Inequalities and Ehrhart \delta-Vectors

# Inequalities and Ehrhart δ-Vectors

A. Stapledon Department of Mathematics, University of Michigan, Ann Arbor, MI 48109, USA
###### Abstract.

For any lattice polytope , we consider an associated polynomial and describe its decomposition into a sum of two polynomials satisfying certain symmetry conditions. As a consequence, we improve upon known inequalities satisfied by the coefficients of the Ehrhart -vector of a lattice polytope. We also provide combinatorial proofs of two results of Stanley that were previously established using techniques from commutative algebra. Finally, we give a necessary numerical criterion for the existence of a regular unimodular lattice triangulation of the boundary of a lattice polytope.

## 1. Introduction

Let be a lattice of rank and set . Fix a -dimensional lattice polytope in and, for each positive integer , let denote the number of lattice points in . It is a result of Ehrhart [5, 6] that is a polynomial in of degree , called the Ehrhart polynomial of . The generating series of can be written in the form

 ∑m≥0fP(m)tm=δP(t)/(1−t)d+1,

where is a polynomial of degree less than or equal to , called the (Ehrhart) -polynomial of . If we write

 δP(t)=δ0+δ1t+⋯+δdtd,

then is the (Ehrhart) -vector of . We will set for and . It is a result of Stanley [18] that the coefficients are non-negative integers. The degree of is called the degree of and is the codegree of . It is a consequence of Ehrhart Reciprocity that is the smallest positive integer such that contains a lattice point in its relative interior (see, for example, [11]). It is an open problem to characterise which vectors of non-negative integers are -vectors of a lattice polytope. Ideally, one would like a series of inequalities that are satisfied by exactly the -vectors. We first summarise the current state of knowledge concerning inequalities and Ehrhart -vectors.

It follows from the definition that and . It is a consequence of Ehrhart Reciprocity that is the number of lattice points in the relative interior of (see, for example, [11]). Since has at least vertices, we have the inequality . We list the known inequalities satisfied by the Ehrhart -vector (c.f. [3]).

 (1) δ1≥δd
 (2) δ0+δ1+⋯+δi+1≥δd+δd−1+⋯+δd−i for i=0,…,⌊d/2⌋−1,
 (3) δ0+δ1+⋯+δi≤δs+δs−1+⋯+δs−i for i=0,…,d,
 (4) if δd≠0 then 1≤δ1≤δi for i=2,…,d−1.

Inequalities (2) and (3) were proved by Hibi in [9] and Stanley in [20] respectively. Both proofs are based on commutative algebra. Inequality (4) was proved by Hibi in [13] using combinatorial methods. Recently, Henk and Tagami [8] produced examples showing that the analogue of (4) when is false. That is, it is not true that for . An explicit example is provided in Example 2.4 below.

We improve upon these inequalities by proving the following result (Theorem 2.19). We remark that the proof is purely combinatorial.

###### Theorem.

Let be a -dimensional lattice polytope of degree and codegree . The Ehrhart -vector of satisfies the following inequalities.

 (5) δ1≥δd,
 (6) δ2+⋯+δi+1≥δd−1+⋯+δd−i for% i=0,…,⌊d/2⌋−1,
 (7) δ0+δ1+⋯+δi≤δs+δs−1+⋯+δs−i for i=0,…,d,
 (8) δ2−l+⋯+δ0+δ1≤δi+δi−1+⋯+δi−l+1 for i=2,…,d−1.
###### Remark 1.1.

Equality can be achieved in all the inequalities in the above theorem. For example, let be a lattice with basis and let be the regular simplex with vertices . In this case, and each inequality above is an equality.

###### Remark 1.2.

Observe that (5) and (6) imply that

 (9) δ1+⋯+δi+δi+1≥δd+δd−1+⋯+δd−i,

for . Since , we conclude that (2) is always a strict inequality. We note that inequality (6) was suggested by Hibi in [13].

###### Remark 1.3.

We can view the above result as providing, in particular, a combinatorial proof of Stanley’s inequality (7).

###### Remark 1.4.

We claim that inequality (8) provides the correct generalisation of Hibi’s inequality (4). Our contribution is to prove the cases when and we refer the reader to [13] for a proof of (4). In fact, in the proof of the above theorem we show that (8) can be deduced from (4), (6) and (7).

In order to prove this result, we consider the polynomial

 ¯δP(t)=(1+t+⋯+tl−1)δP(t)

and use a result of Payne (Theorem 2 [16]) to establish the following decomposition theorem (Theorem 2.14). This generalises a result of Betke and McMullen for lattice polytopes containing a lattice point in their relative interior (Theorem 5 [4]).

###### Theorem.

The polynomial has a unique decomposition

 ¯δP(t)=a(t)+tlb(t),

where and are polynomials with integer coefficients satisfying and . Moreover, the coefficients of are non-negative and, if denotes the coefficient of in , then

 1=a0≤a1≤ai,

for .

Using some elementary arguments we show that our desired inequalities are equivalent to certain conditions on the coefficients of , and (Lemma 2.5) and hence are a consequence of the above theorem.

We also consider the following result of Stanley (Theorem 4.4 [17]), which was proved using commutative algebra.

###### Theorem.

If is a lattice polytope of degree and codegree , then if and only if is a translate of a reflexive polytope.

We show that this result is a consequence of the above decomposition of , thus providing a combinatorial proof of Stanley’s theorem (Corollary 2.18). A different combinatorial proof in the case is provided in [12].

Recent work of Athanasiadis [1, 2] relates the existence of certain triangulations of a lattice polytope to inequalities satisfied by its -vector.

###### Theorem (Theorem 1.3 [1]).

Let be a -dimensional lattice polytope. If admits a regular unimodular lattice triangulation, then

 (10) δi+1≥δd−i for i=0,…,⌊d/2⌋−1,
 (11) δ⌊(d+1)/2⌋≥⋯≥δd−1≥δd,
 (12) δi≤(δ1+i−1i) for i=0,…,d.

As a corollary of our decomposition of , we deduce the following theorem (Theorem 2.20).

###### Theorem.

Let be a -dimensional lattice polytope. If the boundary of admits a regular unimodular lattice triangulation, then

 (13) δi+1≥δd−i
 (14) δ0+⋯+δi+1≤δd+⋯+δd−i+(δ1−δd+i+1i+1),

for .

We note that (13) provides a generalisation of (10) and that (14) may be viewed as an analogue of (12). We remark that the method of proof is different to that of Athanasiadis.

I would like to thank my advisor Mircea Mustaţǎ for all his help. I would also like to thank Sam Payne for some valuable feedback. The author was supported by Mircea Mustaţǎ’s Packard Fellowship and by an Eleanor Sophia Wood travelling scholarship from the University of Sydney.

## 2. Inequalities and Ehrhart δ-Vectors

We will use the definitions and notation from the introduction throughout the paper. Our main object of study will be the polynomial

 ¯δP(t)=(1+t+⋯+tl−1)δP(t).

Since has degree and non-negative integer coefficients, it follows that has degree and non-negative integer coefficients. In fact, we will show that has positive integer coefficients (Theorem 2.14). Observe that we can recover from if we know the codegree of . If we write

 ¯δP(t)=¯δ0+¯δ1t+⋯+¯δdtd,

then

 (15) ¯δi=δi+δi−1+⋯+δi−l+1,

for . Note that and .

###### Example 2.1.

Let be a lattice with basis and let be the standard simplex with vertices . It can be shown that and hence . On the other hand, if is the standard reflexive simplex with vertices and , then . We conclude that does not determine .

###### Remark 2.2.

We can interpret as the Ehrhart -vector of a -dimensional polytope. More specifically, let be the standard reflexive simplex of dimension in a lattice as above. Henk and Tagami [8] defined to be the convex hull in of and . By Lemma 1.3 in [8], is a -dimensional lattice polytope with Ehrhart -vector

Our main objects of study will be the polynomials and in the following elementary lemma.

###### Lemma 2.3.

The polynomial has a unique decomposition

 (16) ¯δP(t)=a(t)+tlb(t),

where and are polynomials with integer coefficients satisfying and .

###### Proof.

Let and denote the coefficients of in and respectively, and set

 (17) ai+1=δ0+⋯+δi+1−δd−⋯−δd−i,
 (18) bi=−δ0−⋯−δi+δs+⋯+δs−i.

We compute, using (15) and since ,

 ai+bi−l =δ0+⋯+δi−δd−⋯−δd−i+1−δ0−⋯−δi−l+δs+⋯+δs−i+l =δi−l+1+⋯+δi=¯δi, ai−ad−i =δ0+⋯+δi−δd−⋯−δd−i+1−δ0−⋯−δd−i+δd+⋯+δi+1 =0, bi−bd−l−i =−δ0−⋯−δi+δs+⋯+δs−i+δ0+⋯+δs−i−1−δs−⋯−δi+1 =0,

for . Hence we obtain our desired decomposition and one easily verifies the uniqueness assertion. ∎

###### Example 2.4.

Let be a lattice with basis and let be the -dimensional lattice polytope with vertices , , , , and . Henk and Tagami showed that (Example 1.1 in [8]). It follows that and . We calculate that and .

We may view our proposed inequalities on the coefficients of the Ehrhart -vector as conditions on the coefficients of , and .

###### Lemma 2.5.

With the notation above,

1. Inequality (2) holds if and only if the coefficients of are non-negative.

2. Inequality (3) holds if and only if the coefficients of are non-negative.

3. Inequality (6) holds if and only if for .

4. Inequality (7) holds if and only if the coefficients of are non-negative.

5. Inequality (8) holds if and only if for .

6. Inequality (9) holds if and only if the coefficients of are positive.

###### Proof.

The result follows by substituting (15),(17) and (18) into the right hand sides of the above statements. ∎

###### Remark 2.6.

The coefficients of are unimodal if . It follows from (17) that for all . Hence the coefficients of are unimodal if and only if for . In Remark 2.17, we show that the coefficients of are unimodal for .

###### Remark 2.7.

A lattice polytope is reflexive if the origin is the unique lattice point in its relative interior and each facet of has the form , for some . Equivalently, is reflexive if it contains the origin in its relative interior and, for every positive integer , every non-zero lattice point in lies on for a unique positive integer . It is a result of Hibi [12] that if and only if is a translate of a reflexive polytope (c.f. Corollary 2.18). We see from Lemma 2.3 that if and only if . Payne and Mustaţǎ gave examples of reflexive polytopes where the coefficients of are not unimodal [15]. Further examples are given by Payne for all [16].

###### Remark 2.8.

It follows from (18) that for all . Hence the coefficients of are unimodal if and only if for . We see from Example 2.4 that the coefficients of are not necessarily unimodal.

Our next goal is to express as a sum of shifted -vectors, using a result of Payne (Theorem 1.2 [16]). We first fix a lattice triangulation of and recall what it means for to be regular. Translate by an element of so that the origin lies in its interior and let denote the fan over the faces of . Then is regular if can be realised as the fan over the faces of a rational polytope. Equivalently, is regular if the toric variety is projective. We may always choose to be a regular triangulation (see, for example, [1]). We regard the empty face as a face of of dimension . For each face of , consider the -vector of ,

 hF(t)=∑F⊆GtdimG−dimF(1−t)d−1−dimG.

We recall the following well-known lemma and outline a geometric proof. We refer the reader to [7] for more details (see also [19]).

###### Lemma 2.9.

Let be a regular lattice triangulation of . If is a face of , then the -vector of is a polynomial of degree with symmetric, unimodal integer coefficients. That is, if denotes the coefficient of in , then for all and . The coefficients satisfy the Upper Bound Theorem,

 hi≤(h1+i−1i),

for .

###### Proof.

As above, translate by an element of so that the origin lies in its interior and let denote the fan over the faces of . For each face of , let denote the smallest linear subspace of containing and let be the complete fan in whose cones are the projections of the cones in containing . We can interpret as the dimension of the cohomology group of the projective toric variety . The symmetry of the follows from Poincaré Duality on , while unimodality follows from the Hard Lefschetz Theorem. The cohomology ring is isomorphic to the quotient of a polynomial ring in variables of degree and hence is bounded by the number of monomials of degree in variables of degree . ∎

Recall that is the smallest positive integer such that contains a lattice point in its relative interior and fix a lattice point in . Let and let denote the projection onto the second factor. We write for the cone over in and for the ray through . For each face of , let denote the cone over and let denote the cone generated by and . The empty face corresponds to the origin and respectively. The union of such cones forms a simplicial fan refining . For each non-zero cone in , with primitive integer generators , we consider the open parallelopiped

 Box(τ)={a1v1+⋯+arvr∣0

and observe that we have an involution

 ι:Box(τ)∩N′→Box(τ)∩N′
 ι(a1v1+⋯+arvr)=(1−a1)v1+⋯+(1−ar)vr.

We also set and . Observe that . For each face of , we define

 BF(t)=∑v∈Box(σF)∩N′tu(v)
 B′F(t)=∑v∈Box(σ′F)∩N′tu(v).

If or then we define or respectively. For example, when is the empty face, and . We will need the following lemma.

###### Lemma 2.10.

For each face of , and .

###### Proof.

Using the involution above,

 tdimF+1BF(t−1)=∑v∈Box(σF)∩N′tdimF+1−u(v)=∑v∈Box(σF)∩N′tu(ι(v))=BF(t).

Similarly,

 tdimF+l+1B′F(t−1)=∑v∈Box(σ′F)∩N′tdimF+1+l−u(v)=∑v∈Box(σ′F)∩N′tu(ι(v))=B′F(t).

Consider any element in and let be the smallest face of such that lies in . Set and let denote the vertices of . Then can be uniquely written in the form

 (19) v={v}+∑(vi,1)∉τ(vi,1)+w,

where lies in , for some subcone of , and is a non-negative integer sum of and . If we write , for some non-negative integers , then

 (20) u(v)=u({v})+dimG−dimF+r∑i=1ai+ar+1l.

Conversely, given in , for some , and a non-negative integer sum of and , then lies in and is the smallest face of such that lies in .

###### Remark 2.11.

With the above notation, observe that for some if and only if lies on a translate of by a non-negative multiple of . Note that if for some (necessarily non-empty) , then and , for any non-negative integer . We conclude that for all faces of if and only if every element in can be written as the sum of an element of and , for some non-negative integers and .

The generating series of can be written as . Payne described this sum by considering the contributions of all in with a fixed . We have the following application of Theorem 1.2 in [16]. We recall the proof in this situation for the convenience of the reader.

###### Lemma 2.12.

With the notation above,

 ¯δP(t)=∑F∈T(BF(t)+B′F(t))hF(t).
###### Proof.

Using (19) and (20), we compute

 ¯δP(t) =(1+t+⋯+tl−1)δP(t)=(1−tl)(1−t)d∑v∈σ∩N′tu(v) =(1−t)d∑F∈T(BF(t)+B′F(t))∑F⊆GtdimG−dimF/(1−t)dimG+1 =∑F∈T(BF(t)+B′F(t))hF(t).

###### Remark 2.13.

We can write . Ehrhart Reciprocity states that, for any positive integer , is times the number of lattice points in the relative interior of (see, for example, [11]). Hence and the generating series of the polynomial has the form .

We will now prove our first main result. When , and the theorem below is due to Betke and McMullen (Theorem 5 [4]). This case was also proved in Remark 3.5 in [21].

###### Theorem 2.14.

The polynomial has a unique decomposition

 ¯δP(t)=a(t)+tlb(t),

where and are polynomials with integer coefficients satisfying and . Moreover, the coefficients of are non-negative and, if denotes the coefficient of in , then

 (21) 1=a0≤a1≤ai,

for .

###### Proof.

Let be a regular lattice triangulation of . We may assume that contains every lattice point of as a vertex (see, for example, [1]). By Lemma 2.12, if we set

 (22) a(t)=∑F∈TBF(t)hF(t)
 (23) b(t)=t−l∑F∈TB′F(t)hF(t),

then . Since contains no lattice points in its relative interior for , if lies in for some face of , then . We conclude that is a polynomial. By Lemma 2.9, the coefficients of are non-negative integers. Since every lattice point of is a vertex of , if lies in for some non-empty face of , then . If we write then Lemma 2.9 implies that for . By Lemma 2.3, we are left with verifying that and . Using Lemmas 2.9 and 2.10, we compute

 tda(t−1)=td∑F∈TBF(t−1)hF(t−1)=∑F∈TBF(t)td−dimF−1hF(t−1)=a(t),
 td−lb(t−1)=td−ltl∑F∈TB′F(t−1)hF(t−1)=t−l∑F∈TB′F(t)td−dimF−1hF(t−1)=b(t).

###### Remark 2.15.

It follows from the above theorem that expressions (22) and (23) are independent of the choice of lattice triangulation and the choice of in .

###### Remark 2.16.

Let be the pyramid over . That is, is the truncation of at level and can be written as

 K:={(x,λ)∈(N×Z)R∣x∈∂(λP),0<λ≤1}∪{0}.

We may view as a polyhedral complex and consider its Ehrhart polynomial and associated Ehrhart -polynomial (p1 [14], Chapter XI [11]). By the proof of Theorem 2.14,

 a(t)/(1−t)d+1=∑F∈T∑\lx@stackrelv∈σ∩N′{v}∈Box(σF)(1+t+⋯+tl−1)tu(v).

It follows from Remark 2.11 that is the generating series of and hence that . With the terminology of [14], is star-shaped with respect to the origin and the fact that , for , is a consequence of Hibi’s results in [14].

###### Remark 2.17.

By Theorem 2.14, and hence the coefficients of are unimodal for (c.f. Remark 2.6).

As a corollary, we obtain a combinatorial proof of a result of Stanley [17]. Recall, from Remark 2.7, that a lattice polytope is reflexive if and only if it contains the origin in its relative interior and, for every positive integer , every non-zero lattice point in lies on for a unique positive integer .

###### Corollary 2.18.

If is a lattice polytope of degree and codegree , then if and only if is a translate of a reflexive polytope.

###### Proof.

Since , we see that if and only if . By Lemma 2.3, we need to show that if and only if is a translate of a reflexive polytope. By Remark 2.11, if and only if every element in can be written as the sum of an element of and , for some non-negative integers and . That is, if and only if is a reflexive polytope. ∎

We now prove our second main result.

###### Theorem 2.19.

Let be a -dimensional lattice polytope of degree and codegree . The Ehrhart -vector of satisfies the following inequalities.

 δ1≥δd,
 δ2+⋯+δi+1≥δd−1+⋯+δd−i for% i=0,…,⌊d/2⌋−1,
 δ0+δ1+⋯+δi≤δs+δs−1+⋯+δs−i for i=0,…,d,
 δ2−l+⋯+δ0+δ1≤δi+δi−1+⋯+δi−l+1 for i=2,…,d−1.
###### Proof.

We observed in the introduction that . By Lemma 2.5, the second inequality is equivalent to , for , and the third inequality is equivalent to for all . Hence these inequalities follow from Theorem 2.14. When , the conditions above imply that for . By Lemma 2.5, this proves the final inequality when . When , the last inequality is Hibi’s result (4). ∎

A lattice triangulation of is unimodular if for every non-empty face of , the cone over in is non-singular. Equivalenty, is unimodular if and only if , for every non-empty face of .

###### Theorem 2.20.

Let be a -dimensional lattice polytope. If admits a regular unimodular lattice triangulation, then

 δi+1≥δd−i
 δ0+⋯+δi+1≤δd+⋯+δd−i+(δ1−δd+i+1i+1),

for .

###### Proof.

From the above discussion, if admits a regular unimodular triangulation then for every non-empty face of . By (22), , where is the empty face of . By Lemma 2.3, and for . The result now follows from the expression (see (17)). ∎

###### Remark 2.21.

When is the regular simplex of dimension , and both inequalities in Theorem 2.20 are equalities.

###### Remark 2.22.

Recall that if is a reflexive polytope then the coefficients of the Ehrhart -vector are symmetric (Remark 2.7). In this case, Theorem 2.20 implies that if admits a regular, unimodular lattice triangulation then the coefficients of the -vector are symmetric and unimodal. Note that if is reflexive then admits a regular unimodular lattice triangulation if and only if admits a regular unimodular lattice triangulation. Hence this result is a consequence of the theorem of Athanasiadis stated in the introduction (Theorem 1.3 [1]). This special case was first proved by Hibi in [10].

###### Remark 2.23.

Recall that and . Hence if and only if . If and admits a regular, unimodular lattice triangulation then, by Theorems 2.19 and 2.20, , for . This implies that for . If, in addition, is reflexive then the coefficients of the -vector are symmetric and hence .

###### Remark 2.24.

Let be a lattice polytope of dimension in and let be the convex hull of and the origin in