Sbornik: Mathematics, 2022, Volume 213, Issue 11, Pages 1530–1558 DOI: https://doi.org/10.4213/sm9727e (Mi sm9727)

DOI: https://doi.org/10.4213/sm9727e

Funding agency	Grant number
Ministry of Science and Higher Education of the Russian Federation	075-15-2022-265
This work was performed at the Steklov International Mathematical Center and supported by the Ministry of Science and Higher Education of the Russian Federation (agreement no. 075-15-2022-265).

Russian version:
Matematicheskii Sbornik, 2022, Volume 213, Number 11, Pages 50–78
DOI: https://doi.org/10.4213/sm9727

Bibliographic databases:

§ 1. Introduction

A lattice triangulation of a (lattice) polygon in $\mathbb{R}^2$ is a triangulation with all vertices in $\mathbb{Z}^2$. As discovered in [3], lattice triangulations are important in algebraic geometry (also see [9]). A lattice triangulation is called primitive (or unimodular) if each triangle is primitive, that is, has the minimum possible area $1/2$ (is a translate of $[(0,0),(x_1,y_1),(x_2,y_2)]$ with $x_1y_2-x_2y_1=1$). We denote the number of primitive lattice triangulations of an $m\times n$ rectangle by $f(m,n)$. Let

was obtained by Anclin [1]. A still better upper bound $c\mkern-2mu<\mkern-2mu4\log_2\frac{1+\sqrt5}2\mkern-2mu=\mkern-2mu\log_2 6.854\mkern-2mu=2.777$ was obtained in [7] and announced in [12]. (I have not seen the manuscript [7] but Professor Welzl has kindly sent me the slides of his talk [13], where the proof of this bound is clearly exposed.)

It is easy to see that

An exact value of $c_3$ in a certain sense is given in Proposition 4.5, where we express $c_3$ in terms of Fredholm’s integral equations for certain generating functions. In particular, Proposition 4.5 provides an algorithm to compute $c_3$ to $n$ digits in polynomial time in $n$. A Mathematica code implementing the main step of this algorithm is presented at the end of § 4.

We have computed $c_3$ with this high precision in hope to find an algebraic equation for it or relate it to some known constants, but we have not succeeded so far.

In § 2.2 we present the results of computations of the exact values of the numbers $f(m,n)$ for some $m$ and $n$. These computations show, in particular, that $c\geqslant c(5,115)= 2.10449551\dots$ .

In § 6 we give an asymptotic upper bound for the number of all (not necessarily primitive) lattice triangulations. However, it seems to be far from optimal.

The rest of this paper (apart from §§ 2.2 and 6) is devoted to the proof of Theorems 1.1 and 1.2. Namely, in § 2.1 recurrence formulae for the number of primitive lattice triangulations of polygons lying in strips of fixed width are presented. They are similar to those in [4] but, in our opinion, are simpler and more efficient. Theorems 1.1 and 1.2 are proved in § 3 and § 4, respectively, using the relations between generating functions coming from the recurrence relations in § 2.1. As mentioned already, $c_3$ is expressed in terms of a Fredholm integral equation for the Cauchy integrals of the generating functions. In § 5 we estimate the approximation error of the most straightforward numerical solution of a Fredholm equation with analytic kernel.

Acknowledgement

I thank the referee for their valuable comments and corrections.

§ 2. Recurrence relations for strips of fixed width

2.1. Recurrence relations

Given a polygon $P\subset\mathbb{R}^2$, the upper part of its boundary is the set $\{(x,y)\in P\mid y'>y\Rightarrow (x,y')\notin P\}$. A vertical side of $P$ is a side of $P$ contained in a line $\{x=x_0\}$. Let $\mathcal T$ be a triangulation of a polygon $P$ in $\mathbb{R}^2$.

We say that $Q$ is a tile of $\mathcal T$ in the following three cases:

• 1) $Q$ is a triangle of $\mathcal T$ without vertical sides;
• 2) $Q$ is a triangle of $\mathcal T$ whose vertical side lies on the boundary of $P$;
• 3) $Q$ is a union of two triangles of $\mathcal T$ which share a common vertical side.

A polygon is called $y$-convex if its intersection with any line $x=\mathrm{const}$ is either an empty set, a point, or a line segment.

Zoom inZoom out

Figure 1

Now we fix an integer $m>0$ and consider primitive lattice triangulations of polygons which lie in the vertical strip $\{0\leqslant x\leqslant m\}$ and are bounded by two graphs of continuous piecewise linear functions.

By analogy with the terminology introduced in [4], § 2.2, we say that $\varphi\colon [0,m]\!\to\!\mathbb{R}$ is an admissible function if it is a continuous piecewise linear function whose graph is a union of line segments with endpoints in $\mathbb{Z}^2$. We fix an admissible function $\varphi_0$ and say that a function $\varphi\colon [0,m]\to\mathbb{R}$ is $\varphi_0$-admissible if it is admissible and $\varphi(x)\geqslant \varphi_0(x)$ for any $x\in[0,m]$. A $\varphi_0$-admissible shape is a polygon $S$ of the form $\{(x,y)\in\mathbb R^2\mid 0\leqslant x\leqslant m,\, \varphi_0(x)\leqslant y\leqslant\varphi(x)\}$ for some $\varphi_0$-admissible function $\varphi$.

As in the above definition of a tile of a triangulation, we say that $Q$ is a primitive lattice tile in the following three cases:

• 1) $Q$ is a primitive lattice triangle without vertical sides;
• 2) $Q$ is a primitive lattice triangle whose vertical side is contained in the boundary of the strip $0\leqslant x \leqslant m$;
• 3) $Q=\Delta_1\cup\Delta_2$, where $\Delta_1$ and $\Delta_2$ are primitive lattice triangles such that $\Delta_1\cap\Delta_2$ is a common vertical side of $\Delta_1$ and $\Delta_2$.

A primitive lattice tile $Q$ is $P$-maximal for a polygon $P$ if $Q\subset P$ and the upper part of the boundary of $Q$ is contained in the upper part of the boundary of $P$. We say that $S'$ is a $\varphi_0$-admissible subshape of a $\varphi_0$-admissible shape $S$ if $S'$ is the closure of $S\setminus(Q_1\cup\cdots\cup Q_n)$, where $Q_1,\dots,Q_n$ are $S$-maximal primitive lattice tiles with pairwise disjoint interiors. Following [4], in this case we set $\#(S',S)=n$. Note that the relation ‘be an admissible subshape’ is not transitive.

Let us denote the number of primitive lattice triangulations of a polygon $P$ by $f^*(P)$. When $P$ lies in the strip $\{0\leqslant x\leqslant m\}$, we also define $f(P)$ as the number of primitive lattice triangulations of $P$ that have no interior edges whose projections onto the $x$-axis are the whole interval $[0,m]$ (we choose simpler notation for a more complicated notion because the numbers $f(P)$ will be used more often than $f^*(P)$).

The following lemma is the inclusion-exclusion formula in our setting. The proof is the same as for Lemma 2.2 in [4].

Example 2.3. Let $m=2$ and $\varphi_0=0$. Given nonnegative integers $a$, $b$ and $c$, let $S_{a,b,c}$ be the $\varphi_0$-admissible shape bounded above by the line segments $[(0,a),(1,b)]$ and $[(1,b),(2,c)]$. Let $f_{a,b,c}=f(S_{a,b,c})$. We also set $f_{a,b,c}=0$ when $\min(a,b,c)<0$. Then (see Figure 2) the recurrence formula in Lemma 2.2 reads

$$ \begin{equation*} f_{a,b,c}= \begin{cases} f_{a-1,b,c}+f_{a,b-1,c}+f_{a,b,c-1}-f_{a-1,b,c-1} &\text{if }(a,b,c)\ne(0,0,0), \\ 1 &\text{if }(a,b,c)=(0,0,0). \end{cases} \end{equation*} \notag $$

Zoom inZoom out

Figure 2

Let $F(x,y,z)=\sum_{a,b,c}f_{a,b,c}x^ay^bz^c$ be the generating function. Then summing the recurrence relations over all triples $(a,b,c)\ne(0,0,0)$ we obtain

$$ \begin{equation*} \begin{aligned} \, &F(x,y,z)-1 = \sum f_{a-1,b,c}x^ay^bz^c+\sum f_{a,b-1,c}x^ay^bz^c+\dotsb \\ &\quad = \sum f_{a,b,c}x^{a+1}y^bz^c+\sum f_{a,b,c}x^ay^{b+1}z^c+\dotsb= F(x,y,z)(x+y+z-xz) \end{aligned} \end{equation*} \notag $$

so that $F(x,y,z)=1/(1-x-y-z+xz)$.

2.2. Some exact values of $f(m,n)$

The recurrence relations in Lemma 2.2 provide an algorithm for the computation of the exact values of $f(m,n)$ for small $m$ and $n$. The algorithm is similar to the one described in [4], § 2.2. We have performed computations using this algorithm, and the reader can see in Table 1 that we have advanced much further in comparison to the computations in [4]. There have been three reasons for this, whose impacts were more or less equal.

The first reason (an obvious one) is that computers have become more powerful. The second reason is that we have used another definition of admissible shapes, which enables us to reduce the amount of memory used with coefficient $3^{m-1}$, which is rather important for $m=9$ (as pointed out in [4], for this kind of algorithm ‘the bottleneck in the computations is always memory’). The third reason is that, instead of long arithmetics, we have used computations modulo different prime numbers and then have recovered the results using the Chinese remainder theorem. This trick has enabled us to ‘convert’ memory into time, the lack of which is not so crucial.

We have computed $f(3,n)$ to $n=600$ and $f(4,n)$ to $n=200$. The exact value of $f(3,600)$ has 1127 digits, and it yields $c_{3,600}=2.07966\dots$ . Comparing this with the limit value $c_3=2.08385$ we see that convergence is very slow. For $m=4$ the last exact value computed is

2.3. Convexity conjecture for the numbers $f(m,n)$

The following conjecture is confirmed by all the computed exact values of the quantities $f(m,n)$ (we set $f(m,0)=1$ by convention).

§ 3. The exact value of $c_2$ (proof of Theorem 1.1)

For $a,c\geqslant 0$, $a\equiv c\mod 2$, let $g^*_{a,c}$ denote the number of primitive lattice triangulations of the trapezoid $T(a,c)$ spanned by $(0,0)$, $(a,0)$, $(1,2)$ and $(1+c,2)$ (if ${a=0}$ or $c=0$, then $T(a,c)$ degenerates to a triangle). When $a\not\equiv c\mod 2$, we set $g^*_{a,c}=0$. We also set $g^*_{0,0}=1$. Let $G^*(x,z)$ be the generating function of the $g^*_{a,c}$:

Zoom inZoom out

Figure 3

The edges of the form $[(k,0),(l,2)]$ of any primitive lattice triangulation cut $T(a,c)$ into smaller trapezoids. These can be transformed into trapezoids $T(a_i,c_i)$ such that $\sum a_i = a$ and $\sum c_i = c$ by uniquely determined lattice automorphisms of the form $(x,y)\mapsto(x+p_iy+q_i,y)$, where $p_i,q_i\in\mathbb{Z}$ (see Figure 3). Hence

It is easy to see (cf. (1)) that the numbers of primitive lattice triangulations of the narrow (that is, of width 1) trapezoids in Figure 3 are binomial coefficients, hence

One can also check that $g_{a,c}=f_{a,(a+c)/2-1,c}$ where the $f_{a,b,c}=f(S_{a,b,c})$ are the numbers discussed in Example 2.3. Hence (cf. Example 2.4)

Lemma 3.2 is proved.

§ 4. Computation of $c_3$ (proof of Theorem 1.2)

4.1. Preparation

For $a,d\geqslant0$ such that $a\not\equiv d+1\mod3$, let $h^*_{a,d}$ be the number of primitive lattice triangulations of the trapezoid $T_3(a,d)$ spanned by $(0,0)$, $(1,3)$, $(1+d,3)$ and $(a,3)$. We set $h^*_{0,0}=1$ and $h^*_{a,d}=0$ for $a\equiv d+1\mod 3$, and we consider the generating function

We illustrate the relation

Zoom inZoom out

Figure 4

Similarly to Lemma 3.1 we have

4.2. Recurrence relations

Using the notation introduced in § 2, set $m=3$, $\varphi_0(x)=\frac13x-1$ and

Zoom inZoom out

Figure 5

In terms of the generating functions the recurrence relations in Lemma 2.2 read (cf. Examples 2.3 and 2.4)

4.3. The equation

We are going to obtain an equation for $\widetilde G(xt^{-1/2},t^{3/2},t^{-1}x)$ by expressing $\widetilde F$ in terms of $\widetilde G$ from (3) and plugging it into (4). To do this we must divide power series by polynomials. However, when some variables appear in powers ranging from $-\infty$ to $+\infty$, the meaning of this division should be clarified. To illustrate the possible ambiguity consider the expression $\operatorname{coef}_{u^{-1}}\bigl[1/(x-uy)\bigr]$. It can be understood either as

The geometric meaning of an exponent of $q$ is twice the signed area of the $\varphi_0$-admissible shape corresponding to the monomial under consideration, that is, $\displaystyle2\int_0^3\varphi(x)\,dx$ where the graph of $\varphi$ is the upper boundary of the shape. One can easily check by hand that

Further, we define $\widetilde F_q$, $\widetilde G_q$ and $\widetilde H_q$ by the same formulae as $\widetilde F $, $\widetilde G $ and $\widetilde H $ in § 4.2, adding the subscript $q$ everywhere. For example,

Let us set

Plugging (3) into (4), denoting the denominator in (6) by $Q_q(x,y,z,w)$ and observing that

We see that the variables $x$ and $q$ are ‘synchronized’ on the right-hand side of the equation obtained: they occur to the same power in each monomial of each power series in this expression. Hence we obtain the following identity in the ring $\mathbb{Z}[t^{\pm1},u^{\pm1}]((x))$:

Having found $g$ from (11), we can compute $\widetilde H(x,x)$. Indeed, by (5) we have

4.4. Computation

In this subsection we study the analytic functions defined by the series discussed in the previous subsection.

By § 4.1 we need to find the smallest positive pole of $H^*(x)$, that is the smallest positive zero $\beta$ of $1-H(x)$. One can check that

Let

The function Psi in the Mathematica system that is presented in Figure 7 computes $\Psi(x,t)$ for $(x,t)\in\Gamma'$ with any prescribed precision.

Notice that one of the functions $u_1(x,t)$ or $u_2(x,t)$ has a branch point at $(x,t)=(1/2,1)$, and hence the functions $\Phi$ and $\Psi$ are also branched at this point. The Laurent-Puiseux expansion of $\Psi(x,1)$ in powers of $s=\sqrt{1/2-x}$ is

Let $x_0^-={16}/{33}$ and $x_0^+={17}/{35}$. We will see below that $x_0\in[x_0^-,x_0^+]$; in fact, $x_0^{\pm}$ are given by initial segments of the continued fraction of $x_0$.

Using the expansion of $\Psi$ at $(1/2,1)$ and computing the values of $\Psi(x,t)$ on a sufficiently dense grid on $\Gamma'$ (using the program presented in Figure 7) one can check that $\Psi$ does not vanish on $\Gamma'\cap\{x<x_0^+\}$ and

Zoom inZoom out

Figure 6.Level curves of $\operatorname{Re}\Psi(x,t)$ for $|t|=1$. The vertical line shown is $x=x_0$ or $x=x_0^+$ (no difference for this resolution).

Applying Lemma 5.2 for an appropriately chosen $h$ to the function $|P(x/4,e^{i\tau},e^{i\theta})|^2$ we find that

Lemmas 4.2 and 4.3 in combination with (11) and (16) imply that the function $g(x,t)$ is analytic in a neighbourhood of $\Gamma'\cap\{x<2^{-3/2}\}$ and satisfies there the condition

Replacing the integrals by integral sums, equation (18) can be solved with any prescribed precision. Then, in view of (13) and (14) we can numerically compute $H(x)$ using the Cauchy integral

In Figure 7 we present a Mathematica function H that computes $H(x)$ with any prescribed precision. The approximation error can be estimated using Lemma 5.4. One can check that the functions $P(x,t,u)$ and $\Psi(x,t)$ do not vanish for $x<x_0^+$, $|u|=1$ and ${10}/{13}<|t|<{13}/{10}$.

Zoom inZoom out

Figure 7.The Mathematica code for the computation of $H(x)$.

In Figures 8 and 9 we show the image of the annulus ${10}/{13}<|t|<{13}/{10}$ under the map $t\mapsto\Psi(x_0,t)$. Thus we can apply the error estimate (27) for $r=10/13$ and therefore for $a=-\log r/(2\pi)=0.04176$.

Zoom inZoom out

Figure 8.A realistic picture of the image of the circles $|t|={10}/{13}$, $|t|=1$ and $|t|={13}/{10}$ under the map $t\mapsto\Psi(x_0,t)$. The left-hand zoom is stretched in the vertical direction.

Zoom inZoom out

Figure 9.A schematic picture of the image of the upper half-annulus $\{\operatorname{Im} t\geqslant0, {10}/{13}\leqslant|t|\leqslant{13}/{10}\}$ under the map $t\mapsto\Psi(x_0,t)$. The lower half-annulus is mapped symmetrically. The shaded part is the domain ${\{|t|<1\}}$.

When estimating the error of $H(x)$ for $x\approx x_0$, in (27) we can set

§ 5. Approximate solutions of Fredholm integral equations with analytic kernels

5.1. Error estimates. Generalities

The notation in this subsection is independent of the notation in the rest of the paper.

5.2. Error estimates for approximate solutions of Fredholm’s equations

Let $\varphi\colon\mathbb{R}\to\mathbb{C}$ be a continuous solution of Fredholm’s integral equation

Given a positive integer $n$, let us see what happens if we replace the integral in (22) by the $n$th integral sum. Namely, consider the vectors $\varphi^{[n]}=(\varphi_1^{[n]},\dots,\varphi_n^{[n]})$ and $f^{[n]}=(f_1^{[n]},\dots,f_n^{[n]})$, and the $n\times n$ matrix $K^{[n]}=(K_{jk}^{[n]})_{jk}$ defined by

We define the norms $\|\cdot\|_p$ on $\mathbb{C}^n$, $1\leqslant p\leqslant\infty$, in the usual way. For a square matrix $A=(a_{jk})_{jk}$ with complex entries we set

5.3. A numerical criterion for the existence and uniqueness of solutions

Here we keep the above assumptions about $K(x,y)$ and $f(x)$, except that we do not assume a priori any longer that equation (22) has a continuous solution $\varphi$.

Let $\mathcal K\colon\mathcal C([0,1])\,{\to}\,\mathcal C([0,1])$ be the Fredholm integral operator with kernel $K(x,y)$, that is, the operator $\varphi\mapsto\psi$ where $\displaystyle\psi(x)=\int_0^1 K(x,y)\varphi(y)\,dy$.

5.4. Analyticity of solutions with respect to a parameter

Let $\Lambda$ be a domain in $\mathbb{C}$, and let $U=\{z\in\mathbb{C}\mid -a\leqslant\operatorname{Im} z\leqslant a\}$, where $a>0$. Let $K(\lambda,x,y)$ be an analytic function in a neighbourhood of $\Lambda\times U^2$ in $\mathbb{C}^3$ and $f(\lambda,y)$ be an analytic function in a neighbourhood of $\Lambda\times U$ in $\mathbb{C}^2$. We assume that $K(\lambda_0,x,y)$ is $(1,1)$-biperiodic and $f(\lambda_0,x)$ is $1$-periodic for any fixed $\lambda_0\in\Lambda$.

For $\lambda\in\Lambda$ let $\mathcal K_\lambda\colon\mathcal C([0,1])\to\mathcal C([0,1])$ be the Fredholm integral operator ${\varphi\mapsto\psi}$, where $\displaystyle\psi(x)=\int_0^1 K(\lambda,x,y)\varphi(y)\,dy$. The next lemma follows directly from Fredholm’s results in his seminal paper [2] (a more general fact was proved in [11]).

§ 6. Nonprimitive lattice triangulations

Denote the number of all (not necessarily primitive) lattice triangulations of the $m\times n$ rectangle by $f^{\mathrm{np}}(m,n)$ and set

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