On Ehrhart polynomials of lattice triangles

The Ehrhart polynomial of a lattice polygon P is completely determined by the pair (b(P),i(P)) where b(P) equals the number of lattice points on the boundary and i(P) equals the number of interior lattice points. All possible pairs (b(P),i(P)) are completely described by a theorem due to Scott. In this note, we describe the shape of the set of pairs (b(T),i(T)) for lattice triangles T by finding infinitely many new Scott-type inequalities.


Introduction
A lattice polygon P ⊆ R 2 is the two-dimensional convex hull of finitely many lattice points, i.e.,points in Z 2 . Two lattice polygons are equivalent if they are mapped onto each other by an affine-linear automorphism of R 2 which maps Z 2 onto Z 2 . Let b = b(P ) (resp. i = i(P )) be the number of lattice points contained in the boundary (resp. in the interior) of P .
The Ehrhart polynomial of P is given by |(kP ) ∩ Z 2 | = a(P )k 2 + b(P ) 2 k + 1 (for k ∈ Z ≥0 ). We refer to the textbook [BR07]. Therefore, the study of Ehrhart polynomials of lattice polygons reduces to the study of the set P of tuples (b(P ), i(P )) for lattice polygons P . In 1976 Scott showed the following result: Theorem 1.1 (Scott). For a lattice polygon P with i(P ) ≥ 1 either (b(P ), i(P )) = (9, 1) or b(P ) ≤ 2i(P ) + 6 holds.
As described in [HS09], this implies a complete description of Ehrhart polynomials of lattice polygons: In this note, we investigate the subset T ⊆ P of tuples (b(T ), i(T )) for lattice triangles T ⊆ R 2 . Since in Ehrhart theory results are often reduced to the case of lattice simplices, we are interested in understanding what this reduction means for the set of Ehrhart polynomials in the simplest case of dimension two. Rather surprisingly, it turns out that the structure of the set T is richer than one might have expected, as the reader can see in Figure 1 and Figure 4. While the structure of T is still to be fully understood, we explain here the appearance of the conspicuous "spikes".
For this, let us introduce the following open affine cones (see Figure 1 and Figure 4).
. It is straightforward to check that the closures of these cones are pairwise disjoint. In this notation, Scott's theorem (Theorem 1.1) is equivalent to the statement (P \ {(9, 1)}) ∩ σ • 1 = ∅. As suggested by Figure 1, the complement of T contains infinitely many more components.
. For c ∈ Z ≥1 let σ c be the translate of the closure of σ • c so its apex is at the origin. As the cones σ c cover the positive orthant, we see that there are no two-dimensional open affine cones in R 2 ≥0 that are disjoint from all of the cones σ • c . Remark 1.4.
(3) The reader might notice many missing lines of slope −1/2 in Figure 4. The reason is that there are only two lattice triangles of prime normalized volume 2a. More precisely, for odd primes p we have T ∩ i = − 1 2 b + p 2 + 1 = (3, p−1 2 ), (p + 2, 0) , for details see [Öbe15]. This follows also from Higashitani's study of lattice simplices with prime normalized volume [Hig14].
(4) From the pictures, it is visible that the points of T in the "spikes" form periodic patterns. It seems to be an interesting open question to make this observation precise.
(5) Scott's inequality (Theorem 1.1) follows from inequalities by del Pezzo and Jung [Jun90,DP87] (see also [Sch00]): any rational projective surface with degree d and sectional genus p > 0 satisfies d ≤ 4p + 4 if (d, p) = (9, 1). Now, Theorem 1.3 can be translated into algebraic geometry as follows: there exists no toric projective surface with Picard number one, degree d, and sectional genus p that satisfies 2(c+1) c (c + p) < d < 2c c−1 p for an integer c ≥ 2. It would be interesting to see whether this is a special case of a more general algebro-geometric statement.
We remark that in an upcoming paper of the first author and Higashitani Theorem 1.3 will be used as the base case of a generalization for lattice simplices of dimension greater than two.

Acknowledgments
This project was originally inspired by an email correspondence with Tyrrell McAllister. The authors would also like to thank Gabriele Balletti for his supply of computational data, e.g., Figure 1 and Figure 4 is due to him, as well as many fruitful discussions. Theorem 1.3 proves a conjecture in the bachelor's thesis of the third author [Öbe15], advised by the second author at Stockholm University. The second author is an affiliated researcher with Stockholm University and partially supported by the Vetenskapsrådet grant NT:2014-3991.

Proof of Theorem 1.3
Our proof uses the ideas of Scott's original proof of Theorem 1.1. For the convenience of the reader we will give complete arguments without assuming prior knowledge of [Sco76]. Let T be a lattice triangle with area a, number of boundary lattice points b, and number of interior lattice points i. We assume that T satisfies for some c ≥ 2 the inequalities We will show that this situation cannot exist.
By replacing T with an equivalent lattice triangle, we may assume that T is contained in a bounding box (i.e.,a rectangle whose edges are parallel to the coordinate axes and that is minimal with respect to the inclusion of T ) with vertical side length p such that p is minimal among all such choices. For an illustration, see Figure 2. Let us note that p equals the lattice width of T . We denote the horizontal side length of the bounding box by p . We observe that necessarily p ≥ p since switching coordinates yields an equivalent triangle. For this, we denote by δ the minimal distance of the x-coordinate of a vertex of T (denoted by v 2 ) on the top edge of the rectangle from the x-coordinate of a vertex of T (denoted by v 1 ) on the bottom edge of the rectangle. By an integral, unimodular shear leaving the horizontal line through the bottom edge of the rectangle invariant, we can achieve δ ≤ p 2 . By possibly flipping along the horizontal or vertical axis, we may also assume that v 2 has x-coordinate greater than or equal to that of v 1 , and the third vertex of T (denoted by v 3 ) has x-coordinate greater than or equal to that of v 2 (recall that p ≥ p > δ). Now, we move the bottom vertex v 1 horizontally to the right until it has the same x-coordinate as that of v 2 . We observe that the area of the obtained triangle is bounded by the area of T (see Figure 3), i.e., where for the second inequality we used δ ≤ p 2 and p ≥ p. This finishes the proof of (3). As each horizontal line between y = 0 and y = p cuts the boundary of T in two points (see Figure 2), we obtain We reformulate the inequality on the right hand side of (2) as 2a c+1 + 2(c + 1) < b. By Pick's Theorem (1), we have a ∈ 1 2 N, and thus the strict inequality becomes 2a+1+2(c+1) 2 c+1 ≤ b. Combining this with (4) gives Let us assume q = 0. Plugging in (3) yields ≤ 2p ⇒ p 2 − 4(c + 1)p + 2 + 4(c + 1) 2 ≤ 0.
Let us translate the left bottom vertex of T into the origin. By applying an integral, unimodular shear leaving the horizontal line through the bottom edge of the rectangle invariant, we can get an equivalent triangle such that the x-coordinate of the top vertex of T is in [0, p). As p ≥ p, this implies q = p , e.g., as in the left example of Figure 2.
A straightforward computation shows that this is only possible for c ≤ 2. As c ≥ 2, we deduce c = 2. Plugging this again into (8) we obtain 28 < q 2 < 36, a contradiction.