Lattice 3-polytopes with few lattice points

LATTICE 3-POLYTOPES WITH FEW LATTICE POINTS

arXiv:1409.6701v2 [math.CO] 26 Mar 2015

´ MONICA BLANCO AND FRANCISCO SANTOS Abstract. This paper is intended as a first step in a program for a full algorithmic enumeration of lattice 3-polytopes. The program is based in the following two facts, that we prove: • For each n there is only a finite number of (equivalence classes of) 3polytopes of lattice width larger than one, where n is the number of lattice points. Polytopes of width one are infinitely many, but easy to classify. • There is only a finite number of 3-polytopes of lattice width larger than one that cannot be obtained by either glueing two smaller ones or by lifting in a very specific manner one of a list of five lattice 2-polytopes. The 3-polytopes in this finite list have at most 11 lattice points. For n = 4, all empty tetrahedra have width one (White). For n = 5 we here show that there are exactly 9 different 3-polytopes of width 2, and none of larger width. Eight of them are the clean tetrahedra previously classified by Kasprzyk and (independently) Reznick. In a subsequent paper, we show that for n = 6 there are 74 polytopes of width 2, two of width 3, and none of larger width.

Contents 1. Introduction 2. Finiteness of lattice d-polytopes via size and width 3. Preliminaries on lattice 3-polytopes 4. A structure theorem for 3-polytopes of size five 5. Classification (of polytopes with five lattice points) References

1 5 8 12 17 23

1. Introduction A lattice polytope is the convex hull of a finite set of points in Zd (or in a d-dimensional lattice). A polytope is d-dimensional if it contains d + 1 affinely independent points. We call size of P its number #(P ∩ Zd ) of lattice points and volume of P its volume normalized to the lattice:   1 ... 1 vol(conv{p1 , . . . , pd+1 }) := det p1 . . . pd+1 2000 Mathematics Subject Classification. 52B10, 52B20. Key words and phrases. Lattice polytopes, unimodular equivalence, lattice points, finiteness. Supported by grants MTM2011-22792 (both authors); BES-2012-058920 of the Spanish Ministry of Science and the European Science Foundation within the ACAT Project (M. Blanco); Alexander von Humboldt Foundation (F. Santos). 1

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´ MONICA BLANCO AND FRANCISCO SANTOS

with pi ∈ Zd . That is, a tetrahedron has volume one if and only if its vertices are an affine lattice basis. Let f : Rd → R be an affine functional such that f (Zd ) ⊂ Z. The integer maxx∈P f (x)−minx∈P f (x) is called the width of P with respect to f . The minimum width among all possible (non-constant) choices of f will be simply called the width of P . Hence, P has width one if its vertices lie in two consecutive parallel hyperplanes of the lattice. Two lattice polytopes P and Q are said Z-equivalent or unimodularly equivalent if there is an affine unimodular transformation t : Rd → Rd that preserves the lattice (f (Zd ) = Zd ) and with t(P ) = Q. We call such a transformation a Z-equivalence. Volume, width, and size are obviously invariant modulo Z-equivalence. We are interested in the complete classification of lattice polytopes in dimension 3, for small size. In dimension 2, once we fix an n ∈ N, there are finitely many Z-equivalence classes of lattice 2-polytopes of size n: For example, using Pick’s formula (1)

vol(P ) = n + i − 2 ≤ 2n − 5,

where i ≤ n − 3 is the number of lattice points in the interior of P , one can show that every polygon of size n is equivalent to one contained in conv({(−2n+6, −2n+ 6), (−2n + 6, 4n − 11), (4n − 11, −2n + 6)}. Another (more efficient) approach is to observe that every 2-polytope of size n + 1 can be obtained by adding to a 2polytope of size n one point at lattice distance one from (at least) one facet. This iterative process should allow the reader to easily check that Figure 1 contains the full list of 2-dimensional polygons with up to five points.

Figure 1. The 2-dimensional polygons with up to five points In dimension 3 the situation is completely different. In particular, there are empty tetrahedra with arbitrarily large volume (see Theorem 3.4). Here, an empty tetrahedron is the same as a 3-polytope of size four: a tetrahedron with integer vertices and no other integer points. Hence, no analogue of Pick’s Theorem is possible and there is an infinite number of Z-equivalence classes of 3-polytopes of each given size n ≥ 4, as was observed in [8]. Although this may seem to imply that there is no point in “classifying” lattice 3-polytopes by their number of lattice points, we show that such a classification does make sense. In fact, in Section 2 we prove that: Theorem 1.1 (Corollary 2.1). For each n ≥ 4, there exist finitely many lattice 3-polytopes of width greater than one and size n.

LATTICE 3-POLYTOPES WITH FEW LATTICE POINTS

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Hence it makes sense to classify, for each size n, separately the 3-polytopes of width one and those of width larger than one. The latter are a finite list. The former are an infinite set, but easy to describe: they consist of two ≤ 2-polytopes of sizes n1 and n2 (n1 + n2 = n) placed on parallel consecutive lattice planes. For each of the two subconfigurations there is a finite number of possibilities and infinitely many ways to “rotate” (in the integer sense, that is, via an element of SL(Z, 2)) one with respect to the other. For example, it is a now classical result that all empty tetrahedra have width one. From this, the classification of the infinitely many types of empty tetrahedra (stated in Theorem 3.4 below) follows easily. Theorem (White [13, Thm. 1]). Every lattice 3-polytope of size four has width one with respect to (at least) one of its three pairs of opposite edges. However, the task of finding the full list of 3-polytopes of a given size n and of width larger than one becomes harder and harder as n grows. In the first two cases (five points and six points) this paper and [2] contain the full classification, which consists, respectively, of 9 and 76 polytopes (see details below). But if one wants to continue this for larger n, some iterative automated process needs to be devised. A first step in this direction is also developed in Section 2, along the following lines. Suppose that a polytope P of width larger than one contains two proper lattice subpolytopes P1 and P2 of width larger than one with P ∩ Z3 = (P1 ∩ Z3 ) ∪ (P2 ∩ Z3 )

and such that (P1 ∩ Z3 ) ∩ (P2 ∩ Z3 ) is still 3-dimensional. Then, we can think of P as being obtained by gluing P1 and P2 , and there are finitely many possible ways of gluing two given polytopes in this fashion. We will refer to a polytope P of width larger than one and that cannot be obtained in this way as minimal if all proper subpolytopes of P have width one, and quasi-minimal otherwise (see precise definitions in Section 2). We prove that: Theorem 1.2 (Lemma 2.3, Corollary 2.6). There exist infinitely many minimal and quasi-minimal lattice 3-polytopes. However, all but finitely many of them project to one of the following polygons in such a way that all of the vertices in the projection have a unique element in the pre-image.

The rest of the paper is devoted to the full classification of lattice 3-polytopes of size five. After introducing some relevant concepts in Section 3, in Section 4 we prove the following structural result: Theorem 1.3 (Theorem 4.1). Let P be a polytope of size five and width larger than one. Let A = {a1 , . . . , a5 } = P ∩ Z3 . Then there exists an integer affine functional f such that, modulo reordering of the points, one of the following happens: (f (ai ))i=1,...,5 = (1, 1, 0, 0, −1) or (f (ai ))i=1,...,5 = (1, 1, 0, 0, −2). That is, the five lattice points are distributed in three parallel hyperplanes in only two possible ways. Once we have this the classification becomes, in a way,

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a 2-dimensional problem and is performed case by case in Section 5. The global result is: Theorem 1.4. There are exactly nine 3-polytopes of size 5 and width two and none of larger width. Eight of them have signature (4, 1) and one has signature (3, 1). Remember that a set A of d+2 points affinely spanning Rd have a unique (modulo a scalar factor) affine dependence. The signature of a d-polytope of size d+2 is (i, j) if this dependence has i positive coefficients and j negative coefficients. Signatures (i, j) and (j, i) are the same, and the five possible signatures of five points in R3 are (4, 1), (3, 2), (2, 2), (3, 1) and (2, 1) (more details in Section 3). More precisely, our classification says: Theorem 1.5 (Theorems 5.1, 5.2, 5.3, 5.4). Every lattice 3-polytope of size 5 is unimodularly equivalent to one of the configurations listed in Table 1. The table is irredundant: configurations in different rows, or configurations obtained for different choices of parameters within each row, are not Z-equivalent. The table includes, apart from a representative for each class, three invariants of the class: its width, its signature and its volume vector. The volume vector is a vector in Z5 recording the volumes of the (perhaps degenerate) tetrahedra spanned by each subset of four of the five points. We give volume vectors with a sign convention that makes them have as many positive and negative entries as given by the signature. Sign. (2, 2) (2, 1)

(3, 2)* (3, 1)*

(4, 1)*

Volume vector (−1, 1, 1, −1, 0) (−2q, q, 0, q, 0) 0 ≤ p ≤ 2q , gcd(p, q) = 1 (−a − b, a, b, 1, −1) 0 < a ≤ b, gcd(a, b) = 1 (−3, 1, 1, 1, 0) (−9, 3, 3, 3, 0) (−4, 1, 1, 1, 1) (−5, 1, 1, 1, 2) (−7, 1, 1, 2, 3) (−11, 1, 3, 2, 5) (−13, 3, 4, 1, 5) (−17, 3, 5, 2, 7) (−19, 5, 4, 3, 7) (−20, 5, 5, 5, 5)

Width 1

Representative (0, 0, 0), (1, 0, 0), (0, 1, 0), (1, 1, 0),(0, 0, 1)

1

(0, 0, 0), (1, 0, 0), (0, 0, 1), (−1, 0, 0),(p, q, 1)

1

(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1),(a, b, 1)

1 2 2 2 2 2 2 2 2 2

(0, 0, 0), (1, 0, 0), (0, 1, 0), (−1, −1, 0),(0, 0, 1) (0, 0, 0), (1, 0, 0), (0, 1, 0), (−1, −1, 0),(1, 2, 3) (0, 0, 0), (1, 0, 0), (0, 0, 1), (1, 1, 1),(−2, −1, −2) (0, 0, 0), (1, 0, 0), (0, 0, 1), (1, 2, 1),(−1, −1, −1) (0, 0, 0), (1, 0, 0), (0, 0, 1), (1, 3, 1),(−1, −2, −1) (0, 0, 0), (1, 0, 0), (0, 0, 1), (2, 5, 1),(−1, −2, −1) (0, 0, 0), (1, 0, 0), (0, 0, 1), (2, 5, 1),(−1, −1, −1) (0, 0, 0), (1, 0, 0), (0, 0, 1), (2, 7, 1),(−1, −2, −1) (0, 0, 0), (1, 0, 0), (0, 0, 1), (3, 7, 1),(−2, −3, −1) (0, 0, 0), (1, 0, 0), (0, 0, 1), (2, 5, 1),(−3, −5, −2)

Table 1. Complete classification of lattice 3-polytopes of size 5.

Let us mention that part of Theorem 1.4 was already known: • Polytopes of signatures (2, 2) and (3, 2) have width 1 by the following result of Howe [12, Thm. 1.3]: Every lattice 3-polytope with no lattice points other than its vertices has width 1. However, the classification of them included in Table 1 is, as far as we know, new. • Polytopes of signature (4, 1), which are the same as “terminal tetrahedra” or “clean tetrahedra with a single interior point”, were classified by

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Kasprzyk [6] and Reznick [10, Thm. 7], who obtained the same list as us. In this sense, Sections 5.3 and 5.4 are only reworking the known classification of them. Still, we have preferred to include them for completeness and because our methods differ from the ones in those papers. Our motivation comes partially from the notion of distinct pair-sums lattice polytopes (or dps polytopes, for short), defined as lattice polytopes in which all the pairwise sums a + b, a, b ∈ P ∩ Zd , are distinct. Equivalently, they are lattice polytopes containing no three collinear lattice points nor the vertices of a non degenerate parallelogram [3, Lemma 1]. They are also the lattice polytopes of Minkowski length equal to one, in the sense of [1]. Dps d-polytopes cannot have two lattice points in the same class modulo (2Z)d , hence they have size at most 2d . For d-polytopes of size d + 2, dps is equivalent to the signature not being (2, 1) nor (2, 2). Hence, the nine polytopes of width greater than one stated in Theorem 1.4 are all dps. Those marked in Table 1 with an * are dps. In a subsequent paper [2] we undertake the complete classification of 3-polytopes of size six and width greater than one, obtaining: Theorem 1.6 ([2]). There are 76 3-polytopes of size 6 and width > 1. Two of them have width 3 and the rest width 2. One and 44 of those, respectively, are dps. In future work we hope to at least classify the dps polytopes of sizes up to 8. Since dps polytopes in dimension three have at most 8 lattice points, this would complete the classification of dps 3-dimensional lattice polytopes, which should help us answer, among others, the following question presented in [3, page 6]: Is every dps 3-polytope a subset of a dps 3-polytope of size 8? Acknowledgment: We thank Bruce Reznick for pointing us to useful references on the topic of this work. 2. Finiteness of lattice d-polytopes via size and width As said in the introduction, for every d ≥ 3 and for every n ≥ d + 1 there are infinitely many classes of lattice d-polytopes of size n [8, Theorem 4]. It is known, however, that this cannot happen if we look only at polytopes containing some interior lattice point: Theorem (Hensley [5, Thm. 3.6]). For each k ≥ 1 there is a number V (k, d) such that no lattice d-polytope with k interior lattice points has volume above V (k, d). Theorem (Lagarias-Ziegler [7, Thm. 2]). For each V ∈ N there is only a finite number of Z-equivalence classes of d-polytopes with volume V or less. Lattice polytopes with no lattice points in their interior are called hollow. For hollow polytopes, although infinitely many for each size, we still have: Theorem (Nill-Ziegler [9, Thm. 1.2]). There is only a finite number of hollow d-polytopes that do not admit a lattice projection onto a hollow (d − 1)-polytope. Combining these three statements with the fact that there is a unique hollow 2-polytope of width larger than one, we get:

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Corollary 2.1. For each n ≥ 4, there are finitely many lattice 3-polytopes of width greater than one and size n. Proof. Once we fix n, every lattice 3-polytope P with n lattice points falls in one of the following (not mutually exclusive) categories: • It is not hollow. In this case Hensley’s Theorem gives a bound for its volume. This, in turn, implies finiteness via the Lagarias-Ziegler Theorem. • It is hollow, but does not project to a hollow 2-polytope. These are a finite family, by the Nill-Ziegler Theorem. • It is hollow, and it projects to a (hollow) 2-polytope of width 1. This implies that P itself has width 1. • It is hollow and it projects to a hollow 2-polytope of width larger than one. The only such 2-polytope is the second dilation of a unimodular triangle. It is easy to check that only finitely many (Z-equivalence classes of) 3polytopes of size n project to it: let P = conv{p1 , ..., pn } be a 3-polytope of size n that projects onto T = conv{(0, 0), (2, 0), (0, 2)}. We must have at least one point projecting to each vertex of T . That is: there are p1 = (0, 0, z1 ), p2 = (2, 0, z2 ) and p3 = (0, 2, z3 ) in P . The unimodular transformation      z3 − z1 z2 − z1 −y (x, y, z) 7→ x, y, z − z1 − x 2 2 allows us to assume that z1 , z2 , z3 ∈ {0, 1}. This implies that P ⊂ T × [1 − n, n], so there is a finite number of possibilities for P .  Still, it is not clear how to find all the (finitely many) 3-polytopes of width larger than one for each given size n. In this paper and the subsequent one [2] we do this for n = 5, 6, but for larger size some automated process would be necessary. We propose one based on the following ideas. For each vertex v in a lattice d-polytope P ⊂ Rd we denote P v := conv((P ∩ d Z ) \ {v}). In the rest of this section we assume P to have width greater than one. Let vert(P ) be the set of all vertices and vert∗ (P ) ⊆ vert(P ) be the set of vertices of P such that P v is either (d − 1)-dimensional or has width one. Definition 2.2. Let P a lattice d-polytope P of width > 1. We say that P is minimal if vert∗ (P ) = vert(P ) and quasi-minimal if # vert∗ (P ) = # vert(P ) − 1. For example, since all 3-polytopes of size 4 have width one, all 3-polytopes of size 5 and width > 1 are minimal. Our interest in (quasi)-minimal polytopes, especially those of dimension three, comes from the following observation: We know that there are finitely many 3polytopes of width larger than one and size bounded by a certain number n. If we can classify all minimal and quasi-minimal ones, then we can easily construct the rest: each of them can be constructed as the (convex hull of the) union of two smaller polytopes of width larger than one. One problem with this approach is that there are infinitely many minimal and quasi-minimal 3-polytopes. Lemma 2.3. There exist infinitely many minimal 3-polytopes and infinitely many quasi-minimal 3-polytopes.

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Proof. The polytope conv{(1, 0, 0), (−1, 0, 0), (0, −1, k), (0, 1, k)} is minimal with k − 1 interior points and volume 4k for all k ∈ N. P (1,0,0) and P (−1,0,0) have width 1 with respect to x. P (0,1,k) and P (0,−1,k) have width 1 with respect to y. Adding the vertex (0, 0, k + 1) the same happens, but now the width of P (0,0,k+1) is two.  However, the infinite families of minimal and quasi-minimal 3-polytopes can easily be understood, thanks to the following structure theorem in arbitrary dimension: Theorem 2.4. Let P be a minimal or quasi-minimal d-polytope. Then one of the following happens: (1) P projects to a minimal or quasi-minimal (d − 1)-polytope P 0 in such a way that every vertex of P 0 has a unique preimage in P and the projection bijects vert∗ (P ) to vert∗ (P 0 ). (2) P has at most 2d + d lattice points. Proof. For each vi ∈ vert∗ (P ) let fi be an affine integer functional giving width one to P v . Two things can happen: (1) If the set {fi : vi ∈ vert∗ (P )} does not linearly span (Rd )∗ , then there is a line where all the fi are constant. Let P 0 = π(P ) be the projection of P in the direction of that line. Since P has width greater than one, so does P 0 . Each vertex of P 0 must be the projection of a unique vertex of P , because π(u) = π(v) implies π(P u ) = π(P v ) = π(P ), which has width two, a contradiction with the fact that all but at most one vertex of P is in vert∗ (P ). Moreover, if π(v) ∈ vert∗ (P 0 ) then P 0π(v) has width one, which implies v P has width one, that is, v ∈ vert∗ (P ). Conversely, if v ∈ vert∗ (P ) then π(P v ) has width one with respect to the functional fi . Thus, π(vert∗ (P )) = vert∗ (P 0 ). Since vert(P 0 ) ⊆ π(vert(P )) and P has at most one vertex not in vert∗ (P ), the same holds for P 0 . That is, P 0 is minimal or quasi-minimal. (2) If the set {fi : vi ∈ vert∗ (P )} linearly spans (Rd )∗ , then assume without loss of generality that f1 , . . . , fd are independent. Then all lattice points of P except for v1 , . . . , vd are contained in the parallelepiped d \

fi−1 ([ki , ki + 1]),

i=1

where [ki , ki + 1], with ki ∈ Z, is the interval containing fi (P vi ). This parallelepiped may not be a lattice polytope, but all its lattice points are vertices. Hence, the set of lattice points in P is contained in the 2d vertices of the parallelepiped plus the d vertices v1 , . . . , vd .  In dimension one the only minimal polytope is a segment of length two, and there is no quasi-minimal one. In dimension two we have: Lemma 2.5. lowing:

• Every minimal 2-polytope is Z-equivalent to one of the fol-

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• Every quasi-minimal 2-polytope is Z-equivalent to one of the following:

Proof. By Theorem 2.4, all minimal or quasi-minimal 2-polytopes with more than six lattice points are quasi-minimal triangles like the two in the statement. An exhaustive search proves the list for lattice polytopes with up to 6 points.  In dimension 3: Corollary 2.6. Let P be a lattice 3-polytope with more than 11 lattice points. (1) If P is quasi-minimal then it projects to one of the following 2-polytopes in such a way that all of the vertices in the projection have a unique element in the pre-image.

(2) If P is minimal then it projects to the second 2-polytope in the figure in such a way that all of the vertices in the projection have a unique element in the pre-image. In particular, these two cases cover all but finitely many minimal or quasi-minimal lattice 3-polytopes. Proof. The last sentence in the statement comes from the fact that there are only finitely many lattice 3-polytopes of width greater than one with 11 lattice points or less (Corollary 2.1). By Theorem 2.4, every minimal or quasi-minimal 3-polytope P with more than 11 lattice points projects to one of the minimal or quasi-minimal 2-polytopes. It cannot project to one of the quasi-minimal triangles, since those have # vert∗ (P ) ≥ 3 > 2 = # vert∗ (P 0 ). In case P is minimal, the lemma also implies that the projection P 0 has the same number of vertices as P (hence, P 0 is not a triangle) and that it is minimal. 

3. Preliminaries on lattice 3-polytopes In this section we review some concepts that are needed in the classification of lattice 3-polytopes. All the contents are either known or their proofs can be considered routine.

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3.1. Volume vectors. Since Z-equivalence preserves volume, the following is a useful invariant for our classification: Definition 3.1. Let A = {p1 , p2 , . . . , pn }, with n ≥ d + 1, be a set of lattice points in Zd . The volume vector of A is the vector n w = (wi1 ...id+1 )1≤i1 d + 1 the affine dependence on {p1 , . . . , pd+1 , pi } (which is encoded in the volume vector of A) allows to write pi as an affine combination of {p1 , . . . , pd+1 }. Since f preserves affine combinations, f (pi ) = qi . This finishes the proof of part (1). For part (2), let Λ(A), Λ(B) ≤ Zd be the affine sublattices spanned respectively by A and B. Since f maps A to B, it maps Λ(A) to Λ(B). The index [Zd : Λ(A)] is the minimal volume (with respect to Zd ) of a basis of Λ(A). Thus the index
[n] divides wI for all I ∈ d+1 , and therefore it divides gcd(wI )I . In particular, if gcd(wI )I = 1, then Λ(A) = Zd = Λ(B). This implies f maps Zd to itself, so it has integer coefficients.  From the volume vector of a point configuration A with n ≥ d + 1 points, we can recover the volume vector of any subconfiguration. Let us look at the case of subconfigurations with d + 2 points. Remember that if d + 2 points {p1 , . . . , pd+2 } affinely span Rd then they have a unique (modulo a scalar factor) affine dependence. The volume vector of d + 2

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points encodes its dependence as follows: let Ik = {1, . . . , d + 2} \ {k} (3)

d+2 X

(−1)k−1 · wIk · pk = 0,

k=1

d+2 X k=1

(−1)k−1 · wIk = 0.

The points with non-zero coefficient in this dependence form a circuit. The signature of the circuit is the pair (i, j) if this dependence has i positive and j negative coefficients. (See more details in [4]). We call signature of the d + 2 points the signature of its (unique) circuit; (i, j) and (j, i) are the same signature. Remark 3.3. For five points A = {p1 , . . . , p5 } ⊂ Rd , the signature (i, j) of A can be (2, 1), (2, 2), (3, 2), (3, 1) or (4, 1), depicted in Figure 2.

(2, 1)

(2, 2)

(3, 2)

(3, 1)

(4, 1)

Figure 2. The five possible signatures (oriented matroids) of five different points in R3 In this situation we modify our sign and order conventions for writing the volume vector, in order to make the signature (and its correspondence to subsets of A) more explicit. More precisely, we take as volume vector for five points p1 , . . . , p5 the vector (v1 , v2 , v3 , v4 , v5 ) where X X vi pi = 0, vi = 0 is the unique affine dependence on A, normalized so that |vi | = vol(conv(A \ {i})). In particular, this way the signature of A equals the number of positive and negative entries in the volume vector. Put differently (see Equation (3)): (v1 , v2 , v3 , v4 , v5 ) = (w2345 , −w1345 , w1245 , −w1235 , w1234 ) where wijkl is as in Equation (2). 3.2. Empty tetrahedra. To review the classification of empty tetrahedra in dimension three one can take as a starting point the fact that every empty tetrahedron has width one with respect to an affine integer functional that is constant in two opposite edges. This easily implies: Theorem 3.4 (Classification of empty tetrahedra, White [13]). Every empty lattice tetrahedron of volume q is unimodularly equivalent to the following T (p, q), for some p ∈ {1, . . . , q} with gcd(p, q) = 1: T (p, q) = conv{(0, 0, 0), (1, 0, 0), (0, 0, 1), (p, q, 1)} Moreover, T (p, q) is Z-equivalent to T (p0 , q) if and only if p0 = ±p±1 (mod q). Definition 3.5. We say that an empty tetrahedron T in a 3-dimensional lattice Λ is of type T (p, q) if there is a lattice isomorphism Λ ∼ = Z3 sending T to T (p, q).

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Among the methods we use to elaborate the classification, we often have a list of tetrahedra and we need to check that they are empty. Of course, one necessary condition is that all facet triangles are empty, which is equivalent to “unimodular in the lattice planes containing them”. Once we know the three points of a unimodular facet of our tetrahedron T , then there is an affine transformation sending these three points to (0, 0, 0), (1, 0, 0) and (0, 1, 0); let (a, b, q) be the image of the fourth point. (Note: the transformation is not unique, but q is unique up to a sign, since it equals the volume of T , and a and b are determined modulo q). Then, our problem reduces to knowing when conv{(0, 0, 0), (1, 0, 0), (0, 1, 0), (a, b, q)} is an empty tetrahedron. The following lemma gives us the answer: Lemma 3.6. The lattice tetrahedron T = conv{(0, 0, 0), (1, 0, 0), (0, 1, 0), (a, b, q)} is empty (with respect to the integer lattice Z3 ) if, and only if, at least one of the following happens: (i) a ≡ 1 (mod q) and gcd(b, q) = 1. (ii) b ≡ 1 (mod q) and gcd(a, q) = 1. (iii) a + b ≡ 0 (mod q) and gcd(a, q) = 1. Proof. Assume without loss of generality that q > 0. T is an empty tetrahedron if, and only if, all its edges are primitive and its width equals one with respect to one of the three pairs of edges. In our case, primitivity of edges is equivalent to (4)

gcd(a, b, q) = gcd(a − 1, b, q) = gcd(a, b − 1, q) = 1

Let us examine the width with respect to the three pairs of edges, depending on the values of a and b. For this, we compute the linear functional taking value 0 on one edge and value 1 on the other. Width is one with respect to that pair of edges if and only if this functional is integer. (i) The functional taking value 0 on the segment (0, 0, 0)(0, 1, 0) and value 1 on (1, 0, 0)(a, b, q) is x + 1−a q z, which is integer if and only if a ≡ 1 (mod q). If this happens, the primitivity conditions (4) become equivalent to gcd(b, q) = 1. (ii) The functional with value 0 on (0, 0, 0)(1, 0, 0) and value 1 on (0, 1, 0)(a, b, q) is y + 1−b q z, which is integer if and only if b ≡ 1 (mod q). If this happens, the primitivity conditions become gcd(a, q) = 1. (iii) The functional with value 0 on (0, 0, 0)(a, b, q) and value 1 on (1, 0, 0)(0, 1, 0) is x + y − a+b q z which is integer if and only if a + b ≡ 0 (mod q). If this happens, the primitivity conditions become gcd(a, q) = 1.  For future reference we include the following statement which can be read as “no vertex of an empty tetrahedron is more special than the others”. Lemma 3.7. Let T (p, q) = conv{p0 = (0, 0, 0), p1 = (1, 0, 0), p2 = (0, 0, 1), p3 = (p, q, 1)} be the standard simplex of type T (p, q), for some 1 ≤ p ≤ q. Then, for each i ∈ {1, 2, 3} there exists a Z-equivalence ti sending pi to p0 and mapping T (p, q) to itself or to T (p0 , q), where p0 ≡ p−1 mod q. Proof. We give the explicit transformation for each case:

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

−1 t1 (x, y, z) =  0 0  0 p t2 (x, y, z) =  q 0  −p0  t3 (x, y, z) = −q 0

    p−1 x 1 q  y  +  0  1 z 0      0 −pp +1 x 0 0 q −q 0  y  +  0  z 1 0 −1      0 pp −1 x p0 1 − p0 q p −q   y  +  q  z 1 0 −1 0 −1 0

Observe that t1 maps T (p, q) to itself, while t2 and t3 map T (p, q) to T (p0 , q).



Remark 3.8. Let us consider, as in the proof of this lemma, the standard tetrahedron of volume q, T (p, q) = {(0, 0, 0), (1, 0, 0), (0, 0, 1), (p, q, 1)}. The transformation t1 in the proof (exchanging (0, 0, 0) ↔ (1, 0, 0) and (0, 0, 1) ↔ (p, q, 1)) is the only unimodular transformation, other than the identity, sending T (p, q) to itself for every p and q. The other 22 affine automorphisms of T (p, q) are automorphisms of Z3 only for particular values of (p, q). This means that the sentence “no vertex of an empty tetrahedron is more special than the others” is not true if we fix a particular class T (p, q) ⊂ Z3 of simplices. If we want to stay within a particular class T (p, q), and in this class p 6≡ p−1 (mod q), then the vertices (0, 0, 0) and (1, 0, 0) are in one orbit of the unimodular automorphism group of T (p, q) and (0, 0, 1) and (p, q, 1) in another.

4. A structure theorem for 3-polytopes of size five In this section we will prove the following theorem (see more precise statements in Theorems 4.4 and 4.5): Theorem 4.1. Let P be a lattice 3-polytope of size 5, and let A = P ∩ Z3 be the set of its lattice points. • If P has signature (2, 2), (2, 1) or (3, 2), then it has width one. • If P has signature (3, 1) or (4, 1), then there exists an affine integer functional taking values (1, 1, 0, 0, h) in A, where h ∈ {−1, −2} and the first four points form the empty subtetrahedron of largest volume. 4.1. A convenient change of coordinates. In order to prove the previous theorem, it will be convenient to make a change of coordinates so that instead of having a tetrahedron of volume q with respect to Z3 we have a tetrahedron whose vertices span Z3 as an affine lattice but we consider it with respect to a finer lattice. A similar transformation achieving this is worked out, for example, in [11]. Theorem 4.2. Let 1 ≤ p ≤ q with gcd(p, q) = 1. The tetrahedron T0 = conv{o = (0, 0, 0), e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1)}, is an empty tetrahedron of type T (p, q) with respect to the lattice Λ(p, q) := h(1/q, −1/q, p/q)i + Z3 .

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Proof. Let T be an empty tetrahedron of type T (p, q) in the lattice Z3 . By Theorem 3.4, we can choose without loss of generality: T = T (p, q) = conv{(0, 0, 0), (1, 0, 0), (0, 0, 1), (p, q, 1)} If q = 1, then Λ(1, 1) = Z3 and the statement is trivial. So assume that q > 1. Let λ be the linear transformation sending T to T0 in the following order: (0, 0, 0) 7→ o,

(1, 0, 0) 7→ e3 ,

(0, 0, 1) 7→ e1 ,

(p, q, 1) 7→ e2 .

3

Let us see that λ is a lattice isomorphism between Z and Λ(p, q). The image Λ of the lattice Z3 by this transformation is clearly a superlattice of 3 Z of index q. Since T has width one with respect to the edges (0, 0, 0)(1, 0, 0) and (0, 0, 1)(p, q, 1), we have that T0 has width one with respect to the functional x + y in the lattice Λ. In other words: Λ ⊂ {(x, y, z) ∈ R3 : x + y ∈ Z}. Let Hi := {(x, y, z) ∈ R3 : x + y = i}, for each i ∈ Z. The different slices Λ ∩ Hi are obtained from one another by integer translation, so understanding one of them is enough to understand Λ. That is, Λ = Z3 + Λ0 , where Λ0 := Λ ∩ H0 = {(x, y, z) ∈ Λ(p, q) : x + y = 0}. We have that Λ0 is a superlattice of Z3 ∩ H0 with index q. Let us consider the rectangle R := conv{(0, 0, 0), (0, 0, 1), (1, −1, 0), (1, −1, 1)},

which is a fundamental rectangle (has area 2) with respect to Z3 ∩ H0 . Then R has area 2q with respect to Λ0 . Again we can say that Λ0 = Z3 + R, hence Λ = Z3 + R. Λ contains no non-integer points on the edges of R, since each primitive integer segment along these lines is a lattice translation of 0e3 or e1 e2 . For the same reason, no two interior points of R can have one coordinate equal, because otherwise we would have a horizontal or vertical lattice segment in Λ of length smaller than one, in contradiction to the fact that oe3 and e1 e2 are primitive in Λ. Hence, by Pick’s formula (1), R contains exactly q − 1 lattice points of Λ0 in its interior. As a conclusion, for each i = 1, . . . , q − 1, R contains exactly one point of Λ with z = i/q and one (which may or may not be the same) with x = −y = i/q. We leave to the reader to check that λ((0, 1, 0)) = (−1/q, 1/q, −p/q), so R contains the point (1/q, −1/q, p/q) ∈ Λ. Since gcd(p, q) = 1, all q − 1 points in R can be obtained by integer translations of the multiples of that point, and the result follows: Λ = Z3 + h(1/q, −1/q, p/q)i = Λ(p, q) 

Observe that if p0 = p−1 (mod q) then (p0 /q, −p0 /q, 1/q) ∈ Λ(p, q). That is: Λ(p, q) = h(1/q, −1/q, p/q)i + Z3 = h(p0 /q, −p0 /q, 1/q)i + Z3 .

We call R(p, q) := conv{(0, 0, 0), (0, 0, 1), (1, −1, 0), (1, −1, 1)} the fundamental rectangle of T (p, q). Observe that all lattice points of Λ(p, q) lie in an integer translation of R(p, q), as illustrated in Figure 3. In Section 4.2 we will need the following result about the fundamental rectangle:

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z

y

x

Figure 3. The standard tetrahedron, the lattice Λ(p, q) and the fundamental rectangle. (The picture is for q = 7 and p = 4) Lemma 4.3. Let q ≥ 2 and let p ∈ {1, . . . , q − 1} with gcd(p, q) = 1. Let p0 := p−1 mod q. Let R(p, q) be the fundamental rectangle in Λ(p, q) and consider the following two triangles contained in it: t1 = conv{(0, 0, 0), (1, −1, 0), (1, −1, 1/2)} and t2 = conv{(0, 0, 0), (1, −1, 0), (1/2, −1/2, 1/2)} (Figure 4). Then, the point (p0 /q, −p0 /q, 1/q) ∈ Λ(p, q) lies in t2 for all (p, q) and it lies in t1 if p ∈ {2, . . . , q−2}.

Figure 4. The triangles t1 and t2 (projected to the plane XZ) Simplices of type T (1, q) are somehow special: the fundamental rectangle has all lattice points in the diagonal, and they have width one with respect to two different pairs of opposite edges. They are called tetragonal in [11]. T (1, 1) and T (1, 2) are even more special: they have width one with respect to any of the three pairs of opposite edges. 4.2. Polytopes of signature (2, ∗). It turns out that every lattice 3-polytope P of size five and signature (2, ∗) has width one: Theorem 4.4. If P is a lattice 3-polytope of size 5 and with signature (3, 2), (2, 2) or (2, 1), then P has width one. Proof. Let A = {p1 , p2 , p3 , p4 , p5 } be the lattice point set of a polytope P = conv A of size five and signature (2, ∗). We assume the points are ordered so that the volume vector v = (v1 , v2 , v3 , v4 , v5 ) of A verifies vi ≤ 0 < v4 ≤ v5 , i = 1, 2, 3. Consider the empty tetrahedron T = conv{p1 , p2 , p3 , p4 }, of type T (p, q) for q = v5 and some p ∈ {1, . . . , q} with gcd(p, q) = 1. By Theorem 4.2, we can consider T to be the standard tetrahedron conv{(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1)} in the lattice Λ(p, q). Moreover, by Lemma 3.7 we can assume p4 = (0, 0, 0) and, by

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symmetry of the conditions so far on the points p1 , p2 and p3 , that p1 = (1, 0, 0), p2 = (0, 1, 0), p3 = (0, 0, 1). Observe now that the barycentric coordinates of p5 with respect to T are   −v1 −v2 −v3 −v4 , , , . v5 v5 v5 v5 By our hypotheses, the first three are non-negative and the last one is between −1 and 0. This implies that p5 lies in the homothetic dilation of factor two of T from point p4 .

z

p3 p4 p2

p1 x

(2, 0, 0)

(1, 1, 0)

(0, 2, 0)

y

Figure 5. The second dilation of an empty tetrahedron That is, p5 ∈ 2T , and since we have the lattice Λ(p, q), this implies that A has width one with respect to the functional x + y unless p5 turns out to be one of the three lattice points in 2T with x + y = 2, namely (2, 0, 0), (1, 1, 0) and (0, 2, 0) (see Figure 5). Let us see what happens in these three cases: • If p5 = (2, 0, 0) then the intersection of conv A with the fundamental rectangle contained in 2T is the triangle conv{(1, 0, 0), (0, 1, 0), (1, 0, 1/2)}. By (a translated version of) Lemma 4.3, for this triangle not to contain additional lattice points of Λ(p, q) we need p = 1. But in this case T has width one also with respect to the functional y + z, and so does P . • The case p5 = (0, 2, 0) is analogous, exchanging the roles of x and y. • If p5 = (1, 1, 0) then the intersection of conv A with the fundamental rectangle contained in 2T is the triangle conv{(1, 0, 0), (0, 1, 0), (1/2, 1/2, 1/2)}. Now Lemma 4.3 implies that this triangle has additional lattice points, no matter the value of p, unless q = 1. But if q = 1 then P has width one with respect to z.  4.3. Polytopes of signature (∗, 1). We assume P to be of signature (4, 1) or (3, 1), since polytopes of other signatures are of width one by Theorem 4.4. In this case: Theorem 4.5. Let P be a lattice 3-polytope of size 5 and signature (4, 1) or (3, 1). Let T be the empty lattice tetrahedron of largest volume q ≥ 1 contained in P .

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Then there exists an affine integer functional taking values 1, 1, 0, 0 in T , and h ∈ {−1, −2} in the fifth point. Moreover, P has symmetric volume vector (q, q, q, q, −4q), if and only if it has signature (4, 1) and h = −2. Proof. Let A = {p1 , p2 , p3 , p4 , p5 } be the lattice points in P = conv A and assume the points are ordered so that the volume vector v = (v1 , v2 , v3 , v4 , v5 ) of A verifies v4 < 0 ≤ vi ≤ v5 , i = 1, 2, 3. That is, point p4 lies in the tetrahedron conv{p1 , p2 , p3 , p5 } = conv(A) (this corresponds to the “1” in the signature) and the (empty) tetrahedron T := conv{p1 , p2 , p3 , p4 } has the maximum volume among the empty tetrahedra in A. Consider T = conv{p1 , p2 , p3 , p4 } to be of type T (p, q) for q = v5 and some p ∈ {1, . . . , q} with gcd(p, q) = 1. Again we can assume without loss of generality that T is the standard tetrahedron conv{p4 = (0, 0, 0), p1 = (1, 0, 0), p2 = (0, 1, 0), p3 = (0, 0, 1)} in the lattice Λ(p, q). The barycentric coordinates of p5 with respect to T are   −v1 −v2 −v3 −v4 , , , . v5 v5 v5 v5 Our hypotheses translate to the first three being in [−1, 0], thus p5 is a point in the negative unit cube [−1, 0]3 (see Figure 6). z

p3 y

p2 p4

p1 x

(−1, −1, 0)

(−1, −1, −1)

Figure 6. The negative unit cube Since the functional x+y is integer on Λ(p, q), its value at p5 is one of {0, −1, −2}. If the value of x + y at p5 is 0, the signature is (2, 1). The result then follows for the functional x + y. Finally, P has symmetric volume vector (q, q, q, q, −4q) if and only if p5 = (−1, −1, −1). This is equivalent to h = −2 and signature (4, 1). 

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5. Classification (of polytopes with five lattice points) In Section 5.1 we completely classify 3-polytopes of size 5 and width one. By Theorem 4.4 this covers all signatures except (3, 1) and (4, 1). Hence, in the rest of the section we only need to analyze the cases allowed by Theorem 4.5. For this we do the following: taking the functional of Theorem 4.5 to be the z-coordinate in the lattice Z3 , we assume without loss of generality that the first four points are p1 = (0, 0, 0), p2 = (1, 0, 0), p3 = (0, 0, 1) and p4 = (p, q, 1) (with gcd(p, q) = 1 and 1 ≤ p ≤ q) since they form an empty tetrahedron, so the point p5 will be (a, b, h) with h = −1, −2. We then look at the possibilities for p5 that make conv{p1 , p2 , p3 , p4 , p5 } not to have extra lattice points. Since all such new lattice points would have to have z = 0 or z = −1, this turns our question into a two-dimensional problem, which we solve case by case. 5.1. Polytopes of width 1. The classification of configurations of width one is easy: Theorem 5.1. Let P be a lattice polytope of size five and width one. Then P is unimodularly equivalent to one of the following: (1) conv{(0, 0, 0), (1, 0, 0), (0, 1, 0), (1, 1, 0), (0, 0, 1)}, of signature (2, 2). (2) conv{(0, 0, 0), (1, 0, 0), (0, 1, 0), (−1, −1, 0), (0, 0, 1)}, of signature (3, 1). (3) conv{(0, 0, 0), (1, 0, 0), (0, 0, 1), (−1, 0, 0), (p, q, 1)}, for some p, q ∈ Z with 0 ≤ p ≤ bq/2c and gcd(p, q) = 1. This is of signature (2, 1). (4) conv{(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1), (a, b, 1)}, for some 0 < a ≤ b, with gcd(a, b) = 1. This is of signature (3, 2). Moreover, two such polytopes are never Z-equivalent to one another. Proof. Width one means the 5 lattice points of P lie in two consecutive lattice planes. Say n0 points are in {z = 0} and 5 − n0 in {z = 1} with n0 ≥ 5 − n0 . That is, n0 ∈ {3, 4}.

• If n0 = 4, then the four points at z = 0 form one of the three 2-dimensional polytopes of size 4 displayed in the right top row of Figure 1, of signatures (3, 1), (2, 2) and (2, 1) respectively, and the position of the fifth point (within the plane z = 1) does not affect the Z-equivalence class of P . The cases (2, 2) and (3, 1) are the configurations of parts (1) and (2). The case (2, 1) is the configuration conv{(0, 0, 0), (1, 0, 0), (−1, 0, 0), (0, 1, 0), (0, 0, 1)} which, via the map (x, y, z) 7→ (x, y, y + z), becomes that of part (3) with (p, q) = (0, 1). • If n0 = 3, then P has three points in the lattice plane z = 0, and two in the next plane z = 1. There are two possibilities: – If the three points at z = 0 are collinear, without loss of generality we can assume they are (−1, 0, 0), (0, 0, 0) and (1, 0, 0). One of the points at z = 1 can be assumed to be (0, 0, 1) and the fifth point has coordinates (p, q, 1) with q 6= 0 (in order to be full-dimensional) and gcd(p, q) = 1 (in order for the edge at z = 1 to be primitive). The map (x, y, z) 7→ (x − ybp/q + 1/2c, y, z) allows us to assume |p| ≤ |q|/2. Reflection of the planes x = 0 and y = 0, if needed, allows us to assume that 0 ≤ p ≤ bq/2c.

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– If the three points at z = 0 are not collinear then they form a unimodular triangle, and without loss of generality we assume they are (0, 0, 0), (1, 0, 0) and (0, 1, 0). One of the points at z = 1 can be assumed to be (0, 0, 1) and the fifth point has coordinates (a, b, 1). By the same argument as before, we need gcd(a, b) = 1 and by symmetries with respect to the triangle at z = 0 we can assume 0 ≤ a ≤ b (details are left to the reader). This configuration has volume vector (−(a + b), a, b, 1, −1), so it is of type (3, 2) unless a = 0 (and hence b = 1 since gcd(0, b) = b). In the case (a, b) = (0, 1) we recover the configuration of part (1), with coordinates x and z exchanged. This finishes the case study, but we still need to check that different configurations in the list are not Z-equivalent. Configurations of different signature are certainly not Z-equivalent. Within those of signature (3, 2), since the volume vector is primitive, Theorem 3.2 says that different values of (a, b) produce inequivalent configurations. In signature (2, 1), however, the volume vector is (q, q, 0, 0, −2q) so, a priori, configurations with different p and the same q could still be Z-equivalent. Let us prove that they are not. For this, let q be fixed and let p, p0 ∈ Z. Let P and P 0 be two of these configurations having (p, q, 1) and (p0 , q, 1) as their fifth point, respectively. All affine transformations that map P to P 0 must preserve the collinearity of the three points at z = 0, so they fix (0, 0, 0) and either fix or exchange (1, 0, 0) and (−1, 0, 0). Similarly, they either fix (0, 0, 1) and send (p, q, 1) to (p0 , q, 1), or they send (p, q, 1) to (0, 0, 1) and (0, 0, 1) to (p0 , q, 1). So we have the following four possibilities:     0 0 (x, y, z) 7→ x + −pq−p y + p0 z, −y + qz, z , (x, y, z) 7→ x + p q−p y, y, z ,     0 0 0 (x, y, z) 7→ −x + p q+p y, y, z , (x, y, z) 7→ −x + p−p y + p z, −y + qz, z . q For any of them to be integer we need p ≡ ±p0 (mod q).



5.2. Configurations of type (3, 1). We can now prove the following: Theorem 5.2. Every polytope P of signature (3, 1) and size 5 has volume vector (−3q, q, q, q, 0) with q ∈ {1, 3} and is unimodularly equivalent to one of (1) conv{(0, 0, 0), (1, 0, 0), (0, 0, 1), (−1, 0, −1), (0, 1, 0)} (of width one) or (2) conv{(0, 0, 0), (1, 0, 0), (0, 0, 1), (−1, 0, −1), (2, 3, 1)} (of width two). Proof. By Theorem 4.5, we have that P consists of the empty lattice subtetrahedron T of largest volume containing two points at each z = 0, 1, and a fifth point at height h ∈ {−1, −2}. Let us first argue that we can assume h = −1 without loss of generality. Notice that the volume vector must be of the form (−3q, q, q, q, 0); that is, all empty subtetrahedra have the same volume. So, for our purposes, it suffices to check that there exists an affine integer functional taking values {1, 1, 0, 0, −1} in the five points. Without loss of generality, our points are: p1 = (0, 0, 0),

p2 = (1, 0, 0),

p3 = (0, 0, 1),

for some 1 ≤ p ≤ q, gcd(p, q) = 1, h = −1, −2.

p4 = (p, q, 1),

p5 = (a, b, h).

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Suppose that h = −2. Four of the points form a (3, 1) circuit. The centroid must be at height z = 0 and, by Lemma 3.7 there is no loss of generality in assuming that p1 is the centroid of p3 , p4 and p5 , so p5 = (−p, −q, −2). If q = 1 = p, then the functional x takes the desired values, so we assume that q > 1. Now, Lemma 3.6 says that in order for the tetrahedron conv{p1 , p2 , p3 , p5 } to be empty we need one of the following conditions: • p = q − 1 and gcd(2, q) = 1. Then take functional x − y. • p = q − 2 and gcd(2, q) = 1. Same, for the functional x − y + z. • −2 ≡ 1 (mod q) and gcd(p, q) = 1. That is, q = 3 and p ∈ {1, 2}. This is a particular case of one of the two above, depending on whether p = 2 or 1. So for the rest of the proof h = −1. That is, p5 = (a, b, −1). In this case, there is no loss of generality in assuming that p1 is the centroid of p2 , p3 and p5 , so p5 = (−1, 0, −1). The intersection of P with z = 0 is then the triangle with vertices     −1 p4 + p5 p−1 q p3 + p5 = , 0, 0 , v = = , ,0 . p2 = (1, 0, 0), 2 2 2 2 2 The question is what possibilities for the third vertex v make this triangle not contain any lattice points other than (0, 0, 0) and
(1, 0, 0). One necessary condition is q q to be odd, because if q is even then p−1 2 , 2 , 0 itself is a lattice point. (Remember that gcd(p, q) = 1). y

p3 p5

p1

z=0

p2

x

Figure 7. The case analysis in the proof of Theorem 5.2. White squares represent the points p1 and p2 of P in the displayed plane z = 0. The gray square is the intersection of p3 p5 with that same plane. Black dots are the lattice points in the plane and black crosses represent the possible intersection points of the edge p4 p5 with the plane z = 0. But other conditions are easy to write. For example, in order for (0, 1, 0) not to be in the triangle, v must be outside the wedge with apex at (0, 1, 0) and rays in the directions of (1, 2, 0) and (−1, 1, 0). (This is the central dark wedge in Figure 7). The same consideration for the other lattice points of the form (k, 1, 0) defines analogous wedges so that at the end the only half-integer points not excluded by the wedges are those with q = 1 or with q = 3 and p = 2 (mod 3). If q = 1 then we get width one, and the configuration is unimodularly equivalent to the first one in

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the statement. If q = 3 then all possibilities for p are unimodularly equivalent to one another. Taking p = 2 we get the second configuration in the statement.  5.3. Non-symmetric configurations of type (4, 1). With similar arguments, but a more involved case study, we get: Theorem 5.3. Apart of those with volume vector of the form (−4q, q, q, q, q), every polytope of size five and signature (4, 1) is Z-equivalent to the one having the points (0, 0, 0), (1, 0, 0), (0, 0, 1) together with one of the following six pairs: • (1, 2, 1) and (−1, −1, −1), volume vector (−5, 1, 1, 1, 2). • (1, 3, 1) and (−1, −2, −1), volume vector (−7, 1, 1, 2, 3). • (2, 5, 1) and (−1, −2, −1), volume vector (−11, 1, 3, 2, 5). • (2, 5, 1) and (−1, −1, −1), volume vector (−13, 3, 4, 1, 5). • (2, 7, 1) and (−1, −2, −1), volume vector (−17, 3, 5, 2, 7). • (3, 7, 1) and (−2, −3, −1), volume vector (−19, 5, 4, 3, 7). Proof. As before, Theorem 4.5 allows us to take the following coordinates for our points: p1 = (0, 0, 0),

p2 = (1, 0, 0),

p3 = (0, 0, 1),

p4 = (p, q, 1),

p5 = (a, b, −1),

for some 1 ≤ p ≤ q, gcd(p, q) = 1. Without loss of generality (by Lemma 3.7) let p1 be the interior point. Then the volume vector of our configuration P = conv {p1 , p2 , p3 , p4 , p5 } is ((a − 2)q − bp, pb − qa, q + b, −b, q)

with sign vector (−, +, +, +, +). Since we assumed the normalized volume of T = conv {p1 , p2 , p3 , p4 } to be the biggest (see Theorem 4.5) we have

(5)

0 < −b < q,

0 < pb − qa ≤ q.

We want to find out what values of a, b, p, q make the intersection of P with {z = 0} not have other lattice points than p1 and p2 . This intersection must now contain p1 in its interior (see Figure 8) and it equals the triangle t with vertices     a b p4 + p5 p+a q+b p3 + p5 = , ,0 , = , ,0 . p2 = (1, 0, 0), 2 2 2 2 2 2 z

p4

p3 z=1 y

z=0

p5

p1

p2

x

z = −1

Figure 8. The setting for the proof of Theorem 5.3

LATTICE 3-POLYTOPES WITH FEW LATTICE POINTS

21

To get more symmetric parameters we set c = a + p and d = b + q > 0. This turns equations (5) into (6)

b < 0 < d,

0 < cb − da ≤ d − b.

Since the rest of the proof happens in the plane z = 0, we drop the last coordinate for every point. We want to study what values of a, b, c, d ∈ Z satisfying equations (6) have (0, 0) and (1, 0) as the only lattice points in the triangle t = conv {(1, 0), (a/2, b/2), (c/2, d/2)}. By symmetry of pairs (a, b) and (c, d), let us further assume that |d| ≤ |b| and observe that, via the transformations (x, y) 7→ (x ± y, y), we are only interested in c modulo d. Let us look at the possible values of (c, d). For d = 1 we take c = 0. For d > 1, taking into account that (c/2, d/2) must be outside the wedge symmetric to the triangle (0, 1)p1 p2 at point (0, 1) we conclude that c/2 6∈ [1 − d/2, 0], which is equivalent to c 6∈ [2 − d, 0]. Thus, the only remaining value for c (mod d) is c = 1. y

z=0

z=0

y

p4 p5 p1

y

z=0 p4 p5

p2 x

p1

p4 p5

p2 x

b = −d = −1

p1 b = −d = −2

p2 x

b = −d = 3

(c, d) = (0, 1)

(c, d) = (1, 2)

(c, d) = (1, 3)

Figure 9. The case analysis in the proof of Theorem 5.3 for the three possibilities of (c, d). White squares represent the points p1 and p2 of P in the displayed plane z = 0. The gray square is the intersection of p4 p5 with that same plane. Black dots are the lattice points in the plane and black crosses represent the possible intersection points of the edge p3 p5 and the plane z = 0.

Figure 9 shows the possibilities for point (a, b) for the first three cases of (c, d), namely (c, d) = {(0, 1), (1, 2), (1, 3)}. The figures must be read in the same way as Figure 7. Each lattice point creates an excluded wedge for the positions of (a, b). The only novelty is that now we have also an excluded (open) half-plane, the one defined by |b| > d, so that the allowed region (the white region in the pictures) gets smaller and smaller and it becomes lattice-point-free (and eventually empty) for d ≥ 4 (picture left to the reader). The crosses in the three pictures (9, 5 and 2, respectively) give a priori 16 possibilities for the points (a, b) and (c, d), that is, for the points p4 and p5 of our original problem. Namely:

´ MONICA BLANCO AND FRANCISCO SANTOS

22

(c, d) (0,1) (0,1) (0,1) (0,1) (0,1)

(a, b) (-1,-1) (-2,-1) (-3,-1) (-1,-2) (-3,-2)

(p, q) (1,2) (2,2) (3,2) (1,3) (3,3)

(c, d) (0,1) (0,1) (0,1) (0,1) (1,2)

(a, b) (-5,-2) (-2,-3) (-5,-3) (-3,-4) (-3,-2)

(p, q) (5,3) (2,4) (5,4) (3,5) (4,4)

(c, d) (1,2) (1,2) (1,2) (1,2) (1,3) (1,3)

(a, b) (-2,-3) (-5,-3) (-3,-4) (-4,-5) (-2,-3) (-3,-4)

(p, q) (3,5) (6,5) (4,6) (5,7) (3,6) (4,7)

Now, these 16 points reduce to only six possibilities by excluding those with gcd(p, q) 6= 1 (which produce extra lattice points at z = 1) or p > q (which are repeated from others). These six are distinguished by having (p, q) in boldface in the table above. To check that they are all inequivalent and that they coincide with those in the statement we simply need to compute their volume vectors. (a, b) (−1, −1) (−1, −2) (−3, −4) (−2, −3) (−4, −5) (−3, −4)

(p, q) (1, 2) (1, 3) (3, 5) (3, 5) (5, 7) (4, 7)

((a − 2)q − bp, pb − qa, q + b, −b, q) (−5, 1, 1, 1, 2) (−7, 1, 1, 2, 3) (−13, 3, 1, 4, 5) (−11, 1, 2, 3, 5) (−17, 3, 2, 5, 7) (−19, 5, 3, 4, 7)

Since the volume vectors are all primitive, they completely characterize the configurations (Theorem 3.2). The representatives in the statement have been chosen to have smaller coordinates.  5.4. Symmetric configurations of type (4, 1). We finally need to deal with configurations having volume vector (−4q, q, q, q, q). Theorem 5.4. Every polytope of size five and signature (4, 1) with a symmetric volume vector (q, q, q, q, −4q), is Z-equivalent to the one having the points (0, 0, 0), (1, 0, 0), (0, 0, 1) together with one of the following two pairs: • (1, 1, 1) and (−2, −1, −2), volume vector (−4, 1, 1, 1, 1). • (2, 5, 1) and (−3, −5, −2), volume vector (−20, 5, 5, 5, 5). Both configurations have width two with respect to the functional x − z. Proof. As before, Theorem 4.5 allows us to take the following coordinates for our points: p1 = (0, 0, 0),

p2 = (1, 0, 0),

p3 = (0, 0, 1),

p4 = (p, q, 1),

p5 = (a, b, −2),

for some 1 ≤ p ≤ q, gcd(p, q) = 1. Without loss of generality (by Lemma 3.7) let p1 be the interior point. Since in these cases the volume vector is (q, q, q, q, −4q) (modulo reordering), then p1 is the baricenter of the other four points and then p5 = (−p − 1, −q, −2). The convex hull of P consists of four thetrahedra glued together, all of normalized volume q; the one we started with and the following three: • T1 = {(0, 0, 0), (1, 0, 0), (0, 0, 1), (−p − 1, −q, −2)}, • T2 = {(0, 0, 0), (1, 0, 0), (p, q, 1), (−p − 1, −q, −2)}, and • T3 = {(0, 0, 0), (0, 0, 1), (p, q, 1), (−p − 1, −q, −2)}. What we need to check is which values of p and q make these three tetrahedra empty. If q = 1 = p, then all tetrahedra are unimodular and therefore empty. Assume q > 1 for the rest of the proof.

LATTICE 3-POLYTOPES WITH FEW LATTICE POINTS

23

Now, Lemma 3.6 says that in order for the tetrahedron T1 = conv{p1 , p2 , p3 , p5 } to be empty we need one of the following conditions: • p = q − 2 and gcd(2, q) = 1. • −2 ≡ 1 (mod q) and gcd(p + 1, q) = 1. That is, q = 3 and p = 1. This is a particular case of the one above. • p = q − 3 and gcd(q − 2, q) = 1. That is, q is odd and p = q − 2 or p = q − 3. In order for the tetrahedron T2 = conv{p1 , p2 , p4 , p5 } to be empty, the same Lemma 3.6 says that we need one of the following conditions: • p = 2 and gcd(2, q) = 1. • −2 ≡ 1 (mod q) and gcd(p − 1, q) = 1. That is, q = 3 and p = 2. This is a particular case of the one above. • p = 3 and gcd(q − 2, q) = 1. That is, q is odd and p = 2 or p = 3.

This implies q = 5 and p ∈ {2, 3}, which makes T3 have width 1 as well: {(0, 0, 0), (0, 0, 1), (2, 5, 1), (−3, −5, −2)} has width one with respect to y − 2x, and {(0, 0, 0), (0, 0, 1), (3, 5, 1), (−4, −5, −2)} has width one with respect to y + z − 2x. A priori, this could lead to two different configurations with q = 5. Not surprisingly, the following matrix represents a Z-equivalence mapping the configuration with p = 2 to the one with p = 3:   1 −1 3  0 −1 5  0 0 1 

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