DGFF 4 – Properties of the Green’s function

I’m at UBC this month for the PIMS probability summer school. One of the long courses is being given by Marek Biskup about the Discrete Gaussian Free Field (notes and outline here) so this seems like a good moment to revive the sequence of posts about the DGFF. Here’s DGFF1, DGFF2, DGFF3 from November.

The first draft of this post was about the maximum of the DGFF in a large box $V_N$, and also about the Green’s function $G^{V_N}(x,y)$, which specifies the covariance structure of the DGFF. This first draft also became too long, so I’m splitting it into two somewhat shorter ones. As we’ll see, some understanding and standard estimates of the Green’s function is enough to say quite a bit about the maximum. In this first post, we’ll explore some ‘low-hanging fruit’ concerning the Green’s function, as defined through a simple random walk, which are useful, but rarely explained in the DGFF literature.

Symmetry of Green’s function

We start with one of these low-hanging fruit. If $G^{V_N}$ is to be a covariance matrix, it has to be symmetric. In the first post, showing that the definition of the DGFF as a random field with given Hamiltonian is equivalent to $\mathcal{N}(0,G^{V_N})$ certainly can be viewed as a proof of symmetry. However, it would be satisfying if there was a direct argument in the language of the definition of the Green’s function.

To make this self-contained, recall the random walk definition of $G^{V_N}(x,y)$. Let $(S_m)_{m\ge 0}$ be simple random walk on $V_N$, and $\mathbb{P}_x,\,\mathbb{E}_x$ denote starting the random walk at $x\in V_N$. As usual, let $\tau_y,\,\tau_A$ denote the hitting time of a vertex y or a set A respectively. Then

$G^{V_N}(x,y):= \mathbb{E}_x \left[ \sum_{m=0}^{\tau_{\partial V_N}}1_{(S_m=y) }\right].$

That is, $G^{V_N}(x,y)$ is the expected number of visits to y by a random walk from x, before it exits $V_N$.

Let’s drop the superscript for now, as everything should hold for a more general subset of the lattice. I don’t think it’s immediately obvious at the level of Markov chains why G(x,y)=G(y,x). In particular, it’s not the case that

$\mathbb{P}_x(\tau_y < \tau_{D^c}) = \mathbb{P}_y(\tau_x <\tau_{D^c}),$

and it feels that we can’t map between paths $x \to \partial D$ and $y\to \partial D$ in a way that preserves the number of visits to y and x, respectively. However, we can argue that for any m

$\mathbb{P}_x(S_m=y, \tau_{D^c}>m) = \mathbb{P}_y(S_m=x, \tau_{D^c}>m),$

by looking at the suitable paths of $(S_m)$. That is, if we have a path $x=S_0,S_1,\ldots,S_m=y$ that stays within D, then the probability of seeing this path starting from x and its reverse direction starting from y are equal. Why? Because

$\mathbb{P}_x(S_0=x,S_1=v_1,\ldots,S_{m-1}=v_{m-1},S_m=y) = \prod_{\ell=0}^{m-1} \frac{1}{\mathrm{deg}(v_\ell)},$

and

$\mathbb{P}_y(S_0=y,S_1=v_{m-1},\ldots,S_{m-1}=v_1, S_m=x) = \prod_{\ell=0}^{m-1} \frac{1}{\mathrm{deg}(v_{m-\ell})} = \prod_{\ell=1}^m \frac{1}{\mathrm{deg}(v_\ell)}.$

Since $D\subset \mathbb{Z}^d$ and x,y are in the interior of D, we must have $\mathrm{deg}(x)=\mathrm{deg}(y)$, and so these two expressions are equal. Summing over all such two-way paths, and then all m gives the result.

Fixing one argument

We now focus on $G^D(\cdot,y)$, where the second argument is fixed. This is the solution to the Poisson equation

$\Delta G^D(\cdot,y) = -\delta_y(\cdot),\quad G^D(x,y)=0,\; \forall x\in \partial D.$

To see this, can use a standard hitting probability argument (as here) with the Markov property. This is harmonic in $D\backslash \{y\}$, and since we know

$G^D(y,y)= \frac{1}{\mathbb{P}_y(\text{RW hits }\partial D\text{ before returning to }y)},$

this uniquely specifies $G^D(\cdot,y)$. Anyway, since harmonic functions achieve their maxima at the boundary, we have $G(y,y)\ge G(x,y)$ for all $x\in D$. We can also see this from the SRW definition as

$G(x,y)=G(y,x) = \mathbb{P}_y (\tau_x < \tau_{\partial D} ) G(x,x) \le G(x,x).$

Changing the domain

Now we want to consider nested domains $D\subset E$, and compare $G^D(\cdot,\cdot)$ and $G^E(\cdot,\cdot)$ on DxD. The idea is that for SRW started from $x\in D$, we have $\tau_{\partial D}\le \tau_{\partial E}$, since one boundary is contained within the other. From this, we get

$G^D(x,y)\le G^E(x,y),\quad \forall x,y\in D,$

and we will use the particular case y=x.

For example, if $x\in V_N$, the box with width N, then the box with width 2N centred on x contains the whole of $V_N$. So, if we set $\bar {V}_{2N}:= [-N,N]^d$, then with reference to the diagram, we have

$G^{V_N}(x,x)\le G^{\bar{V}_{2N}}(0,0),\quad x\in V_N.$

As we’ll see when we study the maximum of the DGFF on $V_N$, uniform control over the pointwise variance will be a useful tool.

Maximising the Green’s function

The idea of bounding $G^{V_N}(x,x)$ by $G^{\bar V_{2N}}(0,0)$ for any $x\in V_N$ is clever and useful. But a more direct approach would be to find the value of x that maximises $G^{V_N}(x,x)$. We would conjecture that when $V_N$ has a central vertex, then this is the maximiser.

We can prove this directly from the definition of the Green’s function in terms of random walk occupation times. Let’s assume we are working with $\bar{V}_N$ for even N, so that 0 is the central vertex. Again, since

$G^D(x,x)=\frac{1}{\mathbb{P}_x(\text{RW hits }\partial D\text{ before returning to }x)},$ (*)

it would suffice to show that this probability is minimised when x=0. This feels right, since 0 is furthest from the boundary. Other points are closer to the boundary in some directions but further in others, so we can’t condition on the maximum distance from its start point achieved by an excursion of SRW (we’re vertex-transitive, so these look the same from all starting points), as even allowing for the four possible rotations, for an excursion of diameter slightly larger than N, starting at the centre is maximally bad.

However, intuitively it does feel as if being closer to the boundary makes you more likely to escape earlier. In fact, with a bit more care, we can couple the SRW started from 0 and the SRW started from $r=(r^x,r^y)\ne 0$ such that the latter always exits first. For convenience we’ll assume also that $r^x,r^y$ are both even.

I couldn’t find any reference to this, so I don’t know whether it’s well-known or not. The following argument involves projecting into each axis, and doing separate couplings for transitions in the x-direction and transitions in the y-direction. We assume WLOG that x is in the upper-right quadrant as shown. Then, let $0=S_0,S_1,S_2,\ldots$ be SRW started from 0, and we will construct $r=R_0,R_1,R_2,\ldots$ on the same probability space as $(S_m)_{m\ge 0}$ as follows. For every m, we set the increment $R_{m+1}-R_m$ to be $\pm(S_{m+1}-S_m)$. It remains to specify the sign, which will be determined by the direction of the S-increment, and a pair of stopping times. The marginal is therefore again an SRW, started from r. Temporarily, we use the unusual notation $S_m= (S^x_m,S^y_m)$ for the coordinates of $S_m$.

So, if $S_{m+1}-S_m=(1,0), (-1,0)$, ie S moves left or right, then we set

$R_{m+1}-R_m = \begin{cases} -(S_{m+1}-S_m) &\quad \text{if }mT^x.\end{cases}$ (*)

where $T^x:= \min\{m\,:\, R^x_m=S^x_m\}$. That is, $R^x$ moves in the opposing direction to $S^x$ until the first time when they are equal (hence the parity requirement), and then they move together. WLOG assume that $r^x>0$. Then suppose $S^x_m=\pm N$ and such m is minimal. Then by construction, if $m\ge T^x$, then $R^x_m=\pm N$ also. If $m, then we must have $S^x_m=-N$, and so since $R^x$‘s trajectory is a mirror image of $S^x$‘s, in fact $R^x_m = N+r^x>N$, so $R^x$ hit +N first. In both cases, we see that $R^x$ hits $\pm N$ at the same time or before $S^x$.

In other words, when $S^x_m$ has non-negative x coordinate, the lazy random walk $R^x$ follows the same trajectory as $S^x$, and when it has negative x coordinate, the $R^x$ mirrors $S^x$. At some time, it may happen that $S^x_m= R^x_m=0$ (recall the parity condition on r). Call this time $T^x$. We then adjust the description of the coupling so that (*) is the mechanism for $m, and then for $m\ge T^x$, we take $S^x_m=R^x_m$.

Similarly, if $S_{m+1}-S_m =(0,1), (0,-1)$, ie S moves up or down, then we set

$R_{m+1}-R_m = \begin{cases} -(S_{m+1}-S_m)&\quad \text{ if }m

with corresponding definition of the stopping time $T^y$.

This completes the coupling, and by considering $T^x\wedge T^y$, we have shown what that the exit time for the walk started from zero dominates the exit time for walk started from r. Recall that so far we are in the case where the box has even width and $r=(r^x,r^y)$ has even coordinates.

This exit time comparison isn’t exactly what we need to compare $G^N(0,0)$ and $G^N(x,x)$. It’s worth remarking at this stage that if all we cared about was the Green’s function on the integer line [-N,N], we would have an easier argument, as by the harmonic property of $G(\cdot,y)$

$G^{[-N,N]}(0,r)=\frac{N-r}{N}G^{[-N,N]}(0,0),$

$G^{[-N,N]}(r,0) = \frac{N}{N+r}G^{[-N,N]}(r,r),$

and so $G(0,0)>G(r,r)$ follows by symmetry. To lift from 1D to 2D directly, we need a bit more than this. It’s possible that S returns in both x- and y- coordinates more often than R, but never at the same time. Fortunately, the coupling we defined slightly earlier does give us a bit more control.

Let $\tau^x(S), \tau^x(R)$ be the first times that $S^x, R^x$ hit $\pm N$. Under this coupling, for any $m\ge 0$

$\mathbb{P}(S^x_m=0, m

since these events are literally equal. Since we showed that $\tau^x(R)\le \tau^x(S)$ almost surely, we can further deduce

$\mathbb{P}(S^x_m=0,m

$=\mathbb{P}(R^x_m=r^x, m

To address the corresponding events for which $m\ge T^x$, we apply the strong Markov property at $T^x$, to obtain SRW $Z_m$ started from r/2, and let $\tau_{-N},\tau_{+N}$ be the hitting times of $-N,+N$ respectively and $\tau_{\pm N}=\tau_{-N}\wedge \tau_{+N}$. It will now suffice to prove that

$\mathbb{P}(Z_m=0, m< \tau_{\pm N}) \ge \mathbb{P}(Z_m=r,m<\tau_{\pm N}),$ (**)

as then we can apply the law of total probability and sum over values of $T^x$ and $m\ge 0$.

To prove this result, we consider the following bijection between trajectories of length m from r/2 to {0,r}. We decompose the trajectories into excursions away from r/2, and then a final meander from r/2 to {0,r} that stays on the same side of r/2. We construct the new trajectory by preserving all the initial excursions, but reversing all the steps of the final meander. So if the original trajectory ended up at 0, the image ends up at r. Trivially, the initial excursions in the image only hit $\pm N$ if the excursions in the original trajectory did this too. But it’s also easy to see, by a similar argument to the coupling at the start of this section, that if the original trajectory ends at r and does not hit $\pm N$, then so does the image. However, the converse is not true. So we conclude (**), and thus

$\mathbb{P}(S_m^x=0) \ge \mathbb{P}(R_m^x=0)$

for all m by combining everything we have seen so far. And so we can now lift to a statement about $S_m$ itself, that is considering both coordinates separately.

The remaining cases for r require a little more care over the definition of $T^x$, though the same projection argument works, for fundamentally the same reason. (Note that in the above argument, if $S^x_m=-N$ and $m, then in fact $R^x_m\ge N+2$, and so it’s not hard to convince yourself that a sensible adjustment to the stopping time will allow a corresponding result with $R^x_m\ge N+1$ in the odd $r^x$ case.) The case for N odd is harder, since in one dimension there are two median sites, and it’s clear by symmetry that we can’t couple them such that RW from one always exits at least as early as RW from the other. However, the distributions of exit times started from these two sites are the same (by symmetry), and so although we can’t find a coupling, we can use similar stopping times to obtain a result in probability.

In the next post, we’ll see how to apply this uniform bound on $G^{V_N}(x,x)$ to control the maximum of the DGFF on $V_N$. In particular, we address how the positive correlations of DGFF influence the behaviour of the maximum by comparison with independent Gaussians at each site.

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 $G=K_n$.