PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 130, Number 12, Pages 3483–3493 S 0002-9939(02)06597-8 Article electronically published on May 9, 2002

ON SYZYGIES OF SEGRE EMBEDDINGS

ELENA RUBEI

(Communicated by Michael Stillman)

Abstract. We study the syzygies of the ideals of the Segre embeddings. Let d N, d 3; we prove that the line bundle (1, ..., 1) on the P 1 ... P 1 ∈ ≥ O × × (d copies) satisfies Property Np of Green-Lazarsfeld if and only if p 3. ≤ Besides we prove that if we have a projective variety not satisfying Property Np for some p, then the product of it with any other projective variety does not satisfy Property Np. From this we also deduce other corollaries about syzygies of Segre embeddings.

1. Introduction Let L be a very ample line bundle on a smooth complex projective variety Y and 0 let ϕL : Y P(H (Y,L)∗) be the map associated to L. Werecallthedefinition → of Property Np of Green-Lazarsfeld, studied for the first time by Green in [Gr1-2] (see also [G-L], [Gr3]): Let Y be a smooth complex projective variety and let L be a very ample line 0 bundle on Y defining an embedding ϕL : Y, P = P(H (Y,L)∗);setS = S(L)= 0 → Sym∗H (L), the homogeneous coordinate ring of the projective space P, and con- 0 n sider the graded S-module G = G(L)= n H (Y,L );letE ∗ 0 En En 1 ... E0 G 0 −→ −→ − −→L −→ −→ −→ be a minimal graded free resolution of G; the line bundle L satisfies Property Np (p N)ifandonlyif ∈ E0 = S,

Ei = S( i 1) for 1 i p. − − ≤ ≤ 0 (Thus L satisfies Property NM0 if and only if Y P(H (L)∗) is projectively normal, ⊂ i.e. L is normally generated; L satisfies Property N1 if and only if L satisfies 0 Property N0 and the homogeneous ideal I of Y P(H (L)∗) is generated by ⊂ quadrics; L satisfies Property N2 if and only if L satisfies Property N1 and the module of syzygies among quadratic generators Qi I is spanned by relations of ∈ the form LiQi =0,whereLi are linear polynomials; and so on.) Now let L = Pn1 ... Pnd (a1, ..., ad), where d, a1, ..., ad,n1, ..., nd are positive integers. AmongP O the papers× × on syzygies in this case we quote [B-M], [Gr1-2], [O-P], [J-P-W], [G-P], [Las], and [P-W]. In the first one, the authors examined the cases in which the resolution is “pure”, i.e. the minimal generators of each module of

Received by the editors December 20, 2000 and, in revised form, July 13, 2001. 2000 Mathematics Subject Classification. Primary 14M25, 13D02.

c 2002 American Mathematical Society

3483

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syzygies have the same degree. We quote the following results from the other papers: Case d = 1, i.e. the case of the Veronese embedding:

Theorem 1 (Green [Gr1-2]). Let a be a positive integer. The line bundle Pn (a) O satisfies Property Na.

Theorem 2 (Ottaviani-Paoletti [O-P]). If n 2, a 3 and the bundle Pn (a) ≥ ≥ O satisfies Property Np,thenp 3a 3. ≤ − Theorem 3 (Josefiak-Pragacz-Weyman [J-P-W]). The bundle Pn (2) satisfies O Property Np if and only if p 5 when n 3 and for all p when n =2. ≤ ≥ (See [O-P] for a more complete bibliography.) Case d =2: Theorem 4 (Gallego-Purnapranja [G-P]). Let a, b 2. The line bundle ≥ P1 P1 (a, b) satisfies Property Np if and only if p 2a +2b 3. O × ≤ − Theorem 5 (Lascoux-Pragacz-Weymann [Las], [P-W]). Let n1,n2 2.Theline ≥ bundle Pn1 Pn2 (1, 1) satisfies Property Np if and only if p 3. O × ≤ Here we consider (1, ..., 1) on P1 ... P1 (d times, for any d). We prove (Section 2): O × × Theorem 6. The line bundle (1, ..., 1) on P1 ... P1 (d times) satisfies Property O × × N3 for any d. Besides we prove (Section 3): Proposition 7. Let X and Y be two projective varieties and let L be a line bundle on X and M alinebundleonY .LetπX : X Y X and πY : X Y Y be the × → × → canonical projections. Suppose L and M satisfy Property N1.Letp 2.IfL does ≥ not satisfy Property Np,thenπ∗ L π∗ M does not satisfy Property Np,either. X ⊗ Y Corollary 8. Let a1, ..., ad be positive integers with a1 a2 ... ad.Suppose ≤ ≤ ≤ k = max i ai =1.Ifk 3 thelinebundle Pn1 ... Pnd (a1, ..., ad) does not { | } ≥ O × × satisfy Property N4 and if d k 2 it does not satisfy Property N2ak+1+2ak+2 2. − ≥ − In particular, from Corollary 8 and Theorem 6, we have:

Corollary 9. Let d 3. The line bundle P1 ... P1 (1, ..., 1) (d times) satisfies ≥ O × × Property Np if and only if p 3. ≤ 2. Proof of Theorem 6 First we have to recall some facts on toric ideals from [St]. k Let k N.LetA = a1, ..., an be a subset of Z . The toric ideal A is defined ∈ { } I as the ideal in C[x1, ..., xn] generated as vector space by the binomials xu1 ... xun xv1 ... xvn 1 n − 1 n n for (u1, ..., un), (v1, .., vn) N ,with i=1,...,n uiai = i=1,...,n viai. ∈ k We have that A is homogeneous if and only if ω Q s.t. ω ai =1 I P ∃P ∈ · i =1, ..., n; the rings C[x1, ..., xn]andC[x1, ..., xn]/ A are multigraded by NA ∀ u1 un I via deg xi = ai; the element x ...x has multidegree b = uiai NA and 1 n i ∈ degree ui = b ω; we define deg b = b ω. i · · P P

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For each b NA,let∆b be the simplicial complex on the set 1, ..., n defined as follows: ∈ { }

∆b = F 1, ..., n : b ai NA { ⊂{ } − ∈ } i F X∈ (thus, by identifying 1, ..., n with A,wehave: { } ∆b = ai , ..., ai , h 1 k i k N,ai ,...,ai A,ai +...+ai =b ∈ 1 k ∈[ 1 k where ai , ..., ai is the simplex generated by ai , ..., ai ). h 1 k i 1 k The following theorem studies the syzygies of the ideal A;itwasprovedby Campillo and Marijuan for k = 1 in [C-M] and by CampilloI and Pison for general k and j = 0 in [C-P]; the following more general statement is due to Sturmfels (Theorem 12.12 p. 120 in [St]).

Theorem 10 (see [St] and also [C-M], [C-P]). Let A = a1, ..., an be a subset of k { } N and A be the associated toric ideal. Let 0 En ... E1 E0 G 0 I → → → → → → be a minimal free resolution of G = C[x1, ..., xn]/ A on C[x1, ..., xn].Eachof I the generators of Ej has a unique multidegree. The number of the generators of multidegree b NA of Ej+1 equalstherankofthej-th reduced homology group ∈ H˜j(∆b, C) of the simplicial complex ∆b. Notation 11. If α is a chain in a topological space, sp(α) will denote the support

of α, i.e. the union of the supports of the simplexes σi s.t. α = i ciσi, ci Z.If X is a simplicial complex, ski(X) will denote the i-skeleton of X. ∈ P Proof of Theorem 6. If we take A = Ad = (1,1, ..., d) i 0, 1 ,wehavethat { 1 |1 ∈{ }} Ad is the ideal of the Segre embedding of P ... P (d times), i.e. the ideal ofI the embedding of P1 ... P1 (d times) by× the line× bundle (1, ..., 1). In fact the ideal of the Segre embedding× × is generated as vector space byO the homogeneous equations of the form

v1,...,d u1 ,...,d x ,..., x ,..., =0 1 d − 1 d 1,...,d 0,1 1,...,d 0,1 Y∈{ } Y∈{ } with u ,..., ,v ,..., N and 1 d 1 d ∈

v1,...,d (1, ..., d)= u1,...,d (1, ..., d).

1,...,d 0,1 1,...,d 0,1 X∈{ } X∈{ } The last condition is equivalent to

v1,...,d (1,1, ..., d)= u1,...,d (1,1, ..., d)

1,...,d 0,1 1,...,d 0,1 X∈{ } X∈{ } (the equality of the first coordinate gives the homogeneity). d In this case ω = ωd =(1, 0, ..., 0) (0 repeated d times) and n =2 . Let b NAd;wehavethatdegb =(=b ω)=k if and only if b is the sum ∈ · d of k (not necessarily distinct) elements of Ad. By identifying the set 1, ..., 2 with A ,wehavethat,ifk =degb,∆ = a {, ..., a }; d b ai ,...,ai Ad,ai +...+ai =b i1 ik 1 k ∈ 1 k h i we say that ai , ..., ai is a degenerate (k 1)-simplex if l, m 1, ..., k with h 1 k i −S ∃ ∈{ } l = m s.t. ai = ai ;thus∆b is equal to the union of the (possibly degenerate) 6 l m (k 1)-simplexes S with vertices in Ad such that the sum of the vertices (with multiplicities)− of S is b.

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By Theorem 10, in order to prove that P1 ... P1 (1, ..., 1) (d times) satisfies N2, O × × we have to prove that H1(∆b)=0foreachb NAd with deg b 4. Analogously ∈ ≥ in order to prove that P1 ... P1 (1, ..., 1) (d times) satisfies N3,wehavetoprove O × × that H2(∆b)=0foreachb NAd with deg b 5. ∈ ≥ The proof is by induction on d. Observe that any b0 NAd+1 with deg b0 = k is b ∈ equal to for some b NAd with deg b = k and for some ε 0, 1, ..., k . Then,  ∈ ∈{ } in order to prove N2 we suppose (by induction) that H1(∆b)=0 b NAd with
∀ ∈ deg b = k, k 4, and we show that H1(∆ b )=0forε 0, ..., k and in order to ≥ (ε) ∈{ } prove N3 we suppose (by induction) that H2(∆b)=0 b NAd with deg b = k, ∀ ∈ k 5, and we show that H2(∆ b )=0forε 0, ..., k . ≥ (ε) ∈{ } Observe that, if ε 0,k (k := deg b), then obviously ∆ b and ∆b are iso- ∈{ } (ε) morphic; besides ∆ b is isomorphic to ∆ b ε 0,...,k (the isomorphism is (k ε) (ε) ∀ ∈{ } given by substituting− 0 with 1 and 1 with 0 in the last coordinate). Thus we may consider only the cases ε 1, ..., [ k ] . ∈{ 2 } First we need some preliminary notation and lemmas.

Notation 12. Let S = a1, ..., ak be a (possibly degenerate) k 1-simplex, ai Ad. Let ε 0, ..., k .Wedenoteh i − ∈ ∈{ } a1 ak Sε0 = , ..., . χ1 χk (χ1,...,χk)s.t.χj 0,1 for j=1,...,k     ∈{ } and exactly ε of χ1[,...,χk are equal to 1

Example 13. Let S = a1,a2,a3,a4 be a (possibly degenerate) tetrahedron, ai h i ∈ Ad.ThesetS10 is the union of the four (possibly degenerate) tetrahedrons a a a a a a a a 1 , 2 , 3 , 4 , 1 , 2 , 3 , 4 , 1 0 0 0 0 1 0 0                 a a a a a a a a 1 , 2 , 3 , 4 , 1 , 2 , 3 , 4 . 0 0 1 0 0 0 0 1                 Thus S10 can be obtained from S by “constructing a tetrahedron on every one of the four faces of S” and considering the union of these four tetrahedrons. The set S20 is the union of the following six (possibly degenerate) tetrahedrons: a a a a a a a a 1 , 2 , 3 , 4 , 1 , 2 , 3 , 4 , 0 0 1 1 0 1 0 1                 a a a a a a a a 1 , 2 , 3 , 4 , 1 , 2 , 3 , 4 , 0 1 1 0 1 0 0 1                 a a a a a a a a 1 , 2 , 3 , 4 , 1 , 2 , 3 , 4 . 1 0 1 0 1 1 0 0                 Then S20 can be obtained from S by “constructing a tetrahedron on every one of the six edges of S” and considering the union of these six tetrahedrons (see Figure 1, representing Sε0 in the case S is not degenerate).

Let b NAd with deg b = k and ε 0, ..., k . Obviously ∈ ∈{ }

∆ b = Sε0 . (ε) S= a1,...,ak with a1+...+ak=b h i [

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S 2 S S S 1 3 a (12) a 3 a 1 a ( 1 ) ( 1 ) ( 3) 0 a1 a4 ( ) ( ) a4 0 a4 0 ( ) 1 a4 (0) a1 (1) a a1 3 a a1 3 a a ( 0 ) 1 3 a1 a ( 0 ) (1 ))(1 a (()) 3 a2 ( 3) 0 0 a ( ) a2 1 2 0 (1 ) a2 ()0 a ( 4) a4 1 (0 ) a4 (1 )

Figure 1.

Note to Figure 1. In the representation of S20 , for the sake of simplicity, we do not a1 a2 a3 a4 a1 a2 a3 a4 represent the tetrahedrons 0 , 1 , 0 , 1 and 1 , 0 , 1 , 0 .







Notation 14. Let b NAd with deg b = k.Forl N,0 l k 1, let ∈ ∈ ≤ ≤ − a a F l(∆ )= i0 , ..., il . b 0 0 a1,...,ak Ad s.t. a1+...+ak=b i0,...,il 1,...,k     ∈ [ [∈{ } l Observe that F (∆b) ∆ b iff k ε l +1. ⊆ (ε) − ≥ The idea of the proof is to consider an l-cycle (for l =1, 2) in ∆ b and to show (ε) l that it is homologous to an l-cycle in F (∆b) and then to show that it is homologous to0byusingthatHl(∆b)=0.

Remark 15. Let S = a1, ..., ak , ai Ad.Ifk 4and1 ε k 2, the set h i al ∈ ≥ ai1 ≤ ai2≤ a−i3 Sε0 contains the cone with vertex 1 on the border of 0 , 0 , 0 for any i1,i2,i3,l 1, ..., k ,withl = ij for j =1, 2, 3. This is true in particular if k 4 ∈{ k } 6



≥ and ε 1, ..., [ 2 ] . ∈{ } al If k 5and1 ε k 3, the set Sε0 contains the cone with vertex 1 on the ≥ ai ≤ ai≤ − border of 1 , ..., 4 for any i1, ..., i4,l 1, ..., k with l = ij for j =1, 2, 3, 4. 0 0 ∈{ } 6
This is true in particular if k 5andε 1, ..., [ k ] .

≥ ∈{ 2 } Definition 16. For any c NAd with deg c = s and ε 1, ..., s , we define Rc,ε to be the following set: ∈ ∈{ }

αi1 αiε 1 αiε αis 1 , ..., − , , ..., − . 1 1 0 0 α1,...,αs Ad i1,...,is 1 1,...,s         ∈ − ∈{ } s.t. α1+[...+αs=c il[=im 6

Lemma 17. Let c NAd with deg c = s. We have that H˜i(∆ c )=0im- (ε 1) ∈ − plies H˜i(Rc,ε)=0if we are in one of the following cases: a) i =0, s 3, ε 1, ..., [ s+1 ] ; b) i =1, s 4, ε 1, ..., [ s+1 ] . ≥ ∈{ 2 } ≥ ∈{ 2 } Proof. Observe that Rc,ε ∆ c .SinceH˜i(∆ c )=0,wehave (ε 1) (ε 1) ⊆ − − i+1 H˜i(sk (∆ c )) = 0. (ε 1) − i+1 i+1 i+1 Obviously sk (Rc,ε) sk (∆ c ). We want to show H˜i(sk (Rc,ε)) = 0. ⊆ (ε 1) i+1 − i+1 Let β be an i-cycle in sk (Rc,ε). Since H˜i(sk (∆ c )) = 0, there exists an (ε 1) −

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i+1 (i +1)-chainη in sk (∆ c )s.t.∂η = β. Suppose sp(η)= Fj,whereFj are (ε 1) j i+1 − i+1 (i +1)-simplexesinsk (∆ c ); consider now an (i +1)-chainψ in sk (Rc,ε) (ε 1) S ˆ −ˆ i+1 ˆ whose support is j Fj,whereFj = Fj if Fj sk (Rc,ε)andFj is a cone on the i+1 ⊆ border of Fj if Fj sk (Rc,ε), in such way that ∂ψ = β (observe that in our S6⊆ cases such cones exist, in fact: Rc,ε is the union of the (possibly degenerate) (s 2)- − simplexes “obtained from the (possibly degenerate) (s 1)-simplexes of ∆ c by − (ε 1) taking off a vertex whose last coordinate is 0”; in the case i = 0 one can check− that the 1-simplexes s.t. the last coordinate of a vertex and the last coordinate of the other vertex are 1, 1 or 1, 0 are contained in Rc,ε, while for a 1-simplex F s.t. the last coordinate of each vertex is 0, there exists a cone, Fˆ,ontheborderofF with i+1 Fˆ Rc,ε,sinces 3; analogously case b)). Thus we proved H˜i(sk (Rc,ε)) = 0. ⊆ ≥ Thus H˜i(Rc,ε)=0.

Proof that P1 ... P1 (1, ..., 1) satisfies property N2. O × × k Lemma 18. Let b NAd, deg b = k, k 4 and ε 1, ..., [ 2 ] . Every 1 cycle γ ∈ 1≥ ∈{ } − in ∆ b is homologous to a 1 cycle in F (∆b) (which is ∆ b since k ε 2). (ε) − ⊆ (ε) − ≥ 1 Proof. Obviously we can suppose sp(γ) sk (∆ b ). The proof is by induction ⊆ (ε) 0 1 on the cardinality of (sp(γ) sk (∆ b )) F (∆b), i.e. we will prove that γ is ∩ (ε) − homologous to a 1-cycleγ ˜ s.t.

0 1 0 1 ]((sp(˜γ) sk (∆ b )) F (∆b)) <]((sp(γ) sk (∆ b )) F (∆b)). ∩ (ε) − ∩ (ε) − a 0 a FirstweremarkthatifP, 1 sk (∆ b )and P, 1 ∆ b ,thenP Rb a,ε ∈ (ε) h i⊆ (ε) ∈ − ( ). ∗ a

In fact, P, ∆ b ,thenP ∆ b a ; we recall that Rb a,ε is 1 (ε) (ε−1) h i⊆ ∈ − −
αi1 αiε 1 αiε αik 2 , ..., − , , ..., − , 1 1 0 0 α1,...,αk 1 Ad i1,...,ik 2 1,...,k 1         − ∈ − ∈{ − } s.t. α1+...[+αk 1=b a il[=im − − 6

i.e. Rb a,ε is the union of the (possibly degenerate) (k 3)-simplexes “obtained − − from the (possibly degenerate) (k 2)-simplexes of ∆ b a by taking off a vertex − (ε−1) whose last coordinate is 0”; then, if the last coordinate of− P is 1, we may conclude at once that P Rb a,ε; also if the last coordinate of P is 0, we may conclude ∈ − that P Rb a,ε, because the number of the vertices whose last coordinate is 0 in ∈ − a (possibly degenerate) (k 2)-simplex of ∆ b a is k 1 (ε 1) 2. − (ε−1) − − − ≥ a 0 1 − Now let 1 (sp(γ) sk (∆ b )) F (∆b), (a Ad). Intuitively we want to ∈ ∩ (ε) − ∈ a modify slightly
γ to obtain a homologous 1-cycleγ ˜ passing through 0 instead of a 1 . a
Let a be the set of all 1-simplexes τ of γ s.t. sp(τ) 1 . We can suppose that
S(1) 3 every 1-simplex τ a be s.t. sp(τ)bea1-simplexof∆b and τ :[0, 1] ∆ b ∈S(1) (
ε) → (ε) be injective. a For τ a ,letτ 0 :[0, 1] ∆ b be a 1-simplex s.t. τ 0()=τ()ifτ() = 1 ∈S(1) → (ε) 6 and τ ()= a if τ()= a ,for 0, 1 .Letα be the 1 cycle ( τ + τ ). 0 0 1 τ a 0
∈{ } ∈S(1) −

P

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By remark ( ), we have that sp(α) C,whereC is the union of the two cones a ∗ a ⊆ 1 ,Rb a,ε and 0 ,Rb a,ε .ObservethatC ∆ b . We state that H1(C)=0: h − i h − i ⊆ (ε) by Theorem 10, since P1 ... P1 (1, ..., 1) (d times) satisfies Property N1 d,we

O × × ∀ have H˜0(∆g)=0 g NAd with deg g 3, d (this can easily be proved directly ∀ ∈ ≥ ∀ ˜ without using that P1 ... P1 (1, ..., 1) satisfies Property N1); then H0(∆ b a )=0; (ε−1) O × × − thus H˜0(Rb a,ε) = 0 by Lemma 17; we have H˜i(C)=H˜i 1(Rb a,ε); thus H1(C)= − − − H˜0(Rb a,ε) = 0. Thus we have that α is homologous to 0. Thus γ is homologous to γ +−α. Obviously the support ofγ ˜ := γ + α can be obtained from sp(γ) by substituting sp(τ)withsp(τ 0) τ a .Then ∀ ∈S(1) 0 1 0 1 ]((sp(˜γ) sk (∆ b )) F (∆b)) <]((sp(γ) sk (∆ b )) F (∆b)); ∩ (ε) − ∩ (ε) − thus we conclude the proof of Lemma 18.

In order to prove that P1 ... P1 (1, .., 1) (d times) satisfies N2 for any d,we O × × suppose (by induction) that H1(∆b)=0 b NAd with deg b = k, k 4, and we k ∀ ∈ ≥ show that H1(∆ b )=0forε 1, ..., [ 2 ] . (ε) ∈{ } Cases ε k 3. We know that every 1-cycle γ in ∆ b is homologous to a ≤ − (ε) 1 2 2 1 cycle in F (∆b) by Lemma 18. Thus, since F (∆b) ∆ b and H1(F (∆b)) = 0 − ⊆ (ε) (because, by induction hypothesis, H1(∆b)=0),wehavethatH1(∆ b )=0. (ε) Cases ε>k 3. These cases are slightly more difficult. By Lemma 18 every 1- − 1 cycle γ in ∆ b is homologous to a 1-cycle γ0 in F (∆b). But in these cases we have (ε) 2 not the inclusion F (∆b) ∆(b), thus we have to conclude the proof in another way. ⊆ ε 2 2 Since H1(F (∆b)) = 0, there exists a 2-chain µ in F (∆b)s.t. ∂µ = γ0.Let 2 sp(µ)= Fi, Fi triangles in F (∆b). Consider a 2-chain ψ in ∆ b whose support i (ε) ˆ ˆ is i Fi,whereS Fi is a cone ∆ b on the border of Fi (there exists by Remark ⊆ (ε) 15), in such way that ∂ψ = γ0;thus[γ0]=0inH1(∆ b ), thus [γ]=0inH1(∆ b ). S (ε) (ε) Thus H1(∆ b )=0. (ε)

Proof that P1 ... P1 (1, ..., 1) satisfies property N3. O × × k Lemma 19. Let b NAd with deg b = k and ε 1, ..., [ 2 ] .Ifk 5, every ∈ ∈{2 } ≥ 2-cycle µ in ∆ b is homologous to a 2-cycle in F (∆b) (which is ∆ b since (ε) ⊆ (ε) k ε 3). − ≥ 2 Proof. Obviously we can suppose that sp(µ) sk (∆ b ). The proof is by in- ⊆ (ε) 0 2 duction on the cardinality of (sp(µ) sk (∆ b )) F (∆b), i.e. we will prove ∩ (ε) − 0 2 that µ is homologous to a 2-cycleµ ˜ s.t. ]((sp(˜µ) sk (∆ b )) F (∆b)) < ∩ (ε) − 0 2 ]((sp(µ) sk (∆ b )) F (∆b)). ∩ (ε) − a 0 a We remark that if P, Q, 1 sk (∆ b )and P, Q, 1 ∆ b ,then P, Q ∈ (ε) h i⊆ (ε) h i⊆ Rb a,ε.() − a

∗ In fact P, Q, ∆ b ;then P, Q ∆ b a .SinceR is the union of 1 − b a,ε h i⊆ (ε) h i⊆ (ε 1) − the (possibly degenerate) (k 3)-simplexes “obtained− from the (possibly degenerate)
− (k 2)-simplexes of ∆ b a by taking off a vertex whose last coordinate is 0” and (ε−1) − −

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since the number of the vertices whose last coordinate is 0 in a (possibly degenerate) (k 2)-simplex of ∆ b a is k 1 (ε 1) 3, we have P, Q Rb a,ε. (ε−1) − − − − − ≥ h i⊆ − Let a be the set of all 2-simplexes τ : (0, 0), (0, 1), (1, 0) ∆ b of µ S(1) h i→ (ε) a s.t. sp(τ) 1 . We can suppose that every 2-simplex τ a be s.t. sp(τ) 3 ∈S(1) 0 be a 2-simplex of ∆ b , τ() sk (∆ b )for (0, 0), (0, 1), (1, 0) and τ :
(ε) ∈ (ε) ∈{ } (0, 0), (0, 1), (1, 0) ∆ b be injective. h i→ (ε) For τ a let τ 0 : (0, 0), (0, 1), (1, 0) ∆ b be a 2-simplex s.t. τ 0()=τ() ∈S(1) h i→ (ε) a a a if τ() = 1 and τ 0()= 0 if τ()= 1 for  (0, 0), (0, 1), (1, 0) .Letα be the 2-cycle6 ( τ + τ ). ∈{ } τ a 0 ∈S
(1) −

By remarkP ( ), we have that sp(α) C,whereC is the union of the two a ∗ a ⊆ cones 1 ,Rb a,ε and 0 ,Rb a,ε .ObservethatC ∆ b . We state that h − i h − i ⊆ (ε) H2(C) = 0: we have already proved that P1 ... P1 (1, ..., 1) satisfies Property N2

O × × i.e. H1(∆g)=0 g with deg g 4; thus H1(∆ b a )=0;thenH1(Rb a,ε)=0by (ε−1) ∀ ≥ − − Lemma 17; we have H˜i(C)=H˜i 1(Rb a,ε); thus H2(C)=H1(Rb a,ε)=0.Thus we have that α is homologous to− 0. Thus− µ is homologous to µ + α−. The support of the cycleµ ˜ := µ + α can be obtained from sp(µ) by substituting sp(τ)withsp(τ 0) τ a .Then ∀ ∈S(1) 0 2 0 2 ]((sp(˜µ) sk (∆ b )) F (∆b)) <]((sp(µ) sk (∆ b )) F (∆b)); ∩ (ε) − ∩ (ε) − thus we conclude the proof of Lemma 19.

In order to prove that P1 ... P1 (1, ..., 1) (d times) satisfies N3 for any d,we O × × suppose (by induction) that H2(∆b)=0 b NAd with deg b = k, k 5andwe k ∀ ∈ ≥ show that H2(∆ b )=0forε 1, ..., [ 2 ] . (ε) ∈{ } Cases ε k 4. We have that every 2-cycle µ in ∆ b is homologous to a 2- ≤ − (ε) 2 3 cycleµ ˜ in F (∆b) by Lemma 19. Since H2(F (∆b)) = 0 (because, by induction 3 hypothesis, H2(∆b)=0),wehavethat[˜µ]=0inH2(F (∆b)) = 0. Since in 3 these cases F (∆b) ∆ b , we may conclude that [µ]=[˜µ]=0inH2(∆ b ), thus ⊆ (ε) (ε) H2(∆ b )=0. (ε) Cases ε>k 4. We have that every 2-cycle µ in ∆ b is homologous to a 2- − (ε) 2 3 cycleµ ˜ in F (∆b) by Lemma 19. Since H2(F (∆b)) = 0 (because H2(∆b)=0), 3 we have that [˜µ]=0inH2(F (∆b)) = 0. But in these cases we have not the 3 3 inclusion F (∆b) ∆ b , thus we may not conclude at once. Since H2(F (∆b)) = 0, ⊆ (ε) 3 there exists a 3-chain ν in F (∆b)s.t. ∂ν =˜µ.Wehavethatsp(ν)= i Fi, Fi 3 ˆ tetrahedrons in F (∆b). Consider a 3-chain ψ in ∆ b whose support is i Fi, (ε) S ˆ where Fi is a cone ∆ b on the border of Fi (there exists by Remark 15), inS such ⊆ (ε) awaythat∂ψ =˜µ;thus[˜µ]=0inH2(∆ b ), thus [µ]=0inH2(∆ b ). Then we (ε) (ε) may conclude that H2(∆ b )=0. (ε) This completes the proof of Theorem 6.

3. Proof of Proposition 7 Let X and Y be two projective varieties and L a very ample line bundle on X 0 and M a very ample line bundle on Y .Let σ0, ..., σk be a basis of H (X, L)and { }

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0 let s0, ..., sl be a basis of H (Y,M); we can suppose y Y s.t. s0(y) =0, { } ∃ ∈ 6 sj(y)=0forj =0;letti,j be the coordinates corresponding to σi sj i,j of the 6 { ⊗ } embedding of X Y by πX∗ L πY∗ M (where π is the projection on )andletti be × ⊗ · · the coordinates corresponding to σ0, ..., σk of the embedding of X by L. { } Remark 20. By setting ti,j =0forj = 0 in an equation of X Y andthentaking off the last index (a 0) of each variable,6 we get an equation of ×X (to prove this, use y).

Remark 21. Let M be a graded module on C[x1, ..., xn] with a minimal set of generators of degree s; then a subset of elements of degree s of M can be extended to a minimal set of generators if and only if these elements are linearly independent on C.

Proof of Proposition 7. Suppose L satisfies Property Np 1 but not Np.Wewantto − show π∗ L π∗ M does not satisfy Property Np; we can suppose π∗ L π∗ M satisfies X ⊗ Y X ⊗ Y Property Np 1.Letlm and qm be the ranks of the m-module of a minimal free − m graded resolution respectively of G(L)andofG(π∗ L π∗ M). Let g j=1,...,l be X ⊗ Y { j } m a minimal set of generators of the m-module Em of a minimal resolution of G(L),

... Em Em 1 ... E0 G(L) 0. → → − → → → → p 1 Since L satisfies Property Np 1 but not Np, there exists a syzygy S of (g1− , ..., p 1 − p 1 p 1 g − ), s.t. S is not generated by linear syzygies of (g − , ..., g − ). Add a 0 to lp 1 1 lp 1 − − the indices of the variables appearing in S and call S˜ the so-obtained vector of ˜ ˜ polynomials; let S0 =(S,0, ..., 0) with 0 repeated qp 1 lp 1 times. Obviously by adding a 0 to the indices of each variable− − appearing− in the equations of X, we get equations of X Y and by adding a 0 to the indices of every variable appearing in the syzygies of ×X we get syzygies of X Y . × m m Add a 0 to the indices of the variables appearing in gj and callg ˜j the so- 1 1 obtained vector of polynomials for j =1, ..., lm;setfj =˜gj for j =1, ..., l1 and m m fj =(˜gj , 0, ..., 0) (0 repeated qm 1 lm 1 times) for j =1, ..., lm and 2 m m − − − ≤ ≤ p 1; fj for j =1, ..., lm are vectors of linear polynomials for 2 m p 1and they− are quadratic if m = 1. Thus, by induction on m and by Remark≤ ≤ 21,− one can m extend this set to a minimal set of generators f j=1,...,q ,ofthem-module of a { j } m minimal resolution of G(π∗ L π∗ M)form p 1 (we recall that we supposed X ⊗ Y ≤ − πX∗ L πY∗ M satisfies Property Np 1);wecandoitinsuchwaythat,whenwe ⊗ − 1 set ti,j =0forj =0,wehavethatfj is zero for j = l1 +1, ..., q1 and the r-th m 6 coordinate of fj is zero for r lm 1 and j = lm +1, ..., qm (we can prove this by induction on m, by using Remark≤ 20− for the case m = 1: it is sufficient to subtract m m linear combination of fj for j =1, ..., lm to fj for j = lm +1, ..., qm). p 1 p 1 ˜ − Obviously S0 is a syzygy of (f1 , ..., fqp− 1 ). − ˜ If πX∗ L πY∗ M satisfies Property Np,thenS0 would be generated by linear ⊗ p 1 p 1 − syzygies of (f1 , ..., fqp− 1 ). − p 1 p 1 ˜ − We state that S0 cannot be generated by linear syzygies of (f1 , ..., fqp− 1 ). In p 1 p 1 − ˜ − fact, if it were, say S0 = α Sα (Sα linear syzygies of (f1 , ..., fqp− 1 )), we set ˜ − ti,j =0forj = 0 in each member of the equality S0 = Sα and, by taking off 6 P α the last index (a 0) of every variable and considering only the first lp 1 coordinates P − of S and Sα, one would obtain that S would be generated by linear syzygies of

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p 1 p 1 (g − , ..., g − ) (observe that by setting t =0forj =0inS and taking the first 1 lp 1 i,j α − p 1 p 1 6 lp 1 coordinates, we get a syzygy of (f1 − , ..., fl − )). − p 1 But S cannot be generated by linear syzygies− by assumption.

The line bundle P1 P1 P1 (1, 1, 1) does not satisfy Property N4; precisely the resolution, with theO notation× × of Introduction, is

0 S( 6) S( 4)9 S( 3)16 S( 2)9 S G 0. → − → − → − → − → → → This has been proved by Barcanescu and Manolache in [B-M] and can be seen also by using the program Macaulay [B-S] (to see only that P1 P1 P1 (1, 1, 1) does O × × not satisfy Property N4, it is sufficient to use the autoduality of the resolution; see [B-M]). From this and from Proposition 7 we deduce that P1 ... P1 (1, ..., 1) (d times) O × × does not satisfy Property N4 for d 3. By also using Gallego-Purnapranja’s ≥ Theorem 4, we deduce that, if a1, ..., ad are integer numbers with a1 a2 ... ad ≤ ≤ ≤ and a1 = ... = ak = 1, the line bundle P1 ... P1 (a1, ..., ad) does not satisfy O × × Property N4 if k 3 and it does not satisfy Property N2ak+1+2ak+2 2 if d k 2. With the same≥ argument as in the Remark in part II, Section 2− of [Gr1-2]− ≥ we deduce Corollary 8.

Acknowledgements I thank G. Ottaviani for several helpful conversations and above all for the valuable suggestion to study syzygies of Segre embeddings. I thank also G. Anzidei and C. Brandigi for some discussions and the referee for some comments.

References

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