*21 August 2017*

**The dimer model part III: The square lattice on a rectangle**

*
This post is part of a series on the dimer model.
This series is a product of a reading project and it does not include original
research. The series closely follows the notes from Richard Kenyon; references that are
used throughout the whole series may be found at the end of the first post.
*

*The dimer model part I: A first introduction**The dimer model part II: Kasteleyn theory**The dimer model part III: The square lattice on a rectangle*

To count the number of tilings of a large graph it is essential to calculate the determinant of a large matrix. This is a hard task in general, but in certain cases some reasoning greatly simplifies the calculations. In this post we show how to do this for the square lattice on a rectangle.

Write $[n]$ for the path graph $\{1,2,...,n\}$. We fix $n$ and $m$ and let $G=G_{[n]\times [m]}$ be the natural rectangular square lattice graph corresponding to the vertex set $[n]\times [m]$. A dimer cover of $G$ corresponds to a domino tiling of a rectangle of width $n$ and height $m$ by $1\times 2$ dominos. The graph $G_{[n]\times [m]}$ is the cartesian product of the path graphs $[n]$ and $[m]$, and thus has adjacency matrix $$A_{[n]\times[m]}=A_{[n]}\otimes I_m + I_n\otimes A_{[m]}.$$ The function on edges $k\equiv 1$ is not a Kasteleyn weighting, since every face has four boundary edges. We let $k$ assign value $i$ to all vertical edges instead, so that the alternating product of $k$ along any square equals $-1$, as desired for faces of four edges. The corresponding Kasteleyn matrix satisfies $$\tilde A_{[n]\times [m]}:=\begin{pmatrix}0&K\\K^t&0\end{pmatrix}=A_{[n]}\otimes I_m + iI_n\otimes A_{[m]}.$$ The eigenvalues of $\tilde A$ are of the form $\lambda_1+i\lambda_2$ where $\lambda_1$ and $\lambda_2$ are eigenvalues of $A_{[n]}$ and $A_{[m]}$ respectively. It is not hard to verify that the $n$ eigenvalues of $A_{[n]}$ are $( 2\cos (j\pi/(n+1)))_{1\leq j\leq n}$ (this calculation is not of particular interest to us, which is why we omit it). The following result follows immediately from above observations and Theorem ii.2.

**Theorem iii.1.**
The square of the number of tilings of $G_{[n]\times [m]}$ equals
\begin{equation}
\label{partitionrectangle}
|\operatorname{Det} K|^2=|\operatorname{Det} \tilde A_{[n]\times [m]}|=\prod_{j=1}^n \prod_{k=1}^m \left|2\cos \frac{j\pi}{n+1}+i2\cos \frac{k\pi}{m+1}\right|.
\end{equation}

**Example iii.2.**
A chessboard can be tiled with dominos in $12,988,816$ different ways.

A dimer tiling of the rectangle can be described by assigning to each white vertex the corresponding black vertex. For most white vertices (those not near the edge of the rectangle) there are four black vertices available: the black vertex must be immediately left, right, above, or below the concerned white vertex. This gives an obvious upper bound on the number of tilings: the number of tilings cannot exceed $4^{|W|}=2^{nm}$. In practice the number of tilings however is much smaller: this is because the condition that the assignment has to be bijective invalidates most maps in $4^W$. We wonder how much freedom we have on average in the assignment of black vertices to each white one, given the condition. In other words, we are interested in the number \begin{equation} C:=\lim_{n,m\to\infty}|V|^{-1}\log Z(n,m)\leq \log 2=\frac{1}{2}\log 4,\label{squarelatticeconstant} \end{equation} where we assume $nm$ to be even and write $Z(n,m)$ for the number of domino tilings of the $n\times m$ rectangle. The appearance of the limit is slightly suggestive in that it is not a priori clear that the limit exists (we show this now). By substituting (\ref{partitionrectangle}) into (\ref{squarelatticeconstant}) we get \begin{align*} C&=\lim_{n,m\to\infty}\frac{1}{2nm}\sum_{j=1}^n \sum_{k=1}^m\log\left( 2\cos \frac{j\pi}{n+1}+i2\cos \frac{k\pi}{m+1}\right)\\ &=\frac{1}{2\pi^2}\int_0^\pi dx\int_0^\pi dy\log(2\cos x+i2\cos y)\\ &=\frac{1}{4\pi^2}\int_0^\pi dx\int_0^\pi dy\log(4\cos^2 x+4\cos^2 y)\\ &=\frac{\log 2}{2}+\frac{1}{4\pi^2}\int_0^\pi dx\int_0^\pi dy\log\left(1+\frac{1}{2}\cos x+\frac{1}{2}\cos y\right) \end{align*} We observe that the integral must be negative by concavity of $\log$, and since we integrate the argument of $\log$ symmetrically around $1$. Hence $C<\frac{1}{2}\log 2$, a much better bound. This means that we have a freedom of less than two black vertices per white vertex in the rectangle. In fact, the integral can be evaluated analytically to find $C=G/\pi$, where $G=1-\frac{1}{9}+\frac{1}{25}-\frac{1}{49}+...$ is Catalan's constant.

Is the constant $C=G/\pi$ universal? It seems independent of the ratio $n/m$ of the sides of the rectangle. What if we had not chosen a rectangular region of the infinite square lattice, but a triangular region, or a circle? Would we get the same constant $C$ in (\ref{squarelatticeconstant}) for those subgraphs as we let the mesh go to zero? It turns out that this depends much on the small-scale boundary that we choose. This is illustrated by the following example: we count the number of dimer tilings of the rectangle $G_{[n]\times [m]}$ (for $n$ even), but now remove all white vertices from the bottom row and all black vertices from the top row. There exists only one dimer tiling, even though the graph does not differ much from the proper rectangle on a macroscopic level. Replacing $Z(n,m)$ by $1$ in (\ref{squarelatticeconstant}) clearly yields $C=0$. It requires some more advanced tools to understand this, and this problem will be addressed in future posts on this website. We can mention here that the value $C=G/\pi$ is universal as an upper bound for such calculations for the square lattice.