A note on the Intersection of Veronese Surfaces

D. Eisenbud, K. Hulek and S. Popescu

February 10, 2003

0 Introduction

The main purpose of this note it to prove the following

Theorem 0.1 Any two Veronese surfaces in ℙ^{5} whose intersection is zerodimensional meet in at most 10 points (counted with multiplicity).

Our initial motivation for this note comes from our paper [EGHPO] where we study linear syzygies of homogeneous ideals generated by quadrics and their restriction to subvarieties of the ambient projective space with known (linear) minimal free resolution. A direct application of the techniques in [EGHPO, Section 3] shows that the homogeneous ideal of a zerodimensional intersection of two Veronese surfaces in ℙ^{5} is 5-regular (see also Lemma 1.1 below), which yields only an upper bound of 12 for its degree, cf. Section 1.

Section 2 analyzes Veronese surfaces on hyperquadrics. The observation that two Veronese surfaces on a smooth hyperquadric Q ⊂ ℙ^{5} meeting in a zero-dimensional subscheme, must meet in a subscheme of length 10 or 6 is classical and goes back to Kummer [Ku] and Reye [Rey] (see also [Jes1] for historical comments): By regarding the smooth hyperquadric Q ⊂ ℙ^{5} as the Plücker embedding of the Grassmannian of lines Gr(ℙ^{1}, ℙ^{3}), a Veronese surface on Q ⊂ ℙ^{5} is, up to duality, the congruence of secant lines to a twisted cubic curve and thus has bidegree (1, 3). More precisely, the congruence has one line passing through a generic point of ℙ^{3} and 3 lines contained in a generic plane. Thus Schubert calculus yields that the possible intersection numbers of two Veronese surfaces on Q are either 10 or 6. See Proposition 2.1 below and the following remark, or the computation of the number of common chords of two space curves in [GH, page 297].

The case of two Veronese surfaces in ℙ^{5} meeting in 10 simple points has also been investigated in relation with association (projective Gale transform) by Coble [Cob], Conner [Con] and others. In [Cob, Theorem 26] Coble claims that 10 points in ℙ^{5} which are associated to the 10 nodes of a symmetroid in ℙ^{3}, the quartic surface defined by the determinant of a symmetric 4 × 4 matrix with linear entries in ℙ^{3}, are the (simple) intersection points of two Veronese surfaces in ℙ^{5} (see Proposition 3.3 below). This is based on Reye’s observation [Rey, page 78-79] that 4 × 4 symmetric matrices with linear entries in ℙ^{3} are actually catalecticant with respect to suitable bases and on the analysis in [EiPo] of the Gale Transform of zerodimensional determinantal schemes. Section 3 contains a modern account of these results.

In Section 4 we briefly discuss which intersection numbers ≤ 10 can actually occur for two Veronese surfaces in ℙ^{5} and in which geometric situation this can happen. For instance, we show that two Veronese surfaces in ℙ^{5} cannot intersect transversally in 9 points, however they may intersect in non-reduced zero-dimensional schemes of this degree.

With the exception of Section 3 all other results in this note are valid in arbitrary characteristic.

1 A reduction step

We shall make essential use of the following lemma whose proof is reminiscent of the linear syzygies techniques used in [EGHPO, Section 3].

Lemma 1.1 If X_{1} and X_{2} are two Veronese surfaces in ℙ^{5} meeting in a zero-dimensional scheme W, then the ideal sheaf I_{W,ℙ^{2}} of W regarded as a subscheme of ℙ^{2} is 5-regular.

Proof. The claim is equivalent to the vanishing h^{1}(I_{W,ℙ^{2}}(4)) = 0. In order to see this we consider the minimal resolution of I_{X_{1}} in ℙ^{5} and restrict it to X_{2}. This yields a complex abutting to I_{W}. Since X_{1} has property N_{p} for all p one immediately computes that h^{1}(I_{W,ℙ^{2}}(4)) = 0. □

As immediate consequence we get a first bound for the number of points where two Veronese surfaces whose intersection is zero-dimensional can meet.

Proposition 1.2 Two Veronese surfaces in ℙ^{5} whose intersection is zerodimensional meet in at most 12 points.

Proof. Let X_{1} and X_{2} be two Veronese surfaces meeting in a zero-dimensional scheme W, and let d = length(W). By Lemma 1.1, W regarded as a subscheme of X_{2} ≅ ℙ^{2} imposes independent conditions on plane quartics, in particular if d ≤ 15. On the other hand X_{2} ⊂ ℙ^{5} is cut out by quadrics scheme-theoretically, and thus the quartics in H^{0}(I_{W,ℙ^{2}}(4)) must cut out W ⊂ ℙ^{2} scheme-theoretically too – in particular there are at least two. If there were only two, then they would form a complete intersection, generating a saturated ideal, and thus W ⊂ ℙ^{2} would be a complete intersection of two plane quartics, which cannot be 5-regular. It follows that h^{0}(I_{W,ℙ^{2}}(4)) ≥ 3, and thus that d ≤ 12. □

Actually the above proof yields also the following estimate

Proposition 1.3 Let X_{1} and X_{2} be two Veronese surfaces in ℙ^{5} meeting in a zero-dimensional scheme W of length d with 10 ≤ d ≤ 12. Then X_{1} ∪ X_{2} lies on at least d − 9 quadrics.

Proof. Let a = h^{0}(I_{X_{1}∪X_{2}}(2)). Then on one hand h^{0}(I_{W,ℙ^{2}}(4)) ≥ h^{0}(I_{X_{2}}(2))− a = 6 − a, on the other hand, by Lemma 1.1, we know that h^{0}(I_{W,ℙ^{2}}(4)) = 15 − d. Combining the two proves the claim of the proposition. □

2 Veronese surfaces on hyperquadrics

In this section we analyze the intersection of two Veronese surfaces which meet in finitely many points, in the case where the two surfaces lie on a common hyperquadric, resp. a pencil of hyperquadrics. We begin with the classical case of congruences of lines:

Proposition 2.1 Assume that X_{1} and X_{2} are Veronese surfaces which meet in finitely many points and assume moreover that there exists a smooth quadric hypersurface Q with X_{1} ∪ X_{2} ⊂ Q. Then X_{1}.X_{2} = 6 or 10.

Proof. Since Q is smooth it is isomorphic to the Grassmannian Gr(ℙ^{1}, ℙ^{3}) and it is well known that H^{4}(Gr(ℙ^{1}, ℙ^{3}), ℤ) = ℤα + ℤβ where α and β are 2-planes. It follows from the double point formula (see for instance [HS]) that every Veronese surface on Gr(ℙ^{1}, ℙ^{3}) has class 3α + β or α + 3β. Since α^{2} = β^{2} = 1 and αβ = 0 it follows that X_{1}X_{2} = 10 or 6 depending on whether X_{1} and X_{2} belong to the same class or not. □

Remark As already mentioned in the introduction, a Veronese surface of class 3α + β on the Grassmannian Gr(ℙ^{1}, ℙ^{3}) ≅ Q ⊂ ℙ^{5} is the congruence of secant lines to a twisted cubic curve (where β is the cycle of lines passing through a point of ℙ^{3} while α is the cycle of lines in a plane). Passing to the dual ℙ^{3} exchanges α and β, so up to duality the same construction accounts also for Veronese surfaces of class α + 3β. It is easy to see, for instance by using Kleiman’s transversality theorem, that both cases described in Proposition 2.1 actually occur.

Proposition 2.2 Assume that X_{1} and X_{2} are Veronese surfaces in ℙ^{5} which meet in finitely many points and assume that there exists a rank 5 hyperquadric Q containing both X_{1} and X_{2}. Then X_{1}.X_{2} = 8.

Proof. We first claim that X_{1} and X_{2} do not pass through the vertex P of the quadric cone Q. Otherwise projection from P would map the Veronese surface to a cubic scroll contained in a smooth hyperquadric Q′ ⊂ ℙ^{4}. But this is impossible, since by the Lefschetz theorem every surface on Q′ is a complete intersection and hence has even degree. Blowing up the point P we obtain a diagram

\xymatrix{Q̃ \ar[r]^{p} \ar[d]^{π} & Q \\ Q′}

where π gives Q̃ the structure of a ℙ^{1}-bundle over Q′. Since X_{1} and X_{2} do not go through the point P they are not blown up and we will, by abuse of notation, also denote their pre-images in Q̃ by X_{1} and X_{2}. The Chow ring of Q̃ is generated by H = p^{∗}(H_{ℙ^{5}}) and H′ = π^{∗}(H_{ℙ^{4}}). Clearly H^{4} = H^{3}H′ = H^{2}(H′)^{2} = H(H′)^{3} = 2 and (H′)^{4} = 0. Let E be the exceptional divisor of the map p. Then E = αH + βH′ and from EH^{3} = 0 and E(H′)^{3} = 2 one deduces α = 1 and β = −1, i.e. E = H − H′. The surfaces X_{i} have class

X_{i} = α_{i}H^{2} + β_{i}HH′ + γ_{i}(H′)^{2}.

From X_{i}H^{2} = deg X_{i} = 4 one computes α_{i} + β_{i} + γ_{i} = 2. Since X_{i} does not meet E we have X_{i}EH′ = 0 and from this one deduces γ_{i} = 0 and hence

X_{i} = α_{i}H^{2} + (2 − α_{i})HH′.

But then X_{i}^{2} = 8 and this proves the claim. □

We analyze next what happens when X_{1} and X_{2} lie on a pencil of hyperquadrics.

Proposition 2.3 Assume that the Veronese surfaces X_{1} and X_{2} meet in finitely many points and assume that they are contained in a pencil of hyperquadrics {λ_{1}Q_{1} + λ_{2}Q_{2} = 0}. Then for a general hyperquadric Q in this pencil

X_{1} ∩ X_{2} ∩ Sing Q = ∅.

Proof. Assume that this is false. Since X_{1} ∩ X_{2} is a finite set it follows that there exists some point P ∈ X_{1}∩X_{2} which is singular for every hyperquadric Q in this pencil. From what we saw in the proof of Proposition 2.2 this also shows that the general quadric in this pencil has rank at most 4 (and at least 3 since both surfaces X_{i} are non-degenerate). Projecting from P maps X_{1} and X_{2} to rational cubic scrolls Y_{1} and Y_{2} in ℙ^{4}, respectively. These cubic scrolls are contained in a pencil of non-degenerate hyperquadrics {λ_{1}Q′_{1} + λ_{2}Q′_{2} = 0} whose general member has rank 3 or 4. For degree reasons this implies Y_{1} = Y_{2}. (Incidentally this also shows that X_{1} and X_{2} are contained in a net of hyperquadrics whose general element has rank 4.)

Let Y be the cone over Y_{1} = Y_{2} with vertex P. We obviously have X_{1}, X_{2} ⊂ Y. We now blow up in P and obtain a diagram

\xymatrix{Ỹ \ar[r]^{p} \ar[d]^{π} & Y ⊂ ℙ^{5} \\ Y_{1}}

where Ỹ is a ℙ^{1}-bundle over Y_{1}. The Picard group of Y_{1} is generated by two elements C_{0} and F with C_{0}^{2} = −1, C_{0}F = 1 and F^{2} = 0. Let F_{1} = π^{∗}C_{0} and F_{2} = π^{∗}F_{1}. Then the Chow group on Ỹ is generated by H = p^{∗}(H_{ℙ^{5}}), F_{1} and F_{2}.

For geometric reasons H^{3} = 3, F_{1}^{2}H = −1, F_{1}F_{2}H = F_{1}H^{2} = F_{2}H^{2} = 1 and F_{2}^{2}H = F_{1}^{2}F_{2} = F_{2}^{2}F_{1} = 0. Let E be the exceptional locus of p and let X̃_{i} denote the strict transforms of the Veronese surfaces. Then π restricted to X̃_{i} defines isomorphisms between X̃_{i} and Y_{1}. Since X_{1} and X_{2} intersect in only finitely many points and since both are contained in the 3-dimensional cone Y it follows that X_{1} ∩ X_{2} = {P }, and from this one concludes that X̃_{1} ∩ X̃_{2} = L where L = E ∩ F_{1} is a projective line, respectively X̃_{1}X̃_{2} = aL for some a ≥ 1. Next we want to determine the class of X̃_{i} in Ỹ. Since the X̃_{i} are sections of the ℙ^{1}-bundle π : Ỹ → Y we find X̃_{i} = H + β_{i}F_{1} + γ_{i}F_{2}; i = 1, 2. Restricting this to E and using that H is trivial on E we immediately find that β_{i} = 1 and γ_{i} = 0, i.e. X̃_{i} = H +F_{1}. But then X̃_{1}X̃_{2} = H^{2} + 2HF_{1} + F_{1}^{2} ≠ aL where the latter inequality can be seen e.g. by intersecting with H. This is a contradiction and the proposition is proved. □

Proposition 2.4 Let X_{1}, X_{2} be two Veronese surfaces in ℙ^{5} intersecting in a finite number of points. If X_{1} and X_{2} are contained in a pencil of hyperquadrics, then X_{1}.X_{2} ≤ 10.

Proof. Let r be the rank of a general element of this pencil of hyperquadrics. If r = 6 or 5 then the assertion follows from Proposition 2.1, or from Proposition 2.2, respectively. On the other hand, since the surfaces X_{i} are nondegenerate we must have r ≥ 3.

We shall first treat the case r = 4. According to Proposition 2.3 we can then choose a rank 4 hyperquadric Q with X_{1} ∩ X_{2} ∩ Sing Q = ∅. Blowing up the singular line L of Q we obtain a diagram

\xymatrix{Q̃ \ar[rr]^{p} \ar[dr]^{π} &  & Q ⊂ ℙ^{5} \\  & Q′ := ℙ^{1} × ℙ^{1}}

where π is the structure map of a ℙ^{2}-bundle. We denote the strict transforms of X_{i} by X̃_{i}, i = 1, 2. Since X_{1} ∩ X_{2} ∩ Sing Q = ∅ we have X̃_{1}.X̃_{2} = X_{1}.X_{2}. Let H = p^{∗}(H_{ℙ^{5}}), let L_{1}, L_{2} denote the rulings of ℙ^{1} ×ℙ^{1} and set F_{i} = π^{∗}L_{i}, i = 1, 2. Then H^{4} = 2, F_{1}H^{3} = F_{2}H^{3} = F_{1}F_{2}H^{2} = 1 and F_{1}^{2} = F_{2}^{2} = 0. Let E be the exceptional locus of p. Its class must be of the form E = H + α_{1}F_{1} + α_{2}F_{2}. From EF_{1}H^{2} = H^{2} = EF_{2}H^{2} = 0 one deduces α_{1} = α_{2} = −1, i.e. E = H − F_{1} − F_{2}.

Now let X be any Veronese surface on Q. We want to determine the possible classes of the strict transform X̃ of X in Q̃. Let X̃ = αH^{2}+β_{1}F_{1}H+ β_{2}F_{2}H + γF_{1}F_{2}. From X̃H^{2} = 4 we obtain 2α + β_{1} + β_{2} + γ = 4. A priory the singular line L can either be disjoint from X, meet it transversally in one point, be a proper secant or a tangent of X. Projection from L shows that only the first and the third of these possibilities can occur.

Assume first that L and X are disjoint. Then π_{|X̃} : X̃ → ℙ^{1} × ℙ^{1} is a 2 : 1 map which shows α = 2. From X̃EF_{1} = X̃EF_{2} = 0 we conclude β_{1} = β_{2} = 0 and hence γ = 0, i.e. X̃ = 2H^{2}.

Assume now that L is a proper secant of X. Blowing up Q along L then blows up X in 2 points and the corresponding exceptional curves are mapped to different rulings in ℙ^{1} × ℙ^{1}. The map π_{|X̃} : X̃ → ℙ^{1} × ℙ^{1} is now birational and hence α = 1. From what we have just said it follows that X̃EF_{1} = X̃EF_{2} = 1 and hence this implies β_{1} = β_{2} = 1. But then γ = 0 and X̃ = H^{2} + HF_{1} + HF_{2}.

Let c_{1} = 2H^{2} and c_{2} = H^{2} + HF_{1} + HF_{2}. The claim of the proposition now follows for the rank 4 case since c_{1}^{2} = c_{2}^{2} = c_{1}c_{2} = 8.

It remains to deal with the case in which the general hyperquadric Q in the pencil has rank 3. By Proposition 2.3 we can again assume that X_{1} ∩ X_{2} ∩ Sing Q = ∅. Projection from the singular locus of Q gives a diagram

\xymatrix{Q̃ \ar[r]^{p} \ar[d]^{π} & Q ⊂ ℙ^{5} \\ C ≅ ℙ^{1}}

where C is a conic section and π : Q̃ → C is a ℙ^{3}-bundle. We denote H = p^{∗}(H_{ℙ^{5}}) and F = π^{∗}(pt). Then H^{4} = 2, H^{3}F = 1 and F^{2} = 0. Let X be any Veronese surface on Q and denote its strict transform on Q̃ by X̃. We first note that X ∩ Sing Q is a finite set. Otherwise X ∩ Sing Q would have to be a conic section and projection from Sing Q would map X onto a plane, not to a conic. Finally let E be the exceptional locus of p. The class of E must be of the form E = H + γF and from EH^{3} = 0 it follows that γ = −1, i.e. E = H − F. Now put X̃ = αH^{2} + βH.F. From X̃H^{2} = 4 one deduces that 2α + β = 4. Since X ∩ Sing Q is finite one must have that X̃EH = 0 and hence β = 0. This shows that X̃ = 2H^{2} and the claim of the proposition follows since X_{1}.X_{2} = X̃_{1}.X̃_{2} = 4H^{4} = 8. □

Remark The above proof shows that if the general element in the pencil of hyperquadrics containing X_{1} and X_{2} has rank 3 or 4, then X_{1}.X_{2} = 8.

3 Catalecticant symmetroids and Veronese surfaces

In this section we prove Coble’s claim [Cob, Theorem 26], mentioned in the introduction, that ten points in ℙ^{5} which are the Gale transform of the nodes of a general quartic symmetroid in ℙ^{3} are the simple intersection points of two Veronese surfaces.

A quartic symmetroid is the quartic surface in ℙ^{3} defined by the determinant of a symmetric 4 × 4 matrix with linear entries in ℙ^{3}; for general choices (of the matrix) the symmetroid has only ordinary double points as singularities (nodes) and their number is 10, by Porteous’ formula. These surfaces are sometimes called Cayley symmetroids, as Cayley initiated their study in [Cay] (cf. [Jes2], but see [Cos] for a modern account of Cayley’s results and much more).

A symmetric matrix whose diagonals are constant is called a catalecticant matrix. Surprisingly enough, it turns out that a symmetric 4×4 matrix with linear entries in ℙ^{3} can always be reduced to a catalecticant form (with respect to suitable bases). This fact goes back to Reye [Rey, page 78-79] and Conner [Con, page 39] and is (re)-proved below.

We will make use of the perfect pairing, called apolarity, between forms of degree n and homogeneous differential operators of order n induced by the action of T = k[∂_{0}, …, ∂_{r}] on S = k[x_{0}, …, x_{r}] via differentiation:

∂^{α}(x^{β}) = α!\binom{β}{α}x^{β−α},

if β ≥ α and 0 otherwise, and where α and β are multi-indices, \binom{β}{α} = ∏ \binom{β_{i}}{α_{i}}, and k is a field of characteristic zero.

Proposition 3.1 The Hessian matrix of a web of quadrics in ℙ^{3} is catalecticant (with respect to a suitable basis) if and only if the quadrics in the web annihilate the quadrics of a twisted cubic curve. (One says in this situation that the web is “orthic” to the twisted cubic curve.)

Proof. Let q : W^{∗} ^{⊂} Sym_{2}V be the web of quadrics on ℙ^{3} = ℙ(V). A twisted cubic C ⊂ ℙ̌^{3} = ℙ(V^{∗}) is defined by its quadrics H^{0}(ℙ̌^{3}, I_{C}(2)). In suitable coordinates, say ∂_{0}, …, ∂_{3}, these are the minors of the matrix

(\begin{matrix}∂_{0} & ∂_{1} & ∂_{2} \\ ∂_{1} & ∂_{2} & ∂_{3} \end{matrix}) .

In terms of the dual coordinates, x_{0}, …, x_{3} of ℙ(V), the web q has the form

a_{0}x_{0}^{2} + a_{4}x_{1}^{2} + a_{7}x_{2}^{2} + a_{9}x_{3}^{2} + 2a_{1}x_{0}x_{1} + 2a_{2}x_{0}x_{2} + 2a_{3}x_{0}x_{3}
+ 2a_{5}x_{1}x_{2} + 2a_{6}x_{1}x_{3} + 2a_{8}x_{2}x_{3}

where a_{0}, a_{1}, …, a_{9} are linear forms in the variables of W. Direct computation shows that a quadric in the web q is annihilated by the equations ∂_{0}∂_{2} − ∂_{1}^{2}, ∂_{0}∂_{3} − ∂_{1}∂_{2}, ∂_{1}∂_{3} − ∂_{2}^{2} if and only if a_{2} = a_{4}, a_{3} = a_{5}, a_{6} = a_{7}. It follows that the web q is orthic to the twisted cubic C iff its Hessian matrix has shape

(\begin{matrix}b_{0} & b_{1} & b_{2} & b_{3} \\ b_{1} & b_{2} & b_{3} & b_{4} \\ b_{2} & b_{3} & b_{4} & b_{5} \\ b_{3} & b_{4} & b_{5} & b_{6} \end{matrix})

where b_{0}, b_{1}, …, b_{6} are linear forms in the variables of W, i.e. it is catalecticant. □

It actually turns out that a 4 × 4 symmetric matrix with linear entries in ℙ^{3} can be represented in two different ways as a catalecticant matrix. Namely

Proposition 3.2 There are exactly two twisted cubic curves whose defining quadrics are annihilated by a general web of quadrics in ℙ^{3}.

Proof. We use the same notation as in the proof of the previous proposition. Namely, let q : W^{∗} ^{⊂} Sym_{2}V denote a general web of quadrics in ℙ^{3} = ℙ(V), and let ℙ̌^{3} = ℙ(V^{∗}) denote the dual space. We also choose coordinates as above such that S = k[x_{0}, …, x_{r}] and T = k[∂_{0}, …, ∂_{r}] are the coordinate rings of ℙ(V) and ℙ(V^{∗}), respectively. Let now q′ : U ⊂ W^{∗} ^{⊂} Sym_{2}(V) be a general subnet. The variety H(q′) of twisted cubics in ℙ^{(}V^{∗}) whose defining quadratic equations are annihilated by the net q′ is the geometric realization of a prime Fano threefold X_{q} of genus 12 (see [Muk1] and [Sch]). Via the apolarity pairing, X_{q} ⊂ Gr(ℙ^{2}, ℙ(U_{q})), where the annihilator U_{q} = (q^{⊥})_{2} ⊂ Sym_{2}V^{∗} is a 7-dimensional vector space. As a subvariety of the Grassmannian the Fano threefold X_{q} is the (codimension 9) common zero-locus of three sections of ∧^{2}E, where E is the dual of the tautological subbundle on Gr(ℙ^{2}, ℙ(U_{q})), cf. [Muk1] or see [Sch, Theorem 5.1] for a complete proof.

The choice of a (general) subnet q′ is equivalent to the choice of a (general) global section of E. But E is a globally generated rank 3 vector bundle whose restriction E = E_{|X_{q}} has third Chern number 2, as an easy direct computation shows. By Kleiman’s transversality theorem the (general) section of E corresponding to q′ must vanish exactly at two (simple) points of X_{q}. These in turn correspond to two twisted cubic curves each of whose defining quadrics are annihilated not only by the net q′, but by the whole web q (the zero locus of a section in E is the special Schubert cycle of subspaces lying in the hyperplane dual to the section). This concludes the proof. □

Let C ⊂ ℙ^{6} be a rational normal sextic curve, and let S = Sec(C) ⊂ ℙ^{6} be its secant variety. S has degree 10, since this is the number of nodes of a general projection of C to a plane. The homogeneous ideal of C is generated by the 2 × 2-minors of either a 3 × 5 or a 4 × 4 catalecticant matrix with linear entries, induced by splittings of O_{ℙ^{1}}(6) as a tensor product of two line bundles of strictly positive degree. Furthermore, it is known that the homogeneous ideal of S = Sec(C) is generated by the 3 × 3 minors of either of the above two catalecticant matrices (see [GP] or [EKS]).

For Π = ℙ^{3} ⊂ ℙ^{6} a general 3-dimensional linear subspace, the linear section Γ = Sec(C) ∩ Π consists of 10 simple points in ℙ^{3} defined by the 3 × 3 minors of a 4 × 4 symmetric (even catalecticant) matrix with linear entries in the variables of Π. Conversely, by Proposition 3.2 above, the set Γ ⊂ ℙ^{3} of 10 nodes of a general quartic symmetroid in ℙ^{3} arises always as a linear section of the secant variety of the rational normal curve in ℙ^{6} (in two different ways). Moreover, it follows that Γ can also be defined by the 3 × 3-minors of each of two different 3 × 5 catalecticant matrices with linear entries in ℙ^{3}. Since the 2 × 2 minors of these catalecticant matrices generate an irrelevant ideal, we may apply [EiPo, Theorem 6.1] (see also [EiPo, Example 6.3] for more details) to obtain the following

Proposition 3.3 (Coble) The Gale transform of the 10 nodes of a general quartic symmetroid in ℙ^{3} are the points of intersection of two Veronese surfaces in ℙ^{5}.

Remark 1) A more careful analysis of the preceding argument shows that the needed generality assumptions on the quartic symmetroid are satisfied if the quartic symmetroid is defined by a regular web of quadrics in ℙ^{3}, see [Cos, Definition 2.1.2].

2) Coble asserts in [Cob] that the converse to Proposition 3.3 should also be true, presumably under suitable generality assumptions. This also relates to the question mentioned in [EiPo] of describing when a collection of 10 points in ℙ^{3} are determinantal.

4 Further results

One may now ask which intersection numbers can actually occur and in which geometric situation this can happen. We are far from having a complete answer to this question, but want to state a number of results in this direction.

We start by considering Veronese surfaces which intersect in 10 points. It is easy to find examples of surfaces X_{1} and X_{2} intersecting transversally in 10 points. For this, one can start with an arbitrary surface X_{1} ⊂ Gr(ℙ^{1}, ℙ^{3}) ⊂ ℙ^{5}. For a general automorphism φ of ℙ^{3} the surface X_{2} = φ(X_{1}) intersects X_{1} transversally by Kleiman’s transversality theorem and since both X_{1} and X_{2} have the same cohomology class we have X_{1}.X_{2} = 10. Actually, we have the following Proposition 4.1 Let X_{1}, X_{2} be two Veronese surfaces in ℙ^{5} intersecting in 10 points. Then X_{1} ∪ X_{2} is contained in a hyperquadric Q and one of the following cases occurs:

(i) Q has rank 6 and X_{1} and X_{2} lie in the same cohomology class,

(ii) The rank of Q is 4 and X_{1} ∩ X_{2} ∩ Sing Q ≠ ∅,

Proof. The existence of Q follows from Proposition 1.3. If Q has rank 6 then X_{1} and X_{2} must have the same class by the proof of Proposition 2.1. The case rank Q = 5 is excluded by Proposition 2.2 and the case of rank Q ≤ 4 and X_{1} ∩ X_{2} ∩ Sing Q = ∅ is excluded by the remark at the end of Section 2. We are now left to exclude the case where rank Q = 3 and X_{1}∩X_{2}∩Sing Q ≠ ∅. We will make use of the diagram and the computations at the end of the proof of Proposition 2.4.

For a Veronese surface X ⊂ Q with rank Q = 3 the class of X̃ in Q̃ equals 2H^{2}. The fibres of the map π|X̃ : X̃ → C ≅ ℙ^{1} are conics and hence this linear system is a subsystem of |2l| on X ≅ ℙ^{2}, i.e. contained in some system of the form |2l − ∑ α_{i}P_{i}|. We have (2l − ∑ α_{i}E_{i})^{2} = 4 − ∑ α_{i}^{2} = 0. This implies either α_{1} = … = α_{4} = 1 or α_{1} = 2. But by the argument in the proof of Proposition 2.4 we already know that X intersects the vertex of Q in a finite non-empty set of points. Since the fibres of the map π|X̃ are conics the first case can only occur if we have the linear system of conics through 4 points in general position. In this case X̃ is mapped to a ℙ^{1} and the general fibre is an irreducible conic, which contradicts what we have. This implies that the linear system is given by |2l − 2P |, in particular X meets the vertex of Q in exactly one point P and that this intersection is not-transversal.

Returning to the case we have to exclude, we can assume that X_{1} is given by the 2 × 2-minors of the matrix

(\begin{matrix}x_{0} & x_{1} & x_{2} \\ x_{1} & x_{3} & x_{4} \\ x_{3} & x_{4} & x_{5} \end{matrix})

and so a typical rank 3 hyperquadric through X_{1} is Q = {x_{0}x_{3} − x_{1}^{2} = 0}. The vertex V of Q is the plane {x_{0} = x_{1} = x_{3} = 0}. The intersection of X_{1} and V is then defined by the ideal (x_{2}x_{4}, x_{2}^{2}, x_{4}^{2}), i.e. the first infinitesimal neighborhood of P. The same holds for the second Veronese surface X_{2} ⊂ Q. Since both surfaces X̃_{1} and X̃_{2} have class 2H^{2} on Q̃, it follows that X̃_{1} and X̃_{2} must meet in 6 other points (different of P). If these points are all in different fibers of π, then X_{1} and X_{2} meet in a (non-reduced) scheme of length 9(= 6 + 3) and thus we got a contradiction. Otherwise, since all the fibers of π|X̃_{i} are conics (whose images in ℙ^{5} already go through the fixed point P), the 2-planes spanned by the fibres of π|X̃_{1} and π|X̃_{2} over some point in C must coincide. But then, by Bezout, the two conic fibres must intersect in 4 points, from which we conclude X_{1}.X_{2} ≥ 11, again impossible. This concludes the proof of the proposition. □

Remark As corollary of the proof of the previous proposition, we find out that two Veronese surfaces in ℙ^{5} may intersect in a non-reduced scheme of length 9. This is indeed the case, for two (general) Veronese surfaces X_{1} and X_{2} lying on a rank 3 hyperquadric Q such that X_{1} ∩ X_{2} ∩ Sing Q consists of a single point P. As above, both surfaces cut out the first infinitesimal neighborhood of P on the vertex of the hyperquadric Q and meet further in 6 simple points. In particular their intersection is non-reduced.

However, we show that

Proposition 4.2 Two Veronese surfaces in ℙ^{5} cannot intersect transversally in 9 points.

Proof. Assume that there exist Veronese surfaces X_{1} and X_{2} in ℙ^{5} such that W = X_{1} ∩ X_{2} is a reduced set of length 9. We consider again W as a subscheme of X_{1} ≅ ℙ^{2} and we recall from Lemma 1.1 that h^{1}(I_{W} (4)) = 0. Equivalently, this means that the linear system δ := |4l − W | on X_{1} ≅ ℙ^{2} has projective dimension 5. It will then be enough to show that the linear system |4l − W | of plane quartics through W contains a smooth curve C, since in this case the restriction of |4l − W | defines, via taking the residual intersection, a g_{7}^{4+ε} on C with ε ≥ 0, thus contradicting Clifford’s theorem.

In order to show the existence of such a C we consider the surface S given by blowing up the set W on X_{1}. It will be enough to check that the linear system |4l−W | is base point free and defines a morphism S → S′ ⊂ ℙ^{5} whose image S′ has no worse than isolated singularities. We want to do this using Reider’s theorem (see [BS, Theorem 2.1] in characteristic 0, and [SB], [Nak], [Ter, Theorem 2.4] in positive characteristic) and for this purpose we write 4l − W = L + K_{δ} where K_{δ} = −3l + W is the canonical divisor on S while L = 7l − 2W. In order to apply a Reider type theorem we need to check that L^{2} ≥ 9 and that L is nef and big. The first is clear since L^{2} = 49−36 = 13. We do not know that L is nef and big, but in the proof of Reider’s theorem (as in the proof of its positive characteristic counterparts, cf [Ter]) this assumption is only used to conclude that h^{1}(K_{δ} +L) = 0, which we already know since h^{1}(I_{W} (4)) = 0 by Lemma 1.1.

We may now argue as follows. If |4l − W | = |K_{δ} + L| is not base-point free, respectively very ample, then there exists a curve a D such that L−2D is ℚ-effective and such that

D^{2} ≥ L.D − k − 1

where k = 0, respectively 1. We write D = al − ∑_{i=1}^{9} b_{i}E_{i}. Since L − 2D must

be ℚ-effective we see immediately that a ≤ 3. For a = 3 we obtain

g − ∑_{i=1}^{9} b_{i}^{2} ≥ 21 − 2 ∑_{i=1}^{9} b_{i} − k − 1

respectively

∑_{i=1}^{9} b_{i}(b_{i} − 1) ≤ −2 + k

which gives a contradiction. For a = 1, 2 the same calculation gives

∑_{i=1}^{9}(b_{i} − 1)^{2} ≤ 4 + k (a = 1),

respectively

∑_{i=1}^{9}(b_{i} − 1)^{2} ≤ k.

On the other hand, since the quadrics through X_{2} cut out X_{2}, then at most 4 points of W ⊂ ℙ^{2} can be collinear and at most 8 points of W can lie on the same conic. This shows that |4l − W | is base point free on S and that S′ has at most isolated singularities, which is our claim, and this concludes the proof of the proposition. □

Remark One may construct pairs of Veronese surfaces on suitable smooth or nodal cubic hypersurfaces in ℙ^{5} which meet in 1, 2, 3, 5 or 6 simple points. It is also possible to check in Macaulay [Mac] that if X ⊂ ℙ^{5} is a Veronese surface and φ is a general linear automorphism of ℙ^{5} fixing m ∈ {1, 2, 3, 5} (general) points on X, then X and φ(X) meet exactly at those m points.

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D. Eisenbud  K. Hulek  S. Popescu
Dept. of Mathematics Inst. für Mathematik  Dept. of Mathematics
UC Berkeley  Univ. Hannover  SUNY Stony Brook
Berkeley CA 94720  D 30060 Hannover  Stony Brook, NY 11794-3651
USA  Germany  USA

de@msri.org         hulek@math.uni-hannover.de  sorin@math.sunysb.edu