)^{\binom{n}{2}-(n-1)} \times (1-p)^{n(N-m)}.$

Here, the first two terms give the probability that the graph structure on {1, …, n} is correct, and the the final term gives the probability of the (independent) event that these vertices are not connected to anything else in the graph. In particular, this has no dependence on the tree structure chosen on [n] (for example, whether it should be a path or a star – both examples of trees). So the conditional distribution is uniform among all trees.

If we work in some limiting regime, where $pn\rightarrow 0$ (for example if n is fixed and $p=\frac{1}{N}\rightarrow 0$), then we can get away asymptotically with less strong conditioning. Suppose we condition instead just that [n] form a component. Now, there are more ways to form a connected graph with one cycle on [n] than there are trees on [n], but the former all require an extra edge, and so the probability that a given one such tree-with-extra-edge appears as the restriction to [n] in G(N,p) is asymptotically negligible compared to the probability that the restriction to [n] of G(N,p) is a tree. Naturally, the local limit of components in G(N,c/N) is a Poisson(c) Galton-Watson branching process, and so this is all consistent with the original construction.

One slightly unsatisfying aspect to this construction is that we have to embed the tree of size [n] within a much larger graph on [N] to see uniform trees. We can’t choose a scaling p=p(n) such that G(n,p) itself concentrates on trees. To guarantee connectivity with high probability, we need to take $p> \frac{\log n}{n}$, but by this threshold, the graph has (many) cycles with high probability.

At this PIMS summer school in Vancouver, one of the courses is focusing on lattice spin models, including the random cluster model, which we now briefly define. We start with some underlying graph G. From a physical motivation, we might take G to be $\mathbb{Z}^d$ or some finite subset of it, or a d-ary tree, or the complete graph $K_N$. As in classical bond percolation (note G(N,p) is bond percolation on $K_N$), a random subset of the edges of G are included, or declared open. The probability of a given configuration w, with e open edges is proportional to

$p^e (1-p)^{|E(G)| - e} q^{k(w)},$ (*)

where the edge-weight $p\in(0,1)$ as usual, and cluster weight $q\in (0,\infty)$, and $k(w)$ counts the number of connected components in configuration w. When q=1, we recover classical bond percolation (including G(N,p) ), while for q>1, this cluster-reweighting favours having more components, and q<1 favours fewer components. Note that in the case $q\ne 1$, the normalising constant (or partition function) of (*) is generally intractable to calculate explicitly.

As in the Erdos-Renyi graph, consider fixing the underlying graph G, and taking $p\rightarrow 0$, but also taking $\frac{q}{p}\rightarrow 0$. So the resulting graph asymptotically ‘wants to have as few edges as possible, but really wants to have as few components as possible’. In particular, 1) all spanning trees of G are equally likely; 2) any configuration with more than one component has asymptotically negligible probability relative to any tree; 3) any graph with a cycle has #components + #edges greater than that of a tree, and so is asymptotically negligible probability relative to any tree.

In other words, the limit of the distribution is the uniform spanning tree of G, and so this (like Aldous-Broder) is a substantial generalisation, which constructs the uniform random tree in the special case where $G=K_n$.

# Real Trees – Root Growth and Regrafting

Two weeks ago in our reading group meeting, Raphael told us about Chapter Five which introduces root growth and regrafting. One of the points of establishing the Gromov-Hausdorff topology in this book was to provide a more natural setting for a discussion of tree-valued processes. Indeed in what follows, one can imagine how to start the construction of a similar process for the excursions which can be used to encode real trees, involving cutting off sub-excursions above one-sided local minima, then glueing them back in elsewhere. But taking account of the equivalence structure will be challenging, and it is much nicer to be able to describe cutting a tree in two by removing a single point without having to worry about quotient maps.

We have seen in Chapter Two an example of a process defined on the family of rooted trees with n labelled vertices which has the uniform rooted tree as an invariant distribution. Given a rooted tree with root p, we choose uniformly at random a vertex p’ in [n] to be the new root. Then if p’=p we do nothing, otherwise we remove the unique first edge in the path from p’ to p, giving two trees. Adding an edge from p to p’ completes the step and gives a new tree with p’ as root. We might want to take a metric limit of these processes as n grows and see whether we end up with a stationary real tree-valued process whose marginals are the BCRT.

To see non-trivial limiting behaviour, it is most interesting to consider the evolution of a particular subtree (which includes the root) through this process. If the vertex chosen for cutting lies in our observed subtree, then the subtree undergoes a prune and regraft operation. On the other hand, if the vertex chosen for cutting does not lie in the subtree, then we do not see any effect of the pruning, except the addition of a new vertex below the original root, which becomes the new root. So essentially, from the point of view of our observed subtree, the root is growing.

Now we can think about interpreting the dynamics of a natural limit process acting on real trees. The key idea is that we don’t change the set on which the tree is supported much, but instead just change the metric. In particular, we will keep the original tree, and add on length at unit rate. Of course, where this length gets added on entirely determines the metric structure of the tree, but that doesn’t stop us giving a simple ‘name’ for the extra length. If we consider a process $X^T$ starting from a particular finite subtree T, then at time t, the tree $X^T_t$ is has vertex set $T \coprod (0,t]$. (Finite subtree here means that it has finite total length.)

Root regrafting should happen at a rate proportional to the total length of the current observed tree. This is reasonable since after all it is supported within a larger tree, so in the discrete case the probability of a prune-regrafting event happening within a given observed subtree is proportional to the number of vertices in that subtree, which scales naturally as length in the real tree limit. It turns out that to get unit rate root growth with $\Theta(1)$ rate prune-regrafting, we should consider subtrees of size $\sqrt{n}$ within a host tree of size n as $n\rightarrow\infty$. We also rescale the lengths by $\frac{1}{\sqrt{n}}$, and time by $\sqrt{n}$ so we actually see prune-regraft events.

Furthermore, if the subtree is pruned, the location of the pruning is chosen uniformly by length of the current observed subtree. So we can view the pruning process as being driven by a Poisson point process with intensity given by the instantaneous length measure of the tree, which at time t has vertex set $T\coprod (0,t]$. It will turn out to be consistent that there is a ‘piecewise isometry’ for want of a better phrase between the metric (and thus length measure) on $X^T_t$ and the canonical induced measure on $T\coprod (0,t]$, so we can describe the instances and locations of the pruning events via a pair of PPPs. The first is supported on $T \times [0,\infty)$, and the second on $\{(t,x): 0 \le x \le t\}$, since we only ‘notice’ pruning at the point labelled x if the pruning happens at some time t after x was created.

If we start from a compact tree T, then the total intensity of this pair is finite up to some time t, and so we have a countable sequence $\tau_0=0<\tau_1<\tau_2<\ldots$ of times for pruning events. It is easy to describe (but a bit messy to notate) the evolution of the metric between these pruning times. Essentially the distance between any pair of points in the observed tree at time $\tau_m$ with root $\rho_{\tau_m}$ is constant between times $\tau_m,\tau_{m+1}$, and new points are added so that the distance between $\rho_{\tau_m}$ and any new point $a\in(\tau_m,\tau_{m+1}]$ is $a-\tau_m$, and everything thing else follows from straightforward consideration of geodesics.

When a pruning event happens at point $x_m$ at time $\tau_m$, distances are preserved within the subtree above $x_m$ in $X^T_{\tau_m -}$, and within the rest of the tree. Again, an expression for the cross distances is straightforward but requires a volume of notation not ideally suited to this medium.

The natural thing to consider is the coupled processes started from different subtrees (again both must contain the original root) of the same host tree. Say $T^1,T^2\le T$, then it is relatively easy to check that $X^{T^1}_t,X^{T^2}_t \le X^T_t \,\forall t$, when we drive the processes by consistent coupled Poisson processes. Furthermore, it is genuinely obvious that the Hausdorff distance between $X^{T^1}_t,X^{T^2}_t$, here viewed as compact subsets of $(X^T_t, d^T_t)$ remains constant during root growth phase.

Less obvious but more important is that the Hausdorff distance decreases during regrafting events. Suppose that just before a regrafting event, the two subtrees are T’ and T”, and the Hausdorff distance between them is $\epsilon$. This Hausdorff distance is with respect to the metric on the whole tree T. [Actually this is a mild abuse of notation – I’m now taking T to be the whole tree just before the regraft, rather than the tree at time 0.]

So for any $a\in T'$, we can choose $b\in T''$ such that $d_T(a,b)\le \epsilon$. This is preserved under the regraft unless the pruning point lies on the geodesic segment (in T) between a and b. But in that case, the distance between a and the pruning point is again at most $\epsilon$, and so after the regrafting, a is at most $\epsilon$ away from the new root, which is in both subtrees, and in particular the regrafted version of T”.

This is obviously a useful first step on the path to proving any kind of convergence result. There are some technicalities which we have skipped over. It is fairly natural that this leads to a Markov process when the original tree is finite, but it is less clear how to define these dynamics when the total tree length is infinite, as we don’t want regrafting events to be happening continuously unless we can bound their net effect in some sense.

Last week, Franz showed us how to introduce the BCRT into matters. Specifically, that BCRT is the unique stationary distribution for this process. After a bit more work, the previous result says that for convergence properties it doesn’t matter too much what tree we start from, so it is fine to start from a single point. Then, the cut points and growth mechanism corresponds very well to the Poisson line-breaking construction of the BCRT. With another ‘grand coupling’ we can indeed construct them simultaneously. Furthermore, we can show weak convergence of the discrete-world Markov chain tree algorithm to the process with these RG with RG dynamics.

It does seem slightly counter-intuitive that a process defined on the whole of the discrete tree converges to a process defined through subtrees. Evans remarks in the introduction to the chapter that this is a consequence of having limits described as compact real trees. Then limitingly almost all vertices are close to leaves, so in a Hausdorff sense, considering only $\sqrt{n}$ of the vertices (ie a subtree) doesn’t really make any difference after rescaling edge lengths. I feel I don’t understand exactly why it’s ok to take the limits in this order, but I can see why this might work after more checking.

Tomorrow, we will have our last session, probably discussing subtree prune-and-regraft, where the regrafting does not necessarily happen at the root.

# Gromov-Hausdorff Distance on Trees

This post continues the exposition of Gromov-Hausdorff distance, as introduced in Chapter Four of Steve Evans’ Probability and Real Trees, which we are reading as a group in the Stats Dept in Oxford at the moment. In this post, we consider applications of the Gromov-Hausdorff distance we have just introduced in the context of trees, viewed as metric spaces.

First we consider a direct application of the previous result, which related the Gromov-Hausdorff distance to the infimum of the distortion across the family of correspondences between the two relevant metric spaces. I found this as Corollary 3.7 in notes by Le Gall and Miermont on scaling limits of random trees and maps, which can be found here. I’m not clear whether there is an original source, but the result is simple enough that probably it does not matter hugely.

Proposition – Given f,g excursions above [0,1], and $T_f, T_g$ the real trees associated with these excursions in the standard way, then

$d_{GH}(T_f,T_g)\le 2 ||f-g||.$

Proof: We construct an appropriate correspondence

$\mathcal {R}=\left\{(a,b) \,:\, \exists t\in[0,1]\text{ s.t. } a=p_f(t), b=p_g(t)\right\}.$

In other words, the trees are defined as projections from [0.1] (with some equivalence structure), so taking pairs of projections from [0,1] gives a natural correspondence. Now, suppose $(p_f(s),p_g(s)), (p_f(t),p_g(t))\in \mathcal {R}$. Then

$d_{T_f}(p_f(s),p_f(t)) = f(s) + f(t) - 2\hat f(s,t),$

where $\hat f(s,t):= \min{r\in[s,t]}f(r)$. Obviously, we can replace f by g to obtain

$d_{T_g}(p_g(s),p_g(t)) = g(s) + g(t) - 2\hat g(s,t).$

By thinking slightly carefully about where the functions achieve their minima compared with where the minimum in the sup norm is achieved, we can conclude that

$|d_{T_f}(p_f(s),p_f(t)) - d_{T_g}(p_g(s),p_g(t)) | \le 4||f-g||,$

and so the result follows from the relation between Gromov-Hausdorff distance and the infimum of distortions over the set of correspondences proved at the end of the previous post.

Gromov-Hausdorff Limits of Real Trees

If we are going to consider the Gromov-Hausdorff topology for limits of tree, we want to be sure that the limit of a sequence of real trees is another real tree. In particular, we want to show that this convergence preserves the property of being a geodesic space.

Theorem 4.19 – Given $X_n$ geodesic spaces, and $X$ complete, such that $X_n \stackrel{d_{GH}}\rightarrow X$, then X is geodesic.

Proof: Here, I’ll go in the opposite order to Evans’ book, as I think it’s easier understand why a special case of this implies the whole result once you’ve actually shown that special case.

Anyway, we want to show that given $x,y\in X$, there is a geodesic segment $x\leftrightarrow y$ in X. We will start by showing that there is a well-defined midpoint of the geodesic, that is a point $z \in X$ such that $d(x,z)=d(y,z)=\frac12 d(x,y)$.

Given arbitrary $\epsilon>0$, we can take n such that $d_{GH}(X,X_n) < \frac{\epsilon}{3}$ and then a correspondence $\mathcal{R}$ between $X,X_n$ with $\mathrm{dis}\mathcal {R}<\frac{2\epsilon}{3}$. Now, by definition of correspondence, we have $x',y'\in X_n$ such that $(x,x'),(y,y')\in\mathcal{R}$. But we do have geodesics in $X_n$, so we can take $z'\in X_n$ the midpoint of geodesic $x'\leftrightarrow y'$. Predictably, we now take $z\in X$ such that $(z,z')\in\mathcal {R}$.

We can show that z is ‘almost’ the midpoint of [x,y] in the sense that

$|d(x,z) - \frac12 d(x,y)| \le |d(x',z')-\frac12 d(x',y')+\frac32 \mathrm{dis}\mathcal{R} \le \epsilon.$

Similarly, we have $|d(y,z)-\frac12 d(x,y)|\le \epsilon$.

Perhaps it’s helpful to think of this point z that we’ve constructed as being like a taut, but not quite rigid string. The midpoint of the string has to be fairly near the midpoint of the endpoints, and in particular, as we let $\epsilon\rightarrow 0$, the z’s we deal with form a Cauchy sequence, and thus converge to some point, which (in a case of poor notation planning) we also call $z \in X$, which is the midpoint of $[x,y]$ as described before.

We can now iterate this iteration, to demonstrate that whenever q is a dyadic rational in [0,1], there exists $z_q \in X$ such that

$d(x,z_q)=q d(x,y),\quad d(y,z_q) = (1-q)d(x,y).$

Again then, if we want the above to hold for some general real r in [0,1], we can approximate r arbitrarily well by dyadic rationals q, and the associated points $z_q$ are Cauchy and thus have a well-defined limit with the required properties. We thus have our geodesic segment $x\leftrightarrow y$.

Rooted Gromov-Hausdorff Distance

In the end, the trees we want to compare might be rooted. For example, we talk about finite trees being invariant under random re-rooting, and we might be interested in similar results for real trees, in particular the BCRT. So we need to compare metric spaces as viewed outwards from particular identified points of each space.

An isometry of rooted spaces must map the root to the root, and so we adjust to obtain rooted Gromov-Hausdorff distance. We might try to consider embeddings into a common space so that the roots are shared, but it will be more convenient to maintain the infimum over metrics on the disjoint union as before. But to ensure the roots are in roughly the ‘same’ place in both set embeddings, we minimise the maximum of the Hausdorff distance between the sets, and the distance between the images of the roots in the common covering space.

Similar results apply as in the unrooted case, and normally the proofs are very similar. As we might expect, when defining a correspondence between rooted spaces, we demand that the pair of roots is one of the roots in the correspondence, and then the same equivalence between the minimal distortion and the rooted G-H distance applies.

Evans shows that $\mathbb{T}^{\mathrm{root}}$, the set of compact rooted trees is complete and separable under the rooted G-H topology. Separability is relatively easy to see. Compact trees have finite $\epsilon$-nets, and there is a canonical way to view the net as the vertices of a finite tree with edge lengths. Approximating these edge lengths by rationals and consider the countable family of isomorphism classes of rooted finite trees gives separability.

If we want, we can also define k-pointed Gromov-Hausdorff distance, where we demand k points in each space are held fixed.

Tree $\eta$-erasure

To show that this is a natural topology to consider for the family of trees, Evans devotes a short section to the operation of $\eta$-erasure, where all points within $\eta$ of a leaf are removed from a given tree.

Formally, $R_\eta$ is a map $\mathbb{T}^{\mathrm{root}}\rightarrow \mathbb{T}^{\mathrm{root}}$ (recalling that these are compact real trees), so that $R_\eta(T)$ consists of the tree

$\{\rho\}\cup\{a \in T \, : \, \exists x\in T,\, a\in[\rho, x],d_T(a,x)\ge \eta\},$

rooted again at $\rho$. We claim that the range of $R_\eta$ is the set of compact rooted trees with a finite number of leaves. In this setting, we want a geodesic definition of leaf, for example a leaf is a point that doesn’t lie in the interior of the (unique) geodesic segment between any other point and the root.

Given a tree T with a finite number of leaves, we can glue disjoint segments of length $\eta$ onto every leaf. Taking $R_\eta$ of this deeper tree will give T. Similarly, suppose $R_\eta(T)$ has infinitely many leaves, which we can label $a_1,a_2,\ldots$. Thus we also have $x_1,x_2,\ldots \in T$ such that $a_i\in[\rho,x_i]$, and the segments $\{[a_i,x_i]\}$ are disjoint, and all have length at least $\eta$, hence T cannot be compact, as it has no finite $\frac{\eta}{2}$ net.

It is clear that for fixed T, the family of these maps applied to T is continuous with respect to $\eta$ in the G-H topology. When $\eta$ is changed a small amount, the amount of extra tree removed is locally small, and so approximating correspondences by points in what is left is absolutely fine. Indeed, the operations $T_{\eta_1}, T_{\eta_2}$ commute to give $T_{\eta_1+\eta_2}$.

We want to show that for fixed $\eta$ this map $R_\eta$ is continuous with respect to G-H topology. Suppose two compact rooted trees S,T are covered by a rooted correspondence $\mathcal R$ with distortion $\epsilon \ll \eta$. We can’t immediately restrict $\mathcal R$ to $R_\eta(S)\times R_\eta(T)$, as it won’t necessarily be a surjection under the projection maps any more.

But we can get around this. Note that if $(x,y)\in\mathcal{R}$, then $|d_S(\rho_S,x) - d_T(\rho_T,y)| \le \epsilon$ by assumption. We will construct a correspondence $\mathcal R'$ on the erased trees as follows. Given $(x,y)\in\mathcal{R}$, if $(x,y)\in R_\eta(S)\times R_\eta(T)$, we keep it, and if $x\not\in R_\eta(S), y\not \in R_\eta(y)$, we throw it away. Suppose we have $(s,t) in \mathcal{R}$ with $s\in R_\eta(S), t\not \in R_\eta(T)$. Let $l_s$ be a leaf of S such that $s \in [\rho_S,l_s]$, and let $\bar t$ be the farthest point from the root of the geodesic $[\rho_T,t]$ restricted to $R_\eta(T)$. So the tree above $\bar t$ includes t and has height at most $\eta$.

In words, if a point appears in a pair in the correspondence but is removed by the erasure, we replace it in the pair with the point closest to the original point that was not removed by the erasure.

It remains to prove that this works. My original proof was short but false, and its replacement is long (and I hope true now), but will postpone writing this down either until another post, or indefinitely. The original proof by Evans and co-authors can be found as the main content of Lemma 2.6 in the original paper [1].

REFERENCES

[1] Evans, Pitman, Winter – Rayleigh processes, real trees, and root growth with regrafting.

# Gromov-Hausdorff Distance and Correspondences

This term, some members of the probability group have been reading Probability and Real Trees, by Steve Evans based on his Saint-Flour course in 2005. A pdf can be found here. This morning was my turn to present, and I gave a precis of Chapter 4, which concerns metrics on metric spaces, a family of tools which will be essential for later chapters which discuss convergence of trees, viewed as metric objects. Hausdorff Distance We start by considering a metric on subsets of a given base space X. The Hausdorff distance between two sets A, B is defined as

$d_H(A,B):=\inf\{ r>0: A\subset U_r(B), B\subset U_r(A)\},$

where $U_r(A):=\{x\in X, d(x,A)< r\}$ consists of set A, and all the points within r of set A. So the Hausdorff distance measures how much we have to fatten each set before it contains the other. Note that if we have a giant set next to a tiny one, we will have to fatten the tiny one a great deal more to achieve this. Sometimes it will be more helpful to think of the following alternative characterisation

$d_H(A,B) = \max\left \{ \sup_{a\in A}\inf_{b\in B} d(a,b), \inf_{a\in A}\sup_{b\in B} d(a,b)\right\}.$

In words, we measure how far away is the point in A farthest from B, and vice versa, and take the larger of the two. The presence of the sups and infs indicates that the inclusion or otherwise of the boundaries of A,B does not affect this distance. In particular, this means that $d_H(A,\bar A) = d_H(A,A^{\circ})$, and so to allow us to call Hausdorff distance a metric, we restrict attention to the closed subsets of X, M(X).

We also observe immediately that if X does not have bounded diameter, then it is possible for the Hausdorff distance between two sets to be infinite. We won’t worry about that for now, because we will mainly be considering host spaces which are compact, for which the following results will be useful.

Lemma 4.4 – If (X,d) is complete, then (M(X),d_H) is also complete.

Proof: Assume we have a sequence of closed sets $S_1,S_2,\ldots \subset X$ which have the Cauchy property under d_H. I’m going to do this slightly differently to Evans, because it’s not the case you can immediately choose $x_n\in S_n$ for each n, such that $d(x_n,x_m)\le d_H(S_n,S_m)$ for all m,n. For an explicit counterexample, see comment from Ed Crane below the article.

Note that if a subsequence of a Cauchy sequence converges to a limit, then the whole sequence converges to that same limit. So we can WLOG replace $S_1,S_2,\ldots$ by some subsequence such that $d_H(S_n,S_{n+1})\le 2^{-n}$. Now it is clear that for any $x_n\in S_n$, there is a choice of $x_{n+1}\in S_{n+1}$ such that $d(x_n,x_{n+1})\le 2^{-n}$ (*). Starting from arbitrary $x_1\in S_1$, we can construct in this manner a sequence $x_1,x_2,\ldots,\in X$ that is Cauchy, and thus has a limit point $x\in X$.

Let $\mathcal{X}$ be the set of sequences $(x_m,x_{m+1},\ldots)$ for some m, with $x_n\in S_n\,\forall n\ge m$, satisfying (*). Now let S be the closure of the set of limit points of this family of sequences, which we have shown is non-empty.

Then for any n, and any $x_n\in S_n$, we can construct such a sequence, and its limit point x, and the triangle inequality down the path $x_n,x_{n+1},\ldots$ gives $d(x_n,S)\le 2^{-(n-1)}$. Furthermore, by construction $S\subset U_{2^{-(n-1)}}(S_n)$, hence it follows that $S_n \stackrel{d_H}\rightarrow S$.

Lemma 4.5 – Given (X,d) compact, (M(X),d_H) is also compact.

Sketch proof: We are working with metric spaces, for which the existence of a finite $\epsilon$-net for every $\epsilon>0$ is equivalent to compactness. [An $\epsilon$-net is a set S of points in the space such that every point of the space lies within $\epsilon$ of an element of S. Thinking another way, it is the set of centres of the balls in a finite covering of the space by $\epsilon$-balls.] It is not too hard to check that if $S_\epsilon$ is an $\epsilon$-net for (X,d), then $\mathcal{P}(S_\epsilon)$ is finite, and an $\epsilon$-net for (M(X),d_H).

Gromov-Hausdorff Distance

So far this is fine, but won’t be useful by itself for comparing how similar two trees are as metric spaces, because we can’t be sure a priori that we can embed them in a common host space. To resolve this, we consider instead the Gromov-Hausdorff distance, which will serve as a distance between metric spaces, even when they are not canonically defined as subsets of a common metric space.

Given X, Y metric spaces, we define

$d_{GH}(X,Y)=\inf_Z \left\{ d_H(X',Y') \, : \, X',Y' \subset (Z,d)\text{ a metric space }, X'\simeq X, Y'\simeq Y\right\}.$

In words, the Gromov-Hausdorff distance between two metric spaces examines the ways to embed them isometrically into a common larger metric space, and gives the minimal Hausdorff distance between them under the class of such embeddings. One issue is that the collection of all metric spaces is not a set. For example, given any set, we can define a metric via the discrete metric, so the collection of metric spaces is at least as large as the collection of all sets, which is not a set. Fortunately, all is not broken, as when we consider a general metric space Z in which we might embed copies of X and Y we are wasting lots of the perhaps very complicated space, because we only need to compare the subsets which are isometric copies of X and Y. So in fact, we lose nothing if we assume that Z is a disjoint union of copies of X and Y, with a metric chosen appropriately. So

$d_{GH}=\inf\left\{ d_H(X,Y) : d\text{ a metric on }X\coprod Y\text{ restricting to }d_X \text{ on }X,\, d_Y\text{ on }Y \right\}.$

In practice though, this is difficult to compute, since the set of things we have to minimize over is complicated. It turns out we can find an equivalent characterisation which will be easier to use in a number of examples, including the case of real trees which is the whole point of the course.

Correspondence and Distortion

We define a correspondence from X to Y to be

$\mathcal{R}\subset X\times Y\text{ s.t. } \pi_X(\mathcal {R}) = X, \, \pi_Y(\mathcal {R}) = Y,$

where $\pi_X,\pi_Y$ are the canonical projection maps from $X\times Y$ into $X,Y$ respectively. So we can think of a correspondence as being something like a matching from X to Y. In a matching, we insist that the projection maps into each set are injections, ie that each element of X (resp Y) can appear in at most one pair, whereas for a correspondence, we demand that the projection maps are surjections, ie that each element of X appears in at least one pair.

Then the distortion of a correspondence

$\mathrm{dis}(\mathcal{R}):= \sup\left\{ |d_X(x,x') - d_Y(y,y')| \,;\, (x,y),(x',y')\in \mathcal{R} \right\}.$

In words, if two sets are non-isomorphic, then a correspondence can’t describe an isometry between the sets, and the distortion is a measure of how far from being an isometry the correspondence is. That is, given a pair of pairs in the correspondence, for an isometry, the distance between the X-elements would be equal to the distance between the Y-elements, and the distortion measures the largest discrepancy between such pairs of pairs over the correspondence.

Theorem 4.11 $d_{GH}(X,Y) = \frac12 \inf_{\mathcal{R}} (\mathrm{dis}\mathcal R)$, where the infimum is taken over all correspondences $\mathcal{R}$ X to Y.

Remark: The RHS of this equivalence can be thought of as the set coupling between X and Y such that the pairs have as equal distances as possible.

Proof: Given an embedding into $X\coprod Y$ with $d_H(X,Y), we have $\mathcal{R}$ with $\mathrm{dis}\mathcal {R}<2r$, by taking:

$\mathcal{R}=\{(x,y): d(x,y)

From the definition of Hausdorff distance, it follows that the for every x, there is a y with d(x,y)<r, and hence the appropriate projection maps are projections.

So it remains to prove that $d_{GH}(X,Y)\le \frac12 \mathrm{dis}\mathcal{R}$. We can define a metric on $X\times Y$ by

$d(x,y)=\int\left\{ d_X(x,x')+d_Y(y,y') + r \,:\, (x',y')\in \mathcal{R} \right\}.$

Then for any $x\in X$, there is $(x,y)\in\mathcal{R}$, and thus $d(x,y)\le r$, and vice versa hence $d_H(X,Y)\le r$.

It only remains to check that this is actually a metric. Let’s take $x,\bar x \in X$, and so

$d(\bar x,y)\le \inf\{ d_X(x,\bar x) + d_X(x,x')+d_Y(y,y')+r \,: \, (x',y')\in\mathcal{R}\},$

so taking $d_X(x,\bar x)$ outside the brackets gives one form of the triangle inequality. We have to check the ‘other combination’ of the triangle inequality. We assume that the infima for $(x,y), (\bar x,y)$ are attained at $(x',y'),(\bar x',\bar y')$ respectively.

$d(x,y)+d(\bar x,y)= 2r+ d_X(x,x')+d_X(\bar x,\bar x') + d_Y(y,y')+d_Y(d,\bar y').$

But we also have $d_X(x',\bar x')-d_Y(y',\bar y')\ge -r$ from the definition of distortion, and so adding these gives the triangle inequality we want, and completes the proof of this theorem.corc

# Enumerating Forests

I’ve just got back from a visit to Budapest University of Technology, where it was very pleasant to be invited to give a talk, as well as continuing the discussion our research programme with Balazs. My talk concerned a limit for the exploration process of an Erdos-Renyi random graph conditioned to have no cycles. Watch this space (hopefully very soon) for a fully rigorous account of this. In any case, my timings were not as slick as I would like, and I had to miss out a chunk I’d planned to say about a result of Britikov concerning enumerating unrooted forests. It therefore feels like an excellent time to write something again, and explain this paper, which you might be able to find here, if you have appropriate journal rights.

We are interested to calculate $a_{n,m}$ the number of forests with vertex set [n] consisting of m unrooted trees. Recall that if we were interested in rooted trees, we could appeal to Prufer codes to show that there are $m n^{n-m-1}$ such forests, and indeed results of Pitman give a coalescent/fragmentation scheme as m varies between 1 and n-1. It seems that there is no neat combinatorial re-interpretation of the unrooted case though, so Britikov uses an analytic method.

We know that

$a_{n,m}= \frac{n!}{m!} \sum_{\substack{k_1+\ldots+k_m=n\\ k_i\ge 1}} \prod_{j=1}^m \frac{k_j^{k_j-2}}{k_j!}.$

To see this, observe that the $k_j$s correspond to the sizes of the m trees in the forest; $\frac{n!}{\prod k_j!}$ gives the multinomial number of ways to assign vertices to the trees; given the labels for a tree of size $k_j$, there are $k_j^{k_j-2}$ ways to make up the tree itself; and $\frac{1}{m!}$ accounts for the fact that the trees have no order.

What we would really like to do is to take the uniform distribution on the set of all labelled trees, then simulate m IID copies of this distribution, and condition the union to contain precisely n vertices. But obviously this is an infinite set, so we cannot choose uniformly from it. Instead, we can tilt so that large trees are unlikely. In particular, for each x we define

$\mathbb{P}(\xi=k) \propto \frac{k^{k-2} x^k}{k!}$,

and define the normalising constant

$B(x):= \sum_{k\ge 1} \frac{k^{k-2}x^k}{k!},$

whenever it exists. It turns out that $x\le e^{-1}$ is precisely the condition for $B(x)<\infty$. Note now that if $\xi_1,x_2,\ldots$ are IID copies of $\xi$, then

$\mathbb{P}(\xi_1+\ldots+\xi_m=n) = \frac{x^n}{B(x)^m} \sum_{k_1+\ldots + k_m=n} \prod_{j=1}^m \frac{k_j^{k_j-2}}{k_j!},$

and so we obtain

$a_{n,m}= \frac{n!}{m!} \frac{B(x)^m}{x^n} \mathbb{P}(\xi_1+\ldots + \xi_m=n).$

So asymptotics for $a_{n,m}$ might follows from laws of large numbers of this distribution $\xi$.

So far, we haven’t said anything about how to choose this value x. But observe that if you want to have lots of trees in the forest, then the individual trees should generally be small, so we take x small to tilt away from a preference for large trees. It turns out that there is a similar interpretation of criticality for forests as for general graphs, and taking x equal to 1/e, its radius of convergence works well for this setting. If you want even fewer trees, there is no option to take x larger than 1/e, but instead one can use large deviations machinery rather than laws of large number asymptotics.

We will be interested in asymptotics of the characteristic function of $\xi$ for x=1/e. In particular $\mathbb{E}[e^{it\xi}]=\frac{B(xe^{it})}{B(x)}$, and it will be enough to clarify the behaviour of this as $t\rightarrow 0$. It’s easier to work with a relation analytic function

$\theta(x)=\sum_{k\ge 1} \frac{k^{k-1}x^k}{k!},$

ie the integral of B. What now feels like a long time ago I wrote a masters’ thesis on the subject of multiplicative coalescence, and this shows up as the generating function of the solutions to Smoluchowski’s equations with monodisperse initial conditions, which are themselves closely related to the Borel distributions. In any case, several of the early papers on this topic made progress by establishing that the radius of convergence is 1/e, and that $\theta(x)e^{-\theta(x)}=x$ everywhere where $|x|\le 1/e$. We want to consider x=1/e, for which $\theta=1$.

Note that $\mathbb{E}\xi = \frac{\theta(x)}{B(x)}$, so we will make progress by relating $B(x),\theta(x)$ in two ways. One way involves playing around with contour integrals in a fashion that is clear in print, but involves quite a lot of notation. The second way is the Renyi relation which asserts that $\theta(x)=B(x)+\frac{\theta(x)^2}{2}$. We will briefly give a combinatorial proof. Observe that after multiplying through by factorials and interpreting the square of a generating function, this is equivalent to

$k^{k-1} = k^{k-2} + \frac12 \sum_{\substack{l+m=k\\l,m\ge 1}} l^{l-1}m^{m-1}\binom{k}{l},$

for all k. As we might expect from the appearance of this equality, we can prove it using a bijection on trees. Obviously on the LHS we have the size of the set of rooted trees on [k]. Now consider the set of pairs of disjoint rooted trees with vertex set [k]. This second term on the RHS is clearly the size of this set. Given an element of this set, join up the two roots, and choose whichever root was not initially in the same tree as 1 to be the new root. We claim this gives a bijection between this set, and the set of rooted trees on [k], for which 1 is not the root. Given the latter, the only pair of trees that leads to the right rooted tree on [k] under this mapping is given by cutting off the unique edge incident to the root that separates the root and vertex 1. In particular, since there is a canonical bijection between rooted trees for which 1 is the root, and unrooted trees (!), we can conclude the Renyi relation.

The Renyi relation now gives $\mathbb{E}\xi = \frac{\theta(x)}{B(x)}=2$ when x=1/e. If we wanted, we could show that the variance is infinite, which is not completely surprising, as the parameter x lies on the radius of convergence of the generating function.

Now, playing around with contour integrals, and being careful about which strands to take leads to the asymptotic as $t\rightarrow 0$

$\mathbb{E}[ e^{it\xi}] = 1+2it + \frac{2}{3}i |2t|^{3/2} (i\mathrm{sign}(t))^{3/2} + o(|t|^{3/2}).$

So from this, we can show that the characteristic function of the rescaled centred partial sum $\frac{\xi_1+\ldots+\xi_N-2N}{bN^{2/3}}$ converges to $\exp(-|t|^{3/2}\exp(\frac{i\pi}{4}\mathrm{sign} t))$, where $b= (32/9)^{1/3}$ is a constant arising out of the previous step.

We recognise this as the characteristic function of the stable distribution with parameters 3/2 and -1. In particular, we know now that $\xi$ is in the domain of attraction for a stable-3/2 distribution. If we wanted a version of the central limit theorem for such partial sums, we could have that, but since we care about the partial sums of the $\xi_i$s taking a specific value, rather than a range of values on the scale of the fluctuations, we actually need a local limit theorem.

To make this clear, let’s return to the simplest example of the CLT, with some random variables with mean $\mu$ and variance $\sigma^2<\infty$. Then the partial sums satisfy

$\mathbb{P}(\mu N + a\sigma\sqrt{N} \le S_N \le \mu_N+b\sigma\sqrt{N}) \rightarrow \int_a^b f_{\mathcal N}(x)dx,$

as $N\rightarrow\infty$. But what about the probability of $S_N$ taking a particular value m that lies between $\mu N+a\sigma \sqrt{N}$ and $\mu N + b\sigma \sqrt{N}$? If the underlying distribution was continuous, this would be uncontroversial – considering the probability of lying in a range that is smaller than the scale of the CLT can be shown in a similar way to the CLT itself. A local limit theorem asserts that when the underlying distribution is supported on some lattice, mostly naturally the integers, then these probabilities are in the limit roughly the same whenever m is close to $\mu N+a\sigma\sqrt{N}$.

In this setting, a result of Ibragimov and Linnik that I have struggled to find anywhere in print (especially in English) gives us local limit theory for integer-supported distributions in the domain of attraction of a stable distribution. Taking p( ) to be the density of this distribution, we obtain

$bm^{2/3}\mathbb{P}(\xi_1+\ldots+\xi_m=n) - p(\frac{n-2m}{b m^{2/3}}) \rightarrow 0$

as $n\rightarrow\infty$, uniformly on any set of m for which $z= \frac{n-2m}{bm^{2/3}}$ is bounded. Conveniently, the two occurrences of b clear, and Britikov obtains

$a_{n,m} = (1+o(1)) \frac{\sqrt{2\pi} n^{n-1/6}}{2^{n-m}(n-m)!} p(\frac{n-2m}{n^{2/3}},$

uniformly in the same sense as before.

# Random Maps 3 – Leaves and Geodesics in BCRT

Recall in the previous two posts, we’ve introduced some of the background to maps on various surfaces. In particular, we’ve introduced the remarkable Cori-Vauquelin-Schaeffer bijection which maps between plane trees labelled with uniform increments and quadrangulations of the sphere, up to some careful fiddling around with rooting and pointing an edge.

We are interested in the case where we choose uniformly a large element from these classes. We want to derive a scaling limit for the uniform planar quadrangulation, and we hope that we will be able to carry some properties of the scaling limit of the labelled trees, which may well be simpler, across the CVS bijection. It is convenient that the vertices of the plane tree become the vertices of the quadrangulation. We are looking to find some sort of metric limit, in the Gromov-Hausdorff sense, and so it will remain to deduce exactly how to use the labelling obtained from the tree to gain information about distances in the (limiting) quadrangulation.

Of course, all of this relies on the fact that there is a nice limit for the ordered plane trees in the first place. Unsurprisingly, it turns out that this is Aldous’s Brownian continuum random tree. The easiest way to see this is to consider the contour process of the ordered plane tree. This is chosen uniformly at random from the set of paths from (0,0) to (2n,0) with increments of size {-1,1} and which stay non-negative. It is thus precisely a simple random walk started at (0,0) conditioned to hit (2n,0) and to be non-negative. Since SRW suitably rescaled converges to Brownian motion, it is unsurprising (but not totally trivial) that this conditioned object converges to a Brownian excursion.

The Brownian excursion can be viewed as a continuous analogue of the contour process for the BCRT, but it is more natural to consider this convergence in the Gromov-Hausdorff topology. In this setting, we say that for a large value of n, the tree is ‘roughly isometric’ to the BCRT in distribution. Here, roughly isometric means the two metric spaces can be embedded isometrically into a common metric space such that they are close together, now in the sense of Hausdorff distance.

At this point, it is worth thinking about this interpretation of the BCRT. We have previously considered this as the scaling limit of a uniformly chosen Cayley tree, that is any unrooted tree on n labelled vertices. Essentially, we are now specifying that the BCRT can carry extra information, namely a root, and geometric information about the order of branches. The root is uncontroversial. Canonically, the root of the BCRT will be at the point associated with time 0 in the driving Brownian excursion. However, we can easily check that the distribution of a uniform rooted plane tree is invariant under re-rooting, and so any argument we have for convergence of the rooted trees to the BCRT will work with the root in a different place. Applying something like a tower law, we conclude that the convergence works when the root is chosen uniformly in the limit.

One potential problem to be discussed is what it means to choose a point uniformly in the limit. We have two possible approaches. One is to consider Lebesgue measure on any path in the BCRT, and glue these together. However, we have a uniform stick-breaking construction of the BCRT, and one consequence of the construction is that the total length of sticks required is infinite, so this won’t work.

The other option is to project Lebesgue measure on [0,1] via the same map that sends points on the Brownian excursion to points in the tree. Note that the so-called real tree is constructed from the excursion by identifying points s and t where f(s)=f(t), and f(x)>f(s) for x in (s,t). But then we might wonder whether this can really be said to be ‘uniform’, since different points in the BCRT will have a different number of pre-images in [0,1]. In fact though, it turns out that in this sense, projected-Leb[0,1]-almost all the points in the BCRT are leaves.

To prove this, naturally we first need to define a leaf, in the setting of these continuum trees. The degree of a vertex is an idea we might keep in mind, but we can’t use this, as we don’t have vertices any more. However, we have a continuous analogue of degree, given by counting the number of connected components remaining after removing a vertex. In particular, we can define the set of leaves as

$\mathcal{L}(\mathcal{T}):=\{x\in\mathcal{T}:\mathcal{T}\backslash \{x\}\text{ is connected}\}.$

We will give a sketch proof of this result about leaves shortly. First, we clarify some notation, and consider properties of geodesics (shortest-length paths) in the tree.

Define $\check{f}(u,v):=\min_{x\in[u\wedge v,u\vee v]} f(x)$ to be the minimum value attained by f between u and v. Consider the value x at which this minimum is attained. Then, projecting onto the tree, p(x) is the ‘most recent common ancestor’ of points u and v. We can make this a bit more precise by considering geodesics in the tree starting at the root. Analogous to the unique path property in a discrete tree, in this continuous setting there is a unique path from the root to any given point, along which the height is strictly increasing. This is not surprising. It follows from one of the definitions of a real tree that the length of the path from p(0) to p(s) should be f(s), and so there is a unique isometric embedding of [0,f(s)] into $\mathcal{T}_f$ which starts at 0 and ends at p(s). Anyway, under this $p(\check{f}(s,t))$ gives the point at which the geodesics from p(s) to p(0) and from p(t) to p(0) meet.

Furthermore, we can now describe the distance in the tree between p(s) and p(t). This is given by

$d_f(s,t):= f(s)+f(t)-2\check{f}(s,t),$

and with the geodesic picture, it is easy to see why. Consider the point x at which $\check{f}(s,t)$ achieves the minimum. As we have said, this lies on the geodesics from p(s) to 0 and p(t) to 0, and paths between points are unique, so removing point x disconnects p(s) and p(t) in the tree. So we need to concatenate the geodesic from p(s) to p(x) and from p(x) to p(t). But these are subsets of the two geodesics discussed, and their respective lengths are $f(s)-\check{f}(s,t)$ and $f(t)-\check{f}(s,t)$.

We can now give a sketch proof the result that almost all the support of $\lambda_f$, the projection on Lebesgue measure from [0,1] onto $\mathcal{T}_{f}$ is on $\mathcal{T}(\mathcal{T}_f)$.

Given $s,t\in[0,1]$, suppose we are removing p(s), and this separates p(t) from the root, which is canonically p(0). Without loss of generality, take t>s. Now suppose that $\check{f}(s,t), and that, as before, this infimum is attained at $x\in[s,t]$. Then the geodesic from p(0) to p(t) will pass through p(x), but not through p(s), so in particular, removing p(s) cannot disconnect p(t) from the root.

Thus, p(s) is not a leaf if and only if there exists some small window [s,t] such that $f(s)\le f(x),\;\forall x\in[s,t]$. By Blumenthal’s 0-1 law, for fixed x, this happens with probability 0 if f is Brownian motion. Here, f is not Brownian motion, but a Brownian excursion with length 1. However, Blumenthal’s 0-1 law depends on the instantaneous behaviour after time s, ie the sigma field $\mathcal{F}_s^+$. So, for $s\in(0,1)$, the value of a Brownian at time 1 is independent of this sigma field, so if we imagine Brownian excursion as a ‘conditioned’ Brownian motion, this conditioning should have no effect on the conclusion of this corollary to Blumenthal’s 0-1 law.

This is not a formal argument, but it sketches why with probability 1, p(s) is a leaf for each $s\in(0,1)$, from which the result follows.

# Random Maps 2 – The Schaeffer Bijection

As indicated at the end of the previous post, our aim is to find a natural bijection between the set of pointed, rooted quadrangulations with n faces, and some set of objects based on decorating rooted plane trees with n edges in some fashion. Unlike our previous example, the construction of this bijection is definitely not trivial. It seems like a foolish ambition to explain this without several pictures, so I’m going to focus on some aspects of the analysis which I found challenging, rather than the construction itself.

Anyway, we don’t yet know what the extended set of trees should be. We need an extra factor of $3^n$, so it is natural to consider adding some sort of labelling of the tree, where for each non-root vertex in turn there are three options. So, given a rooted tree T, we label the vertices such that the root has label 0, and if a parent vertex has label k, any offspring has label k-1, k or k+1. Such a labelling is called admissable, and $\mathbb{T}_n$ is the set of rooted plane trees with n edges and an admissable labelling.

We now demonstrate how to construct an element of $\mathcal{Q}_n$ from an element of $\mathbb{T}_n$. Various authors had considered this problem to various extents, and so what follows is known as the Cori-Vauquelin-Schaeffer bijection, at least in this course.

Consider a contour exploration of the tree. That is, start out at the root and at all times take first-edge you encounter going clockwise from your current direction. When you arrive at a leaf, you will indeed therefore immediately retrace your most recent step. The key property is that you traverse each edge exactly twice, and so we may think of the tree as having 2n oriented edges. It is more useful to think about corners. A corner is the directed arc (WLOG clockwise) between adjacent edges at a vertex. There is a natural bijection between corners and directed edges, by looking anti-clockwise from the tail of the edge. So the contour process explored the directed edges in some order, and hence explores the corners of the tree. One thing I found confusing initially was switching between considering vertices and corners. I feel in retrospect that the only reason we need the vertices themselves is to induce the labelling onto the corners. These are the only thing we will use in the construction.

As we trace out the contour process, naturally we see different labels. We define the successor of a corner with label k to be the next corner seen in the contour process (taken modulo 2n if necessary) with label k-1. Note that any corner on a vertex with minimal label will not have a successor. To counter this, we add a new vertex, suggestively called $v_*$, with a single corner (ie no edges yet) and denote this corner to be the successor of the corners in the original tree with minimal label.

To construct our quadrangulation, we simply join up every corner with its successor corner. Note that if you are thinking of the successor of a corner as a vertex (rather than as a corner) you will get in trouble here, as it might be several ways to draw this arc.

The red arcs and vertex v* are added to form the quadrangulation. Note the blue angles indicate the three corners around the vertex labelled -1.

It is not obvious that it is possible to do this so that the arcs do not overlap. However, by considering the label process as you explore via the contour process, it becomes clear that you can discount the possibility of any overlaps one by one. This applies equally to pairs of new arcs overlapping, as well as new arcs overlapping with edges of the original tree. In any case, we remove the edges of the original tree to obtain the quadrangulation.

Note that when you move from any corner of a vertex with label k to its successor, then to the successor of its successor and so on, the labels are decreasing, so eventually you must end up at a corner with minimal label, and hence at $v_*$. We conclude that the graph of arcs is connected. It remains to show that it is a quadrangulation.

This is rather fiddly to do without a diagram. Note first that whenever we have a directed edge in the tree going from label k to label k-1, then this edge essentially becomes an arc of the quadrangulation. We show that the edge oriented in the other direction, called say e, induces three further arcs of a quadrangle. So e goes from label k-1 to k. Consider the corners before and following e in the contour exploration, which is a corner around the vertex with label k. The successor of the corner after e is a corner with label k-1, and this has a successor with label k-2. By construction, this must also be the successor of the corner before e. Why? Well as we traverse the contour beyond e, the first appearance of label k-1 must happen before the first appearance of label k-2, as the increments can only be in {-1,0,1}. This gives us the three further arcs. Note also that the 2-colouring of the quadrangulation is given by the parity of the tree-labelling.

I was bothered about what happens if two vertices with label k-1 are in fact the same. This would happen if, for example, the vertex labelled k is a leaf. Then, at least two of the corners around the single vertex with label k-1 have the same corner as successor. A naïve attempt at drawing the resulting arcs did not give a quadrangle. The key observation is that you have to draw the arcs in the direction of the contour process. So in this case, the arc from the corner before edge e will loop all the way around the vertex with label k, so it contains the other two relevant arcs on its way to the vertex with label k-2, giving us the ‘pacman’ quadrangle discussed earlier.

The other case we have to check is when our base edge joins two vertices with label k. Then the other two vertices of the face will have label k-1. This is similar to the above, and slightly easier.

As a preliminary to checking that we can invert this construction, we observe that the vertices of the quadrangulation are the vertices of the original tree plus $v_*$, and furthermore, the labels in the tree are given by the graph distance from $v_*$ in the quadrangulation, with a constant added uniformly so that the root vertex has label 0.

At this point, we observe that in the construction, we didn’t specify how to choose the rooted edge of the quadrangulation. Canonically, we take it to be the arc between the first corner of the root in the contour process, and its successor. However, we can orient it in either direction, giving us the extra factor of 2 we were looking for.

Returning to the inverse, it is clear what to do when we see a quadrangle corresponding to the second case above – namely put an edge between the two vertices with label k. In the case where the face has labels {k,k-1,k-1,k-2} it is less obvious. Note though that by starting at the first corner of the root, which is identified by the rooted edge in the quadrangulation, we can recover the contour process from the arcs of the quadrangulation, and the labels. So when we see such a face, we can use this information to choose which of the (k-1)-labelled vertices to join to the vertex with label k.

Anyway, now we are convinced that this bijection works, the next stage is to apply it to gain extra information about a uniformly-chosen large quadrangulation. We can view the vertices as being those of a large uniform plane tree, and the labels as given by a random walk along this large tree. We might expect to see this labelling structure converge to something that looks like Brownian motion indexed by a Brownian continuum random tree, in a sense to be made more precise. And the labelling is not merely a decoration in the quadrangulation, since it specifies the distance to the identified point $v_*$. In particular, this gives a bound on the distance between any two vertices in the quadrangulation, eg two vertices chosen uniformly at random. In fact, by looking more carefully at the scaling limit of the uniform tree’s contour process, we can say rather more than that.

# Random Maps 1 – Towards the Schaeffer Bijection

I have spent the past ten days in Saint Flour, an inaccessible but picturesque town in the rolling hills of Cantal, in the middle of France, and venue for perhaps the most notable summer school in probability. My highlight has been the course ‘Aspects of Random Maps’ delivered by Gregory Miermont, and I thought I should write a few posts about points of interest encountered during the lectures and private study.

A map is an embedding of a connected graph onto a surface. We typically do not care about the nature of this embedding up to homeomorphisms of the surface which preserve orientations of the map. One advantage for doing this is that the set of maps now might be countable, and the set of maps with n edges might be finite. This can be proved by considering a map to be obtained by glueing together polygonal faces. Some potential glueings are impossible, and some are equivalent, but for a fixed number of edges, the number of such sets of polygons and a possible glueing is finite. In fact we can be much more precise than this about how to describe precisely the legal glueing through a triple of permutations, but I won’t discuss this here.

I haven’t yet given a complete definition of a map. We want a typical large map, that is a map with a large number of vertices and edges, to be topologically roughly the same as the surface it is embedded into. In particular, the map needs to encode the geometric features of the surface. So a small triangle on the surface of a torus should not be considered a map. To rigorise this, we demand that any face of a map should be a topological disc, in particular, it should be simply connected. Since the torus itself is not simply connected, this excludes our triangle example. Note a single vertex on a torus is also excluded.

Although it goes against the usual order of definitions, it might be helpful to think of a map as an embedding which satisfies Euler’s formula: V – E + F = 2 -2g, where g is the genus of the surface. For a connected planar graph, induction is on the number of edges and vertices is the typical way to prove this result. The inductive step works the same on a more general surface, but it is less clear what the base case should be. Another consequence of the definition is that we should work on the sphere rather than the plane. From now on, this is our surface of interest.

We begin by considering $\mathcal{M}_n$ to be the family of rooted plane maps with n edges. The root is a distinguished oriented edge. Our aim is to count the size of this set.

Before doing this, we digress onto the topic of rooted plane trees. Note that any (rooted) tree in the classical sense is planar, but in a rooted plane tree, we also specify the geometric ordering of the offspring. For example, if the root has two offspring, of which one has precisely one offspring and the other has none, we consider these as two separate cases.

So now, if we denote by $a_k$ the number of plane trees with k edges, we can define a generating function via $A(z):=\sum_{k\ge 0} a_k z^k$. If the root vertex v has no offspring, this gives one possibility corresponding to k=0. Otherwise, there is a well-defined left-most offspring of the root, called u. Then u and its descendents form a plane tree, and v and its descendents apart from those through u also form a plane tree. So after accounting for the edge between u and v, we obtain

$A(z)=1+zA(z)^2,$

whenever A(z) is defined. We now can apply whichever is our favourite method of showing that this is the generating function of the Catalan numbers, $a_k=\frac{1}{k+1}\binom{2k}{k}$.

There is a more complicated version of this generating function argument due to Tutte that allows us to enumerate $\mathcal{M}_n$. It is convenient to work with a second variable in the generating function that encodes the degree of the root face. The resulting equation of generating functions is less well-known but using the Lagrange inversion formula gives the explicit expression

$|\mathcal{M}_n|=\frac{2}{n+2}\cdot \frac{3^n}{n+1}\binom{2n}{n}.$

Although there are extra terms, this motivates seeking a bijection between maps, and some version of rooted plane trees, perhaps decorated with some extra information. As in many cases, this will turn out to be possible. The bijection we end up with will not just help us enumerate the maps, but will also allow us to control a lot more information about distances in the map, which will be particularly useful when we try to take limits.

The first observation is that given a map, we can construct a dual map, by placing fresh vertices somewhere in the middle of each face, and joining a pair of these if the corresponding faces in the original graph share an edge.

Alternatively, we can place the same fresh vertices in the middle of each face, then join each new vertex to an original vertex, if that original vertex lies on the face corresponding to the new vertex. If you focus in on an original edge, it is clear that it is now surrounded by a ‘diamond’ (if you’ve drawn the diagram in a natural way) of new edges. Removing the original edges thus leaves us with a quadrangulation. This procedure is called the ‘trivial bijection’ between $\mathcal{M}_n$ and $\mathcal{Q}_n$, the family of rooted quadrangulations with n faces. Note that the root in such a quadrangulation is an identified directed edge, rather than a vertex. We haven’t yet specified how to describe the root of the resulting quadrangulation. It suffices to take the first new edge which lies clockwise of the root edge in the original graph, seen from the ‘tail’ of the root, which is of course oriented.

In this, and the bijections which follow, the natural questions to ask are: a) is the inverse obvious? and b) what happens to self-loops and isthmuses? Here, the inverse really is obvious. Any quadrangulation is bipartite, hence two-colourable, so we need to fix one colour and join the two vertices of that colour within each face to recover the original graph. The root tells us which colour we need to take. As for the second question, first we should say that an isthmus is an edge which has the same face on both sides. This causes no problems in this particular bijection. For the self-loops, we get a sort of Pacman-like quadrangle, with two ‘outer-edges’ between the same two vertices, and an edge between one of the outer vertices and some internal vertex. This edge contributes twice to the degree of the face.

The upshot of this is that for a simple enumeration, it suffices to prove that $|\mathcal{Q}_n|=\frac{2}{n+2}\cdot \frac{3^n}{n+1}\binom{2n}{n}$. This may not look like we have achieved much, but we can now apply Euler’s formula to any quadrangulation in this set to deduce that the number of vertices present is n+2. If we consider the set $\mathcal{Q}_n^*$, where now we identify a particular vertex $v_*$ in the quadrangulation, it suffices to prove that $|\mathcal{Q}_n^*|=2.3^n \cdot a_n$, where $a_n$ is the nth Catalan number as before. Now we have the most efficient setup to look for a bijection with some type of decorated plane tree as discussed before.

# Multitype Branching Processes

One of the fundamental objects in classical probability theory is the Galton-Watson branching process. This is defined to be a model for the growth of a population, where each individual in a generation gives birth to some number (possibly zero) of offspring, who form the next generation. Crucially, the numbers of offspring of the individuals are IID, with the same distribution both within generations and between generations.

There are several ways one might generalise this, such as non-IID offspring distributions, or pairs of individuals producing some number of offspring, but here we consider the situation where each individual has some type, and different types have different offspring distributions. Note that if there are K types, say, then the offspring distributions should now be supported on $\mathbb{Z}_{\ge 0}^K$. Let’s say the offspring distribution from a parent of type i is $\mu^{(i)}$.

The first question to address is one of survival. Recall that if we want to know whether a standard Galton-Watson process has positive probability of having infinite size, that is never going extinct, we only need to know the expectation of the offspring distribution. If this is less than 1, then the process is subcritical and is almost surely finite. If it is greater than 1, then it is supercritical and survives with positive probability. If the expectation is exactly 1 (and the variance is finite) then the process is critical and although it is still almost surely finite, the overall population size has a power-law tail, and hence (or otherwise) the expected population size is infinite.

We would like a similar result for the multitype process, saying that we do not need to know everything about the distribution to decide what the survival probability should be.

The first thing to address is why we can’t just reduce the multitype change to the monotype setting. It’s easiest to assume that we know the type of the root in the multitype tree. The case where the type of the root is random can be reconstructed later. Anyway, suppose now that we want to know the offspring distribution of a vertex in the m-th generation. To decide this, we need to know the probability that this vertex has a given type, say type j. To calculate this, we need to work out all the type possibilities for the first m generations, and their probabilities, which may well include lots of complicated size-biasing. Certainly it is not easy, and there’s no reason why these offspring distributions should be IID. The best we can say is that they should probably be exchangeable within each generation.

Obviously if the offspring distribution does not depend on the parent’s type, then we have a standard Galton-Watson tree with types assigned in an IID manner to the realisation. If the types are symmetric (for example if M, to be defined, is invariant under permuting the indices) then life gets much easier. In general, however, it will be more complicated than this.

We can however think about how to decide on survival probability. We consider the expected number of offspring, allowing both the type of the parent and the type of the child to vary. So define $m_{ij}$ to be the expected number of type j children born to a type i parent. Then write these in a matrix $M=(m_{ij})$.

One generalisation is to consider a Galton-Watson forest started from some positive number of roots of various types. Suppose we have a vector $\nu=(\nu_i)$ listing the number of roots of each type. Then the expected number of descendents of each type at generation n is given by the vector $\nu M^n$.

Let $\lambda$ be the largest eigenvalue of M. As for the transition matrices of Markov chains, the Perron-Frobenius theorem applies here, which confirms that, because the entries of M are positive, the eigenvalue with largest modulus is simple and real, and the associated eigenvector has entirely positive entries. [In fact we need a couple of extra conditions on M, including that it is possible to get from any type to any other type – we say irreducible – but that isn’t worth going into now.]

So in fact the total number of descendents at generation n grows like $\lambda^n$ in expectation, and so we have the same description of subcriticality and supercriticality. We can also make a sensible comment about the left-$\lambda$-eigenvector of M. This is the limiting proportion of the different types of vertices.

It’s a result (eg. [3]) that the height profile of a depth-first search on a standard Galton-Watson tree converges to Brownian Motion. Another way to phrase this is that a GW tree conditioned to have some size N has the Brownian Continuum Random Tree as a scaling limit as N grows to infinity. Miermont [4] proves that this result holds for the multitype tree as well. In the remainder of this post I want to discuss one idea along the way to the proof, and one application.

I said initially that there wasn’t a trivial reduction of a multitype process to a monotype process. There is however a non-trivial embedding of a monotype process in a multitype process. Consider all the vertices of type 1, and all the paths between such vertices. Then draw a new tree consisting of just the type 1 vertices. Two of these are joined by an edge if there is no other type 1 vertex on the unique path between them in the original tree. If that definition is confusing, think of the most sensible way to construct a tree on the type 1 vertices from the original, and you’ve probably chosen this definition.

There are two important things about this new tree. 1) It is a Galton-Watson tree, and 2) if the original tree is critical, then this reduced tree is also critical. Proving 1) is heavily dependent on exactly what definitions one takes for both the multitype branching mechanism and the standard G-W mechanism. Essentially, at a type 1 vertex, the number of type 1 descendents is not dependent on anything that happened at previous generations, nor in other branches of the original tree. This gives IID offspring distributions once it is formalised. As for criticality, we note that by the matrix argument given before, under the irreducibility condition discussed, the expectation of the total population size is infinite iff the expected number of type 1 vertices is also infinite. Since the proportion of type 1 vertices is given by the first element of the left eigenvector, which is positive, we can make a further argument that the number of type 1 vertices has a power-law tail iff the total population size also has a power-law tail.

I want to end by explaining why I was thinking about this model at all. In many previous posts I’ve discussed the forest fire model, where occasionally all the edges in some large component are deleted, and the component becomes a set of singletons again. We are interested in the local limit. That is, what do the large components look like from the point of view of a single vertex in the component? If we were able to prove that the large components have BCRT as the scaling limit, this would answer this question.

This holds for the original random graph process. There are two sensible ways to motivate this. Firstly, given that a component is a tree (which it is with high probability if its size is O(1) ), its distribution is that of the uniform tree, and it is known that this has BCRT as a scaling limit [1]. Alternatively, we know that the components have a Poisson Galton-Watson process as a local limit by the same argument used to calculate the increments of the exploration process. So we have an alternative description of the BCRT appearing: the scaling limit of G-W trees conditioned on their size.

Regarding the forest fires, if we stop the process at some time T>1, we know that some vertices have been burned several times and some vertices have never received an edge. What is clear though is that if we specify the age of each vertex, that is, how long has elapsed since it was last burned; conditional on this, we have an inhomogeneous random graph. Note that if we have two vertices of ages s and t, then the probability that there is an edge between them is $1-e^{-\frac{s\wedge t}{n}}$, ie approximately $\frac{s\wedge t}{n}$. The function giving the probabilities of edges between different types of vertices is called the kernel, and here it is sufficiently well-behaved (in particular, it is bounded) that we are able to use the results of Bollobas et al in [2], where they discuss general sparse inhomogeneous random graphs. They show, among many other things, that in this setting as well the local limit is a multitype branching process.

So in conclusion, we have almost all the ingredients towards proving the result we want, that forest fire components have BCRT scaling limit. The only outstanding matter is that the Miermont result deals with a finite number of types, whereas obviously in the setting where we parameterise by age, the set of types is continuous. In other words, I’m working hard!

References

[1] Aldous – The Continuum Random Tree III

[2] Bollobas, Janson, Riordan – The phase transition in inhomogeneous random graphs

[3] Le Gall – Random Trees and Applications

[4] Miermont – Invariance principles for spatial multitype Galton-Watson trees

# Discontinuous Phase Transitions

Yesterday, Demeter Kiss from Cambridge gave a seminar in Oxford about a model for self-destructive percolation on $\mathbb{Z}^2$ that had implications for the (non-)existence of an infinite-parameter forest fire model on the same lattice. I enjoyed talking about this and his recent work on the related model of frozen percolation on $\mathbb{Z}^2$. Considering these models in the lattice setting present a whole range of interesting geometric challenges that are not present in the mean-field case that has mainly occupied my research direction so far.

The afternoon’s discussion included lots of open problems about percolation. Several of these are based around continuity of the phase transition, so I thought I would write a quite post about some simple examples of this, and one example where it does not hold.

A helpful base example is bond percolation on the lattice $\mathbb{Z}^2$. Here, we specify some probability p in [0,1], and we declare edges of the lattice open with probability p, independently of each other. We then consider the graph induced by the open edges. We say that percolation occurs if the origin is contained in an infinite open component. The terminology arises from the interpretation as fluid being added at the origin and flowing down open edges. We define $\theta(p)$ to be the probability that the origin is in an infinite component when the parameter is p. By translation-invariance, we can get some sort of 0-1 law, to conclude that there is an infinite component somewhere in the system with probability either 0 or 1, depending on whether $\theta(p)$ is positive or zero. Indeed, we can further show that if it is positive, then with probability 1 there is a unique infinite component.

We define the critical probability $p_c:= \inf\{\theta(p)>0\}$. A question worth asking is then, what is $\theta(p_c)$? In some examples, we can find $p_c$, but we cannot prove that $\theta(p)$ is continuous around $p_c$. In the case of $\mathbb{Z}^2$ this is known, and it is known from work of Kesten that $p_c=1/2$. See below for a plot of $\theta(p)$ in this setting (obtained from this blog, though possibly originating elsewhere).

The aim is to find an example where we do not have such a continuous phase transition. The original work on frozen percolation took place on trees, and one of Kiss’s results is confirms that these show qualitatively different phenomena to the same process on the lattice. In some sense, trees lie halfway between a lattice and a mean-field model, since there is often some independence when we look down the tree from a given generation, if it is well-defined to use such language.

Anyway, first we consider percolation on an infinite regular rooted k-ary tree. This means we have a root, which has k children, each of which in turn has k children, and so on. As before we consider bond percolation with parameter p. In this setting, we have a language to describe the resulting open component of the root. The offspring distribution of any vertex in the open component is given by Bin(k,p) independently of everything else, so we can view this component as the realisation of a Galton-Watson tree with this offspring distribution. This distribution has finite mean kp, and so we can state explicitly when the survival probability is positive. This happens when the mean is greater than 1, ie p>1/k.

For our actual example, we will consider the survival probability, but the technicalities are easier to explain if we look at the extinction probability, now using the language of branching processes. Suppose the offspring distribution has pgf given by

$f(x)=p_0+p_1x+p_2x^2+\ldots.$

Then the extinction probability q satisfies f(q)=q. I want to pause to consider what happens if this equation has multiple solutions. Indeed, in most interesting cases it will have multiple solutions, since f(1) will always be 1 if it is a non-defective offspring distribution. It is typically cited that: the extinction probability q is the smallest solution to this equation. I want to discuss why that is the case.

To approach this, we have to consider what extinction means. It is the limit in the event sense of the events {we are extinct after n generations}. Let the probabilities of these events be $q_n$, so $q_0=0$. Then by a straightforward coupling argument, we must have

$0=q_0\le q_1\le q_2 \le\ldots\le q:= \lim q_n \le 1.$

But, by the same generating function argument as before, $q_{n+1}=f(q_n)\ge q_n$. So if we split [0,1] into regions A where $f(x)\ge x$ and B where $f(x), all the $(q_n)$s must occur in the former, and so since it is closed, their limit must be in A also. Note that if f(x) intersects x lots of times, then region A is not necessarily connected. In the diagram below, in moving from $q_n$ to $q_{n+1}$ we might jump across part of B.

This is bad, as we are trying to prove that q is the right boundary of the connected component of A containing 0. But this cannot happen, as f is monotonic. So if one of the roots of f(x)=x in between the hypothesised $q_n is called z, then $f(q_n)< f(z)=z < q_{n+1}$, a contradiction.

Ok, so now we are ready to consider our counterexample to continuity over the percolation threshold. See references for a link to the original source of this example. We have to choose a slightly more complicated event than mere survival or extinction. We consider bond percolation as before on the infinite ternary tree, where every vertex has precisely 3 offspring. Our percolation event is now that the root is the root of an infinite binary tree. That is, the root has at least two children, each of which have at least two children, each of which, and so on.

If we set this probability equal to q, and the probability of an edge being open equal to p, then we have the recurrence:

$q=3p^2(1-p)q^2+p^3[3q^2(1-q)+q^3].$

The first term corresponds to the root having two open edges to offspring, and the second to the root having all three open edges to offspring. After manipulating, we end up with

$q\left[2p^3q^2-3p^2q+1\right]=0.$

We are therefore interested in roots of the quadratic lying between 0 and 1. The discriminant can be evaluated as

$\Delta=p^3(9p-8),$

and so there are no real roots where p<8/9. But when p=8/9, we have a repeated root at q=27/32, which is obviously not zero!

This equation is qualitatively different to the previous one for the extinction probability of a Galton-Watson tree. There, we had a quadratic, with one root at 1. As we varied p, the other root moved continuously from greater than one to less than one, so it passed through 1, giving continuity at the critical probability. Here, we have a cubic, again with one root at 1. But now the other roots are complex for small p, meaning that the local minimum of the cubic lies above the x-axis. As p gets to the critical value, it the local minimum passes below the x-axis, and suddenly we have a repeated root, not at zero.

I would like to have a neat probabilistic heuristic for this result, without having to make reference to generating functions. At the moment, the best I can come up with is to say that the original problem is simple, in the sense that the critical probability is as small as it could be while still making sense in expectation. To be concrete, when the mean of the offspring generation is less than 1, the expected size of the nth generation tends to zero, so there certainly could not be positive probability of having an infinite component.

Whereas in the binary tree example, we only require p=2/3 to have, in expectation, the right number of open edges to theoretically allow an infinite binary tree. If we think of percolation as a dynamic process by coupling in p, essentially as we move from p=2/3 to p=8/9 we need to add enough edges near the origin to be able to take advantage of the high density of edges available far from the origin. The probability of this working given you start from n vertices grows much faster (as n grows) than in the original problem, so you might expect a faster transition.

This is so content-free I’m reluctant even to call it a heuristic. I would be very interested to hear of any more convincing argument for this phenomenon!

REFERENCES

Dekking, Pakes – On family trees and subtrees of simple branching processes (link)