Shifted symmetric $δ$-vectors of convex polytopes

0.16 MB·

by  Akihiro Higashitani

SHIFTED SYMMETRIC δ-VECTORS OF CONVEX POLYTOPES
AKIHIROHIGASHITANI
0
1
Abstract. Let P ⊂ RN be an integral convex polytope of dimension d and δ(P) =
0
2 (δ0,δ1,...,δd) its Ehrhart δ-vector. It is known that Pij=0δd−j ≤ Pij=0δj+1 for each
n 0 ≤ i ≤ [(d − 1)/2]. A δ-vector δ(P) = (δ0,δ1,...,δd) is called shifted symmetric
a if Pij=0δd−j = Pij=0δj+1 for each 0 ≤ i ≤ [(d − 1)/2], i.e., δd−i = δi+1 for each
J 0 ≤ i ≤ [(d−1)/2]. In this paper, some properties of integral convex polytopes with

9 shiftedsymmetricδ-vectorswillbestudied. Moreover,asanaturalfamilyofthose,(0,1)-
1 polytopeswillbeintroduced. Inaddition,shiftedsymmetricδ-vectorswith(0,1)-vectors
] are classified when Pdi=0δi ≤5.
O
C
.
h Introduction
t
a
m An integral convex polytope is a convex polytope any of whose vertices has integer
coordinates. Let P ⊂ RN be an integral convex polytope of dimension d and
[

2 i(P,n) = |nP ∩ZN|, n =1,2,3,....
v
2 Here nP = {nα : α∈ P} and|X| is thecardinality of a finiteset X. Thesystematic study
8 of i(P,n) originated in thework of Ehrhart[3], whoestablished the following fundamental
1
properties:
1
.
0 (0.1) i(P,n) is a polynomial in n of degree d. (Thus in particular i(P,n) can be defined
1 for every integer n.)
9
(0.2) i(P,0) = 1.
0
: (0.3) (loi de r´eciprocit´e) (−1)di(P,−n) =|n(P −∂P)∩ZN| for every integer n > 0.
v
i We say that i(P,n) is the Ehrhart polynomial of P. We refer the reader to [10, pp.
X
235–241] and [4, Part II] for the introduction to the theory of Ehrhart polynomials.
r
a
We define the sequence δ ,δ ,δ ,... of integers by the formula
0 1 2
∞ ∞
(1) (1−λ)d+1 1+ i(P,n)λn = δ λi.
i
" #
n=1 i=0
X X
It follows from the basic fact (0.1) and (0.2) on i(P,n) together with a fundamental result
on generating function ([10, Corollary 4.3.1]) that δ = 0 for every i> d. We say that the
i
sequence
δ(P) = (δ ,δ ,...,δ )
0 1 d
which appears in Eq.(1) is the δ-vector of P. Alternate names of δ-vector are for example
Ehrhart δ-vector, Ehrhart h-vector or h∗-vector.
Thus δ = 1 and δ = |P ∩ZN|−(d+1). Let ∂P denote the boundary of P and
0 1
i∗(P,n) = |n(P −∂P)∩ZN|, n = 1,2,3,....
1991 Mathematics Subject Classification. Primary: 52B20 Secondary: 13F20.
1
By using (0.3) one has
∞ d δ λi+1
(2) i∗(P,n)λn = i=0 d−i .
(1−λ)d+1
n=1 P
X
In particular,
δ = |(P −∂P)∩ZN|.
d
Hence δ ≥ δ . Moreover, each δ is nonnegative ([11]). In addition, if (P −∂P)∩ZN is
1 d i
nonempty, then one has δ ≤ δ for every 1 ≤ i≤ d−1 ([5]).
1 i
When d = N, the leading coefficient ( d δ )/d! of i(P,n) is equal to theusual volume
i=0 i
of P ([10, Proposition 4.6.30]). In general, the positive integer vol(P) = d δ is said to
P i=0 i
be the normalized volume of P.
P
It follows from Eq.(2) that
max{i :δ 6= 0} = d+1−min{i :i(P −∂P)∩ZN 6= ∅}.
i
Recently, δ-vectors of integral convex polytopes have been studied intensively. (For
example, see [9],[13] and [14].)
There are two well-known inequalities of δ-vectors. Let s = max{i : δ 6= 0}. Stanley
i
[12] shows the inequalities
(3) δ +δ +···+δ ≤δ +δ +···+δ , 0≤ i ≤ [s/2]
0 1 i s s−1 s−i
by using the theory of Cohen–Macaulay rings. On the other hand, the inequalities
(4) δ +δ +···+δ ≤ δ +δ +···+δ +δ , 0 ≤ i≤ [(d−1)/2]
d d−1 d−i 1 2 i i+1
appear in [5, Remark (1.4)]. The above two inequalities are generalized in [13].
A δ-vector δ(P) = (δ ,δ ,...,δ ) is called symmetric if the equalities hold in Eq. (3)
0 1 d
for each 0 ≤ i ≤ [s/2], i.e., δ = δ for each 0 ≤ i ≤ [s/2]. The δ-vector δ(P) of
i s−i
P is symmetric if and only if the Ehrhart ring [4, Chapter X] of P is Gorenstein. A
combinatorial characterization for the δ-vector to be symmetric is studied in [6] and [8].
We say that a δ-vector δ(P) = (δ ,δ ,...,δ ) is shifted symmetric if the equalities hold
0 1 d
in Eq. (4) for each 0 ≤ i ≤ [(d−1)/2], i.e., δ = δ for each 0 ≤ i ≤ [(d−1)/2]. It
d−i i+1
seems likely that an integral convex polytope with a shifted symmetric δ-vector is quite
rare. Thus it is reasonable to sutudy a property of and to find a natural family of integral
convex polytopes with shifted symmetric δ-vectors. In section 2, some characterizations
of an integral convex polytope with a shifted symmetric δ-vector are given. Concretely, in
Theorem 2.1, it is shown that integral convex polytopes with shifted symmetric δ-vectors
have a special property. Moreover, as a generalization of an integral convex polytope with
a shifted symmetric δ-vector, an integral simplicial polytope any of whose facet has the
normalized volume 1 is considered in section 2. In section 3, a family of (0,1)-polytopes
withshiftedsymmetricδ-vectors ispresented. Theseshiftedsymmetricδ-vectors are(0,1)-
vectors. Inaddition, by usingthoseexamples, weclassify shiftedsymmetricδ-vectors with
(0,1)-vectors when d δ ≤ 5 in section 4.
i=0 i
2
P
1. Review on the computation of the δ-vector of a simplex
We recall from [4, Part II] the well-known combinatorial technique how to compute the
δ-vector of a simplex.
• Given an integral d-simplex F ⊂ RN with the vertices v ,v ,...,v , we set F =
0 1 d
(α,1) ∈ RN+1 : α ∈ F . And ∂F = (α,1) ∈RN+1 : α∈ ∂F is its boundary.
• Let C(F) = C = {rβ : β ∈ F,0 ≤ r ∈ Q} ⊂ RN+1. Its boundary is ∂Ce =
(cid:8) (cid:9) (cid:8) (cid:9)
e
rβ : β ∈ ∂F,0 ≤ r ∈ Q .
e e
• Lnet S (resp. S∗) be the seot of all points α ∈ C∩ZN+1 (resp. α ∈ (C−∂C)∩ZN+1)
of the form αe = d r (v ,1), where each r ∈ Q with 0 ≤ r < 1 (resp. with
i=0 i i i i
0 < r ≤ 1).
i
P
• The degree of an integer point (α,n) ∈ C is deg(α,n) := n.
Lemma 1.1. (a) Let δ be the number of integer points α ∈ S with degα= i. Then
i
∞
δ +···+δ λd
1+ i(F,n)λn = 0 d .
(1−λ)d+1
n=1
X
(b) Let δ∗ be the number of integer points α∈ S∗ with degα= i. Then
i
∞ δ∗λ+···+δ∗ λd+1
i∗(F,n)λn = 1 d+1 .
(1−λ)d+1
n=1
X
(c) One has δ∗ = δ for each 1≤ i ≤ d+1.
i (d+1)−i
We say thataδ-vector δ(P) = (δ ,δ ,...,δ )is shifted symmetric if δ = δ for each
0 1 d d−i i+1
0 ≤ i ≤ [(d−1)/2]. Since δ = δ , an integral convex polytope with a shifted symmetric
d 1
δ-vector is always a d-simplex.
The followings are some examples of a simplex with a shifted symmetric δ-vector.
Let e denote the ith canonical unit coordinate vector of Rd.
i
Examples 1.2. (a) We define v ∈ Rd,i = 0,1,...,d by setting v = e for i = 1,...,d
i i i
and v = (−e,...,−e), where e is a nonnegative integer. Let P = conv{v ,v ,...,v }.
0 0 1 d
Then one has vol(P) = ed+1 by using an elementary linear algebra. When e = 0, it is
clear that δ(P) = (1,0,0,...,0). When e is positive, we know that
d
j (e−j)d+1
(v ,1)+ (v ,1) = (j −e,j −e,...,j −e,1)
i 0
ed+1 ed+1
i=1
X
and 0 < j , (e−j)d+1 < 1 for every 1 ≤ j ≤ e. Then Lemma 1.1 says that δ ,δ ≥ e.
ed+1 ed+1 1 d
Since δ ≥ δ for 1≤ i ≤ d−1 and vol(P) =ed+1, we obtain δ(P) = (1,e,e,...,e).
i 1
(b) Let d ≥ 3. We define v ∈ Rd,i = 0,1,...,d by setting v = e for i = 1,...,d and
i i i
v = (e,...,e), where e is a positive integer. Let P = conv{v ,v ,...,v }. Then one has
0 0 1 d
vol(P) = ed−1 by using an elementary linear algebra. And we know that
d
j (e−j)d−1
(v ,1)+ (v ,1) = (e−j,e−j,...,e−j,1)
i 0
ed−1 ed−1
i=1
X
3
and 0 < j ,(e−j)d−1 < 1 for every 1 ≤ j ≤ e−1. Thus δ ,δ ≥ e−1 by Lemma 1.1. In
ed−1 ed−1 1 d
addition we know that
d
ke+j (e−j)d−1−k
(v ,1)+ (v ,1) = (e−j,e−j,...,e−j,k+1)
i 0
ed−1 ed−1
i=1
X
and 0 < ke+j, (e−j)d−1−k < 1 for every 0 ≤ j ≤ e − 1 and 1 ≤ k ≤ d − 2. Hence
ed−1 ed−1
δ(P) = (1,e−1,e,e,...,e,e−1).
2. Some characterizations of an integral convex polytope with a shifted
symmetric δ-vector
In the first half of this section, two results of an integral convex polytope with a shifted
symmetric δ-vector are given. And in the latter half of this section, we generalize an
integral convex polytope with a shifted symmetric δ-vector.
Theorem 2.1. Let P be a d-simplex whose vertices are v ,v ,...,v ∈Rd and S (S∗) the
0 1 d
set which appears in section 1. Then the following conditions are equivalent:
(i) δ(P) is shifted symmetric;
(ii) the normalized volume of all facets of P is equal to 1;
(iii) each element (α,n) ∈S\{(0,...,0,0)} has a unique expression of the form:
d
(5) (α,n) = r (v ,1) with 0 < r < 1 for j = 0,1,...,d,
j j j
j=0
X
where α∈ Zd and n ∈ Z.
Proof. ((i) ⇔ (iii)) If each element (α,n) ∈ S\{(0,...,0,0)} has the form (5), each
element (α′,n′) ∈ S∗\{( d v ,..., d v ,d + 1)} also has the same form (5). This
j=0 j j=0 j
implies that δ(P) is shifted symmetric. On the other hand, suppose that δ(P) is shifted
P P
symmetric. Let min{i : δ 6= 0,i > 0} = s and δ = m (6= 0). Then one has d +
i 1 s1 1
1− max{i : δ 6= 0} = s and both S and S∗ have the m elements with degree s . If
i 1 1 1
an element (α′,s ) ∈ S∗ does not have the form (5), there is 0 ≤ j ≤ d with r = 1,
1 j
say, r = r = ··· = r = 1 and 0 < r ,r ,...,r < 1. Then S has an element
0 1 a a+1 a+2 d
(α′ −v −v −··· −v ,s −a −1) 6= (0,...,0,0), a contradiction. Thus each element
0 1 a 1
(α′,s ) ∈ S∗ has the form (5). If we set min{i : δ 6= 0,i > s } = s , the same discussions
1 i 1 2
can be done as above. Thus each element (β′,s ) ∈ S∗ has the form (5). Hence each
2
element (α′,n′)∈ S∗\{( d v ,..., d v ,d+1)} has the form (5), that is to say, each
j=0 j j=0 j
element (α,n) ∈ S\{(0,...,0,0)} has the form (5).
P P
((ii) ⇔ (iii)) Let δ(P) = (δ ,δ ,...,δ ) ∈ Zd+1 be the δ-vector of P and δ(F) =
0 1 d
(δ′,δ′,...,δ′ )∈ Zd theδ-vector ofafacetF ofP. Thenonehasδ′ ≤ δ for0≤ i ≤ d−1.
0 1 d−1 i i
If there is a facet F with vol(F) 6= 1, say, its vertices are v ,v ,...,v , there exists an
0 1 d−1
element (α,n) ∈ S withα = d−1r v +0·v and n >0. Thisimplies that thereexists an
j=0 j j d
element of S\{(0,...,0,0)} which doesnothave theform (5). Ontheother hand,suppose
P
that there exists an element (α,n) ∈ S\{(0,...,0,0)} which does not have the form (5),
i.e., (α,n) = d r (v ,1) and there is 0 ≤ j ≤ d with r = 0, say, r = 0. Then the
j=0 j j j d
normalized volume of the facet whose vertices are v ,v ,...,v is not equal to 1. (cid:3)
0 1 d−1
P
4
Theorem 2.2. Let P be a d-simplex. If vol(P) = p with a prime number p and
min{i :δ 6= 0,i >0} = d+1−max{i :δ 6= 0},
i i
then δ(P) is shifted symmetric.
Proof. Theelements of S formacyclic groupof primeorder. Thenevery non-zeroelement
of S is a generator. Thus it can be considered whether S\{(0,...,0,0)} is disjoint from
∂S, which case satisfies the condition of Theorem 2.1 (iii), or S is contained in a facet of
∂S, where ∂S is a cyclic group generated by the vertices of a facet of P. In the latter
case, let x ∈ S be an element of the maximal degree deg(x), and let −x denote its inverse.
Since S is contained in ∂S, deg(x)+deg(−x) < d+1, which contradicts the assumption.
Therefore, since Theorem 2.1, δ(P) is shifted symmetric. (cid:3)
Recall that an integral convex polytope with a shifted symmetric δ-vector is always
a simplex. Then we expand the definition of shifted symmetric to an integral simplicial
polytope. WestudyanintegralsimplicialpolytopeP anyofwhosefacethasthenormalized
volume 1. When P is a simplex, its δ-vector is shifted symmetric by Theorem 2.1.
Leth(∆(P)) = (h ,h ,...,h )denotetheh-vector oftheboundarycomplex ofP. (See,
0 1 d
[4, Part I].) Then the following is a well-known fact about a lower bound of the h-vector
for a simplicial (d−1)-sphere.
Lemma 2.3. ([1],[2]) The h-vector h(∆(P)) = (h ,h ,...,h ) of a simplicial (d − 1)-
0 1 d
sphere satisfies h ≤ h for every 1 ≤ i ≤ d−1.
1 i
Now,allofh-vectors ofsimplicial(d−1)-spheressatisfyingthelowerbound,i.e., h = h
1 i
for every 1 ≤ i ≤ d−1, are given by h-vectors of the boundary complexes of simplicial
polytopes any of whose facet has the normalized volume 1. In fact,
Theorem 2.4. For an arbitrary positive integer h , there exists a d-dimensional simplicial
1
polytope P any of whose facet has the normalized volume 1 and whose h-vector of the
boundary complex coincides with (1,h ,...,h ,1) ∈ Zd+1.
1 1
Proof. Let d = 2. A convex polygon is always simplicial. And each facet of an integral
polygonhasthenormalizedvolume1ifandonlyifthereisnointeger pointinitsboundary
except its vertices. Hence, foran arbitrary positive integer h , we can say thatthere exists
1
an integral polygon with h +2 vertices any of whose facet has the normalized volume 1.
1
We assume when d ≥ 3. Let P be the d-dimensional integral convex polytope whose
vertices v ∈ Rd, i = 0,1,...,d+n, are of the form:
i
(0,...,0) for i= 0,
v = e for i= 1,2,...,d,
i i


(c ,...,c ,j) for i= d+1,d+2,...,d+n,
j j


 (n−j)(j−1)
where n is a positive even number, j = i−d and c = n+ .
j 2
5
First step. We prove that P is a simplicial convex polytope. We define the {(n+1)(d−
1)+2} convex hulls by setting
F := conv{v ,v ,...,v ,v ,...,v } for i= 1,...,d,
i 0 1 i−1 i+1 d
F′ := conv{v1,...,vd−1,vd+1},

Gi,j := conv{v1,...,vi−1,vi+1,...,vd−1,vd+j,vd+j−1} for i = 1,...,d−1,j = 2,...,n,

G′ := conv{v ,...,v ,v ,...,v ,v ,v } for i = 1,...,d−1,
i 1 i−1 i+1 d−1 d d+n



and the followings are the equations of the hyperplanes containing the above convex hulls:
H ⊃ F : −x = 0 for i= 1,...,d,
i i i
H′ ⊃F′ : d−1x −(n(d−1)−1)x = 1,
k=1 k d

I ⊃ G :c′ i−1 x −(1−(d−2)c′)x +c′ d−1 x +(j − n+2)x = c′
 i,j i,jPj k=1 k j i j k=i+1 k 2 d j
for i =1,...,d−1,j = 2,...,n, where c′ = ((j −1)c −jc )= j2−j+n,
P j P j j−1 2
Ii′ ⊃ Gi′ : n ik−=11xk −(n(d−1)−1)xi +n dk=i+1xk = n for i = 1,...,d−1.



We prove thaPt these {(n+1)(d−1)+2} coPnvex hulls are all facets of P. If we write

H ⊂ Rd for the hyperplane defined by the equation a x +···+a x = b, then we write
1 1 d d
H(+) ⊂ Rd for the closed half-space defined by the inequality a x +···+a x ≤ b.
1 1 d d
• Let P = conv{v ,...,v }. Then one has P = ( d H(+))∩(x +···+x ≤
n+1 0 d n+1 i=1 i 1 d
1).
T
• Let P = conv{P ∪{v }} for k = n,n−1,...,1. Then it can beshown easily
k k+1 d+k
that
d d−1
(+) (+)′ (+)
P = ( H )∩( I )∩( I )∩(kx +···+kx −(c (d−1)−1)x ≤ k).
k i i i,j 1 d−1 k d
i=1 i=1 1≤i≤d−1
\ \ \
k+1≤j≤n
Then one has P = conv{P ∪{v }} = P. Thus we obtain the following equality:
1 2 d+1
d d−1
P = ( H(+))∩( I(+)′)∩( I(+))∩H′(+).
i i i,j
i=1 i=1 1≤i≤d−1
\ \ \
2≤j≤n
Hence we can say that F ,F′,G and G′ are all facets of P and they are (d−1)-simplices.
i i,j i
Second step. We prove that the normalized volume of each facet of P is equal to 1.
One has vol(F ) = 1 for i = 1,...,d since vol(P ) = 1 and one has vol(G′) = 1 for
i n+1 i
1 ≤ i ≤ d−1 since conv{v ,...,v ,v } is a simplex with a shifted symmetric δ-vector
1 d d+n
by Examples 1.2(b). And one has vol(F′) = 1 since vol(conv{v ,v ,...,v ,v }) = 1.
0 1 d−1 d+1
When we consider vol(G ), we are enough to prove that vol(G )= 1 by the symmetry.
i,j d−1,j
For a (d−1)-simplex G , we consider the elements of the set S which appears in
d−1,j
section 1:
(α ,...,α ,r)= r (v ,1)+···+r (v ,1)+r (v ,1)+r (v ,1),
1 d 1 1 d−2 d−2 d−1 d+j d d+j−1
where (α ,...,α ,r)∈ Zd+1 and 0 ≤ r < 1 for 0 ≤ i≤ d. Then one has
1 d i
(α ,...,α )= (r +r c +r c ,...,r +r c +r c ,r c +r c ,r j+r (j−1)).
1 d 1 d−1 j d j−1 d−2 d−1 j d j−1 d−1 j d j−1 d−1 d
6
Since r = α −α ∈ Z, we obtain r = 0. Similary we obtain r = ··· = r = 0.
1 1 d−1 1 2 d−2
Hence we can rewrite (α ,...,α ,r) =r (v ,1)+r (v ,1). It then follows that
1 d d−1 d+j d d+j−1
vol(G )= vol(conv{v ,v }) = vol(conv{(c ,j),(c ,j −1)})
d−1,j d+j d+j−1 j j−1
Since r j +r (j −1) ∈ Z and r +r ∈ Z, one has r = r = 0. This implies that
d−1 d d−1 d d−1 d
vol(conv{(c ,j),(c ,j −1)}) = 1. Thus vol(G ) = 1.
j j−1 d−1,j
Third step. By the first step and the second step, P is a d-dimensional simplicial
polytope with {(n+1)(d−1)+2} facets and (d+n+1) vertices any of whose facet has
the normalized volume 1. Hence, by Lemma 2.3, one has h(∆(P)) = (1,n+1,...,n+1,1)
for a positive even number n. Thus, when h is odd and h ≥ 3, we know that there
1 1
exists a simplicial polytope with h(∆(P)) = (1,h ,...,h ,1) any of whose facet has the
1 1
normalized volume 1. When h = 1, it is clear that h(∆(P )) = (1,1,...,1). When h
1 n+1 1
is even and h ≥ 2, let P′ =conv{v ,...,v }. Then we can verify that P′ is a simplicial
1 1 d+n
polytope with h(∆(P′)) = (1,n,...,n,1) any of whose facet has the normalized volume
1. (cid:3)
3. A family of (0,1)-polytopes with shifted symmetric δ-vectors
In this section, a family of (0,1)-polytopes with shifted symmetric δ-vectors is studied.
Weclassify completely theδ-vectors ofthosepolytopes. Moreover, weconsiderwhenthose
δ-vectors are both symmetric and shifted symmetric.
Let d = m+n with positive integers m and n. We study the δ-vector of the integral
convex polytope P ⊂ Rd whose vertices are of the form:
e +e +···+e i = 1,...,d,
i i+1 i+m−1
(6) v =
i
((0,...,0) i = 0,
where e =e .
d+i i
The normalized volume of P is equal to the absolute value of the determinant of the
circulant matrix
v
1
.
(7) (cid:12) .. (cid:12).
(cid:12) (cid:12)
(cid:12)v (cid:12)
(cid:12) d(cid:12)
(cid:12) (cid:12)
This determinant (7) can be calculated e(cid:12)asi(cid:12)ly. In fact,
(cid:12) (cid:12)
Proposition 3.1. When (m,n) = 1, the determinant (7) is equal to ±m. And when
(m,n) 6= 1, the determinant (7) is equal to 0. Here (m,n) is the greatest common divisor
of m and n.
A proof of this proposition can be given by the formula of the determinant of the
circulant matrix. Thus one has vol(P) = m when (m,n) = 1.
In this section, we assume only the case of (m,n) = 1. Hence (m,d) = 1.
For j = 1,2,...,d−1, let q be the quotient of jm divided by d and r its remainder
j j
i.e., one has the equalities
jm = q d+r for j = 1,2,...,d−1.
j j
7
It then follows from (m,d) = 1 that
0 ≤ q ≤ m−1,1 ≤ r ≤ d−1
j j
and
rj 6=rj′ if j 6= j′
for every 1 ≤ j,j′ ≤ d−1. In addition, for k = 1,2,...,m−1, let j ∈ {1,2,....d−1} be
k
the integer with r = k, i.e., one has the equalities
jk
j m = q d+r = q d+k for k = 1,2,...,m−1.
k jk jk jk
Then q > 0. Thus one has
jk
1 ≤q ,r ≤ m−1
jk jk
for every 1 ≤ k ≤m−1.
For an integer a, let a denote the residue class in Z/dZ.
Theorem 3.2. Let P be the integral convex polytope whose vertices are of the form (6)
and δ(P) =(δ ,δ ,...,δ ) its δ-vector. For each 1≤ i ≤ d, one has im ∈ {1,2,...,m−1}
0 1 d
if and only if one has δ = 1. Moreover, δ(P) is shifted symmetric, i.e., δ = δ for
i i+1 d−i
each 0≤ i ≤ [(d−1)/2].
Proof. By using the above notations, we obtain
q r
jk {(v ,1)+(v ,1)+···+(v ,1)}+ jk(v ,1) = (q ,...,q ,j ) ∈ Zd+1
m 1 2 d m 0 jk jk k
and 0 < qjk, rjk < 1 for every 1 ≤ k ≤ m−1. Then Lemma 1.1 guarantees that one has
m m
δ ≥ 1 for k = 1,...,m−1. Considering d δ = m by Proposition 3.1, it turns out
jk i=0 i
that δ(P) coincides with
P
1 i = 0,j ,j ,...,j ,
1 2 m−1
δ =
i
(0 otherwise.
Now im ∈ {1,2,...,m−1}isequivalentwithi ∈ {j ,...,j }. Thereforeonehasδ = 1
1 m−1 i
if and only if im ∈ {1,2,...,m−1} for each 1 ≤ i≤ d.
In addition, by virtue of Theorem 2.1, δ(P) is shifted symmetric, as required.
Corollary 3.3. Let P be the integral convex polytope whose vertices are of the form (6)
and δ(P) = (δ ,δ ,...,δ ) its δ-vector. Then δ(P) is symmetric, i.e., δ = δ for each
0 1 d i s−i
0 ≤ i≤ [s/2] if and only if one has d ≡ m−1 (mod m).
Proof. Letpbethequotientofddividedbymandritsremainder,i.e.,onehasd= mp+r.
And let j = min{j ,j ,...,j }. On the one hand, one has j m = d+t. On the other
t 1 2 m−1 t
hand, one has (p+1)m = d+m−r and 1≤ m−r ≤ m−1. It then follows from Theorem
3.2 that p+1 = j = min{i :δ 6= 0,i > 0}. Hence d−p = s = max{i : δ 6= 0} since δ(P)
t i i
is shifted symmetric.
When d≡ m−1 (mod m), i.e., r = m−1, we can obtain the equalities
d−p = mp+r−p = mp+m−1−p = (m−1)(p+1).
8
In addition, for nonnegative integers l(p+1), l = 1,2,...,m−1, the following equalities
hold:
l(p+1)m = l(mp+m)= l(mp+m−1)+l = ld+l = l ∈ {1,2,...,m−1}.
Thus it turns out that δ(P) coincides with
1 i= 0,p+1,2(p+1),...,(m−1)(p+1),
δ =
i
(0 otherwise,
by Theorem 3.2. It then follows that
δ = δ = δ = 1
k(p+1) (m−1−k)(p+1) s−k(p+1)
for every 0 ≤ k ≤m−1 and
δ = δ = 0
i s−i
for every 0 ≤ i ≤ s with i 6= k(p+1), k = 0,1,...,m−1. These equalities imply that
δ(P) is symmetric.
Suppose that δ(P) is symmetric. Our work is to show that r = m−1. Then one has
δ = δ = δ =δ = 1.
0 s d−p (m−1)(p+1)
Since δ(P) is also shifted symmetric, one has δ = δ . Hence one has δ =
(m−1)(p+1) p+1 p+1
δ = δ = ··· = δ =1 since δ(P) is both symmetric and shifted
(m−2)(p+1) 2(p+1) [(m−1)/2](p+1)
symmetric. When m is odd, one has d−p = m−1(p+1) since δ(P) is symmetric. Thus
2 2
r = m−1. When m is even, one has d+1 = m(p+1) since δ(P) is shifted symmetric.
2 2
Thus r = m−1.
Therefore δ(P) is symmetric if and only if d ≡ m−1 (mod m). (cid:3)
4. Classifications of shifted symmetric δ-vectors with (0,1)-vectors
In this section, we will classify all the possible shifted symmetric symmetric δ-vectors
with (0,1)-vectors when d δ ≤ 5 by using the examples in the previous section.
i=0 i
In [7], the possible δ-vectors of integral convex polytopes are classified completely when
P
d δ ≤ 3.
i=0 i
LPemma 4.1. Let d ≥ 3. Given a finite sequence (δ ,δ ,...,δ ) of nonnegative integers,
0 1 d
where δ = 1 and δ ≥ δ , which satisfies d δ ≤ 3, there exists an integral convex
0 1 d i=0 i
polytope P ⊂ Rd of dimension d whose δ-vector coincides with (δ ,δ ,...,δ ) if and only
0 1 d
P
if (δ ,δ ,...,δ ) satisfies all inequalities (3) and (4).
0 1 d
As an analogy of Lemma 4.1, we classify shifted symmetric symmetric δ-vectors with
(0,1)-vectos when d δ = 4 or 5.
i=0 i
Now, in what follows, a sequence (δ ,δ ,...,δ ) with each δ ∈ {0,1}, where δ = 1,
0 1 d i 0
P
which satisfies all inequalities (3) and all equalities (4) together with d δ = 4 or 5 will
i=0 i
be considered.
P
At first, we consider the case of d δ = 4. Let δ = δ = δ = 1 with 1 ≤ m <
i=0 i m1 m2 m3 1
m < m ≤ d. Let p = m −1, p = m −m −1, p = m −m −1 and p = d−m .
2 3 1 1 2 2 1 3 3 2 4 3
P
9
By δ = δ for 0 ≤ i≤ [(d−1)/2], one has p = p and p = p . Moreover, by (3), one
i+1 d−i 1 4 2 3
has p ≥ p . Thus,
1 2
(8) p ≥ p ≥ 0, 2p +2p = d−3.
1 2 1 2
Our work is to construct an integral convex polytope P with dimension d whose δ-vector
coincides with
δ(P) =(1,0,...,0,1,0,...,0,1,0,...,0,1,0,...,0)
p1 p2 p2 p1
for an arbitrary integer 1 ≤ m| {z< }m |< {mz }≤ d|sa{tzisfy}ing|th{ezco}nditions (8). When
1 2 3
p = p = 0, it is easy to construct it by Examples 1.2 (a). When p = p > 0, if we set
1 2 1 2
d = 4p +3 and m = 4, then the δ-vector of the integral convex polytope whose vertices
1
are of the form (6) coincides with δ(P) = (1,0,...,0,1,0,...,0,1,0,...,0,1,0,...,0) by
p1 p1 p1 p1
virtue of Corollary 3.3.
| {z } | {z } | {z } | {z }
Lemma 4.2. Let d = 4k +3, l ≥ 1 and d′ = d+2l. There exists an integral simplex
P ⊂ Rd′ of dimension d′ whose δ-vector coincides with
(9) (1,0,...,0,1,0,...,0,1,0,...,0,1,0,...,0) ∈Zd′+1.
k+l k k k+l
Proof. When l = 1,|if{zwe}set|d {=z 4}k +|5 a{nzd}m =| 4{,zth}en the δ-vector of the integral
convex polytope whose vertices are of the form (6) coincides with (9). When l ≥ 2, let
v′,v′,...,v′ ∈ R4k+2l+3 be the vertices as follows:
0 1 4k+2l+3
(v ,1,1,...,1,1), i = 1,
1
 2l−2
(v ,0,1,...,0,1), i = 2,3,
 i | {z }


v′ =  2l−2
i (v ,0,0,...,0,0), i = 4,...,4k+5,
 i | {z }

2l−2
ei, | {z } i = 4k+6,...,4k+2l+3,


(0,...,0), i = 0,






where v ,...,v are of the form (6) with d = 4k + 5 and m = 4. Then a simple
1 4k+5
computation enables us to show that
A
*
(cid:12) (cid:12)
v′ (cid:12) (cid:12)
1 (cid:12) (cid:12)
(cid:12) ... (cid:12)= (cid:12)(cid:12) (cid:12)(cid:12) =4,
(cid:12) (cid:12) (cid:12) 1 (cid:12)
(cid:12)v′ (cid:12) (cid:12) 0 (cid:12)
(cid:12) 4k+2l+3(cid:12) (cid:12) ... (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) 1(cid:12)
(cid:12) (cid:12)
(cid:12) 10 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)