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arXiv:1107.4862v1 [math.CO] 25 Jul 2011

EHRHART POLYNOMIALS OF INTEGRAL SIMPLICES

WITH PRIME VOLUMES

AKIHIRO HIGASHITANI

Abstract. For an integral convex polytope P ⊂ RNof dimension d, we call δ(P) =

(δ0,δ1,...,δd) its δ-vector and vol(P) =?d

we will establish the new equalities and inequalities on δ-vectors for integral simplices

whose normalized volumes are prime. Moreover, by using those, we will classify all the

possible δ-vectors of integral simplices with normalized volume 5 and 7.

i=0δi its normalized volume. In this paper,

Introduction

One of the most fascinating problems on enumerative combinatorics is to characterize

the δ-vectors of integral convex polytopes.

Let P ⊂ RNbe an integral convex polytope of dimension d, which is a convex polytope

any of whose vertices has integer coordinates. Let ∂P denote the boundary of P. Given

a positive integer n, we define

i(P,n) = |nP ∩ ZN|,i∗(P,n) = |n(P \ ∂P) ∩ ZN|,

where nP = {nα : α ∈ P} and |X| is the cardinality of a finite set X. The enumerative

function i(P,n) is called the Ehrhart polynomial of P, which was studied originally in the

work of Ehrhart [1]. The Ehrhart polynomial has the following fundamental properties:

• i(P,n) is a polynomial in n of degree d. (Thus, in particular, i(P,n) can be defined

for every integer n.)

• i(P,0) = 1.

• (loi de r´ eciprocit´ e) i∗(P,n) = (−1)di(P,−n) for every integer n > 0.

We refer the reader to [2, Part II] and [7, pp. 235–241] for the introduction to the theory

of Ehrhart polynomials.

We define the sequence δ0,δ1,δ2,... of integers by the formula

?∞

n=0

(1 − λ)d+1

?

i(P,n)λn

?

=

∞

?

i=0

δiλi. (1)

Then, from a fundamental result on generating function ([7, Corollary 4.3.1]), we know

that δi= 0 for every i > d. We call the integer sequence

δ(P) = (δ0,δ1,...,δd),

2000 Mathematics Subject Classification: Primary 52B20; Secondary 52B12.

Keywords: Integral simplex, Ehrhart polynomial, δ-vector.

The author is supported by JSPS Research Fellowship for Young Scientists.

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which appears in (1), the δ-vector of P. In addition, by the reciprocity law, one has

∞

?

n=1

i∗(P,n)λn=

?d

i=0δd−iλi+1

(1 − λ)d+1

.

The δ-vector has the following fundamental properties:

• δ0= 1 and δ1= |P ∩ ZN| − (d + 1).

• δd= |(P \ ∂P) ∩ ZN|. Hence, we have δ1≥ δd.

• Each δiis nonnegative ([8]).

• If (P \ ∂P) ∩ ZNis nonempty, then one has δ1≤ δifor every 1 ≤ i ≤ d − 1 ([3]).

• When d = N, the leading coefficient (?d

?d

Recently, the δ-vectors of integral convex polytopes have been studied intensively. For

example, see [6], [10] and [11].

There are two well-known inequalities on δ-vectors. Let s = max{i : δi?= 0}. One is

i=0δi)/d! of i(P,n) is equal to the usual

volume of P ([7, Proposition 4.6.30]). In general, the positive integer vol(P) =

i=0δiis said to be the normalized volume of P.

δ0+ δ1+ ··· + δi≤ δs+ δs−1+ ··· + δs−i,0 ≤ i ≤ ⌊s/2⌋, (2)

which is proved in Stanley [9], and another one is

δd+ δd−1+ ··· + δd−i≤ δ1+ δ2+ ··· + δi+ δi+1,0 ≤ i ≤ ⌊(d − 1)/2⌋, (3)

which appears in Hibi [3, Remark (1.4)].

When?d

pletely ([4, Theorem 5.1]) by (2) and (3) together with an additional condition. Further-

more, by the proofs of [5, Theorem 0.1] and [4, Theorem 5.1], we know that all the possible

δ-vectors can be realized as the δ-vectors of integral simplices when?d

further classifications of the δ-vectors with?d

straints on δ-vectors for integral simplices whose normalized volumes are prime numbers.

The following theorem is our main result of this paper.

i=0δi≤ 3, the above inequalities (2) and (3) characterize the possible δ-vectors

completely ([5]). Moreover, when?d

i=0δi= 4, the possible δ-vectors are determined com-

i=0δi≤ 4. However,

unfortunately, it is not true when?d

δ-vectors of integral simplices. In this paper, in particular, we establish some new con-

i=0δi= 5. (See [4, Remark 5.2].) Therefore, for the

i=0δi≥ 5, it is natural to investigate the

Theorem 0.1. Let P be an integral simplex of dimension d and δ(P) = (δ0,δ1,...,δd)

its δ-vector. Suppose that?d

Then,

i=0δi= p is an odd prime number. Let i1,...,ip−1 be the

positive integers such that?d

(a) one has

i=0δiti= 1 + ti1+ ··· + tip−1with 1 ≤ i1≤ ··· ≤ ip−1≤ d.

i1+ ip−1= i2+ ip−2= ··· = i(p−1)/2+ i(p+1)/2≤ d + 1;

(b) one has

ik+ iℓ≥ ik+ℓ for 1 ≤ k ≤ ℓ ≤ p − 1 with k + ℓ ≤ p − 1.

We prove Theorem 0.1 in Section 1 via the languages of elementary group theory.

As an application of Theorem 0.1, we give a complete characterization of the possible

δ-vectors of integral simplices when?d

i=0δi= 5 and 7.

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Theorem 0.2. Given a finite sequence (δ0,δ1,...,δd) of nonnegative integers, where δ0=

1 and?d

ik+ iℓ≥ ik+ℓfor 1 ≤ k ≤ ℓ ≤ 4 with k + ℓ ≤ 4, where i1,...,i4are the positive integers

such that?d

Theorem 0.3. Given a finite sequence (δ0,δ1,...,δd) of nonnegative integers, where δ0=

1 and?d

and ik+iℓ≥ ik+ℓfor 1 ≤ k ≤ ℓ ≤ 6 with k+ℓ ≤ 6, where i1,...,i6are the positive integers

such that?d

By virtue of Theorem 0.1, the “Only if” parts of Theorem 0.2 and 0.3 are obvious. A

proof of the “If” part of Theomre 0.2 is given in Section 2 and that of Theorem 0.3 is

given in Section 3.

Finally, we note that we cannot characterize the possible δ-vectors of integral simplices

with higher prime normalized volumes only by Theorem 0.1. In fact, since the volume of

an integral convex polytope containing a unique integer point in its interior has an upper

bound, if p is a sufficiently large prime number, then the integer sequence (1,1,p − 3,1)

cannot be a δ-vector of some integral simplex of dimension 3, although (1,1,p − 3,1)

satisfies all the conditions of Theorem 0.1.

i=0δi= 5, there exists an integral simplex P ⊂ Rdof dimension d whose δ-vector

coincides with (δ0,δ1,...,δd) if and only if i1,...,i4satisfy i1+ i4= i2+ i3≤ d + 1 and

i=0δiti= 1 + ti1+ ··· + ti4with 1 ≤ i1≤ ··· ≤ i4≤ d.

i=0δi= 7, there exists an integral simplex P ⊂ Rdof dimension d whose δ-vector

coincides with (δ0,δ1,...,δd) if and only if i1,...,i6satisfy i1+i6= i2+i5= i3+i4≤ d+1

i=0δiti= 1 + ti1+ ··· + ti6with 1 ≤ i1≤ ··· ≤ i6≤ d.

1. A proof of Theorem 0.1

The goal of this section is to give a proof of Theorem 0.1.

First of all, we recall from [2, Part II] the well-known combinatorial technique how to

compute the δ-vector of an integral simplex.

Given an integral simplex F in RNof dimension d with the vertices v0,v1,...,vd, we

set

? F =?(α,1) ∈ RN+1: α ∈ F?,

Clearly, we have i(F,n) = i(? F,n) for all n. Let

which is an integral simplex in RN+1of dimension d with the vertices (v0,1),(v1,1),...,(vd,1).

C(? F) = {rβ : β ∈? F,0 ≤ r ∈ Q}.

i(F,n) =

Then one has

???

?

(α,n) ∈ C(? F) : α ∈ ZN????.

Each rational point α ∈ C(? F) has a unique expression of the form α =?d

where ri∈ Q with 0 ≤ ri< 1. We define the degree of α =?d

i=0ri(vi,1) with

i=0ri(vi,1),0 ≤ ri∈ Q. Let S be the set of all points α ∈ C(? F)∩ZN+1of the form α =?d

with deg(α) =?d

Lemma 1.1. Let δibe the number of integer points α ∈ S with deg(α) = i. Then,

i=0ri(vi,1) ∈ C(? F) ∩ ZN+1

i=0ri, i.e., the last coordinate of α.

δ(F) = (δ0,δ1,...,δd).

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Notice that the elements of S form an abelian group with a unit (0,...,0) ∈ S. For

α and β in S with α =

i=0ri(vi,1) and β =

0 ≤ ri,si< 1, we define the operation in S by setting α⊕β :=?d

in order to distinguish the operation in S from the usual addition, we use the notation ⊕,

which is not a direct sum.)

?d

?d

i=0si(vi,1), where ri,si ∈ Q with

i=0{ri+si}(vi,1), where

{r} = r−⌊r⌋ denotes the fractional part of a rational number r. (Throughout this paper,

We prove Theorem 0.1 by using the above notations.

A proof of Theorem 0.1. Let v0,v1,...,vdbe the vertices of the integral simplex P and S

the group appearing above. Then, since vol(P) = p is prime, it follows from Lemma 1.1

that the order of S is also prime. In particular, S is a cyclic group.

(a) Write gi1,...,gip−1∈ S \{(0,...,0)} for (p−1) distinct elements with deg(gij) = ij

for 1 ≤ j ≤ p − 1, that is, S = {(0,...,0),gi1,...,gip−1}. Then, for each gij, there

exists its inverse −gijin S \ {(0,...,0)}.

gij=?d

gi′

Let −gij= gi′

j. If gijhas the expression

q=0rq(vq,1), where rq∈ Q with 0 ≤ rq< 1, then its inverse has the expression

j=?d

deg(gij) + deg(gi′

q=0

q=0{1 − rq}(vq,1). Thus, one has

j) =

d

?

(rq+ {1 − rq}) ≤

d

?

q=0

(rq+ 1 − rq) = d + 1

for all 1 ≤ j ≤ p − 1.

For j1,j2∈ {1,...,p−1} with j1?= j2, let gij1=?d

write gij2and gi′

q=0r(1)

q (vq,1) and gij2=?d

q=0r(2)

q (vq,1).

Since S is a cyclic group with a prime order, gij1generates S, which implies that we can

j2as follows:

gij2= gij1⊕ ··· ⊕ gij1

?

???

t

,gi′

j2= gi′

j1⊕ ··· ⊕ gi′

??

j1

?

?

t

for some integer t ∈ {2,...,p − 1}. Thus, we have

d

?

q=0

(r(2)

q

+ {1 − r(2)

q}) = deg(gij2) + deg(gi′

j2)

= deg(gij1⊕ ··· ⊕ gij1

?

?

be p. Hence, if 0 < r(1)

0 < {t(1 − r(1)

obviously, if r(1)

q

r(1)

??

?

?

t

) + deg(gi′

j1⊕ ··· ⊕ gi′

??

j1

?

?

t

) =

d

?

q=0

({tr(1)

q} + {t(1 − r(1)

q)}).

Moreover, gij1⊕ ··· ⊕ gij1

??

p

= (0,...,0) holds. Thus, we have {pr(1)

q } = 0 for all 0 ≤ q ≤ d.

Again, since p is prime, it follows that the denominator of each rational number r(1)

< 1 (resp. 0 < {1 − r(1)

q )} < 1), so r(1)

q

+ {1 − r(1)

= {1 − r(1)

q } = {tr(1)

q )} = 0. Thus, deg(gij1)+deg(gi′

q

must

qq } < 1), then 0 < {tr(1)

q } = {tr(1)

q } = 0, then {tr(1)

q } < 1 (resp.

q } + {t(1 − r(1)

q } = {t(1 − r(1)

q )} = 1. In addition,

q )} = 0, so r(1)

q

+ {1 −

q }+{t(1−r(1)

j1) = deg(gij2)+deg(gi′

j2), i.e.,

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ij1+ i′

j1= ij2+ i′

j2. Hence, we obtain

i1+ i′

1= ··· = i(p−1)/2+ i′

(p−1)/2(= i(p+1)/2+ i′

(p+1)/2= ··· = ip−1+ i′

p−1) ≤ d + 1.

Our work is to show that i′

First, we consider i′

i′

1= im< ip−1. Thus, it follows that

j= ip−jfor all 1 ≤ j ≤ (p − 1)/2.

1. Suppose that i′

1?= ip−1. Then, there is m ∈ {1,...,p − 2} with

ip−1+ i′

p−1= i1+ i′

1= i1+ im< i1+ ip−1≤ i′

p−1+ ip−1,

a contradiction.

gi′

done. Hence, i′

Therefore, we obtain the desired conditions

Thus, i′

1must be ip−1.

2among {i2,...,ip−2}. Then, the same discussion can be

2= ip−2. Similarly, we have i′

Next, we consider i′

2. Since gi′

2?= gi1and

2?= gip−1, we may consider i′

3= ip−3,...,i′

(p−1)/2= i(p+1)/2.

i1+ ip−1= i2+ ip−2= ··· = i(p−1)/2+ i(p+1)/2≤ d + 1.

(b) Write gi1,...,giℓ∈ S \ {(0,...,0)} for ℓ distinct elements with deg(gij) = ij for

1 ≤ j ≤ ℓ. Let A = {gi1,...,giℓ}. Then there are k distinct elements hi1,...,hikin

A with deg(hij) = ij for 1 ≤ j ≤ k satisfying |A| + |B| = k + ℓ ≤ p − 1, where B =

{hi1,...,hik} ⊂ A. Moreover, for each g ∈ A ⊕ B = {a ⊕ b : a ∈ A,b ∈ B}, g satisfies

deg(g) ≤ ik+ iℓ. In fact, for gij∈ A and hij′∈ B, if they have the expressions

gij=

d

?

q=0

rq(vq,1) and hij′=

d

?

q=0

r′

q(vq,1),

where rq,r′q∈ Q with 0 ≤ rq,r′q< 1, then one has

deg(gij⊕ hij′) =

d

?

q=0

{rq+ r′

q} ≤

d

?

q=0

(rq+ r′

q) = ij+ ij′ ≤ ik+ iℓ.

Now, Lemma 1.2 below guarantees that there exist at least k elements in A ⊕ B \ A ∪

{(0,...,0)}. In addition, each gijin A satisfies deg(gij) ≤ iℓ≤ ik+ iℓ. Thus, we can

say that there exist at least (k + ℓ) distinct elements in S \ {(0,...,0)} whose degrees

are at most ik+ iℓ. From the definition of i1,...,ip−1, this means that ik+ iℓ≥ ik+ℓ, as

desired.

?

Lemma 1.2. Let G be a group with prime order p, where its operation is denoted by +,

and let G∗= G \ {0}, where 0 is the unit of G. We choose two subsets (not subgroups) A

and B of G satisfying B ⊂ A ⊂ G∗and |A| + |B| ≤ p − 1 and we set C = G∗\ A. Then

one has

|(A + B) ∩ C| ≥ |B|, (4)

where A + B = {a + b : a ∈ A,b ∈ B}.

Proof. Let A = {a1,...,aℓ} and B = {b1,...,bk}. We show the assertion by induction on

k.

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