Page 1

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.

1

Page 2

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.

2

Page 3

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).

3

Page 4

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.,

4

Page 5

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.

5

Page 6

First, we consider k = 1, i.e., B = {b1}. Then, ℓ + 1 ≤ p − 1. For 1 ≤ i ≤ ℓ, let

ai+ b1= a′

i∈ G. Then we have

?

i=1

(0 + b1) +

ℓ

?

i=1

(ai+ b1) =0 +

ℓ

?

ai

?

+ b1+ ··· + b1

?

2,...,a′

???

ℓ+1

= b1+

ℓ

?

i=1

a′

i.

If we suppose that A + {b1} ⊂ A ∪ {0}, then we have {b1,a′

b1,a′

1,a′

2,...,a′

ℓ} ⊂ A ∪ {0}. Since

1,a′

2,...,a′

ℓare distinct, one has {b1,a′

1,a′

ℓ} = A∪{0}. Thus, b1+ ··· + b1

?

???

ℓ+1

∈ G cannot

= 0

from the above equality. However, since |G| is prime and l+1 < p, b1+ ··· + b1

?

???

ℓ+1

be 0, a contradiction. Hence, A+{b1} ?⊂ A∪{0}, which implies that |(A+{b1})∩C| ≥ 1.

Next, we consider k ≥ 2. Let B′= {b1,...,bk−1}. Then, by the hypothesis of induction,

one has |(A+B′)∩C| ≥ k−1. When |(A+B′)∩C| > k−1, the assertion holds. Thus, we

assume that |(A+B′)∩C| = k −1. Let (A+B′)∩C = {c1,...,ck−1}, where c1,...,ck−1

are (k − 1) distinct elements, A′= A ∪ {c1,...,ck−1} and C′= G∗\ A′. Then, again

by the hypothesis of induction, one has |(A′+ {bk}) ∩ C′| ≥ 1. This implies that there

exists at least one element ckin C′such that a + bk= ckfor some a ∈ A′. When a ∈ A,

then ck ∈ (A + B) ∩ C′, which says that the assertion holds. Hence, we assume that

a ∈ {c1,...,ck−1}, say, a = c1.

Now, again by the hypothesis of induction, it is easy to see that we have the following

equalities by renumbering c1,...,ck−1∈ (A + B′) ∩ C if necessary:

where ai1,...,aik−1∈ A. Suppose that the inequality

c1= ai1+ b1,

c2= ai2+ b2,

...

ck−1= aik−1+ bk−1,

(5)

|(A + B) ∩ C| ≥ k(6)

is not satisfied. From (5), one has

ck= a + bk= c1+ bk= ai1+ b1+ bk.

Set c′

ck∈ (A+B)∩C′, which means that (6) holds. When c′

one has c′

1∈ (A+B)∩C′, which also means that (6) holds. Moreover, c′

ck?= b1. In addition, c′

say, c′

1= c2. Then, again from (5),

1= ai1+ bk. When c′

1∈ A, since c′

1+ b1∈ A + B and c′

1+ b1= ck∈ C′, one has

1∈ C′, since c′

1cannot be 0 since

1∈ {c2,...,ck−1},

1= ai1+bk∈ A+B,

1cannot be c1since b1?= bk. Hence, it must be c′

ck= c1+ bk= c2+ b1= ai2+ b2+ b1.

Set c′

be 0, c1and c2. Hence, it must be c′

discussions, we obtain

2= ai2+ b1. Similarly, when c′

2∈ A or c′

2∈ {c3,...,ck−1}, say, c′

2∈ C′, (6) holds. Moreover, c′

2= c3. By repeating these

2cannot

ck= c1+ bk= c2+ b1= ··· = ck−1+ bk−2= aik−1+ bk−1+ bk−2.

6

Page 7

Set c′

k−1= aik−1+ bk−2. However, we have

c′

k−1?∈ A ∪ C′∪ {0,c1,c2,...,ck−1} = G,

a contradiction. Thus, the inequality (6) must be satisfied.

Therefore, we obtain the required inequality (4).

?

Remark 1.3. (a) When i1+ip−1= ··· = i(p−1)/2+i(p+1)/2= d+1, the δ-vector is shifted

symmetric. Shifted symmetric δ-vectors are studied in [6]. Moreover, the theorem [6,

Theorem 2.3] says that if i1+ip−1= d+1, then we have i1+ip−1= ··· = i(p−1)/2+i(p+1)/2=

d + 1.

(b) The inequalities i1+iℓ≥ iℓ+1are not new. In fact, for example, when i1< ··· < ip−1,

by (2), one has

δ0+ ··· + δi1≤ δip−1+ ··· + δip−1−i1.

Thus, we obtain ip−1− i1≤ ip−2, i.e., i1+ ip−2≥ ip−1. Similarly, one has

δ0+ ··· + δi2≤ δip−1+ ··· + δip−1−i2.

Thus, we obtain ip−1− i2 ≤ ip−3. Since i1+ ip−1 = i2+ ip−2, this is equivalent to

i1+ ip−3≥ ip−2. In the same way, we can obtain all inequalities i1+ iℓ≥ iℓ+1. On the

other hand, when k ≥ 2, there are many new inequalities.

2. The possible δ-vectors of integral simplices with?d

In this section, we give a proof of the “If” part of Theorem 0.2, i.e., we classify all the

possible δ-vectors of integral simplices whose normalized volume is 5.

Let (δ0,δ1,...,δd) be a nonnegative integer sequence with δ0 = 1 and?d

positive integers such that?d

i1+ i2≥ i3). From the conditions δ0= 1,?d

(i) (1,0,...,0,4,0,... ,0);

(ii) (1,0,...,0,2,0,... ,0,2,0,... ,0);

(iii) (1,0,...,0,1,0,... ,0,2,0,... ,0,1,0,... ,0);

(iv) (1,0,...,0,1,0,... ,0,1,0,... ,0,1,0,... ,0,1,0,... ,0).

i=0δi= 5

i=0δi = 5

which satisfies i1+ i4= i2+ i3≤ d + 1, 2i1≥ i2and i1+ i2≥ i3, where i1,...,i4are the

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

i1+i4= i2+i3, we notice that i1+i3≥ i4(resp. 2i2≥ i4) is equivalent to 2i1≥ i2(resp.

i=0δi= 5 and i1+ i4= i2+ i3, the possible

sequences are only the following forms:

Our work is to find integral simplices whose δ-vectors are of the above forms.

To construct integral simplices, we define the following integer matrix, which is called

the Hermite normal form:

7

A5(d1,...,d4) =

1

...

1

∗∗ ···5

1

...

1

,(7)

Page 8

where there are djj’s among the ∗’s for j = 1,...,4 and the rest of the entries are all 0.

Then, clearly, it must be dj≥ 0 and d1+ ··· + d4≤ d − 1. By determining d1,...,d4, we

obtain an integer matrix A5(d1,...,d4) and we define the integral simplex P5(d1,...,d4)

from the matrix as follows:

P5(d1,...,d4) = conv({(0,...,0),v1,...,vd}) ⊂ Rd,

where viis the ith row vector of A5(d1,...,d4). The following lemma enables us to compute

δ(P5(d1,...,d4)) easily.

Lemma 2.1 ([4, Corollary 3.1]). If δ(P5(d1,...,d4)) = (δ0,δ1,...,δd), then we have

d

?

i=0

δiti= 1 +

4

?

i=1

t1−si,

where

si=

i

5−

4

?

j=1

?ij

5

?

dj

,for i = 1,...,4.

2.1. The case (i). Let i1= i2= i3= i4= i. Thus, one has i − 1 ≥ 0 and 2i − 2 ≤ d − 1

from our conditions. Hence, we can define P5(0,i − 1,i − 1,0). Then, by Lemma 2.1,

δ(P5(0,i − 1,i − 1,0)) coincides with (i) since s1= s2= s3= s4= −i + 1.

2.2. The case (ii). Let i1= i2= i and i3= i4= j. Thus, one has 2i ≥ j, 2j −2i−2 ≥ 0

and i + j − 2 ≤ d − 1. Hence, we can define P5(0,i,2i − j,2j − 2i − 2) and its δ-vector

coincides with (ii) since s1= s2= −j + 1 and s3= s4= −i + 1.

2.3. The case (iii). Let i1= i,i2= i3= j and i4= k. Thus, one has 2i ≥ j, 3j−3i−2 ≥ 0

and 2j−2 ≤ d−1. Hence, we can define P5(0,2i−j,i,3j−3i−2) and its δ-vector coincides

with (iii) since s1= −2j + i + 1 = −k + 1, s2= s3= −j + 1 and s4= −i + 1.

2.4. The case (iv). In this case, one has 2i1≥ i2,i1+ i2≥ i3,i2+ 2i3− 3i1− 2 ≥ 0 and

i2+i3−2 ≤ d−1. Hence, we can define P5(0,2i1−i2,i1+i2−i3,i2+2i3−3i1−2) and its

δ-vector coincides with (iv) since s1= i1−i2−i3+1 = −i4+1,s2= −i3+1,s3= −i2+1

and s4= −i1+ 1.

Remark 2.2. (a) The classification of the case (iv) is essentially given in [6, Lemma 4.3].

(b) The inequalities 2i1≥ i2and i1+i2≥ i3can be obtained from (2) as we mentioned in

Remark 1.3 (b). Thus, the possible δ-vectors of integral simplices with normalized volume

5 can be essentially characterized only by Theorem 0.1 (a) and the inequalities (2).

3. The possible δ-vectors of integral simplices with?d

In this section, similarly to the previous one, we give a proof of the “If” part of Theorem

0.3, i.e., we classify all the possible δ-vectors of integral simplices whose normalized volume

is 7.

Let (δ0,δ1,...,δd) be a nonnegative integer sequence with δ0 = 1 and?d

2i2≥ i4, where i1,...,i6are the positive integers such that?d

i=0δi= 7

i=0δi = 7

which satisfies i1+ i6 = i2+ i5 = i3+ i4 ≤ d + 1, i1+ il ≥ il+1 for 1 ≤ l ≤ 3 and

i=0δiti= 1 + ti1+ ··· + ti6

8

Page 9

with 1 ≤ i1≤ ··· ≤ i6≤ d. Since i1+ i6= i2+ i5= i3+ i4, we need not consider the

inequalities i1+ i4≥ i5, i1+ i5≥ i6, i2+ i3≥ i5, i2+ i4≥ i6and 2i3≥ i6. From the

conditions δ0= 1,?d

(i) (1,0,...,0,6,0,... ,0);

(ii) (1,0,...,0,3,0,... ,0,3,0,... ,0);

(iii) (1,0,...,0,1,0,... ,0,4,0,... ,0,1,0,... ,0);

(iv) (1,0,...,0,2,0,... ,0,2,0,... ,0,2,0,... ,0);

(v) (1,0,...,0,1,0,... ,0,2,0,... ,0,2,0,... ,0,1,0,... ,0);

(vi) (1,0,...,0,2,0,... ,0,1,0,... ,0,1,0,... ,0,2,0,... ,0);

(vii) (1,0,...,0,1,0,... ,0,1,... ,0,2,0,... ,0,1,0,... ,0,1,0,... ,0);

(viii) (1,0,...,0,1,0,... ,0,1,... ,0,1,0,... ,0,1,0,... ,0,1,0,... ,0,1,0,... ,0).

i=0δi= 7 and i1+ i6= i2+ i5= i3+ i4, the possible sequences are

only the following forms:

In the same way as the previous section, we define the following integer matrix:

A7(d1,...,d6) =

1

...

1

∗∗···7

1

...

1

,(8)

where there are djj’s among the ∗’s for j = 1,...,6 and the rest of the entries are all 0.

Then it must be dj≥ 0 and d1+ ··· + d6≤ d − 1. By determining d1,...,d6, we obtain

the integral simplex

P7(d1,...,d6) = conv({(0,...,0),v1,...,vd}) ⊂ Rd,

where viis the ith row vector of A7(d1,...,d6). Similarly, the following lemma enables us

to compute δ(P7(d1,...,d6)) easily.

Lemma 3.1 ([4, Corollary 3.1]). If δ(P7(d1,...,d6)) = (δ0,δ1,...,δd), then we have

d

?

i=0

δiti= 1 +

6

?

i=1

t1−si,

where

si=

i

7−

6

?

j=1

?ij

7

?

dj

,for i = 1,...,6.

3.1. The case (i). Let i1= ··· = i6= i. Thus, one has i − 1 ≥ 0 and 2i − 2 ≤ d − 1

from our conditions. Hence, we can define P7(0,0,i − 1,i − 1,0,0). Then, by Lemma 3.1,

δ(P7(0,0,i − 1,i − 1,0,0)) coincides with (i) since s1= ··· = s6= −i + 1.

3.2. The case (ii). Let i1 = ··· = i3 = i and i4 = ··· = i6 = j.

j−i ≥ 0,2i ≥ j,2j −2i−2 ≥ 0 and i+j−2 ≤ d−1. Hence, we can define P7(0,j −i,2i−

j,2i−j,0,2j −2i−2) and its δ-vector coincides with (ii) since s1= s2= s3= −j +1 and

s4= s5= s6= −i + 1.

9

Thus, one has

Page 10

3.3. The case (iii). Let i1 = i,i2 = ··· = i5 = j and i6 = k. Thus, one has i +

j ≥ k,k − j ≥ 0,k − i − 1 ≥ 0,i − 1 ≥ 0 and i + k − 2 ≤ d − 1.

define P7(i + j − k,k − j,k − i − 1,0,0,i − 1) and its δ-vector coincides with (iii) since

s1=−4i+j−4k

77

−2i−3j−2k

77

Hence, we can

+1 = −j+1,s2=−i+2j−8k

+ 1 = −j + 1,s5=−6i−2j+k

+1 = −k+1,s3=−5i+3j−5k

+ 1 = −i + 1 and s6=−3i−j−3k

7

+1 = −j+1,s4=

+ 1 = −j + 1.

7

3.4. The case (iv). Let i1 = i2 = i,i3 = i4 = j and i5 = i6 = k. Thus, one has

i − 1 ≥ 0,i + j ≥ k,3k − 3j − 1 ≥ 0 and 2i − 2j + 2k − 2 = i + k − 2 ≤ d − 1. Hence,

we can define P7(0,0,i − 1,i + j − k,0,3k − 3j − 1) and its δ-vector coincides with (iv)

since s1 = s2 = −i + 2j − 2k + 1 = −k + 1,s3 = s4 = −i + j − k + 1 = −j + 1 and

s5= s6= −i + 1.

3.5. The case (v). Let i1= k1,i2= i3= k2,i4= i5= k3and i6= k4. Thus, one has

2k1≥ k2,k2− k1≥ 0,k1+ k2≥ k3,2k3− 2k1− 2 ≥ 0 and k2+ k3− 2 ≤ d − 1. Hence, we

can define P7(0,2k1− k2,0,k2− k1,k1+ k2− k3,2k3− 2k1− 2) and its δ-vector coincides

with (v) since s1= k1− k2− k3+ 1 = −k4+ 1,s2= s3= −k3+ 1,s4= s5= −k2+ 1 and

s6= −k1+ 1.

3.6. The case (vi). Let i1= i2= k1,i3= k2,i4= k3and i5= i6= k4. Thus, one has

k3−k2−1 ≥ 0,k1+k2≥ k3,2k1≥ k3,k2+2k3−3k1−1 ≥ 0 and k2+k3−2 ≤ d−1. Hence,

we can define P7(0,k3−k2−1,k1+k2−k3,2k1−k3,0,k2+2k3−3k1−1) and its δ-vector

coincides with (vi) since s1= s2= k1−k2−k3+1 = −k4+1,s3= −k3+1,s4= −k2+1

and s5= s6= −k1+ 1.

3.7. The case (vii). Let i1= k1,i2= k2,i3= i4= k3,i5= k4and i6= k5. Thus, one

has 2k1≥ k2,k1+k2≥ k3,k2−k1≥ 0,3k3−2k1−k2−2 ≥ 0 and 2k3−2 ≤ d−1. Hence,

we can define P7(0,0,2k1− k2,k1+ k2− k3,k2− k1,3k3− 2k1− k2− 2) and its δ-vector

coincides with (vii) since s1= k1− 2k3+ 1 = −k5+ 1,s2= k2− 2k3+ 1 = −k4+ 1,s3=

s4= −k3+ 1,s5= −k2+ 1 and s1= −k1+ 1.

3.8. The case (viii). In this case, one has i1+ i2≥ i3,2i2≥ i4,i3+ 2i4− 2i1− i2− 2 ≥

0,2i1≥ i2,i1+i3≥ i4and i3+i4−2 ≤ d−1. Hence, we can define P7(0,i1+i2−i3,i1+

i3− 2i2,0,2i2− i4,i3+ 2i4− 2i1− i2− 2) if i1+ i3≥ 2i2and P7(0,2i1− i2,0,2i2− i1−

i3,i1+ i3− i4,i3+ 2i4− 2i1− i2− 2) i1+ i3≤ 2i2. Moreover, each of δ-vectors of them

coincides with (viii) since s1= i1− i3− i4+ 1 = −i6+ 1, s2= i2− i3− i4+ 1 = −i5+ 1,

s3= −i4+ 1, s4= −i3+ 1, s5= −i2+ 1 and s6= −i1+ 1.

Remark 3.2. When we discuss the cases of (vi) and (viii), we need the new inequality

2i2≥ i4. In fact, for example, the sequence (1,0,2,0,1,1,0,2,0) cannot be the δ-vector

of an integral simplex, although this satisfies i1+ il≥ il+1,l = 1,...,3. Similarly, the

sequence (1,0,1,1,0,1,0,1,0,1,1,0) also cannot be the δ-vector of an integral simplex,

although this satisfies i1+ il≥ il+1,l = 1,...,3.

References

[1] E. Ehrhart, “Polynˆ omes Arithm´ etiques et M´ ethode des Poly` edres en Combinatoire,” Birkh¨ auser,

Boston/Basel/Stuttgart, 1977.

10

Page 11

[2] T. Hibi, “Algebraic Combinatorics on Convex Polytopes,” Carslaw Publications, Glebe NSW, Aus-

tralia, 1992.

[3] T. Hibi, A lower bound theorem for Ehrhart polynomials of convex polytopes, Adv. in Math. 105

(1994), 162 – 165.

[4] T. Hibi, A. Higashitani and N. Li, Hermite normal forms and δ-vector, to appear in J. Comb. Theory

Ser. A, also available at arXiv:1009.6023v1.

[5] T. Hibi, A. Higashitani and Y. Nagazawa, Ehrhart polynomials of convex polytopes with small volume,

European J. Combinatorics 32 (2011), 226–232.

[6] A. Higashitani, Shifted symmetric δ-vectors of convex polytopes, Discrete Math. 310 (2010), 2925–

2934.

[7] R. P. Stanley, “Enumerative Combinatorics, Volume 1,” Wadsworth & Brooks/Cole, Monterey, Calif.,

1986.

[8] R. P. Stanley, Decompositions of rational convex polytopes, Annals of Discrete Math. 6 (1980), 333 –

342.

[9] R. P. Stanley, On the Hilbert function of a graded Cohen–Macaulay domain, J. Pure and Appl. Algebra

73 (1991), 307 – 314.

[10] A. Stapledon, Inequalities and Ehrhart δ-vectors, Trans. Amer. Math. Soc. 361 (2009), 5615–5626.

[11] A. Stapledon, Additive number theorem and inequalities in Ehrhart theory, arXiv:0904.3035v1

[math.CO].

Akihiro Higashitani, Department of Pure and Applied Mathematics, Graduate School of

Information Science and Technology, Osaka University, Toyonaka, Osaka 560-0043, Japan

E-mail address: a-higashitani@cr.math.sci.osaka-u.ac.jp

11