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 réciprocité) (−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)

Shifted symmetric $δ$-vectors of convex polytopes

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