Random three dimensional partitions

Secret Blogging Seminar 2018-03-12

Back in graduate school, I read a beautiful paper of Kenyon, Okounkov and Sheffield. It started with the following physical story.

Electron microscope image of a salt crystal

This is the corner of a crystal of salt, as seen under an electron microscope. (I took the image from here, unfortunately I couldn’t find better information about the sourcing.) As you can see, the corner is a bit rounded, where some of the molecules have rubbed away. They ask the question: “What is the shape of that rounded corner?”

The molecules of a salt crystal form a cubical lattice. We can index them as $\{ (i,j,k) \in \mathbb{Z}_{\geq 0}^3 \}$ . But it’s not the ones from the interior that are missing – if $(i_1, j_1, k_1)$ is missing and $i_2 \leq i_1$ , $j_2 \leq j_1$ and $k_2 \leq k_1$ , then $(i_2, j_2, k_2)$ is missing as well.

A finite subset $P$ of $\mathbb{Z}_{\geq 0}^3$ with the property that $(i_1,j_1,k_1) \in P$ implies $(i_2, j_2, k_2) \in P$ for $i_2 \leq i_1$ , $j_2 \leq j_1$ , $k_2 \leq k_1$ is what I’ll call a three dimensional partition. (Here I am deliberately rebelling against the awful classical terminology, which is a “plane partition“.) The opening of Kenyon, Okounkov and Sheffield’s paper discusses the question: “What is the shape of a random three dimensional partition of size $N$ ?”

It was only after I had worked through this paper, and its sequels 1 2 fairly carefully that I realized they hadn’t actually answered their motivating question. They certainly implied what the answer should be, and they laid out all the necessary tools, but they never came back and said “let’s do the salt crystal example”.

In this post, I want to lay out in outline how this question is answered. The previous posts on Legendre transforms in statistical mechanics, and on random partitions, were meant as warm ups, where it is easier to make complete arguments.

Let me acknowledge right out that I am making no attempt at rigor here. What I want to do is sketch the argument, and hope this encourages some of you to read the amazing sequence of papers I have linked to.

In the previous posts, our first step was to analyze partitions of a given slope. Namely, we showed that there are roughly $\exp(n h(p))$ partitions that fit in a $(pn) \times ((1-p) n)$ box, where $h(p) = -p \log p - (1-p) \log (1-p)$ . We would now like to similarly analyze three dimensional partitions with a given slope.

We could specify a normal vector $(p,q,r)$ and ask for the partition to have slope roughly $(p,q,r)^{\perp}$ , but there is an approach which turns out to be equivalent and is a bit easier to formulate. Let $(p,q,r)$ be positive real numbers with $p+q+r=1$ . The boundary of a plane partition is made up of squares which lie either in the $(x,y)$ plane, the $(x,z)$ plane or the $(y,z)$ plane. Let us suppose there were some function $h(p,q,r)$ such that the number of partitions with some specified boundary, using roughly $pN$ squares of the first type, $qN$ of the second type and $rN$ of the third type, is roughly $\exp(N h(p,q,r))$ . We would like to know what this $h$ function is.

I said “some specified boundary”. What boundary shall we use? As is often the case in mathematical physics, the nicest thing to do is to use “periodic boundary conditions” — which is to say, to avoid the issue of boundaries by wrapping the problem on a torus.

A perspective drawing can give us inspiration. The boundary of a three dimensional partition looks like a tiling of the plane by rhombi with angles $60^{\circ}$ and $120^{\circ}$ . The three planes which squares can lie in turn into the three possible orientations of the rhombi (colored red, black and blue below).

(Image taken from Eventually Almost Everywhere, who has further nice discussion of the relation between rhombus tilings and plane partitions.)

Tile the plane by equilateral triangles in the standard manner and then quotient that plane by some lattice to produce a torus. Let that torus contain $N$ upward pointing and $N$ downward pointing triangles; so we can hope to tile it with $N$ rhombi.

Let $m(a,b,c)$ be the number of such tilings which use $a$ , $b$ and $c$ rhombi of the three potential types. So our goal is to determine $\lim_{N \to \infty} \frac{1}{N} \log m(pN,qN,rN)$ .

As in the case of two dimensional partitions, the best strategy is to form a generating function. Let $Z(u,v,w) = \sum_{a+b+c=N} m(a,b,c) u^a v^b w^c$ .

There is an amazing explicit formula for this generating function. To tantalize you, I will state it without proof, and simply point you to the search term “Kasteleyn’s method” if you want to learn more.

Let us suppose our torus has a fundamental domain which is $n \times n$ , so $N = n^2$ , and let $n$ be even. Set

$\displaystyle{ K(u,v,w) = \prod_{a=0}^{n-1} \prod_{b=0}^{n-1} (u+e^{2 \pi i a/n} v + e^{2 \pi i b/n} w) }$ .

We have (up to possible sign errors on my part)

$\displaystyle{Z(u,v,w) = }$

$\displaystyle{\frac{ K(-u,v,w) + K(u,-v,w) + K(u,v,-w) - K(-u,-v,-w) }{2}}$ .

Asymtotically, all four terms contributing to $Z$ are about the same size, so $|Z(u,v,w)| \approx |K(u,v,w)|$ . And how big is $K$ ? We can approximate that sum by an integral:

$\displaystyle{ \log K(u,v,w) \approx n^2 \int_{\alpha=0}^{1} \int_{\beta=0}^{1} \log | u+ve^{2 \pi i \alpha} + w e^{2 \pi i \beta} |}$ .

Set $R(u,v,w) = \int_{\alpha=0}^{1} \int_{\beta=0}^{1} \log | u+ve^{2 \pi i \alpha} + w e^{2 \pi i \beta} |$ . This is known as the “Rankin function”. Then, as before, we conclude that $\sigma$ is the Legendre transform of the Rankin function. And we conclude that a random three dimensional partition has the shape $R(e^{Cx}, e^{Cy}, e^{Cz}) = 1$ for some constant $C$ .

Something interesting happens. Suppose that $u>v+w$ . So $u+\zeta+\eta$ cannot be zero for any complex numbers $\zeta$ and $\eta$ with $|\zeta| \leq v$ , $|\eta| \leq w$ . So $\log |u+\zeta+\eta|$ is a harmonic function when we restrict $\zeta$ and $\eta$ to those discs. The average of a harmonic function over the boundary of a disc is the value at the center of the disc. So $R(u,v,w)$ simply equals $u$ when $u < v+w$ .

Thus, the surface $R(e^{Cx}, e^{Cy}, e^{Cz})=1$ contains, in particular, the planar region $x=0$ , $e^{Cx} <e^{Cy}+e^{Cz}$ , and similar planar regions in the $y=0$ and $z=0$ planes. This is a major difference between the two dimensional and three dimensional limit shapes — no part of the two dimensional limit curve is contained in the coordinate axes. While I'm not sure how seriously this picture should really be taken, it might do a bit to explain why those salt crystals above look so cubical.

There is another very important difference between the two and three dimensional pictures. In the two dimensional case, the only solutions of the variational problem were of the form $e^{Cx}+e^{Cy} = 1$ . But, in the three dimensional case, $R(e^{Cx}, e^{Cy}, e^{Cz})=1$ is only one of many solutions to the resulting PDE. There is so much more to say about all of this. If you want to read more, I refer you to Chapter One of Kenyon and Okounkov.