Unimodular covers of 3-dimensional parallelepipeds and Cayley sums

We show that the following classes of lattice polytopes have unimodular covers, in dimension three: the class of parallelepipeds, the class of centrally symmetric polytopes, and the class of Cayley sums $\text{Cay}(P,Q)$ where the normal fan of $Q$ refines that of $P$. This improves results of Beck et al.~(2018) and Haase et al.~(2008) where the last two classes were shown to be IDP.


Introduction
A lattice polytope P ⊂ R d has the integer decomposition property if for every positive integer n, every lattice point p ∈ nP ∩ Z d can be written as a sum of n lattice points in P .We abbreviate this by saying that "P is IDP".Being IDP is interesting in the context of both enumerative combinatorics (Ehrhart theory) and algebraic geometry (normality of toric varieties).It falls into a hierarchy of several properties each stronger than the previous one; see, e.g., [2, Section 2.D], [7, Sect.1.2.5], [10, p. 2097], [11, p. 2313].Let us here only mention that P has a unimodular triangulation ⇒ P has a unimodular cover ⇒ P is IDP.
Remember that a unimodular triangulation is a triangulation of P into unimodular simplices, and a unimodular cover is a collection of unimodular simplices whose union equals P .
Oda ([12]) posed several questions regarding smoothness and the IDP property for lattice polytopes.Following [6,16], we say that a pair (P, Q) of lattice polytopes has the integer decomposition property, or that the pair (P, Q) is IDP, if A lattice polytope Q is called smooth if it is simple and the primitive edge directions at every vertex form a linear basis for the lattice; equivalently, if the projective toric variety defined by the normal fan of Q is smooth.The following versions of Oda's questions are now considered conjectures [8,11], and they are open even in dimension three: Conjecture 1.1.
(2) (Related to problems 1, 3, 4, 6 in [12]) Every pair (P, Q) of lattice polytopes with Q smooth and the normal fan of Q refining that of P is IDP.
When the normal fan of a polytope Q refines that of another polytope P , as in the second conjecture, we say that P is a weak Minkowski summand of Q, since this is easily seen to be equivalent to the existence of a polytope P such that P + P = kQ for some dilation constant k > 0. This property has the following algebraic implication for the projective toric variety X Q : P is a weak Minkowski summand of Q if and only if the Cartier divisor defined by P on X Q is numerically effective, or "nef" (see [3,Cor. 6.2.15,Prop. 6.3.12],but observe that what we here call "weak Minkowski summand" is simply called "Minkowski summand" there).
Motivated by these and other questions, several authors have studied the IDP property for different classes of lattice polytopes.For example, very recently Beck et al. [1] proved that all smooth centrally symmetric 3-polytopes are IDP.More precisely, they show that any such polytope can be covered by lattice parallelepipeds and unimodular simplices, both of which are trivially IDP.In Section 2 we show: Theorem 1.2.Every 3-dimensional lattice parallelepiped has a unimodular cover.
This, together with the mentioned result from [1], gives: The two-dimensional case of Conjecture 1.1( 2) is known to hold, with three different proofs by Fakhruddin [5], Ogata [13] and Haase et al. [8].This last one actually shows that smoothness of Q is not needed.In dimension three, however, the conjecture fails without the smoothness assumption.Indeed, if we let P = Q be any non-unimodular empty tetrahedron, then P is obviously a weak Minkowski summand of Q but the pair (P.Q) is not IDP.By an empty tetrahedron we mean a lattice tetrahedron containing no lattice points other than its vertices (see the proof of Lemma 2.2 for a classification of them).
An alternative approach to Conjecture 1.1(2) is via Cayley sums, which we discuss in Section 3. Recall that the Cayley sum of two lattice polytopes P, Q ⊂ R d is the lattice polytope We normally require Cay(P, Q) to be full-dimensional (otherwise we can delete coordinates) but this does not need P and Q to be full-dimensional.It only requires the linear subspaces parallel to them to span R d .
As we note in Proposition 3.1, if the Cayley sum of P and Q is IDP then the pair (P, Q) is IDP.In particular, the following statement from Section 3 is stronger than the afore-mentioned result of [5,8,13]: Theorem 1.6.Let Q be lattice polygon, and P a weak Minkowski summand of Q.Then the Cayley sum Cay(P, Q) has a unimodular cover.
This has the following two corollaries, also proved in Section 3. A prismatoid is a polytope whose vertices all lie in two parallel facets.A polytope has width 1 if its vertices lie in two consecutive parallel lattice hyperplanes.Observe that this is the same as being (SL(Z, d)-equivalent to) a Cayley sum.
A special case of the latter are integer dilations of empty tetrahedra.That their dilations have unimodular covers is [15,Cor. 4.2] (and is also implicit in [9]).
We believe that the 3-polytopes in all these statements have unimodular triangulations, but this remains an open question.
Acknowledgements: We would like to thank Akiyoshi Tsuchiya, Spencer Backman, and Johannes Hofscheier for posing these questions to us and Christian Haase for helpful discussions.

Parallelepipeds
The main tool for the proof of Theorem 1.2 is what we call the parallelepiped circumscribed to a given tetrahedron, defined as follows: Definition 2.1.Let T be a tetrahedron with vertices p 1 , p 2 , p 3 , and p 4 .Consider the points , and let C(T ) is a parallelepiped with facets conv (p i , p j , q k , q l ) for all choices of {i, j, k, l} = [4].We call it the parallelepiped circumscribed to T .For each i ∈ [4], let T i = conv (q i , p j , p k , p l ), with {i, j, k, l} = [4]; we call these T i the corner tetrahedra of C(T ).Together with T they triangulate C(T ).
Modulo an affine transformation, the situation of T and C(T ) is exactly that of the regular tetrahedron inscribed in a cube; see Figure 1.Lemma 2.2.Let T = conv {p 1 , p 2 , p 3 , p 4 } be an empty lattice tetrahedron that is not unimodular.Let C(T ) be the parallelepiped circumscribed to T and let T 1 , T 2 , T 3 and T 4 be the corresponding corner tetrahedra in C(T ).Then, every T i contains at least one lattice point different from {p 1 , . . ., p 4 }.
Lemma 2.3.Let P be a lattice parallelepiped and let T ⊂ P be a tetrahedron.Then, at least one of the four corner tetrahedra T i of the circumscribed parallelogram C(T ) is fully contained in P .
Proof.Let us denote the vertices of T by p 1 , p 2 , p 3 , p 4 and the vertices of C(T ) not in T by q 1 , q 2 , q 3 , q 4 , with the conventions of Definition 2.1.
We call band any region of the form f −1 ([α, β]) for some functional f ∈ (R 3 ) * and closed interval [α, β] ⊂ R. We claim that any band containing T must contain at least three of the q i s.This claim implies that the parallelepiped P , which is the intersection of three bands, contains at least one of the q i s and hence it fully contains the corresponding T i .
The translation of vector 1 2 (p 1 + p 2 − p 3 − p 4 ) sends q 1 , q 2 , p 3 , p 4 to p 2 , p 1 , q 4 , q 3 (in this order).By applying this to inequality (3), we obtain so that q 3 , q 4 ∈ B. This finishes the proof of the claim, and of the lemma.
Corollary 2.4.Let T be an empty lattice tetrahedron contained in a lattice parallelepiped P .Then, T can be covered by unimodular tetrahedra contained in P .
Proof.We proceed by induction on the (normalized) volume of T , which is a positive integer.If this volume equals 1 then T is unimodular and there is nothing to prove, so we assume T is not unimodular.Let p 1 , p 2 , p 3 , p 4 denote the vertices of T .
Lemma 2.3 guarantees that one of the corner tetrahedra T i of the parallelepiped C(T ) is contained in P .Without loss of generality, suppose T 4 = conv (p 1 , p 2 , p 3 , q 4 ) is in P .By Lemma 2.2, we know that T 4 contains a lattice point other than the p i s, which we denote by u.Then S = conv (T ∪{u}) can be triangulated in two different ways: S = T ∪ T 4 , where T 4 = conv (p 1 , p 2 , p 3 , u) ⊆ T 4 and S = S 1 ∪ S 2 ∪ S 3 , with Each of the tetrahedra S i has lattice volume strictly smaller than T because, for each i, p i is the unique point of C(T ) maximizing the distance to the opposite facet conv (p j , p k , p l ) of T .Thus, S 1 , S 2 and S 3 cover T and have volume strictly smaller than T .The S i may not be empty, but we can triangulate them into empty tetrahedra, which by inductive hypothesis they can be covered unimodularly.
Proof of Theorem 1.2.Arbitrarily triangulate the parallelepiped into empty lattice tetrahedra and apply Corollary 2.4 to these tetrahedra.
Let us say that a lattice 3-polytope P has the circumscribed parallelepiped property if it satisfies the conclusion of Lemma 2.3: "for every empty tetrahedron T contained in P at least one of the four corner tetrahedra in C(T ) is contained in P ".If this holds then P has a unimodular cover, since then the proofs of Corollary 2.4 and Theorem 1.2 work for P .Hence, a positive answer to the following question would imply that every smooth 3-polytope has a unimodular cover, which in turn implies Conjecture 1.1(1) in dimension three.
(5) Indeed, in both polytopes the only lattice points are the six vertices and the origin.The point (1, 1, 1) lies in the second dilation but is not the sum of two lattice points in the polytope.Hence, they are not IDP, which implies they do not admit unimodular covers.

Cayley sums
Let P and Q be two lattice polytopes in R d .We do not require them to be fulldimensional, but we assume their Minkowski sum is.Remember that the Minkowski sum P + Q and the Cayley sum of P and Q are defined as: The so-called Cayley Trick is the isomorphism 2 Cay(P, Q) ∩ (R d × {1}) ∼ = P + Q, which easily implies: The Cayley Trick also provides the following canonical bijections: polyhedral subdivisions of Cay(P, Q) ↔ mixed subdivisions of P + Q triangulations of Cay(P, Q) ↔ fine mixed subdivisions of P + Q unimodular simplices in Cay(P, Q) ↔ unimodular prod-simplices in P + Q.
See [4] for more details on the Cayley Trick and on triangulations and polyhedral subdivisions of polytopes.In fact these bijections can be taken as definitions of the objects in the right-hand sides.In particular, we call prod-simplices in P + Q the Minkowski sums T 1 + T 2 where T 1 ⊂ P and T 2 ⊂ Q are simplices with complementary affine spans.A prod-simplex is unimodular if the edge vectors from a vertex of T 1 and from a vertex of T 2 form a unimodular basis.
We now turn our attention to d = 2, in order to prove Theorem 1.6.A triangulation of Cay(P, Q) ⊂ R 3 consists of tetrahedra of types (1, 3), (2, 2) and (3,1), where the type denotes how many vertices they have in P and in Q. Empty tetrahedra of types (1, 3) or (3, 1), which are Cayley sums of a triangle in P and a point in Q, or viceversa, are automatically unimodular.The case that we need to study are therefore tetrahedra of type (2, 2), which are Cayley sums of a segment p ⊂ P and a segment q ⊂ Q.The following lemma, whose proof we postpone to Section 4, is crucial to understand how to unimodularly cover these tetrahedra.We use the Lemma 3.2.Let Q be a two-dimensional lattice polytope and P a weak Minkowski summand of it.Let p = [p 1 , p 2 ] ⊂ P and q = [q 1 , q 2 ] ⊂ Q be two primitive and non-parallel lattice segments, and − → p and − → q be the lines spanned by them.If the parallelogram p + q is not unimodular, then at least one of the regions contains a lattice point.See Figure 2. Proof.The proof is by induction on the normalized volume of T , which we assume to be at least 2. This implies that T is of type (2, 2), since empty tetrahedra of types (1, 3) and (3, 1) are unimodular.Thus, T is the Cayley sum of primitive segments p = [p 1 , p 2 ] ⊂ P and q = [q 1 , q 2 ] ⊂ Q.Let u be the lattice point whose existence is guaranteed by Lemma 3.2.Assume (the other case is similar) that and call t the triangle t = conv (u, p 1 , p 2 ) ⊂ P .Let us denote ũ, p1 , p2 , q1 , q2 the points corresponding to u, p 1 , p 2 , q 1 , q 2 in Cay(P, Q).That is, pi = p × {0}, qi = p × {1}, and ũ = u × {0}.Observe that the assumption u ∈ ((p 1 , p 2 ) + − → q implies that of the segments [u, q i ] crosses the triangle conv (p 1 , p 2 , q j ), where {i, j} = {1, 2}, see Figure 3.
The tetrahedra Cay(t, {q j }) and Cay(t, {q i }) are unimodular, which implies that T = Cay(p, q) has volume equal to the sum of the volumes of Cay([p 1 , u], q) and Cay([p 2 , u], q).In particular, we have covered T by the three tetrahedra in T − , Figure 3. [u, q 2 ] intersects conv (p 1 , p 2 , q 1 ) which are of smaller volume and hence have unimodular covers by inductive assumption.
Proof of Theorem 1.6.Arbitrarily triangulate Cay(P, Q) into empty lattice tetrahedra and apply Corollary 3.3 to these tetrahedra.
Let us now show how to derive Corollaries 1.7 and 1.8 from this theorem.Prismatoids were defined in [14] as polytopes whose vertices all lie in two parallel facets.In particular, a lattice prismatoid is any d-polytope SL(Z, d)-equivalent to one of the form conv (Q where Q 1 , Q 2 are lattice (d − 1)-polytopes and k ∈ Z >0 .This is almost a generalization of Cayley sums, which would be the case k = 1, except the definition of prismatoid requires Q 1 and Q 2 to be full-dimensional, while the Cayley sum only requires this for Proposition 3.4.Let Q 1 , Q 2 be two lattice polygons and consider the prismatoid ) is a lattice polygon then P has a unimodular cover.
Proof.The condition that P ∩ (R 2 × {1}) is a lattice polygon implies the same for P ∩ (R 2 × {i}), for every i.Indeed, the condition implies that every edge of Cay(P, Q) of the form [u × {0}, v × {k}] has a lattice point in R 2 × {i}, and hence it has a lattice point in P ∩ (R 2 × {i}), for every i.
Observe that for every i ∈ {1, . . ., k − 1} the intersection P ∩ (R 2 × {i}) has the same normal fan as Q 1 + Q 2 .Thus, each slice is a Cayley polytope.For i ∈ {2, . . ., k − 1}, both bases have the same normal fan (and therefore each is a weak Minkowski summand of the other); for i ∈ {1, k} one base is a weak Minkowski summand of the other.We can therefore apply Theorem 1.6 to each slice and combine the covers thus obtained to get a unimodular cover of P .
Proof of Corollaries 1.7 and 1.8.In both cases the polytope under study satisfies the hypotheses of Proposition 3.4: in Corollary 1.7, the smoothness of the prismatoid implies that every edge of the form [u × {0}, v × {k}] has lattice points in all slices.In Corollary 1.8, since P has width one,

Proof of Lemma 3.2
Let f q be the primitive lattice functional constant on q and f p the one constant on p.We assume that f q (p 1 ) < f q (p 2 ) and f p (q 1 ) < f p (q 2 ).
Observe that in the strip q + − → p , there is a unique lattice point on the line f q (x) = −1; indeed, since q is primitive, the only way that in the strip there could be two lattice points on f q (x) = −1 is if they were on the boundary of the strip, which would however imply that p + q is a unimodular paralellogram, against our assumptions.Since translating the polytopes by lattice vectors will not result in any loss of generality, we can assume that p 1 is that unique lattice point.That is, f q (p 1 ) = −1, or equivalently, the triangle conv (q 1 , q 2 , p 1 ) is unimodular.Similarly, the unique lattice point in the strip on the line f q (x) = 1 is then q 1 + q 2 − p 1 .
Proof of Lemma 3.2.Suppose by contradiction that there is no lattice point as described in the lemma.In particular, no lattice point on the boundary of Q can be in the interior of the strip q + − → p .Thus the boundary of Q contains two primitive segments which each have one vertex on each side of the strip q + − → p ; , with b and t crossing the strip in H 1 and H 2 respectively and the convention that f p (b 2 ) > f p (b 1 ) and f p (t 2 ) > f p (t 1 ).This readily implies The same holds for P and the strip p + − → q , and we call the segments = [l 1 , l 2 ] and r = [r 1 , r 2 ], with and r crossing the strip p + − → q in V 1 and V 2 respectively.
The only difference is that in the case that P is one dimensional we have = r = p.Again we have f q (l 1 ) ≤ f q (p 1 ), f q (l 2 ) ≥ f q (p 2 ), f q (r 1 ) ≤ f q (p 1 ), f q (r 2 ) ≥ f q (p 2 ).(7) Observe that a priori one of l and r can coincide with p, if this is on the boundary of P , and similarly one of t, b might be q, if this is on the boundary of Q.Each inequality is strict, unless the segment in question coincides with p or q.
Proof.The inequality ≥ w follows in each case from (7) and (6).
If one of the inequalities, say the one for , is not strict, then has one endpoint on each of the boundary lines of (p+ − → q ).Unless = p, one of the endpoints of is not an endpoint of p, say l 1 = p 1 .Thus the triangle T = conv (p 2 , p 1 , l 1 ) is contained in P and its edge [p 1 , l 1 ] is an integer dilation of q.Since width fq (T ) = w ≥ 2, T must contain a lattice point in the interior of the strip.
Claim 4.2.f q (b 2 − b 1 ) and f q (t 2 − t 1 ) are non-zero and have the same sign.That is, f q achieves its maximum over b and over t on the same halfplane V 1 or V 2 .
Proof.Suppose by contradiction that the maximum of f q on t lies in V 1 and that the maximum on b lies in V 2 .Then Q ∩ V 2 is contained in the open strip {−1 < f q (x) < w − 1}, of width w.This cannot contain a translated copy of r, since width fq (r) ≥ w, see Figure 5.This is a contradiction, since P is a Minkowski summand of Q and therefore Q must have an edge parallel to t.We assume w.l.o.g. that the maximum on t (and hence on b) is achieved in V 2 , that is to say, f p and f q increase in the same direction along t (and hence along b).Claim 4.3.Assume w.l.o.g. that b and t either are parallel or their affine spans cross in V 2 .Then, (1) The intersection of Q with any line parallel to p in V 2 has width w.r.t.f q strictly smaller than w.(2) f p (r 2 ) > f p (r 1 ), that is, f p achieves its maximum over r in H 2 .
Proof.Both t and b must intersect p, otherwise p 1 or p 2 are the lattice points we are looking for in Q.Their intersections with p are thus endpoints of a segment of width w.r.t f q less than w, the width of p. Since t and b cross in V 2 , the same is true for any segment parallel to p contained in Q ∩ V 2 .
For part (b), If f p (r 2 ) ≤ f p (r 1 ), it would be impossible to fit a translated copy r of r in the correct side of Q: r would need to lie inside the triangle delimited by the affine line t and the inequalities f q (x) ≥ f q (r 1 ), f p (x) ≤ f p (r 1 ).However, this triangular region has width less than w w.r.t.f q , by combining part (a) with the fact that f p and f q increase in the same direction along t, see Figure 6.
The last two claims can be summarized as saying that in the pictures b, t and r have positive slope.Observe that this implies that q is not in the boundary of Q and p = r, so both P and Q are full dimensional.Let g be the primitive lattice functional constant on [p 1 , q 2 ] (and therefore constant also on [q 1 , q 1 + q 2 − p 1 ]).By the assumption on p 1 , the values of g on these segments differer by 1.We choose the sign of g so that g([p 1 , q 2 ]) = g([q 1 , q 1 + q 2 − p 1 ]) + 1.
Proof.Since b and t must respectively separate p 1 and q 1 + q 2 − p 1 from the other two vertices of the parallelogram conv (q 1 , p 1 , q 2 , q 1 + q 2 − p 1 ), they must respectively intersect its (parallel) edges [p 1 , q 2 ] and [q 1 , q 1 + q 2 − p 1 ], which implies the stated inequalities for b and t.The same argument applied to the parallelogram conv (p 1 , q 2 , p 2 , p 1 + p 2 − p 2 ), yields the inequalities for and r.We are now ready to show a contradiction.Since the normal fan of Q refines that of P , Q must have an edge r which is a translated copy of r.Let r 1 and r 2 be its endpoints.Now consider the lattice line d through r 1 parallel to [p 1 , q 2 ], that is, g is constant on d.Let d be the parallel line defined by g(d ) = g(d) + 1.
Consider the segment s contained in r 1 + − → p with endpoints s 1 = r 1 on d and s 2 on d .Since t separates q 1 and q 1 + q 2 − p 1 and g decreases from t 1 to t 2 (by Claim 4.4), the inequality g(x) < g(d ) holds on Q ∩ V 2 , and in particular for r 2 .Since r 2 is a lattice point, g(r 2 ) ≤ g(d) = g(r 1 ), which contradicts Claim 4.4).

Figure 1 .
Figure 1.In red we have a tetrahedron T , in black its circumscribed parallelepiped C(T ), and in blue the corner simplex T 4 .
following conventions: if a, b are points, we denote by [a, b] and (a, b) respectively the closed and open segments with endpoints a, b.Given a segment s = [a, b], we denote the vector − → s := b − a and the line spanned by − → s by − → s .

Figure 4 .
Figure 4. Setup for the proof of Lemma 3.2

Figure 5 .
Figure 5. Illustration of the proof of Claim 4.2