On the Hilbert Function of Partially General Unions of Double Points

Abstract

We study the Hilbert function of a union of a fixed set of double points and a prescribed number of general double points. Our main results are for double points of Veronese varieties and of Segre–Veronese varieties. Among the Segre–Veronese varieties, we study the ones with all factors of dimension one and the ones with two factors, one of them of dimension one. We give many examples with exceptional or controlled behavior for a small number of double points.

1. Introduction

A classical topic of last century algebraic geometry was the study of the secant varieties ${\sigma }_x\left(X\right)\subseteq {\mathbb{P}}^r$ of an embedded projective variety $X\subset {\mathbb{P}}^r$ [1,2,3]. The key information about them is the dimension of these secant varieties. This topic was revived, because for certain embedded varieties, it gave important results for applied mathematics, roughly speaking related to tensors [2,3,4,5,6,7].

Our main results are about the Veronese embeddings of projective spaces and on the Segre–Veronese varieties. Veronese embeddings of projective spaces are related to the additive decomposition of forms. The dimension of a secant variety of a Segre–Veronese embedding is related to the dimension of the set of all partially symmetric tensors with a fixed format and a fixed partially symmetric tensor rank. Knowing these dimensions is very useful, because a standard way to handle a tensor (or a received packet of data) is to approximate it with a low rank one.

The Terracini Lemma relates this dimension to the dimension of the linear span of a “general” union of double points (ref. [1], Cor. 10), i.e., to the following interpolation problem.

For any irreducible quasi-projective variety W, let $W_{\mathrm{reg}}$ denote the set of its smooth points. For any positive integer x, let $S(W_{\mathrm{reg}},x)$ denote the set of all subsets of $W_{\mathrm{reg}}$ with cardinality x. For all $p\in W_{\mathrm{reg}}$, let $(2p,W)$ or just $2p$ denote the closed subscheme of W with ${\left({\mathcal{I}}_{p,W}\right)}^2$ as its ideal scheme. The scheme $(2p,W)$ is a connected zero-dimensional scheme of degree $1+\dim W$. For all $S\in S(W_{\mathrm{reg}},x)$, set $(2S,W):={\cup }_{p\in S}(2p,W)$. Often we write $2S$ if W is clear in a statement. The set $S(W_{\mathrm{reg}},x)$ is an irreducible variety of dimension $x\dim X$, and hence, we are allowed to speak about its general element. The Terracini Lemma gives dim ${\sigma }_x\left(X\right)=\dim ⟨2S⟩$ for a general $S\in S(X_{\mathrm{reg}},x)$, where $⟨⟩$ denote the linear span. The case $x=1$ of it explains the usefulness of $(2p,W)$: if W is embedded in a projective space ${\mathbb{P}}^r$, then $⟨(2p,W)⟩\subset {\mathbb{P}}^r$ is the Zariski tangent space of W at p.

In this paper we start with a fixed $A\in S(X_{\mathrm{reg}},a)$ and give (in some cases) dim$⟨2A\cup 2S⟩$ for a general $S\in S(X_{\mathrm{reg}},x-a)$. The integer $\dim ⟨2A\cup 2S⟩$ is the dimension of the $(x-a)$-secant variety linear projection from the linear space $⟨2A⟩$.

For the Veronese embedding of ${\mathbb{P}}^n$, we prove the following result.

Theorem 1.

Fix positive integers $a,x$ and $A\in S({\mathbb{P}}^n,a)$. Fix an integer $d\ge \max \{5,2a-1\}$ and a general $S\in S({\mathbb{P}}^n\setminus A,x)$. Let $V(n,d)$ be the vector space of degree d forms in $n+1$ variables. Set $V(n,d)(-2S-2A):=\{f\in V(n,d)\mid f_{|2S\cup 2A}=0\}$. Then,

$$\dim V(n,d)(-2S-2A)=\max \{0,\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-(n+1)(a+x)\}.$$

We will prove Theorem 1 using sheaf cohomology. In terms of sheaf cohomology, Theorem 1 is equivalent to the following statement.

Theorem 2.

Fix positive integers $a,x$ and $A\in S({\mathbb{P}}^n,a)$. Fix an integer $d\ge \max \{5,2a-1\}$ and a general $S\in S({\mathbb{P}}^n\setminus A,x)$. Then, either $h^1\left({\mathcal{I}}_{2S\cup 2A}\left(d\right)\right)=0$ or $h^0\left({\mathcal{I}}_{2S\cup 2A}\left(d\right)\right)=0$.

It is interesting also to handle $2S\cup A$ instead of $2S\cup 2A$. It corresponds to computing the dimension of a secant variety of the image of X by the linear projection from $⟨A⟩$ (see Proposition 1 for the case of Veronese embeddings).

We use the following fundamental theorem proven by J. Alexander and A. Hirschowitz [8,9,10,11].

Theorem 3

(Alexander–Hirschowitz). Fix positive integers n, d and x. Take a general set $S\in S({\mathbb{P}}^n,x)$. Then, either $h^1\left({\mathcal{I}}_{2S}\left(d\right)\right)=0$ or $h^0\left({\mathcal{I}}_{2S}\left(d\right)\right)=0$, except in the following cases:

1. 

$n>1$ and $d=2$;

2. 

$(n,d)\in \left\{\right(2,4),(3,4),(4,3),(4,4\left)\right\}$.

This is the sheaf-cohomology version of the Alexander–Hirschowitz theorem. In the language of Theorems, 1 and 3 are stated in the following way.

Theorem 4

(Alexander–Hirschowitz). Fix positive integers n, d and x. Take a general set $S\in S({\mathbb{P}}^n,x)$. Then, $\dim V(n,d)(-2S)=\max \{0,\left(\genfrac{}{}{0pt}{}{n+d}{n}\right)-x(n+1)\}$, except in the following cases:

1. 

$n>1$ and $d=2$;

2. 

$(n,d)\in \left\{\right(2,4),(3,4),(4,3),(4,4\left)\right\}$.

Remark 1.

In the statement of Theorem 3, the values of $h^0\left({\mathcal{I}}_{2S}\left(d\right)\right)$ and $h^1\left({\mathcal{I}}_{2S}\left(d\right)\right)$ in the listed exceptional cases were known before [8], and they are listed in [8,9,10,11], except one case solved in [12]. There are many proofs of the Alexander–Hirschowitz theorem [11,13,14,15].

We recall that the Segre–Veronese varieties are the embeddings by complete linear systems of multiprojective spaces. We study this problem for two types of multiprojective spaces, ${\mathbb{P}}^m\times {\mathbb{P}}^1$ and ${\left({\mathbb{P}}^1\right)}^k$, $k\ge 2$.

We use the automorphism group of the multiprojective space to drastically simplify the proofs, reducing to the case of the “worst A”. We think that this reduction is also useful to test computationally for many other interpolation problems. The word “worst” is used, because if this element A of $S(X_{\mathrm{reg}},a)$ has good interpolation, then all $B\in S(X_{\mathrm{reg}},a)$ have good interpolation by the semicontinuity theorem for cohomology (ref. [16], Th. III.12.8). For subsets of ${\mathbb{P}}^n$, the worst ones are the one collinear (Lemma 3). We are able to describe and use the next worst ones when all but one of the points are collinear (Lemma 4). For the multiprojective spaces and their Segre–Veronese embeddings, there is no single notion of “bad”. We use the one with respect to the first factor (Remark 7). These lemmas and remarks may be used for interpolations with other zero-dimensional schemes, not just double points.

We raise the following interpolation problem.

Question: Fix positive integers a and x. Take an integral and non-degenerate variety $X\subset {\mathbb{P}}^r$ and an integral subvariety $T\subset X$ such that $T\cap X_{\mathrm{reg}}\ne \varnothing$. Compute $\dim ⟨2A\cup 2S⟩$ and $\dim ⟨(2A,T)\cup 2S⟩$ for a general $A\in S(T_{\mathrm{reg}}\cap X_{\mathrm{reg}},a)$ and a general $S\in S(X_{\mathrm{reg}},x)$.

It may be easier when T is a hypersurface of X. We use it in one case (Lemma 10) to prove all the main theorems of Section 6. The question arises when we use the so-called Differential Horace Lemma according to J. Alexander and A. Hirschowitz (Lemma 2) and an inductive procedure using the hypersurface T. In [11], there is a case in which higher codimension is needed and handled.

In Section 2, we fix the notation and give the preliminary remarks and lemmas used in the paper. Most proofs later are inductive, the induction step being handled using in a tricky way the Differential Horace Lemma.

Section 3 contains many lemmas on the Veronese embeddings and a result (Proposition 1) about the Hilbert function of $A\cup 2S$, S general in $S({\mathbb{P}}^n,x)$. At the beginning of the section, we describe our method (Remark 6).

Section 4 contains the proof of Theorem 2.

Section 5 and Section 6 are for Segre–Veronese varieties. In Section 5, we give the notation and background and prove the case ${\mathbb{P}}^m\times {\mathbb{P}}^1$ (Theorem 5 for $2A\cup 2S$, Proposition 3 for $A\cup 2S$) using induction on the integer m. Section 6 contains four major results for the embeddings of ${\left({\mathbb{P}}^1\right)}^k$, $k\ge 3$, (Theorems 6–9), the case $k=2$ being proven in Section 5. At the beginning of Section 5, we describe the set-up of the Segre–Veronese varieties and give three possible extensions for the interested readers.

Section 7 contains examples for ${\mathbb{P}}^n$ and ${\mathbb{P}}^1\times {\mathbb{P}}^1$ for up to 4 points. We also consider multiple points of multiplicity $m>2$.

In the last section, we propose the following extension of the question: instead of T, we take a flag $T_1⊊T_2⊊\cdots ⊊T_s⊊X$ and prescribe the number of points contained in each stratum $T_{i+1}\setminus T_i$.

We thank the referees for helpful feedback, which led to some changes in the exposition.

2. Preliminary Results

We work over an algebraically closed field $\mathbb{K}$ of characteristic 0.

Let M be an integral projective variety, D an effective Cartier divisor of M and $Z\subset M$ a zero-dimensional scheme. The residual scheme ${\mathrm{Res}}_D\left(Z\right)$ of Z with respect to D is the zero-dimensional subscheme of M with ${\mathcal{I}}_{Z,M}:{\mathcal{I}}_{D,M}$ as its ideal sheaf.

We have ${\mathrm{Res}}_D\left(Z\right)\subseteq Z$ and $\mathrm{deg}\left(Z\right)=\mathrm{deg}(Z\cap D)+\mathrm{deg}\left({\mathrm{Res}}_D\left(Z\right)\right)$. If $Z=A\cup B$ with $A\cap B=\varnothing$, then ${\mathrm{Res}}_D\left(Z\right)={\mathrm{Res}}_D\left(A\right)\cup {\mathrm{Res}}_D\left(B\right)$. For all line bundles R on M, we have an exact sequence

$$0\to {\mathcal{I}}_{{\mathrm{Res}}_D\left(Z\right)}\otimes R(-D)\to {\mathcal{I}}_Z\otimes R\to {\mathcal{I}}_{Z\cap D,D}\otimes R_{|D}\to 0$$

(1)

For all $p\in M_{\mathrm{reg}}$, let $(2p,M)$ denote the zero-dimensional subscheme of M with ${\left({\mathcal{I}}_{p,M}\right)}^2$ as its ideal sheaf. We have ${(2p,M)}_{\mathrm{red}}=\left\{p\right\}$ and $\mathrm{deg}\left(\right(2p,M\left)\right)=\dim M+1$. For all finite set $S\subset M_{\mathrm{reg}}$, set $(2S,M):={\cup }_{p\in S}(2p,M)$, with the convention $(2\varnothing ,M)=\varnothing$.

For any quasi-projective variety W and all positive integers x, let $S(W,x)$ denote the set of all subsets of W with cardinality x. Set $S(W,0)=\{\varnothing \}$.

Remark 2.

Let $A\subset B\subset {\mathbb{P}}^r$ be zero-dimensional schemes. We have $⟨A⟩\subseteq ⟨B⟩$ and $\dim ⟨B⟩\le \dim ⟨A⟩+\mathrm{deg}\left(B\right)-\mathrm{deg}\left(A\right)$. Thus, if B is linearly independent, then A is linearly independent, while if $⟨A⟩={\mathbb{P}}^r$, then $⟨B⟩={\mathbb{P}}^r$.

Remark 3.

Let $X\subset {\mathbb{P}}^r$ be an integral and non-degenerate variety. Fix a closed subscheme $B\subset X$ and a positive integer x. Take a general $S\in S(X\setminus B_{\mathrm{red}},x)$. Then we have

$$\dim ⟨B\cup S⟩=\min \{r,\dim ⟨B⟩+x\}.$$

The following observation is a cohomological version of Remark 3.

Remark 4.

Let X be an integral projective variety, L a line bundle on X, B a closed subscheme of X and x a positive integer. Let S be a general element of $S(X\setminus B,x)$. Then

$$h^0({\mathcal{I}}_{B\cup S}\otimes L)=\max \{0,h^0({\mathcal{I}}_B\otimes L)-x\},$$

$$h^1({\mathcal{I}}_{B\cup S}\otimes L)=h^1({\mathcal{I}}_B\otimes L)+\max \{0,x-h^0({\mathcal{I}}_B\otimes L)\}.$$

Remark 5.

Fix an integral and non-degenerate variety $X\subset {\mathbb{P}}^r$ and a linear subspace $M\subset {\mathbb{P}}^r$. Set $m:=\dim M$. If $m=r-1$, then $⟨M\cup \left\{p\right\}⟩={\mathbb{P}}^r$. Now assume $m\le r-2$. Let ${\mathcal{l}}_M:{\mathbb{P}}^r\setminus M\to {\mathbb{P}}^{r-m-1}$ denote the linear projection from M. Since X is non-degenerate, $X⊈M$. Hence, $\mu :={\mathcal{l}}_{M|X\setminus M}X\setminus M\to {\mathbb{P}}^{r-m+1}$ is a morphism. Since X is non-degenerate, $\mu (X\setminus M\cap X)$ spans ${\mathbb{P}}^{r-m-1}$. Hence, μ is not a constant map. Since we are in characteristic zero and $\dim X\ge 1$, the differential of μ is non-zero at a general $p\in X_{\mathrm{reg}}$. Hence $\dim ⟨M\cup v⟩=m+2$ for a general connected zero-dimensional subscheme v of $X_{\mathrm{reg}}$ such that $\mathrm{deg}\left(v\right)=2$.

Lemma 1.

Fix an integral projective variety X of positive dimension, L a line bundle on X and D an effective irreducible Cartier divisor of X such that $h^1\left(L(-D)\right)=0$. Let $Z\subset X$ be a zero-dimensional scheme such that $h^1({\mathcal{I}}_Z\otimes L)=0$. Set $a:=h^0({\mathcal{I}}_Z\otimes L)$. Take a positive integer x and a general $S\in S(D,x)$. Assume $h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(Z\right)}\otimes L(-D))\le \max \{0,a-x\}$. Then $h^0({\mathcal{I}}_{Z\cup S}\otimes L)=\max \{0,a-x\}$ and $h^1({\mathcal{I}}_{Z\cup S}\otimes L)=\max \{0,x-a\}$.

Proof. 

Remark 4 gives $h^1\left(L\right)=0$. Take a maximal $S^{\prime }\subseteq S$ with the property that we have $h^0({\mathcal{I}}_{Z\cup S^{\prime }}\otimes L)=\max \{0,a-\#S^{\prime }\}$ (we allow the case $S^{\prime }=\varnothing$). Assume $S^{\prime }\ne S$ (and hence, assume $h^0({\mathcal{I}}_{Z\cup S^{\prime }}\otimes L)>0$) and take $p\in S\setminus S^{\prime }$. Note that ${\mathrm{Res}}_D(Z\cup S^{\prime })={\mathrm{Res}}_D\left(Z\right)$. Since p is general in D, the maximality of $S^{\prime }$ implies that D is the base locus of $|{\mathcal{I}}_{Z\cup S^{\prime }}\otimes L|$. Hence, we obtain $h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(Z\right)}\otimes L(-D))=a-\#S^{\prime }$, a contradiction.   □

The following result according to J. Alexander and A. Hirschowitz is called the Differential Horace Lemma. It was proven and used in this generality in [8] and in their previous joint papers.

Lemma 2

(Differential Horace Lemma). Let M be an integral projective variety, R a line bundle on M, $Z\subset M$ a zero-dimensional scheme and D an integral effective Cartier divisor. Fix an integer $x>0$, a general $S\in S(M,x)$ and a general $A\in S(D,x)$. Then for all $i\in \mathbb{N}$, we have

$$h^i(M,{\mathcal{I}}_{Z\cup (2S,M)}\otimes R)\le h^i({\mathcal{I}}_{{\mathrm{Res}}_D\left(Z\right)\cup A}\otimes R_{|D})+h^i({\mathcal{I}}_{Z\cup (2S,M)}\otimes R(-D))$$

The case $S=\varnothing$ of Lemma 2 is called the Horace Lemma.

3. Veronese Embeddings

Remark 6.

The Alexander–Hirschowitz theorem was proven by double induction on the dimension, n, of the projective space and the degree, d, of the forms considered for the interpolation problem “x general double points”. The main difficulty is that the cases with very low d are exceptional. If for some d, one knows the theorem in ${\mathbb{P}}^{n-1}$ and ${\mathbb{P}}^n$ for the degree d, then it would be not too difficult to prove it in ${\mathbb{P}}^n$ for forms of degree $d+1$. For our interpolation problems in our main theorems we are not (by our choice) in the exceptional range. We show some exceptional cases in Section 7. Section 7 shows that our tools work also to discover exceptional cases and prove their exact defect (the integer $h^1\left({\mathcal{I}}_{2A\cup 2S}\left(d\right)\right)$). In Section 3 and Section 4, we reduce the case of $2A\cup 2S$ for d to a case for $2S^{\prime }$, $S^{\prime }$ general for a lower $d^{\prime }$ such that $(n,d^{\prime },\#S^{\prime })$ is not in the list of exceptional cases. It is a key part of a less than 80 page long manuscript to use as much as possible the statement of the Alexander–Hirschowitz theorem. We first show that instead of an arbitrary $A\in S({\mathbb{P}}^n,a)$, it is sufficient to handle all $B\in S({\mathbb{P}}^n,a)$ contained in a line L: if $2B\cup 2S$ is not exceptional in degree d, then $2A\cup 2S$ is not exceptional in degree d (Lemmas 3 and 4). Points B contained in a line L are easy to control, because for the Differential Horace Lemma, we take a hyperplane $H\supseteq L$, and in the residual, the multiplicity of each point decreases by 1. Hence, in 2 steps, we go from $2B$ to ∅. This observation may be used for many other interpolation problems. Instead of $2A$, we may take multiple points of arbitrary multiplicity in the sense of Notation 1. See [17] for the Differential Horace Lemma for multiple points of multiplicity $m>2$. Our proofs are in some sense from d to a lower $d^{\prime }$. Instead of the set (A or B) of cardinality a which we had at level d, we have the set ∅ at level $d^{\prime }$. To do that (with $\#S^{\prime }$ as small as needed), we use not only the Differential Horace Lemma but Lemma 1, which drastically reduces the degree of the zero-dimensional scheme that we need to control in degree $d-1$ and $d-2$. See [4,18,19] for some uses of this approach to the Differential Horace Lemma, refs. [18,19] not being restricted to the case of Segre–Veronese varieties. The case $A\cup 2S$ is carried out in Proposition 1.

See (ref. [16] Figure 11, p. 260) as a much more complicated example of linear projection as a flat limit of a family of automorphisms. This example motivated the following two lemmas.

Lemma 3.

Fix positive integers a, x, d and $A\in S({\mathbb{P}}^n,a)$. Then, there is $B\in S({\mathbb{P}}^n,x)$ such that B is contained in a line, and for a general $S\in S(X\setminus A,x)$, we have:

1. 

$h^1\left({\mathcal{I}}_{2A\cup 2S}\left(d\right)\right)\le h^1\left({\mathcal{I}}_{2B\cup 2S}\left(d\right)\right)$;

2. 

$h^0\left({\mathcal{I}}_{2A\cup 2S}\left(d\right)\right)\le h^0\left({\mathcal{I}}_{2B\cup 2S}\left(d\right)\right)$;

3. 

$h^1\left({\mathcal{I}}_{A\cup 2S}\left(d\right)\right)\le h^1\left({\mathcal{I}}_{2B\cup 2S}\left(d\right)\right)$;

4. 

$h^0\left({\mathcal{I}}_{A\cup 2S}\left(d\right)\right)\le h^0\left({\mathcal{I}}_{2B\cup 2S}\left(d\right)\right)$.

Proof. 

Projectively equivalent zero-dimensional schemes have the same Hilbert function, i.e., for all $d\in \mathbb{N}$, the vector spaces of degree d forms vanishing on them have the same dimension. By the semicontinuity theorem for cohomology (ref. [16], Th. III.12.8), it is sufficient to prove that there is $B\in S(L,a)$, which is a flat limit of a family of elements of $S({\mathbb{P}}^n,a)$, all of them projectively equivalent to B. Fix a line $L\subset {\mathbb{P}}^n$. Take a general codimension 2 linear subspace V of ${\mathbb{P}}^n$. Since V is general, $A\cap V=\varnothing$, $L\cap V=\varnothing$ and $⟨L\cup V⟩={\mathbb{P}}^n$. Thus, we may take a homogeneous system of coordinates $x_0,\dots ,x_n$ such that $V=\{x_0=x_1=0\}$ and $L=\{x_2=\cdots =x_n=0\}$. Let $h:{\mathbb{P}}^n\setminus V\to {\mathbb{P}}^1$ denote the linear projection for V. We identify L with the target of h. Since V is general, V does not intersect any line spanned by 2 points of A. Hence, $\#h\left(A\right)=a$. Set $B:=h\left(A\right)\in S(L,a)$. For any non-zero constant c, let $h_c$ denote the linear automorphism of ${\mathbb{P}}^n$ defined by the formula $h_c\left([x_0:x_1:x_2:\cdots :x_n]\right)=[x_0:x_1:c{x}_2:\cdots :c{x}_n]$. Each $h_c\left(A\right)$ is projectively equivalent to A. If we take $c=0$ in the formula for $h_c$, we obtain h. Since $\#h\left(A\right)=a$, B is the flat limit of the family ${\left\{h_c\left(A\right)\right\}}_{c\ne 0}$.   □

Lemma 4.

Take a finite set $S\subset {\mathbb{P}}^n$, $n\ge 2$, $p\in S$ and a line $L\subset {\mathbb{P}}^n$ such that $p\notin L$. Set $A:=S\setminus \left\{p\right\}$. Assume that no line spanned by 2 points of A contains p. Then, there is $B\subset L$ such that $\#B=\#S^{\prime }$, and $\left\{p\right\}\cup B$ is the flat limit of a family of subsets of ${\mathbb{P}}^n$ projectively equivalent to S.

Proof. 

Take a general codimension 2 linear subspace V of ${\mathbb{P}}^n$ containing p.

Let $h:{\mathbb{P}}^n\setminus V\to {\mathbb{P}}^1$ denote the linear projection for V. We identify L with the target of h. Take a homogeneous system of coordinates $x_0,\dots ,x_n$ such that $V=\{x_0=x_1=0\}$, $p=[0:0:\cdots :1]$ and $L=\{x_2=\cdots =x_n=0\}$. Set $B:=h\left(A\right)$. Since no line spanned by 2 points of A contains p, and V is a general codimension 2 linear space containing p, V does not intersect any line spanned by 2 points of A. Hence, $\#B=\#A$. For any non-zero constant c, let $h_c$ denote the linear automorphism of ${\mathbb{P}}^n$ defined by the formula $h_c\left([x_0:x_1:x_2:\cdots :x_n]\right)=[x_0:x_1:c{x}_2:\cdots :c{x}_n]$. Each $h_c\left(A\right)$ is projectively equivalent to B. Note that if we take $c=0$ in the definition of the automorphism $h_c$, we obtain the linear projection h. Since $\#h\left(A\right)=a$, B is the flat limit of the family ${\left\{h_c\left(A\right)\right\}}_{c\ne 0}$. Since $h_c\left(p\right)=p$, $\left\{p\right\}\cup B$ is the flat limit of the family ${\{\left\{p\right\}\cup h_c\left(A\right)\}}_{c\ne 0}$, concluding the proof.   □

Proposition 1.

Fix integers $a\ge 3$, $n\ge 2$ and $d\ge \max \{a-1,5\}$. Take $A\in S({\mathbb{P}}^n,a)$ and a general $S\in S({\mathbb{P}}^2\setminus A,x)$. If $(n,d,a)=(2,4,5)$, assume $x\ne 5$. Then, either $h^1\left({\mathcal{I}}_{2S\cup A}\left(d\right)\right)=0$ or $h^0\left({\mathcal{I}}_{2S\cup A}\left(d\right)\right)=0$.

Proof. 

By Lemma 3, it is sufficient to prove the case in which A is contained in a line L. We will prove (inductively) a stronger result in which we allow as A any degree a zero-dimensional subscheme of L.   □

Set $y_{t,n}:=\lfloor (\left(\genfrac{}{}{0pt}{}{n+t}{n}\right)-a)/(n+1)\rfloor$ and $x_{t,n}:=\lceil (\left(\genfrac{}{}{0pt}{}{n+d}{n}\right)-a)/(n+1)\rceil$, where t is a positive integer. By Remark 2 to prove Proposition 1 for A and d and all x, it is sufficient to prove that $h^1\left({\mathcal{I}}_{2A\cup 2E}\left(d\right)\right)=0$ for a general $E\in S({\mathbb{P}}^n\setminus A,y_{d,n})$, and $h^0\left({\mathcal{I}}_{A\cup 2F}\left(d\right)\right)=0$ for a general $F\in S({\mathbb{P}}^n\setminus A,x_{d,n})$.

Take $z_{d,n}\in \{y_d,x_{d,n}\}$ and a general $S\in S({\mathbb{P}}^2\setminus L,z_{d,n})$.

(a) Assume $n=2$. First, assume $d=a-1$. Recall that S is general. Since $h^i(L,{\mathcal{I}}_{A,L}(a-1))=0$, $i=0,1$, ${\mathrm{Res}}_L\left(A\right)=\varnothing$ and $S\cap L=\varnothing$, the residual exact sequence of L gives $h^i\left({\mathcal{I}}_{2S\cup A}\left(d\right)\right)=h^i\left({\mathcal{I}}_{2S}(d-1)\right)$, $i=0,1$. Since we excluded the case $(d,x)=(4,5)$, it is sufficient to use the Alexander–Hirschowitz theorem (Theorem 3). Now assume $d\ge a$ and that for all $a^{\prime }\le a$, the proposition is true for the integer $d-1$.

(a1) Assume $d-a$ odd. Since $3y_{d,2}\ge \left(\genfrac{}{}{0pt}{}{d+2}{2}\right)-a-2$ and $d\ge a+1$, we have the inequality $y_{d,2}\ge (d+1-a)/2$. We specialize S to $S_1\cup S_2$ with $S_2$ a general element of $S(L,(d+1-a)/2)$ and $S_2$ a general element of $S({\mathbb{P}}^2,z_{d,2}-(d+1)/2)$. Note that we have $L\cap (A\cup 2S_1\cup 2S_2)=A\cup (2S_2,L)$ and ${\mathrm{Res}}_L(A\cup 2S_1\cup 2S_2)=2S_1\cup S_2$. Since $h^i(L,{\mathcal{I}}_{A\cup (2S_2,L)}\left(d\right))=0$, $i=0,1$, the residual exact sequence of L shows that to prove the proposition in degree d, it is sufficient to prove that either $h^1\left({\mathcal{I}}_{2S_1\cup S_2}(d-1)\right)=0$ (case $z_{d,2}=y_{d,2}$) or $h^0\left({\mathcal{I}}_{2S_1\cup S_2}(d-1)\right)=0$ (case $z_{d,2}=x_{d,2}$). Since $\#S_2=(d-1)/2\le d-1$, we may use the inductive assumption, unless $d-1=4$. Assume $d=5$. Since $d-a$ is odd, we obtain $a=4$, $\#S_2=1$, $y_{5,2}=5$ and $x_{5,2}=6$. Take the case $z_{5,2}=y_{5,2}$. We have $h^1\left({\mathcal{I}}_{2S_1}\left(4\right)\right)=0$ by the Alexander–Hirschowitz theorem, and hence, $h^0\left({\mathcal{I}}_{2S_1}\left(4\right)\right)=3$. Since ${\mathbb{P}}^2$ is homogeneous and we may take $S_2$ as general after fixing $S_1$, we obtain $h^0\left({\mathcal{I}}_{2S_1\cup S_2}\left(4\right)\right)=2$, i.e., $h^1\left({\mathcal{I}}_{2S_1\cup S_2}\left(4\right)\right)=0$. Now, assume $z_{5,2}=x_{5,2}$. By the Alexander–Hirschowitz theorem, we have $h^1\left({\mathcal{I}}_{2S_1}\left(4\right)\right)=h^0\left({\mathcal{I}}_{2S_1}\left(4\right)\right)=1$ with $|{\mathcal{I}}_{2S_1}\left(4\right)|$ being the double of the conic containing $S_1$. Hence, $h^0\left({\mathcal{I}}_{2S_2\cup S_1}\left(4\right)\right)=0$ for a general $S_2$.

(a2) Assume $d-a$ even. Fix a general $B_1\subset L$ such that $\#B_2=(d-a)/2$, a general $p\in L\setminus (A\cup B_1)$ and a general $B_2\in S({\mathbb{P}}^2,z_{d,2}-1-(d-a)/2)$. Note that $h^i(L,{\mathcal{I}}_{A\cup (2B_2,L)\cup \left\{p\right\},L}\left(d\right))=0$, $i=0,1$.

By the Differential Horace Lemma (Lemma 2), to prove the theorem in degree d, it is sufficient to prove that either $h^1\left({\mathcal{I}}_{2B_1\cup B_2\cup (2p,L)}(d-1)\right)=0$ (case $z_{d,2}=y_{d,2}$) or $h^0\left({\mathcal{I}}_{2B_1\cup B_2\cup (2p,L)}(d-1)\right)=0$ (case $z_{d,2}=x_{d,2}$). It is for using induction on d that we allowed an arbitrary zero-dimensional scheme contained in a line. We have $\mathrm{deg}(B_2+(2p,L))=(d-a)/2+2\le d$. Hence, we may use the inductive assumption, unless $d=5$. Since $a\ge 3$ and $d-a$ is even, we obtain $a=3$, and hence, $y_{5,2}=x_{5,2}=6$. We use a different proof but first note that the proof just given for the general case would give $h^1\left({\mathcal{I}}_{A\cup 2S}\left(5\right)\right)=h^0\left({\mathcal{I}}_{A\cup 2S}\left(5\right)\right)\le 1$. Take a smooth conic C such that $\mathrm{deg}(C\cap A)=1$. We specialize S to the union $E_1\cup \left\{q\right\}$ with $E_1$, the union of 5 general points of C and q general in ${\mathbb{P}}^2$. Since $C\cong {\mathbb{P}}^1$ and $h^0(C,{\mathcal{O}}_C\left(5\right))=11$, we obtain $h^i(C,{\mathcal{I}}_{(2E_1,C)\cup A\cap C,C}\left(5\right))=0$, $i=0,1$. Note that $h^0\left({\mathcal{I}}_{E_1\cup 2q\cup {\mathrm{Res}}_L\left(A\right)}\left(3\right)\right)=h^1\left({\mathcal{I}}_{E_1\cup 2q\cup {\mathrm{Res}}_L\left(A\right)}\left(3\right)\right)$. Thus, by the residual exact sequence of C, it is sufficient to prove that $h^0\left({\mathcal{I}}_{E_1\cup 2q\cup {\mathrm{Res}}_L\left(A\right)}\left(3\right)\right)=0$. Assume $h^0\left({\mathcal{I}}_{E_1\cup 2q\cup {\mathrm{Res}}_L\left(A\right)}\left(3\right)\right)=1$. Since $\mathrm{deg}(C\cap A)=1$, $\mathrm{deg}(C\cap {\mathrm{Res}}_L\left(A\right))\le 1$. The residual exact sequence of C gives $h^1\left({\mathcal{I}}_{E_1\cup {\mathrm{Res}}_L\left(A\right)}\left(3\right)\right)=0$. Thus, $h^0\left({\mathcal{I}}_{E_1\cup {\mathrm{Res}}_L\left(A\right)}\left(3\right)\right)=3$. The linear system $|{\mathcal{I}}_{E_1\cup {\mathrm{Res}}_L\left(A\right)}\left(3\right)|$ has dimension 2, and all its elements are singular at q. Since q may be a general point of ${\mathbb{P}}^2$, the Bertini theorem gives a contradiction.

(b) Assume $n>2$ and that the proposition is true in lower dimensional projective spaces. Take a hyperplane $H\supset L$. In step (b1), we take $z_{d,n}=y_{d,n}$. In step (b2), we take $z_{d,n}=y_{d,n}$. Set $e_{d,n-1}:=\left(\genfrac{}{}{0pt}{}{n+d-1}{n-1}\right)-n{y}_{d,n-1}$. Note that $0\le e_{d,n-1}\le n-2$. We use the following numerical claims.

Claim 1.

We have $y_{d,n}\ge y_{d,n-1}+n-1$.

Proof of Claim 1.

Assume $y_{d,n}\le y_{d,n-1}+n-2$. Multiplying this inequality by $(n+1)n$ and using that $(n+1)y_{d,n}\ge \left(\genfrac{}{}{0pt}{}{n+d}{n}\right)-n$ and $n{y}_{d,n-1}\le \left(\genfrac{}{}{0pt}{}{n+d}{n}\right)$, we obtain

$$n\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-n^2\le (n+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n-1}}\right)+n(n+1)(n-2)$$

(2)

We have $n\left(\genfrac{}{}{0pt}{}{n+d}{n}\right)-(n+1)\left(\genfrac{}{}{0pt}{}{n+d-1}{n-1}\right)=(d-1)\left(\genfrac{}{}{0pt}{}{n+d-1}{n-1}\right)$. Hence, (2) is equivalent to

$$(d-1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n-1}}\right)\le n^3-2n$$

(3)

The left hand side of (3) is an increasing function of d. Thus, to prove Claim 1 by contradiction, it is sufficient to prove that (3) fails for $d=5$. Take $d=5$. The left hand side of (3) is $4(n+4)(n+3)(n+2)(n+1)>n^3-2n$ for all $n\ge 3$.   □

Claim 2.

We have $y_{d,n}-y_{d,n-1}\le y_{d-1,n}-1$.

Proof of Claim 2.

Assume $y_{d,n}\ge y_{d,n-1}+y_{d-1,n}$. We multiply this inequality by $n(n+1)$ and then use that $n(n+1)y_{d,n}\le n\left(\genfrac{}{}{0pt}{}{n+d}{n}\right),$

$$n(n+1)y_{d,n-1}\ge (n+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n-1}}\right)-(n-1)(n+1),$$

and $n(n+1)y_{d-1,n}\ge n\left(\genfrac{}{}{0pt}{}{n+d-1}{n}\right)-n^2$. We have $\left(\genfrac{}{}{0pt}{}{n+d}{n}\right)-\left(\genfrac{}{}{0pt}{}{n+d-1}{n-1}\right)=\left(\genfrac{}{}{0pt}{}{n+d-1}{n-1}\right)$. We obtain the inequality

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n-1}}\right)\le (n-1)(n+1)+2n^2-n-1$$

(4)

The left hand side of this inequality is an increasing function of d. Since $d\ge 5$ and $n\ge 3$ (4) gives $(n+4)(n+3)(n+1)n\le 3n^2-n-1$. This contradiction proves Claim 2.

(b1) Take $z_{n,d}=y_{n,d}$. Claim 1 gives $y_{d,n}\ge y_{d,n-1}+e_{d,n-1}$. Take a general set $S_1\in S({\mathbb{P}}^n,y_{d,n}-y_{d,n-1}-e_{d,n-1})$ and a general $(S_2,S_3)\in S(H,y_{d,n-1})\times S(H,e_{d,n-1})$. Since Proposition 1 is proven for H, Remarks 3 and 4 give $h^i(H,{\mathcal{I}}_{(2S_2,H)\cup S_3,H}\left(d\right))=0$, $i=0,1$. Hence, the Differential Horace Lemma (Lemma 2) shows that to prove the proposition for n, d and $x:=y_{d,n-1}$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{2S_1\cup S_2\cup (2S_3,H)}(d-1)\right)=0$. Note that $\mathrm{deg}(2S_1\cup S_3\cup (2S_2,H))=\left(\genfrac{}{}{0pt}{}{n+d-1}{n}\right)-(\left(\genfrac{}{}{0pt}{}{n+d}{n}\right)-(n+1)y_{d,n})\le \left(\genfrac{}{}{0pt}{}{n+d-1}{n}\right)$.   □

Claim 3.

We have $h^1\left({\mathcal{I}}_{2S_1\cup (2S_3,H)}(d-1)\right)=0$.

Proof of Claim 3.

Since $(2S_3,H)\subset 2S_3$, Remark 2 gives that it is sufficient to prove that $h^1\left({\mathcal{I}}_{2S_1\cup 2S_3}(d-1)\right)=0$. Note that $S_1\cup S_3$ does not depend on A. Since $e_{d,n-1}\le n-1$ and any $n-1$ points of ${\mathbb{P}}^n$ are contained in a hyperplane, $h^1\left({\mathcal{I}}_{2S_1\cup 2S_2}(d-1)\right)=0$ if and only if $h^1\left({\mathcal{I}}_{2E}(d-1)\right)=0$ for a general $E\in S({\mathbb{P}}^n,y_{d,n}-y_{d,n-1})$. Since $y_{d,n}-y_{d,n-1}\le y_{d-1,n}$ (Claim 2) and $d\ge 5$, it is sufficient to quote the Alexander–Hirschowitz theorem, unless $d=5$ and $3\le n\le 4$. In these cases, we use that $y_{d,n}-y_{d,n-1}\le y_{d-1,n}-1$ (Claim 2).   □

We have ${\mathrm{Res}}_H(2S_1\cup S_2\cup (2S_3,H))=2S_1$ and $\mathrm{deg}(2S_1\cup S_3\cup (2S_2,H)\le y_{d-1,n}$. Using Claim 3 and Lemma 1 to prove that $h^1\left({\mathcal{I}}_{2S_1\cup S_2\cup (2S_3,H)}(d-1)\right)=0$, it is sufficient to prove that $h^0\left({\mathcal{I}}_{2S_1}(d-2)\right)\le \max \{0,\left(\genfrac{}{}{0pt}{}{n+d-1}{n}\right)-\mathrm{deg}\left(2S_1\right)-y_{d,n-1}-n{e}_{d,n-1}\}$. Assume for the moment that $(n,d)\ne (4,5)$. Since $(n,d-2)$ is not in the Alexander–Hirschowitz list of exceptional cases, $h^0\left({\mathcal{I}}_{2S_1}(d-2)\right)=\max \{0,\left(\genfrac{}{}{0pt}{}{n+d-2}{n}\right)-\mathrm{deg}\left(2S_1\right)\}$. Since

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-2}{n}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-2}{n-1}}\right),$$

for $(n,d)\ne (4,5)$, it is sufficient to check the inequality $y_{d,n-1}+e_{d,n-1}\le \left(\genfrac{}{}{0pt}{}{n+d-2}{n-1}\right)$. We have $n{y}_{d,n-1}+e_{d,n-1}=\left(\genfrac{}{}{0pt}{}{n+d-1}{n-1}\right)$ and

$$n\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-2}{n-1}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n-1}}\right)={\displaystyle \frac{(n+d-2)!}{(n-1)!d!}}(nd(d-2)-n-d-1).$$

Thus, it is sufficient to check the inequality $(n-1)e_{d,n-1}\le \frac{(n+d-2)!}{(n-1)!d!}(nd(d-2)-n-d-1)$. Since $e_{d,n-1}\le n-1$, it is sufficient to check the inequality

$${(n-1)}^2\le {\displaystyle \frac{(n+d-2)!}{(n-1)!d!}}(nd(d-2)-n-d-1)$$

(5)

Since $n+d-1\ge d+1$, the right hand side of (5) is an increasing function of d. For $d=5$, the inequality (5) is the inequality

$${(n-1)}^2\le (n+3)(n+2)(n+1)n(14n-6)/120,$$

which is true for all $n\ge 3$.

Now, assume $n=4$ and $d=5$. Since $d\ge a-1$, we obtain $3\le a\le 6$. We have $y_{5,4}=24$, $x_{5,4}=25$. If $3\le a\le 4$, then $y_{5,3}=13$ and $e_{5,3}=4-a$. If $5\le a\le 6$, then $y_{5,3}=12$ and $e_{5,3}=8-a$. It is sufficient that for a general $E\in S({\mathbb{P}}^4,7)$, we have $h^0\left({\mathcal{I}}_{2E}\left(3\right)\right)=1$ by the Alexander–Hirschowitz theorem.

(b2) Now, assume $z_{d,n}=x_{d,n}$. By step (b1), we may assume $y_{d,n}<z_{d,n}$. Thus, we have $z_{d,n}=y_{d,n}+1$. We specialize S to $S_3\cup S_4$ with $S_3$ a general element of $S({\mathbb{P}}^n,y_{d,n}-y_{d,n-1})$ and $S_4$ a general union of $y_{d,n-1}+1$ points of H. We have $y_{d,n-1}+1\ge x_{d,n-1}$. The inductive assumption gives $h^0(H,{\mathcal{I}}_{A\cup (2S_4,H)}\left(d\right))=0$. Hence, to prove Proposition 1 in this case, it is sufficient to prove that $h^0\left({\mathcal{I}}_{2S_3\cup S_4}(d-1)\right)=0$. Since $d-1>2$, the the triple $(n,d-1,\#S_3)$ is not in the Alexander–Hirschowitz list, and hence,

$$h^0\left({\mathcal{I}}_{2S_3}(d-1)\right)=\max \{0,\left({\displaystyle \genfrac{}{}{0pt}{}{d+n}{n}}\right)-(n+1)(y_{d,n}-y_{d,n-1})\}.$$

Then, we use Lemma 1.

4. Proof of Theorem 2

In this section, we prove Theorem 2.

For all integers a, n and d such that $a\ge 0$, $n\ge 1$ and $d\ge a-1$, set

$$y_{d,n,a}:=\lfloor (\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-a)/(n+1)\rfloor ,x_{d,n,a}:=\lceil (\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-a)/(n+1)\rceil$$

and $e_{d,n,a}:=\left(\genfrac{}{}{0pt}{}{n+d}{n}\right)-a-(n+1)y_{d,n,a}$. We have $y_{d,n,a}\le x_{d,n,a}\le y_{d,n,a}+1$, $0\le e_{d,n,a}\le n$, $(n+1)y_{d,n,a}+a+e_{d,n,a}$. Thus, we have $y_{d,1,a}=\lfloor (d+1-a)/2\rfloor$, $e_{d,1,a}=0$ if $d-a$ is odd and $e_{d,1,a}=1$ if $d-a$ is even.

For all integers a, n and d such that $a\ge 0$, $n\ge 1$ and $d\ge 2a-1$, set

$u_{d,n,a}:=\lfloor \left(\genfrac{}{}{0pt}{}{n+d}{n}\right)/(n+1)\rfloor -a$, $v_{d,n,a}:=\lceil \left(\genfrac{}{}{0pt}{}{n+d}{n}\right)/(n+1)\rceil -a$ and

$$f_{d,n,a}:=\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)-(n+1)a-(n+1)u_{d,n,a}.$$

We have $u_{d,n,a}\le v_{d,n,a}\le u_{d,n,a}+1$, $0\le f_{d,n,a}\le n$, $(n+1)u_{d,n,a}+(n+1)a+f_{d,n,a}$. Thus, we have $y_{d,1,a}=\lfloor (d+1-a)/2\rfloor$, $f_{d,1,a}=0$ if d is odd, $f_{d,1,a}=1$ if d is even, while we have $u_{d,n,a}=y_{n,d,0}-a=u_{n,d,0}-a$, $v_{d,n,a}=v_{n,d,0}-a=v_{n,d,0}-a$ and $f_{d,n,a}=f_{d,n,0}$.

If a is fixed (in a statement or a proof), then we write $y_{n,d}$, $x_{n,d}$, $v_{n,d}$, $e_{n,d}$ and $f_{n,d}$ instead of $y_{n,d,a}$, $x_{n,d,a}$, $v_{n,d,a}$, $e_{n,d,a}$ and $f_{n,d,a}$. For a fixed a, the integers $y_{n,d}$ and $x_{n,d}$ are the ones used in the proof of Lemma 1. We fix a and $ϵ\in \{1,2\}$. For all $n\ge 1$, we use the integer $y_{n,d}\left(ϵ\right):=y_{n,d}-ϵ$, $x_{n,d}\left(ϵ\right):=x_{n,d}-ϵ$ and $e_{d,n}\left(ϵ\right):=e_{d,n}$.

Remark 7.

Let $X\subset {\mathbb{P}}^r$ be an integral and non-degenerate surface. Fix a linear space $M\subseteq {\mathbb{P}}^r$. Then $\dim ⟨M\cup v⟩=\min \{r,2+\dim M\}$ for a general connected degree 2 zero-dimensional scheme whose reduction is contained in $X_{\mathrm{reg}}$ (Remark 5).

Remark 8.

Fix $(n,d,a)$ and assume $f_{d,n,a}\le 2$. By Remark 7, to prove Theorem 2 for the triple $(n,d,a)$, it is sufficient to prove it for $x=u_{d,n}$.

Remark 9.

Assume $a=3$. Take a line $L\subset {\mathbb{P}}^n$, $A\subset L$ such that $\#A=3$ and a hyperplane $H\supseteq L$. By Remark 8, if $f_{d,n}\le 2$ to prove the case for this triple $(n,d,A)$, it is sufficient to check the case $x=u_{d,n}$. Take a general $S\in S({\mathbb{P}}^n,u_{d,n})$.

(a) Take $n=2$ and $d=5$. We have $u_{5,2}=4$ and $f_{5,2}=0$. Since $h^i(L,{\mathcal{I}}_{(2A,L),L}\left(5\right))=0$, $i=0,1$, to prove Theorem 2 in this case, it is sufficient to use that $h^1\left({\mathcal{I}}_{2S\cup A}\left(4\right)\right)=0$ (Lemma 1).

(b) Take $n=2$ and $d=6$. We have $u_{6,2}=6$ and $f_{6,2}=1$. Take a smooth plane cubic C such that $C\cap L=A$ and degenerate S to a general $S^{\prime }\in S(C,6)$. Note that we have $\mathrm{deg}((2S^{\prime }\cup 2A)\cap C))=18=\mathrm{deg}\left({\mathcal{O}}_C\left(6\right)\right)$. For a general $S^{\prime }$, the scheme $(2S^{\prime }\cup 2A)\cap C$ is not the complete intersection of C with a degree 6 plane curve. Since C is an elliptic curve, $h^i(C,{\mathcal{I}}_{(2S^{\prime }\cup 2A)\cap C,C}\left(6\right))=0$, $i=0,1$. By the residual exact sequence of C, it is sufficient to prove that $h^0\left({\mathcal{I}}_{A\cup S_1}\left(3\right)\right)=0$. This is is true according to the Bezout theorem and the fact that for a general $S_1$ the set $A\cup S_1$ is not a complete intersection of C with another plane cubic.

(c) Take $n=3$ and $d=5$. We have $u_{5,3}=11$ and $f_{5,3}=0$. We degenerate S to $S_1\cup S_2$ with $S_1$ general in $S({\mathbb{P}}^3,5)$ and $S_2$ general in $S(H,4)$. Part (a) gives $h^i(H,{\mathcal{I}}_{(2A,H)\cup (2S_2,H}\left(5\right))=0$. Hence, it is sufficient to prove that $h^i\left({\mathcal{I}}_{2S_1\cup A\cup S_2}\left(4\right)\right)=0$. Proposition 1 gives $h^1\left({\mathcal{I}}_{2S_1\cup A}\left(4\right)\right)=0$. By Lemma 1, it is sufficient to use that $h^i\left({\mathcal{I}}_{2S_1}\left(3\right)\right)=0$, $i=0,1$.

(d) Take $n=3$ and $d=6$. We have $u_{6,3}=18$ and $f_{6,3}=0$. We take a general $S^{\prime }\in S({\mathbb{P}}^3,11)$ a general $S^{″}\in S(H,6)$ and a general $q\in H$. Step (b) gives the equalities $h^i(H,{\mathcal{I}}_{(2S^{″},H)\cup (2A,H)\cup \left\{q\right\},H}\left(6\right))=0$, $i=0,1$. Hence, the Differential Horace Lemma (Lemma 2) applied to q shows that it is sufficient to prove the quality $h^i({\mathcal{I}}_{2S^{\prime }\cup A\cup S^{″}\cup (2q,H)}\left(5\right))=0$. Since q is general in a general plane containing L, $S^{\prime }\cup \left\{q\right\}$ may be seen as a general element of $S({\mathbb{P}}^3,12)$. Proposition 1 gives $h^1({\mathcal{I}}_{2S^{\prime }\cup 2q\cup A}\left(5\right))=0$. Hence, $h^1({\mathcal{I}}_{2S^{\prime }\cup (2q,H)\cup A}\left(5\right))=0$ (Remark 2). Since $S^{″}$ is general in H, to conclude this case, it is sufficient to observe that $h^0\left({\mathcal{I}}_{2S^{\prime }}\left(4\right)\right)=0$ (use for $x\ne 9$ the case $(n,d,x)=(3,4,x)$ is not defective).

(e) Take $n=4$ and $d=5$. We have $u_{5,4}=22$ and $f_{5,4}=1$. We degenerate S to $S^{\prime }\cup S^{″}$ with $S^{\prime }$ general in $S({\mathbb{P}}^4,11)$ and $S^{″}$ general in $S(H,11)$. Part (c) gives the equalities $h^i(H,{\mathcal{I}}_{(2A,H)\cup (2S^{\prime },H),H}\left(5\right))=0$, $i=0,1$. Hence, it is sufficient to prove the equality $h^1\left({\mathcal{I}}_{2S^{\prime }\cup A\cup S^{\prime \prime }}\left(4\right)\right)=0$. Proposition 3 gives $h^1\left({\mathcal{I}}_{2S^{\prime }\cup A}\left(4\right)\right)=0$, and hence, $h^0\left({\mathcal{I}}_{2S^{\prime }\cup A}\left(4\right)\right)=12$. By Lemma 1, it is sufficient to use that $h^0\left({\mathcal{I}}_{2S^{\prime }}\left(3\right)\right)=0$.

(f) Take $n=4$ and $d=6$. We have $u_{6,4}=39$ and $f_{6,4}=0$. We take a general $S^{\prime }\in S({\mathbb{P}}^4,21)$ and a general $S^{″}\in S(H,18)$. Part (d) gives $h^i\left(H,{\mathcal{I}}_{(2A,H)\cup (2S^{″},H),H}\left(6\right)\right)=0$. By the Horace Lemma, it is sufficient to prove that $h^i\left({\mathcal{I}}_{2S^{\prime }\cup A\cup S^{″}}\left(5\right)\right)=0$. Proposition 1 gives $h^1\left({\mathcal{I}}_{2S^{\prime }\cup A}\left(5\right)\right)=0$, i.e., $h^0\left({\mathcal{I}}_{2S^{\prime }\cup A}\left(5\right)\right)=18$. Using Lemma 1 to conclude, it is sufficient to use that $h^0\left({\mathcal{I}}_{2S^{\prime }}\left(4\right)\right)=0$.

Lemma 5.

Assume $a\ge 3$ and $d\ge 2a-1$. Then,

$$(n+1)(v_{d,n}-u_{d,n-1}-f_{d,n-1}+f_1)\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d-2}{n}}\right).$$

Proof. 

We have $(n+1)v_{d,n-1}\le \left(\genfrac{}{}{0pt}{}{n+d}{n}\right)-a+n$. Since $n{u}_{d,n-1}+f_{d,n-1}=\left(\genfrac{}{}{0pt}{}{n+d-1}{n-1}\right)-a$ and $0\le f_{d,n-1}\le n-1$, $n(u_{d,n-1}+f_{d,n-1})\ge \left(\genfrac{}{}{0pt}{}{n+d-1}{n-1}\right)$. Note that $n\left(\genfrac{}{}{0pt}{}{n+d}{n}\right)-n\left(\genfrac{}{}{0pt}{}{n+d-1}{n-1}\right)=n\left(\genfrac{}{}{0pt}{}{n+d-1}{n}\right)$ and that $n\left(\genfrac{}{}{0pt}{}{n+d-1}{n}\right)=d\left(\genfrac{}{}{0pt}{}{n+d-1}{n-1}\right)(d-1)$. Hence, to prove that

$$n(n+1)(v_{d,n}-u_{d,n-1}-f_{d,n-1}+f_1)\le n\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-2}{n}}\right),$$

it is sufficient to observe that $(d-1)\left(\genfrac{}{}{0pt}{}{n+d-1}{n-1}\right)\ge n^2+a$ for all $d\ge \max \{5,2a-1\}$ and all $n\ge 2$.   □

Proof of Theorem 2.

Since any 2 points of ${\mathbb{P}}^n$ are projectively equivalent, the Alexander–Hirschowitz theorem (Theorem 3) shows that it is sufficient to use the case $a\ge 3$, assuming $d\ge 2a-1\ge 5$. The integer a is fixed, and hence, in this proof, $u_{d,n}:=u_{d,n,a}$, and so on.

By Lemma 3, we may assume that A is contained in a line L. To prove the theorem for fixed A, n and d and all positive integers x, it is sufficient to prove that $h^1\left({\mathcal{I}}_{2A\cup 2E}\left(d\right)\right)=0$ for a general $E\in S({\mathbb{P}}^n\setminus A,u_{d,n})$ and $h^0\left({\mathcal{I}}_{2A\cup 2F}\left(d\right)\right)=0$ for a general $F\in S({\mathbb{P}}^n\setminus A,v_{d,n})$. Fix a hyperplane $H\supseteq L$. Take $z_{d,n}\in \{u_{d,n},v_{d,n}\}$ and a general $S\in S({\mathbb{P}}^n,z_{d-n})$. Since S is general, $S\cap H=\varnothing$. Note that the theorem is true for $n=1$. Hence, even in the case $n=2$, we may assume that the theorem is true in lower dimensional projective spaces. Hence, we may apply Proposition 3 and the theorem in H, even if $n=2$, and hence, $H=L$.

Fix an integer $d\ge 2a-1$. Recall that $0\le f_{d,n-1}\le n-1$. Until step (d), we assume $z_{d,n}\ge u_{d,n-1}+f_{d,n-1}$. Set $f:=f_{d,n-1}$.

Take a general $(S_1,S_2,S_3)\in S({\mathbb{P}}^n,z_{d,n}-u_{d,n-1}-f)\times S(H,u_{d,n-1})\times S(H,f)$.

By the inductive assumption on the dimension of the projective space, we have $h^j(H,{\mathcal{I}}_{(2A,H)\cup (2S_2,H)\cup S_3,H}\left(d\right))=0$, $j=0,1$. By the Differential Horace Lemma, to conclude the proof in this case, it is sufficient to prove that $h^i\left({\mathcal{I}}_{2S_1\cup S_2\cup A\cup (2S_3,H)}(d-1)\right)=0$.

(a) In this step, we prove that $h^1\left({\mathcal{I}}_{2S_1\cup A\cup (2S_3,H)}(d-1)\right)=0$.

Set $e_1:=\lfloor ((\genfrac{}{}{0pt}{}{d+n-2}{n-1})-a-u_{d,n-1}-nf)/n\rfloor$ and $f_1:=\left(\genfrac{}{}{0pt}{}{n+d-2}{n-1}\right)-a-u_{d,n-1}-n{e}_1$. We have $0\le f_1\le n-1$.

We assume for the moment that $z_{d,n}\ge u_{d,n-1}+f+e_1+f_1$, leaving the case $z_{d,n}<u_{d,n-1}+f+e_1+f_1$ to step (c). Take a general $(S_4,S_5)\in S(H,e_1)\times S(H,f_1)$. We specialize $e_1$ of the points of $S_1$ to different points of $S_4$ and apply the Differential Horace Lemma (Lemma 2) to the points of $S_5$. By the inductive assumption, we have $h^j(H,{\mathcal{I}}_{A\cup (2S_3,H)\cup (2S_4,H)\cup S_5,H}(d-1))=0$, $j=0,1$. By the Differential Horace Lemma, to prove that $h^1\left({\mathcal{I}}_{2S_1\cup A\cup (2S_3,H)}(d-1)\right)=0$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{2S_1\cup S_4\cup (2S_5,H)}(d-2)\right)=0$.

(a1) In this step, we prove that $h^1\left({\mathcal{I}}_{2S_1\cup (2S_5,H)}(d-1)\right)=0$. By Remark 2, it is sufficient to prove that $h^1\left({\mathcal{I}}_{2S_1\cup 2S_5}(d-2)\right)=0$. We have $\#(S_1\cup S_5)=z_{d,n}-u_{d,n-1}-f+f_1$. Assume for the moment that $(n,d-2,\#(S_1\cup S_5))$ is not in the list of the exceptional cases of the Alexander–Hirschowitz theorem (Theorem 3). Since $z_{d,n}\le v_{n,d}$, to prove that $h^1\left({\mathcal{I}}_{2S_1\cup (2S_5,H)}(d-1)\right)=0$, it is sufficient to use that Lemma 5 gives the inequality

$$(n+1)(v_{d,n}-u_{d,n-1}-f+f_1)\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n}}\right).$$

Now, assume that $(n,d-2,\#(S_1\cup S_5))$ is in the list of the exceptional cases of the Alexander–Hirschowitz theorem. Since $d\ge 2a-1$ with $a\ge 3$, we obtain $a=3$ and

$$(n,d)\in \left\{\right(2,6),(3,6),(4,5),(4,6\left)\right\}.$$

These cases are carried out in Remark 9.

(a2) In this step, we prove that $h^i\left({\mathcal{I}}_{2S_1\cup A\cup (2S_3,H)}(d-1)\right)=0$. Step (a1) gives $h^0\left({\mathcal{I}}_{2S_1\cup A\cup (2S_3,H)}(d-1)\right)=\left(\genfrac{}{}{0pt}{}{n+d-1}{n}\right)-\mathrm{deg}\left(2S_1\right)-a-nf$. By Lemma 1 and step (a1), to prove the theorem in this case, it is sufficient to prove that

$$h^0\left({\mathcal{I}}_{2S_1}(d-2)\right)\le \max \{0,\left(\genfrac{}{}{0pt}{}{n+d-1}{n}\right)-\mathrm{deg}\left(2S_1\right)-a-nf-u_{d,n-1}\}.$$

Assume for the moment that $(n,d-2,\#S_1)$ is not in one of the exceptional cases of the Alexander–Hirschowitz theorem. We have $h^0\left({\mathcal{I}}_{2S_1}(d-2)\right)=\left(\genfrac{}{}{0pt}{}{n+d-2}{n}\right)-\mathrm{deg}\left(2S_1\right)$. Hence, it is sufficient to use that $f\le n$ and $\left(\genfrac{}{}{0pt}{}{n+d-2}{n-1}\right)\ge a+n^2$.

Since $d\ge 5$, the exceptional cases for $(n,d-2)$ are settled in Remark 9.

(b) In this step, we conclude the proof of Theorem 2, proving the equality $h^i\left({\mathcal{I}}_{2S_1\cup S_2\cup (2S_3,H)\cup A}(d-1)\right)=0$. Step (a) gives

$$h^1\left({\mathcal{I}}_{2S_1\cup (2S_3,H)}(d-1)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+d-1}{n}}\right)-\mathrm{deg}\left(2S_1\right)-a-nf.$$

Since $S_2$ is general in H and $\#S_2=u_{d,n-1}$, Lemma 1 gives that it is sufficient to prove that $h^0\left({\mathcal{I}}_{2S_1}(d-2)\right)\le \max \{0,\left(\genfrac{}{}{0pt}{}{n+d-1}{n}\right)-\mathrm{deg}\left(2S_1\right)-a-nf-u_{d,n-1}\}$. Remember that $d\ge 2a-1\ge 5$. Hence, if $(n,d-2)$ is in the exceptional list of the Alexander–Hirschowitz theorem, we use Remark 9. If $(n,d-2)$ is not in this list, then $h^0\left({\mathcal{I}}_{2S_1}(d-2)\right)=\max \{0,\left(\genfrac{}{}{0pt}{}{n+d-2}{n}\right)-\mathrm{deg}\left(2S_1\right)\}$. Hence, it is sufficient to prove that $a+nf+u_{d,n-1}\le \left(\genfrac{}{}{0pt}{}{n+d-2}{n-1}\right)$. We have $n(u_{n,d-1}+f)=\left(\genfrac{}{}{0pt}{}{n+d-1}{n-1}\right)-na$. Hence, it is sufficient to prove that $a+(n^2-1)f\le n\left(\genfrac{}{}{0pt}{}{n+d-2}{n-1}\right)-\left(\genfrac{}{}{0pt}{}{n+d-1}{n-1}\right)$. Since $f\le (n-1)$, it is sufficient to prove that $a+(n^2-1)(n-1)\le n\left(\genfrac{}{}{0pt}{}{n+d-2}{n-1}\right)-\left(\genfrac{}{}{0pt}{}{n+d-1}{n-1}\right)$. Call $t(n,d,a)$ the difference between the right hand side and the left one of the inequality we want to prove. We have $t(2,d,a)=d-1-a-3$, which is $\ge 0$, except in the case $a=3$, $n=2$, $d\le 6$ settled in Remark 9. We have $t(3,d,a)=d^2-1-a-16$. Since $a\ge 3$ and $d\ge 2a-1$, $t(3,d,a)>0$, we have $t(4,d,a)=(d+2)(d+1)(d-1)/2-16-a>0$. Call $w(n,d,a)$ the left hand side of the inequality we want to prove, and $q(n,d,a)$ its right hand side. We have $q(n,d,a)=\frac{(n+d-2)!(nd-n-d+1)}{(n-1)!d!}$. We have $w(n+1,d,a)/w(n,d,a)\le {(n+1)}^3/n^3$. We have $q(n+1,d,a)/q(n,d,a)=\frac{(n+d-1)\left(\right(n+1)d-n-d)}{n(nd-n-d+1)}$. We have $\left(\right(n+1)d-n-d)=(n-1)d$. For a fixed n, the rational number $q(n+1,d,a)/q(n,d,a)$ increases with d. Hence, it is sufficient to use that $q(n+1,5,a)/q(n,5,a)>5/4$.

(c) Assume $e+f\le z_{d,n}<e+f+e_1+f_1$. We take $f_2:=\max \{0,z_{d,n}-e-f\}$ and $e_2:=\min \{e_1,z_{d,n}-f_1\}$, $\#S_5=f_2$ and $\#S_4=e_2$ and $\#S_3=\max \{0,z_{d,n}-e_2-f_2\}$.

(d) Assume $z_{d,n}<e+f$. We take $f^{\prime }:=\max \{0,z_{d,n}-e\}$ and $S_1=\varnothing$.   □

5. Segre–Veronese Varieties

In this section, we fix integers $k\ge 2$ and $n_i>0$, $i=1,\dots ,k$. Set $n:=n_1+\cdots +n_k$ and $Y:={\mathbb{P}}^{n_1}\times \cdots {\mathbb{P}}^{n_k}$. We have $\dim Y=n$, and Y is the multiprojective space associated to tensors of format $(n_1+1)\times \cdots \times (n_k+1)$. Write $G:=\mathrm{Aut}\left({\mathbb{P}}^{n_1}\right)\times \cdots \times \mathrm{Aut}\left({\mathbb{P}}^{n_k}\right)$. We have $G\subset \mathrm{Aut}\left(Y\right)$ and $g^{*}\left({\mathcal{O}}_Y(x_1,\dots ,x_k)\right)\cong {\mathcal{O}}_Y(x_1,\dots ,x_k)$ for all $(x_1,\dots ,x_k)\in {\mathbb{Z}}^k$.

Remark 10.

We discussed our approach in Remark 6. For Segre–Veronese varieties, the reduction is from an arbitrary $A\in S(Y,a)$ to a specific $B\in S(Y,a)$ contained in a curve L of multidegree ${ϵ}_1$ (Lemma 6). This is a key step because using the Differential Horace Lemma twice with respect to hypersurfaces containing L, we pass from A or B to ∅. To avoid several very dense pages, we reduce to results known in the literature, (ref. [4] Cor. 3.14) for this section, ref. [7] for the next one. The interested reader may use the published literature in a similar way in the following very interesting settings:

1. 

$k=2$, arbitrary biprojective space, bidegree $(d_1,d_2)$ with $d_1\ge 3$ and $d_3\ge 3$ [6]. A few cases with $d_1=2$ and $d_2\ge 5$ are in [19].

2. 

$k\ge 3$, arbitrary multiprojective space, multidegree $(d_1,\dots ,d_k)$ with $d_1\ge 3$, $d_2\ge 3$, $d_i\ge 2$ for all $i>2$ [5,19]. A few cases with $d_k=1$ are in [5].

3. 

$k=2$, $d_1=1$. This part is wide open even for $A=\varnothing$ [20,21,22].

Lemma 6.

Take $A\in S(Y,a)$ such that $\#{\pi }_1\left(A\right)=a$. Fix $p_i\in {\mathbb{P}}^{n_i}$, $2\le i\le k$, and a line $R\subseteq {\mathbb{P}}^{n_1}$. Set $L:=R\times p_2\times \cdots \times p_k$. Then there is $B\subset L$ such that $\#B=a$, and B is a flat limit of set G-equivalents to A.

Proof. 

The case $k=1$ (omitted in the statement) is true by the proof of Lemma 3. Hence, it is sufficient to find an irreducible family of elements of G that for $i=2,\dots ,k$ sends the i-th coordinate of all points of A to a family with $p_i$ as its limit, preserving the condition that the first coordinate of all points of A are distinct. For $i=2,\dots ,k$, take a general hyperplane $H_i\subset {\mathbb{P}}^{n_i}$. Since $H_i$ is general, we have $p_i\notin H_i$. For $i=2,\dots ,k$, let $Y\left(H_i\right)$ denote the multiprojective space with $H_i$ as its i-th factor and ${\mathbb{P}}^{n_j}$ as its j-th factor for all $j\in \{1,\dots ,k\}\setminus \left\{i\right\}$. Set $\tilde{Y}:={\cup }_{i=2}^{k}Y\left(H_i\right)$. Let $\mathcal{l}:Y\setminus \tilde{Y}\to {\mathbb{P}}^{n_1}\times p_2\times \cdots \times p_k$ be defined by the formula $\mathcal{l}\left((x_1,\dots ,x_k)\right)=(x_1,p_1,\dots ,p_k)$. Since each $H_i$ is general, $A\cap \tilde{Y}=\varnothing$, and hence, ℓ is defined at each point of A. Since $\#{\pi }_1\left(A\right)=a$, we have $\#\mathcal{l}\left(A\right)=a$. Write $\mathcal{l}=({\mathcal{l}}_1,\dots ,{\mathcal{l}}_k)$ with ${\mathcal{l}}_1:{\mathbb{P}}^{n_1}\to {\mathbb{P}}^{n_1}$ the identity map and ${\mathcal{l}}_i:{\mathbb{P}}^{n_i}\setminus H_i\to p_i$ the linear projection from $H_i$ for $i=2,\dots ,k$. Fix $i\in \{2,\dots ,k\}$. In the proof of Lemma 3 (taking a system of homogeneous coordinates of ${\mathbb{P}}^{n_i}$ with $p_i$ a coordinate points), we proved that ${\mathcal{l}}_i$ is a flat limit of a family ${\left\{g{\left(i\right)}_c\right\}}_{c\in \mathbb{K}\setminus \left\{0\right\}}$ of automorphisms of ${\mathbb{P}}^{n_i}$. Write $g{\left(1\right)}_c:={\mathcal{l}}_1$ for all $c\in \mathbb{K}\setminus \left\{0\right\}$. Note that $G=\mathrm{Aut}\left({\mathbb{P}}^{n_1}\right)\times \cdots \times \mathrm{Aut}\left({\mathbb{P}}^{n_k}\right)$. Thus, for all $c\in \mathbb{K}\setminus \left\{0\right\}$ the automorphism $g_c:=(g{\left(1\right)}_c,\dots ,g{\left(k\right)}_c)$ is an element of G. By construction, the set $\mathcal{l}\left(A\right)$ is the flat limit of the family ${\left\{g_c\left(A\right)\right\}}_{c\in \mathbb{K}\setminus \left\{0\right\}}$. If $n_1=1$, the proof is over. If $n_1>1$, we apply Lemma 3 to the first factor of Y.   □

Lemma 7.

Fix positive integers $d_1,\dots ,d_k$, a and x. Take $A\in S(Y,a)$ such that $\#{\pi }_1\left(A\right)=a$. Fix $p_i\in {\mathbb{P}}^{n_i}$, $2\le i\le k$ and a line $R\subseteq {\mathbb{P}}^{n_1}$. Set $L:=R\times p_2\times \cdots \times p_k$. Then there is $B\in S(L,a)$ such that for a general $S\in S(Y,x)$, we have $h^0\left({\mathcal{I}}_{2B\cup 2S}(d_1,\dots ,d_k)\right)\ge h^0\left({\mathcal{I}}_{2A\cup 2S}(d_1,\dots ,d_k)\right)$ and $h^0\left({\mathcal{I}}_{B\cup 2S}(d_1,\dots ,d_k)\right)\ge h^0\left({\mathcal{I}}_{A\cup 2S}(d_1,\dots ,d_k)\right)$.

Proof. 

Use the semicontinuity theorem for cohomology (ref. [16], Th. III.12.8) and Lemma 6.   □

Remark 11.

Take $a=1$ and a positive integer x. Fix k, $n_i$ and $d_i$, $1\le i\le k$. Take a general element $(S,S^{\prime })$ of the set $S(Y,x)\times S(Y,x+1)$. Since G is transitive, we have the equality $h^0\left({\mathcal{I}}_{2A\cup 2S}(d_1,\dots ,d_k)\right)=h^0\left({\mathcal{I}}_{2S^{\prime }}(d_1,\dots ,d_k)\right)$ and the equality

$$h^0\left({\mathcal{I}}_{A\cup 2S}(d_1,\dots ,d_k)\right)=\max \{0,h^0\left({\mathcal{I}}_{2S}(d_1,\dots ,d_k)\right)-1,0\}.$$

Hence, it is important to know when a secant variety of a Segre–Veronese variety is defective. Many cases are known [4,5,6,7]. We list the following 2 cases that will be needed in this paper. We call $h^0$ the integer $h^0\left({\mathcal{I}}_{2S}(d_1,\dots ,d_k)\right)$ for a general $S\in S(Y,x)$ and

$$\delta :=\mathrm{deg}\left(2S\right)-h^0\left({\mathcal{O}}_Y(d_1,\dots ,d_k)\right)+h^0=h^0\left({\mathcal{I}}_{2S}(d_1,\dots ,d_k)\right).$$

(a) Assume $n_i=1$ for all i. With no loss of generality, we may assume $d_1\le \dots \le d_k$. By [7], the 4 exceptional cases are as follows:

1. 

$k=2$, $d_1=2$, $d_2=2b$ with b a positive integer, $x=2b+1$, $h^0=1$, $\delta =1$;

2. 

$k=3$, $d_1=d_2=1$, $d_3=2b$ with b a positive integer, $x=2b+1$, $h^0=1$, $\delta =1$;

3. 

$k=3$, $d_1=d_2=d_3=2$, $x=7$, $h^0=1$, $\delta =2$;

4. 

$k=4$, $d_1=d_2=d_3=d_4=1$, $x=3$, $h^0=2$, $\delta =1$.

(b) Assume $k=2$, $n_2=1$. If $d_2\ge 3$, then the associated Segre–Veronese variety is defective if and only if $d_1=2$ and $d_2$ is even (ref. [4], Cor. 3.14).

Remark 12.

Take $a=2$. The group G acts transitively on the subset of $S(Y,2)$ formed by 2 points not contained in a curve of multidegree ${ϵ}_1$. Hence, for $A\in S(Y,2)$ not contained in a curve of multidegree ${ϵ}_1$, we have $h^0\left({\mathcal{I}}_{A\cup 2S}(d_1,\dots ,d_k)\right)=\max \{0,h^0\left({\mathcal{I}}_{2S}(d_1,\dots ,d_k)\right)-2\}$ and $h^0\left({\mathcal{I}}_{2A\cup 2S}(d_1,\dots ,d_k)\right)=h^0\left({\mathcal{I}}_{2S^{\prime }}(d_1,\dots ,d_k)\right)$ for a general $S\in S(Y,x)$ and a general $S^{\prime }\in S(Y,x+2)$.

Remark 13.

The group G acts transitively on the set of all curves $L\subset Y$ of multidegree ${ϵ}_1$. Fix any such curve L and call $G_L$ the set of all $g\in G$ such that $g\left(L\right)=L$. The group $G_L$ acts 3-transitively on L. Thus, if $a\le 3$ and $A,B\in S(L,a)$, then $h^0\left({\mathcal{I}}_{A\cup 2S}(d_1,\dots ,d_k)\right)=h^0\left({\mathcal{I}}_{B\cup 2S}(d_1,\dots ,d_k)\right)$ and $h^0\left({\mathcal{I}}_{2A\cup 2S}(d_1,\dots ,d_k)\right)=h^0\left({\mathcal{I}}_{2B\cup 2S}(d_1,\dots ,d_k)\right)$. Taking B general in $S(L,a)$, we obtain

$$h^0\left({\mathcal{I}}_{A\cup 2S}(d_1,\dots ,d_k)\right)=\max \{h^0\left({\mathcal{I}}_{2S}(d_1,\dots ,d_k)\right)-a,h^0\left({\mathcal{I}}_{L\cup 2S}(d_1,\dots ,d_k)\right)\}.$$

Remark 14.

Set $Y:={\mathbb{P}}^1\times {\mathbb{P}}^1$, i.e., take $k=2$ and $n_1=n_2=1$. Fix positive integers a, $d_1$, $d_2$, x such that $d_1\ge \max \{1,a-1\}$. The triple $(d_1,d_2,1)$ is not defective by Remark 11 and the fact that in case $k=2$ of part (a) of Remark 11, all triples have defect at most 1. Now assume $a\ge 2$, $d_1=a-1$ and A contained in a curve L of multidegree ${ϵ}_1$. We have $L\cong {\mathbb{P}}^1$ and $\mathrm{deg}\left({\mathcal{O}}_L(x_1,x_2)\right)=x_1$ for all $(x_1,x_2)\in {\mathbb{Z}}^2$. For all integers $x_1$, $x_2$, all zero-dimensional schemes $Z\subset Y$, we have the exact sequence

$$0\to {\mathcal{I}}_{{\mathrm{Res}}_L\left(Z\right)}(x_1,x_2-1)\to {\mathcal{I}}_Z(x_1,x_2)\to {\mathcal{I}}_{Z\cap L,L}(x_1,x_2)\to 0$$

(6)

We have $h^0\left({\mathcal{I}}_A(d_1,d_2)\right)=h^0\left({\mathcal{O}}_Y(d_1,d_2-1)\right)$ by the long cohomology exact sequence associated to (6). Hence, the associated linear system $|{\mathcal{I}}_{A\cup 2S}(d_1,d_2)|$ is defective for a general $S\in S(Y,x)$ if and only if either $d_2=1$ and $2x\le d_1$ or $d_2=3$, $d_1=a-1$ is even and $x=a$ or $d_1=2$, $a=3$, $d_2$ is odd, $d_2\ge 3$ and $x=d_2$.

Remark 15.

Take an integral and non-degenerate variety $X\subset {\mathbb{P}}^r$, an integral curve $C\subset X$ and integers $a\ge 0$, $b\ge 0$. Let $Z\subset C_{\mathrm{reg}}$ be a general union of a points and b double points of $C_{\mathrm{reg}}$. By Remark 5 applied to $C\subseteq ⟨C⟩$, we have $\dim ⟨Z⟩=\min \{a+2b-1,\dim ⟨C⟩\}$.

Lemma 8.

Fix integers $a\ge 1$, $d_1\ge \max \{2a-2,2\}$, $d_2\ge 2$ and $x>0$. Take $A\in S(Y,a)$ such that $\#{\pi }_1\left(A\right)=a$ and a general $S\in S(Y,x)$. Then, either $h^0\left({\mathcal{I}}_{A\cup 2S}(d_1,d_2)\right)=0$ or $h^1\left({\mathcal{I}}_{A\cup 2S}(d_1,d_2)\right)=0$.

Proof. 

By Remark 7, it is sufficient to use the case in which A is contained in a curve $L\subset Y$ of multidegree ${ϵ}_1$.

The case $a=1$ is true by Remark 11 and the fact that in part (a), case $k=2$, of Remark 11, the defect is at most 1.

Thus, we may assume $a\ge 2$. Set

$$y:=\lfloor (d_1+1)(d_2+1)/3\rfloor -a,ϵ:=(d_1+)(d_2+1)-a-3y.$$

Since $(d_1+1)(d_2+1)-3y-3a\le 2$, to prove the lemma for a fixed triple $(d_1,d_2,A)$ and all integers x, it is sufficient to prove that $h^1\left({\mathcal{I}}_{A\cup 2S}(d_1,d_2)\right)=0$ for a general $S\in S(Y,y)$ (Remarks 3 and 7).

Since $a\ge 2$, $2a-2\ge a$ with equality if and only if $a=2$.

Assume $a=2$. By Remark 13, to prove the lemma, it is sufficient to prove the inequality $h^0\left({\mathcal{I}}_{L\cup 2S}(d_1,d_2)\right)\le h^0\left({\mathcal{I}}_{2S}(d_1,d_2)\right)-2$, i.e., $h^0\left({\mathcal{I}}_{2S}(d_1,d_2-1)\right)\le h^0\left({\mathcal{I}}_{2S}(d_1,d_2)\right)-2$, and that $h^1\left({\mathcal{I}}_{2S}(d_1,d_2-1)\right)=0$. Note that $h^0\left({\mathcal{O}}_Y(d_1,d_2)\right)-h^0\left({\mathcal{O}}_Y(d_1,d_2-1)\right)=d_1+1\ge 3$. First, assume $d_2\ge 3$. Since $d_2\ge 3$, ${\mathcal{O}}_Y(d_1,d_2)$ is not defective if $d_1\ge 3$, and hence, we conclude by the case $k=2$ of Remark 11. Now, assume $d_1=2$. Note that we have $y=d_2+1-\lceil a/3\rceil =d_2$ and $ϵ=1$. We use that in the exceptional cases with $k=2$ of Remark 11, we have $\delta =1$. Now, assume $d_2=2$. Since $a=2$, we have $y=d_1$ and $ϵ=1$. We have $h^0\left({\mathcal{I}}_{2S}(d_1,2)\right)=3$ and $h^0\left({\mathcal{I}}_{2S}(d_1,1)\right)=0$.

Assume $a>2$. Set $e:=\lfloor (d_1+1-a)/2\rfloor$ and $f:=d_1+1-a-2e$. We have $0\le f\le 1$.

Assume for the moment $y\ge e+f$.

Take a general $(S_1,S_2,S_3)\in S(Y,y-e-f)\times S(L,e)\times S(L,f)$ with the convention $S_1=\varnothing$ if $y-e-f=0$, $S_2=\varnothing$ if $e=0$ and $S_3=\varnothing$ if $f=0$. Note that we have $\mathrm{deg}((2S_2,L)\cup S_3\cup A)=d_1+1$. By the Differential Horace Lemma (Lemma 2), it is sufficient to prove that $h^1\left({\mathcal{I}}_{2S_1\cup S_2\cup (2S_3,L)}(d_1,d_2-1)\right)=0$.

Observation 1: Since $a+2e+f=d_1+1$ and $3y+a+ϵ=(d_1+1)(d_2+1)$, we have $\mathrm{deg}(2S_1\cup S_2\cup (2S_3,L))=d_2(d_1+1)-ϵ$.

(a) In this step, we prove that $h^1\left({\mathcal{I}}_{2S_1}(d_1,d_2-1)\right)=0$. We have $0\le ϵ\le 2$ and $f\le 1$. We have $\mathrm{deg}\left(2S_1\right)=3y-3e-3f=(d_1+1)(d_2+1)-a-ϵ-3e-3f$ and $a+2e+f=d_1+1$. Hence, $3y\le d_1(d_2+1)$ if and only if either $e+ϵ\ge f$ or $e=0$, $f=1$ and $ϵ>0$. We have $e=0$ and $f=1$ if and only if $a=d_1$, which is false because $a\ge 3$ and $d_1\ge 2a-2$. Hence, $h^1\left({\mathcal{I}}_{2S_1}(d_1,d_2-1)\right)=0$ if $(d_1,d_2-1,y)$ is not in the list of Remark 11, part (a) for $k=2$. Since $d_1\ge 4$, the triple $(d_1,d_2-1,y)$ may be in the list only if $d_2-1=2$ and $d_1$ is even, and in that case, the defect is at most 1. To conclude, it would be sufficient to have $ϵ+e-f>0$. For $d_1=4$, we have $a=3$, and hence, $e=1$ and $f=0$. For $d_1=6$, we have $a=4$, $e=f=1$, $y=6$ and $ϵ=2$. For $d_1\ge 8$, we have $e\ge 2>f$.

(b) By Lemma 1, Remark 15 and Observation 1 to conclude for $y\ge e+f$, it is sufficient to prove that $h^0\left({\mathcal{I}}_{2S_1}(d_1,d_2-2)\right)\le \max \{0,d_2(d_1+1)-3y-e-2f\}$.

First, assume $d_2=2$. We have $y=d_1+1-\lceil a/3\rceil$ and $ϵ=3\lceil a/3\rceil -a$. Since $S_1$ is general, $h^0\left({\mathcal{I}}_{2S_1}(d_1,0)\right)=\max \{0,d_1+1-2y\}$, we have $d_1+1-2y\le 0$, because $2y=2d_1+2-2\lceil a/2\rceil$, $d_1\ge 2a-2$ and $a\ge 3$.

Now, assume $d_2\ge 3$ and that $(d_1,d_2-2)$ is not defective. We obtain

$$h^0\left({\mathcal{I}}_{2S_1}(d_1,d_2-2)\right)=\max \{0,(d_1+1)(d_2-1)-3y\},$$

concluding this case.

Now, assume $d_2\ge 3$ and that $(d_1,d_2-2,y)$ is defective. By Remark 11 and the inequality $d_1\ge 3$, we have $d_2=4$, $d_1\ge 4$, and $d_1$ is even. Moreover, the defect is 1 only if $y=d_1+1$. We have $3y+2a+ϵ=5d_1+5$ with $0\le ϵ\le 2$. We obtain $3+2a+ϵ=5d_1+5$ with $d_1\ge 2a-2$ and $a\ge 3$, a contradiction.

(c) Now, assume $y\le e+f-1$. We take $S_3=\varnothing$, $S_2$ of cardinality $\min \{e,y\}$ and $S_1$ of cardinality $y-\min \{e,y\}$.   □

Proposition 2.

Set $Y:={\mathbb{P}}^1\times {\mathbb{P}}^1$, i.e., take $k=2$ and $n_1=n_2=1$. Fix positive integers a, $d_1$, $d_2$, x such that $d_1\ge \max \{3,2a-1\}$ and $d_2\ge 3$. Take $A\in S(Y,a)$ such that $\#{\pi }_1\left(A\right)=a$ and a general $S\in S(Y,x)$. Then, either $h^0\left({\mathcal{I}}_{A\cup 2S}(d_1,d_2)\right)=0$ or $h^1\left({\mathcal{I}}_{A\cup 2S}(d_1,d_2)\right)=0$.

Proof. 

By Remark 7, it is sufficient to use the case in which A is contained in a curve $L\subset Y$ of multidegree ${ϵ}_1$.

The case $a=1$ is true by Remark 11 and in particular (part (a), case $k=2$ of Remark 11) for $d_1\ge 3$ and $d_2\ge 3$, no case is defective. Thus, we may assume $a\ge 2$. Set $y:=\lfloor (d_1+1)(d_2+1)/3\rfloor -a$ and $ϵ:=(d_1+1)(d_2+1)-3a-3y$. Since $ϵ\le 2$, to prove the lemma for a fixed triple $(d_1,d_2,A)$ and all integers x, it is sufficient to prove that $h^1\left({\mathcal{I}}_{2A\cup 2S}(d_1,d_2)\right)=0$ for a general $S\in S(Y,y)$ (Remarks 3 and 4).

First, assume $d_1=2a-1$. Since $\mathrm{deg}(L\cap 2A)=2a$, $\mathrm{deg}\left({\mathcal{O}}_L(d_1,d_2)\right)=d_1$ and $L\cong {\mathbb{P}}^1$, the long cohomology exact sequence of (6) gives $h^i\left({\mathcal{I}}_{2A\cup 2S}(d_1,d_2)\right)=h^i\left({\mathcal{I}}_{A\cup 2S}(d_1,d_2-1)\right)$. Hence, Lemma 8 concludes this case. Thus, we may assume $d_1\ge 2a$.

Set $e:=\lfloor (d_1+1-2a)/2\rfloor$ and $f:=d_1+1-2a-2e$. We have $0\le f\le 1$. Until step (c), we assume $y\ge e+f$. Take a general $(S_1,S_2,S_3)\in S(Y,y-e-f)\times S(L,e)\times S(L,f)$ with the convention $S_1=\varnothing$ if $y-e-f=0$, $S_2=\varnothing$ if $e=0$ and $S_3=\varnothing$ if $f=0$. We have $\mathrm{deg}((2S_2,L)\cup S_3\cup (2A,L))=d_1+1$. By the Differential Horace Lemma (Lemma 2), it is sufficient to prove that $h^1\left({\mathcal{I}}_{2S_1\cup S_2\cup (2S_3,L)}(d_1,d_2-1)\right)=0$.

Observation 1: Since $2a+2e+f=d_1+1$ and $3y+3a+ϵ=(d_1+1)(d_2+1)$, we have $\mathrm{deg}(2S_1\cup S_2\cup (2S_3,L)\cup A)=d_2(d_1+1)-ϵ$.

(a) In this step, we prove that $h^1\left({\mathcal{I}}_{2S_1\cup A}(d_1,d_2-1)\right)=0$. Since $d_2-1\ge 2$, Lemma 8 gives that it is sufficient to prove that $3(y-e-f)+a\le (d_1+1)d_2$. Observation 1 gives $3(y-e-f)+a=d_2(d_1+1)-ϵ-e$.

(b) In this step, we conclude the case $y\ge e+f$. By step (a), we have the equality $h^0\left({\mathcal{I}}_{2S_1\cup A}(d_1,d_2-1)\right)=d_2(d_1+1)-a-3(y-e-f)$. By step (a), Observation 1 and Lemma 1, it is sufficient to prove that

$$h^0\left({\mathcal{I}}_{2S_1}(d_1,d_2-2)\right)\le \max \{0,d_2(d_1+1)-a-3y+2e+f\}.$$

Since $S_1$ is general, this inequality is obvious if $(d_1,d_2-2,y-e-f)$ is not defective. Since $d_1\ge 3$ and $d_2\ge 3$, $(d_1,d_2-2,x)$ is defective if and only if $d_2=4$, $d_1$ is even and $x=d_1+1$, and in that case the defect is 1 and occurs only if $h^0\left({\mathcal{I}}_{2S}(d_1,d_2)\right)=1$ (Remark 11). Assume $d_2=4$. Since $h^0\left({\mathcal{O}}_Y(d_1,3)\right)-h^0\left({\mathcal{O}}_Y(d_1,1)\right)$, it is sufficient to prove that $a+e+2f\le d_1$. This inequality is obvious, because $f\le 1$, $a\ge 3$ and $2a+2e+f=d_1+1$.

(c) Assume $y\le e+f-1$. We take $S_3=\varnothing$, $S_2$ of cardinality $\min \{y,e\}$ and $S_1$ of cardinality $y-\max \{e,y\}$.   □

Remark 16.

Theorem 5 and Proposition 3 are not optimal because we made the assumption $d_1\ge 5$ to use Theorem 2. To use the case with $d_1\in \{3,4\}$, one needs to first obtain all exceptional cases for $d=3,4$ for the Veronese variety. By [4], Table 1 at p. 91, we need $d_1\ge 3$ and $d_2\ge 3$ to avoid all exceptional cases for the case $a\le 1$.

Theorem 5.

Fix integers $a\ge 0$, $d_1\ge \max \{4,2a-1\}$, $d_2\ge 3$ and $m\ge 1$. Set $Y:={\mathbb{P}}^m\times {\mathbb{P}}^1$. Take $A\in S(Y,a)$ such that $\#{\pi }_1\left(A\right)=a$ and a general $S\in S(Y,x)$. Then, either $h^0\left({\mathcal{I}}_{A\cup 2S}(d_1,d_2)\right)=0$ or $h^1\left({\mathcal{I}}_{A\cup 2S}(d_1,d_2)\right)=0$.

Lemma 9.

Fix integers $a\ge 0$, $d_1\ge 3+a$, $d_2\ge 3$ and $x>0$. Take $A\in S(Y,a)$ such that $\#{\pi }_1\left(A\right)=a$ and a general $S\in S(Y,x)$. Then, either $h^0\left({\mathcal{I}}_{A\cup 2S}(d_1,d_2)\right)=0$ or $h^1\left({\mathcal{I}}_{A\cup 2S}(d_1,d_2)\right)=0$.

Proof. 

By Remark 7, it is sufficient to use the case in which A is contained in a curve $L\subset Y$ of multidegree ${ϵ}_1$.

The case $a\le$ is true by Remark 11 and the fact that in part (a), case $k=2$, because $d_1\ge 3$ and $d_2\ge 3$. Thus, we may assume $a\ge 2$. Set $y:=\lfloor (d_1+1)(d_2+1)/3\rfloor -a$ and $ϵ:=(d_1+)(d_2+1)-a-3y$. Since $(d_1+1)(d_2+1)-3y-3a\le 2$, to prove the lemma for a fixed triple $(d_1,d_2,A)$ and all integers x, it is sufficient to prove that $h^1\left({\mathcal{I}}_{A\cup 2S}(d_1,d_2)\right)=0$ for a general $S\in S(Y,y)$ (Remark 7).

Assume $a=2$. By Remark 13, it is sufficient to prove two statements, the inequality $h^0\left({\mathcal{I}}_{L\cup 2S}(d_1,d_2)\right)\le h^0\left({\mathcal{I}}_{2S}(d_1,d_2)\right)-2$, i.e., $h^0\left({\mathcal{I}}_{2S}(d_1,d_2-1)\right)\le h^0\left({\mathcal{I}}_{2S}(d_1,d_2)\right)-2$, and that $h^1\left({\mathcal{I}}_{2S}(d_1,d_2-1)\right)=0$. Note that $h^0\left({\mathcal{O}}_Y(d_1,d_2)\right)-h^0\left({\mathcal{O}}_Y(d_1,d_2-1)\right)=d_1+1\ge 3$. First, assume $d_2\ge 3$. Since $d_2\ge 3$, ${\mathcal{O}}_Y(d_1,d_2)$ is not defective by the inequality $d_1\ge 3$, hence, we conclude by the case $k=2$ of Remark 11. Assume $a>2$. Set $e:=\lfloor (d_1+1-a)/2\rfloor$ and $f:=d_1+1-a-2e$. We have $0\le f\le 1$.

Assume for the moment $y\ge e+f$.

Take a general $(S_1,S_2,S_3)\in S(Y,y-e-f)\times S(L,e)\times S(L,f)$ with the convention $S_1=\varnothing$ if $y-e-f=0$, $S_2=\varnothing$ if $e=0$ and $S_3=\varnothing$ if $f=0$.

We have $\mathrm{deg}((2S_2,L)\cup S_3\cup A)=d_1+1$. By the Differential Horace Lemma (Lemma 2), it is sufficient to prove that $h^1\left({\mathcal{I}}_{2S_1\cup S_2\cup (2S_3,L)}(d_1,d_2-1)\right)=0$.

Observation 1: Since $a+2e+f=d_1+1$ and $3y+a+ϵ=(d_1+1)(d_2+1)$, we have $\mathrm{deg}(2S_1\cup S_2\cup (2S_3,L))=d_2(d_1+1)-ϵ$.

(a) In this step, we prove that $h^1\left({\mathcal{I}}_{2S_1}(d_1,d_2-1)\right)=0$. The only difference with respect to the proof of Lemma 6 is that to exclude that $(e,f)=(0,1)$, it is sufficient to use that $a+2e+f=d_1+1$ with $d_1\ge 3$.

(b) By Lemma 1, Remark 15 and Observation 1, to conclude for $y\ge e+f$, it is sufficient to prove that $h^0\left({\mathcal{I}}_{2S_1}(d_1,d_2-2)\right)\le \max \{0,d_2(d_1+1)-3y-e-2f\}$. We have $d_2\ge 3+a\ge 5$, excluding almost all cases. The remaining ones are trivial by Remark 5.

(c) Now assume $y\le e+f-1$. We take $S_3=\varnothing$, $S_2$ of cardinality $\min \{e,y\}$ and $S_1$ of cardinality $y-\min \{e,y\}$.   □

Proposition 3.

Fix a positive integer m and set $Y:={\mathbb{P}}^m\times {\mathbb{P}}^1$. Take integers $a\ge 0$, x, $d_1\ge 3+a$ and $d_2\ge 3$. Take $A\in S(Y,a)$ such that $\#{\pi }_1\left(A\right)=a$ and a general $S\in S(Y,x)$. Then, either $h^0\left({\mathcal{I}}_{A\cup 2S}(d_1,d_2)\right)=0$ or $h^1\left({\mathcal{I}}_{A\cup 2S}(d_1,d_2)\right)=0$.

Proof. 

Since the case $m=1$ is true by Lemma 9, we assume $m\ge 2$ and use induction on m. By Remark 7, it is sufficient to use the case in which A is contained in a curve $L\subset Y$ of multidegree ${ϵ}_1$. We have $\dim Y=m+1$ and $h^0\left({\mathcal{O}}_Y(d_1,d_2)\right)=\left(\genfrac{}{}{0pt}{}{m+d_1}{m}\right)(d_2+1)$.

Set $y:=\lfloor ((\genfrac{}{}{0pt}{}{m+d_1}{m})(d_2+1)-a)/(m+2)\rfloor$, $x_1:=\lceil ((\genfrac{}{}{0pt}{}{m+d_1}{m})(d_2+1)-a)/(m+2)\rceil$ and $ϵ\left(\genfrac{}{}{0pt}{}{m+d_1}{m}\right)(d_2+1)-a-(m+2)y$. By Remark 2, to prove the proposition for a fixed $(Y,d_1,d_2,A)$ and all x, it is sufficient to prove it for $x=y$ and $x=x_1$. In steps (a) and (b), we use the case $z=y$. In step (c), we handle the case $x=x_1$. Take a general $S\in S(Y,y)$.

By Remark 11 and [4], Cor 3.14, the case $a=1$ is true. Hence, we may assume $a\ge 2$. Since $A\subset L$, there is $M\in |{\mathcal{I}}_A(1,0)|$. We have $M\cong {\mathbb{P}}^{m-1}\times {\mathbb{P}}^1$ and $a\le \left(\genfrac{}{}{0pt}{}{m+d_1-1}{m-1}\right)(d_2+1)$. Set $e:=\lfloor ((\genfrac{}{}{0pt}{}{m+d_1-1}{m-1})(d_2+1)-a)/(m+1)\rfloor$ and $f:=\left(\genfrac{}{}{0pt}{}{m+d_1-1}{m-1}\right)(d_2+1)-(m+1)e-a$. We have $0\le f\le m$.

Observation 1: Assume $a=2$. If $\#{\pi }_2\left(A\right)$, then the set A is in the open orbit of $S(Y,2)$ for the action of the group G considered in Remark 11. Hence, this case follows for the non-defectivity of Y for the linear system $|{\mathcal{O}}_Y(t_1,t_2)$ for all $t_1\ge 3$, $t_2\ge 3$ (ref. [4], Cor. 3.14).

(a) Assume $y\ge e+f$.

Take a general $(S_1,S_2,S_3)\in S(Y,y-e-f)\times S(M,e)\times S(M,f)$. Note that $\mathrm{deg}(A\cup (2S_2,M)\cup S_3)=\left({\displaystyle \genfrac{}{}{0pt}{}{m+d_1-1}{m-1}}\right)(d_2+1)$. By the inductive assumption on m, we have $h^i(M,{\mathcal{I}}_{A\cup (2S_2,M)\cup S_3,M}(d_1,d_2))=0$, $i=0,1$. We specialize e of the points of S to different points of $S_2$ and apply the Differential Horace Lemma to each point of S. We find that to prove that $h^1\left({\mathcal{I}}_{A\cup 2S}(d_1,d_2)\right)=0$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{2S_1\cup S_2\cup (2S_3,M)}(d_1-1,d_2)\right)=0$.

(a1) In this step, we prove that $h^1\left({\mathcal{I}}_{A^{\prime }\cup 2S_1\cup (2S_3,M)}(d_1-1,d_2)\right)=0$. By Remark 2, it is sufficient to prove that $h^1\left({\mathcal{I}}_{2S_1\cup 2S_3}(d_1-1,d_2)\right)=0$. Note that here, A does not appear, except that M is a hyperplane containing L. We take as M a general hyperplane containing L. The group $\mathrm{Aut}\left(M\right)$ acts transitively on the set of of all lines of M. Hence, $S_3$ may be considered as a general subset of f points of M. Taking $S_1$ general, we find that $2S_1\cup 2S_3$ has the bigraded Hilbert function of the union of $y-e$ general double points of Y. By [4], Cor. 3.14, to prove that $h^1\left({\mathcal{I}}_{2S_1\cup 2S_3}(d_1-1,d_2)\right)=0$, it is sufficient to prove that $(m+2)(y-e)\le \left(\genfrac{}{}{0pt}{}{m+d_1-1}{m-1}\right)$. Since $\mathrm{deg}(A\cup (2S_2,M)\cup S_3)=\left(\genfrac{}{}{0pt}{}{m+d_1-1}{m-1}\right)(d_2+1)$ and $(m+2)y+a=\left(\genfrac{}{}{0pt}{}{m+d_1}{m}\right)-ϵ$, we have $(m+2)(y-e)=f-ϵ-e+\left(\genfrac{}{}{0pt}{}{m+d_1-1}{m-1}\right)(d_2+1)$. Hence, it is sufficient to prove that $e\ge f$. Assume $e\le f-1$. By the definition of e as a floor, we obtain $\left(\genfrac{}{}{0pt}{}{m+d_1-1}{m-1}\right)\le 1+m+(f-1)m$. Since $f\le m$, we obtain $\left(\genfrac{}{}{0pt}{}{m+d_1-1}{m-1}\right)\le 1+m^2$. It is sufficient to use that $d_1\ge 3+a\ge 5$.

(a2) Step (a1) gives

$$h^0\left({\mathcal{I}}_{2S_1\cup (2S_3,M)}(d_1-1,d_2)\right)=(d_2+1)\left({\displaystyle \genfrac{}{}{0pt}{}{m+d_1-1}{m}}\right)-(m+2)(y-e-f)-(m+1)f.$$

By Lemma 2, to prove that $h^1\left({\mathcal{I}}_{2S_1\cup S_2\cup (2S_3,M)}(d_1-1,d_2)\right)=0$, it is sufficient to prove that

$$h^1\left({\mathcal{I}}_{2S_1}(d_1-2,d_2)\right)\le (d_2+1)\left({\displaystyle \genfrac{}{}{0pt}{}{m+d_1-1}{m}}\right)-(m+2)(y-e-f)-(m+1)f-e.$$

Since $\mathrm{deg}(2S_1\cup S_2\cup (2S_3,M))=h^0\left({\mathcal{O}}_Y(d_1-1,d_2)\right)-ϵ$ with $ϵ\ge 0$, to prove that $h^1\left({\mathcal{I}}_{2S_1\cup S_2\cup (2S_3,M)}(d_1-1,d_2)\right)=0$, it is sufficient to prove that

$$h^0\left({\mathcal{I}}_{2S_1}(d_1-2,d_2)\right)\le \max \{0,d_2+1)\left({\displaystyle \genfrac{}{}{0pt}{}{m+d_1-1}{m}}\right)-(m+2)(y-e-f)-(m+1)f-e\}.$$

Since $d_1-2\ge 3$, $d_2\ge 3$ and $S_1$ is general, we may apply [4], Cor. 3.14, and find that either $h^0\left({\mathcal{I}}_{2S_1}(d_1-2,d_2)\right)=0$ or $h^1\left({\mathcal{I}}_{2S_1}(d_1-2,d_2)\right)=0$.

Hence, we may assume that $h^1\left({\mathcal{I}}_{2S_1}(d_1-2,d_2)\right)=0$, i.e., we may assume that $h^0\left({\mathcal{I}}_{2S_1}(d_1-2,d_2)\right)=h^0\left({\mathcal{O}}_Y(d_1-2,d_2)\right)-2S_1$.

Note that $h^0\left({\mathcal{O}}_Y(d_1-1,d_2)\right)-h^0\left({\mathcal{O}}_Y(d_1-2,d_2)\right)=\left(\genfrac{}{}{0pt}{}{m+d_1-2}{m-1}\right)(d_2+1)$. Thus, it is sufficient to check that $e\ge f$. Assume $e\le f-1$. We find $a+f^2\ge \left(\genfrac{}{}{0pt}{}{m+d_1-1}{m-1}\right)(d_2+1)$ with $d_2\ge 3$, $d_1\ge 3+a$ and $f\le m$, a contradiction.

(b) Assume $y<e+f$. We modify step (a), taking instead of the triple $(y-e-f,e,f)$, the following triple $(y^{\prime },e^{\prime },f^{\prime })$ with $y^{\prime }=0$, $f^{\prime }:=\max \{0,f-(e+f-y)\}$ and $e^{\prime }:=y-f^{\prime }$.

(c) In this step, we handle the case $x=x_1$. We may assume $x_1\ne y$, i.e., $x_1=y+1$. We take the same e, f, $S_2$ and $S_3$, but here, $S_1$ has cardinality $x_1-e-f$. Assume for the moment $x_1\ge e+f$. In step (a1), the only modification arises if $h^1\left({\mathcal{I}}_{2S_1\cup 2S_3}(d_1-1,d_2)\right)>0$. Assume $h^1\left({\mathcal{I}}_{2S_1\cup 2S_3}(d_1-1,d_2)\right)>0$. By [4], Cor. 3.14, we obtain $h^0\left({\mathcal{I}}_{2S_1\cup 2S_3}(d_1-1,d_2)\right)=0$. Since $S_2$ is general in M, it is sufficient to use that $e\ge f$.

If $x_1<e+f$, the proof of step (b) works verbatim.   □

Proof of Theorem 5.

The cases with $a\le 1$ are true by [4], Cor. 3.14, and Remark 11. Thus, we may assume $a\ge 2$, and hence, $d_1\ge 6$. The case $m=1$ is true by Proposition 3. Thus, we assume $m\ge 2$ and use induction on m.   □

We have $\dim Y=m+1$ and $h^0\left({\mathcal{O}}_Y(d_1,d_2)\right)=\left(\genfrac{}{}{0pt}{}{m+d_1}{m}\right)(d_2+1)$.

Set $y:=\lfloor (\genfrac{}{}{0pt}{}{m+d_1}{m})(d_2+1)/(m+2)\rfloor -a$, $x_1:=\lceil (\genfrac{}{}{0pt}{}{m+d_1}{m})(d_2+1)/(m+2)\rceil -a$ and $ϵ\left(\genfrac{}{}{0pt}{}{m+d_1}{m}\right)(d_2+1))-(m+2)(y+a)$. By Remark 2, to prove the proposition for a fixed $(Y,d_1,d_2,A)$ and all x, it is sufficient to prove it for $x=y$ and $x=x_1$. In steps (a) and (b), we use the case $x=y$. In step (c), we use the case $x=x_1$. Take a general $S\in S(Y,y)$.

By Remark 7, it is sufficient to use the case in which A is contained in a curve $L\subset Y$ of multidegree ${ϵ}_1$. Take a general $M\in |{\mathcal{I}}_L(1,0)|$. We have $\dim M=m$ and $h^0(M,{\mathcal{O}}_M(d_1,d_2))=\left(\genfrac{}{}{0pt}{}{m+d_1-1}{m-1}\right)(d_2+1)$. Set $e:=-a+\lfloor (\genfrac{}{}{0pt}{}{m+d_1-1}{m-1})(d_2+1)/(m+1)\rfloor$ and $f:=\left(\genfrac{}{}{0pt}{}{m+d_1-1}{m-1}\right)(d_2+1)-(m+1)(e+a)$. We have $0\le f\le m$ and

$$2a+(m+1)e+f=(d_2+1)\left({\displaystyle \genfrac{}{}{0pt}{}{m+d_1-1}{m-1}}\right)$$

(7)

(a) Assume $y\ge e+f$.

Take a general $(S_1,S_2,S_3)\in S(Y,y-e-f)\times S(M,e)\times S(M,f)$. Note that $\mathrm{deg}(2A\cup (2S_2,M)\cup S_3)=\left(\genfrac{}{}{0pt}{}{m+d_1-1}{m-1}\right)(d_2+1)$. By the inductive assumption on m, we have $h^i(M,{\mathcal{I}}_{2A\cup (2S_2,M)\cup S_3,M}(d_1,d_2))=0$, $i=0,1$. We specialize e of the points of S to different points of $S_2$ and apply the Differential Horace Lemma (Lemma 2) to each point of S. We find that to prove that $h^1\left({\mathcal{I}}_{2A\cup 2S}(d_1,d_2)\right)=0$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{2S_1\cup A\cup S_2\cup (2S_3,M)}(d_1-1,d_2)\right)=0$.

Set $g:=\lfloor ((\genfrac{}{}{0pt}{}{m+d_1-2}{m-1})-a-e-(m+1)f)/(m+1)\rfloor$ and

$$h:=\left({\displaystyle \genfrac{}{}{0pt}{}{m+d_1+m-2}{m-1}}\right)(d_2+1)-a-mf-(m+1)g.$$

We have $0\le h\le m$ and

$$a+e+(m+1)f+(m+1)g+h=(d_2+1)\left({\displaystyle \genfrac{}{}{0pt}{}{m+d_1-2}{m-1}}\right)$$

(8)

Until step (a3), we assume $y\ge e+f+g+h$. Take a general

$$(S_4,S_5,S_6)\in S(Y,y-e-f-g-h)\times S(Y,g)\times S(Y,h).$$

Note that $\mathrm{deg}((2S_5,M)\cup (2S_3,M)\cup A\cup S_6))=h^0(M,{\mathcal{O}}_M(d_1-1,d_2))$ and that $d_1-1\ge 3+a$. Hence, Proposition 3 gives $h^i(M,{\mathcal{I}}_{(2S_5,M)\cup (2S_3,M)\cup A\cup S_6}(d_1-1,d_2))=0$, $i=0,1$. We specialize g of the points of $S_1$ to different points of $S_5$ and apply the Differential Horace Lemma to each point of $S_6$. Since $h^i(M,{\mathcal{I}}_{(2S_5,M)\cup (2S_3,M)\cup A\cup S_6}(d_1-1,d_2))=0$, $i=0,1$, to prove that $h^1\left({\mathcal{I}}_{2S_1\cup A\cup S_2\cup (2S_3,M)}(d_1-1,d_2)\right)=0$, it is sufficient to prove that

$$h^1\left({\mathcal{I}}_{2S_4\cup S_5\cup (2S_6,M)}(d_1-2,d_2)\right)=0.$$

(a1) Assume $y\ge e+f+g+h$. In this step, we prove the vanishing of the cohomology group $H^1\left({\mathcal{I}}_{2S_4\cup (2S_6,M)}(d_1-2,d_2)\right)=0$. Note that here, A does not appear, except that M is a hyperplane containing L. We take as M a general hyperplane containing L. The group $\mathrm{Aut}\left(M\right)$ acts transitively on the set of of all lines of ${\mathbb{P}}^m$. Hence, $S_3$ may be considered as a general subset of f points of M. Taking $S_1$ general, we find that $2S_4\cup 2S_6$ has the bigraded Hilbert function of the union of $y-e-f-g$ general double points of Y. By [4], Cor. 3.14, to prove that $h^1\left({\mathcal{I}}_{2S_4\cup 2S_6}(d_1-2,d_2)\right)=0$, it is sufficient to prove that $(m+2)(y-e-f-g)\le \left(\genfrac{}{}{0pt}{}{m+d_1-1}{m-1}\right)$.

Hence, $h^1\left({\mathcal{I}}_{2S_4\cup S_5\cup (2S_6,M)}(d_1-2,d_2)\right)=0$.

(a2) In this step, we assume $y\ge e+f+g+h$. In this step, we prove that $h^1\left({\mathcal{I}}_{2S_4\cup S_5\cup (2S_6,M)}(d_1-2,d_2)\right)=0$. Note that $\mathrm{deg}(S_5\cup (2S_6,M))\le \left(\genfrac{}{}{0pt}{}{m+d_1-3}{m-1}\right)(d_2+1)$.

Observation 1: For all integers $d\ge 3$, we have

$$\left({\displaystyle \genfrac{}{}{0pt}{}{m+d-2}{m-1}}\right)/\left({\displaystyle \genfrac{}{}{0pt}{}{m+d-3}{m-1}}\right)=(m+d-2)/(d-1).$$

Claim 4.

We have $\mathrm{deg}(S_5\cup (2S_6,M))\le \left(\genfrac{}{}{0pt}{}{m+d_1-3}{m-1}\right)(d_2+1)$.

Proof of Claim 4.

We have $\mathrm{deg}(S_5\cup (2S_6,M))=g+(m+1)h$.

Assume $g+(m+1)h-1\ge \left(\genfrac{}{}{0pt}{}{m+d_1-3}{m-1}\right)(d_2+1)$, and hence,

$$(m+1)g+{(m+1)}^{2}h-m-1\ge (m+1)(d_2+1)\left({\displaystyle \genfrac{}{}{0pt}{}{m+d_1-3}{m-1}}\right).$$

(9)

Set $\alpha :=\left(\genfrac{}{}{0pt}{}{m+d_1-3}{m-1}\right)$. Multiplying (8) by (m + 1), subtracting (9) and using Observation 1, we obtain

$$-(m+1)a-(m+1)e+(m^2+2m)h-m-1\ge \alpha (d_2+1){\displaystyle \frac{(m+d-2)(m+1)}{d-1}}$$

(10)

Since $h\le m$, we obtain

$$-(m+1)a-(m+1)e+m^3+2m^2-m-1\ge \alpha (d_2+1){\displaystyle \frac{(m+d-2)(m+1)}{d-1}}$$

(11)

To obtain a contradiction, it is sufficient to use that $d_1\ge 5$, $d_2\ge 3$ and that (7) $e\ge m^2/3$.   □

Claim 1 and Lemma 1 show that to prove that $h^1\left({\mathcal{I}}_{2S_4\cup S_5\cup (2S_6,M)}(d_1-2,d_2)\right)=0$, it is sufficient to prove that

$$h^0\left({\mathcal{I}}_{2S_4}(d_1-3,d_2)\right)\le \max \{0,h^0\left({\mathcal{O}}_Y(d_1-2,d_2)\right)-\mathrm{deg}(2S_4\cup S_5\cup (2S_6,M))\}.$$

This inequality is true because $S_4$ is general and ${\mathcal{O}}_Y(d_1-3,d_2)$ is not defective (ref. [4], Cor. 1.3).

(a3) Assume $e+f\le y<e+f+g+h$. We modify the proof of step (a2), taking $(g^{\prime },h^{\prime })$ instead of $(g,h)$ with $h^{\prime }:=\max \{0,h-(e+f+g+h-y)\}$ and $\#S_5=g^{\prime }=y-h^{\prime }$.

(b) Assume $y<e+f$. We modify step (a1), taking instead of the previous sets $S_1$, $S_2$, $S_3$ the sets $S_1,S_2,S_3$ with $S_1=\varnothing$, $\#S_3=\max \{0,f-(e+f-y)\}$ and $\#S_2=y-\#S_3$.

(c) Take $x=x_1$. We may assume $x_1\ne y$, i.e., $x_1=y+1$. We take the same e, f, g and h$S_2$, $S_3$, $S_5$ and $S_6$, but here, $\#S_1=x_1-e-f$ and $\#S_6=x_1-e-f-g-h$.

6. Segre–Veronese Varieties with One-Dimensional Factors

In this section, we prove the following theorem and several related results.

Theorem 6.

Set $Y:={\left({\mathbb{P}}^1\right)}^k$, $k\ge 3$. Fix positive integers a, x, $d_i$, $1\le i\le k$ such that $d_1\ge \max \{5,2a-1\}$, $d_i\ge 3$. Take $A\in S(Y,a)$ such that $\#{\pi }_1\left(A\right)=a$ and a general $S\in S(Y,x)$. Then, either $h^0\left({\mathcal{I}}_{A\cup 2S}(d_1,\dots ,d_k)\right)=0$ or $h^1\left({\mathcal{I}}_{A\cup 2S}(d_1,\dots ,d_k)\right)=0$.

To prove Theorem 6 (or similar statements) using a proof by induction, the following result may be very useful.

Theorem 7.

Set $Y:={\left({\mathbb{P}}^1\right)}^k$, $k\ge 3$. Fix positive integers a, x and $d_i$, $1\le i\le k$ such that $d_1\ge \max \{3,a-1\}$, $d_2\ge 3$ and $d_i\ge 2$ for all $3\le i\le k$. Take $A\in S(Y,a)$ such that $\#{\pi }_1\left(A\right)=a$ and a general $S\in S(Y,x)$. Then, either $h^0\left({\mathcal{I}}_{A\cup 2S}(d_1,\dots ,d_k)\right)=0$ or $h^1\left({\mathcal{I}}_{A\cup 2S}(d_1,\dots ,d_k)\right)=0$.

To prove these theorems, we introduce the following notation and make the following remarks and results. Set $\underline{d}:=(d_1,\dots ,d_{k-1})\in {\mathbb{N}}^{k-1}$. We fix positive integers $t_1,\dots ,t_k-1$ such that $t_1\ge a-1$ and set $\underline{t}=(t_1,\dots ,t_{k-1})$. In the proofs of Theorem 6 and 7, we take $t_i=d_i$ for all $i<k$.

In Theorems 6 and 7, we assumed $k\ge 3$. We quote Proposition 3 and Theorem 5 for the case $k=2$ and use induction on the integer k.

Remark 17.

By [7] and the homogeneity of Y, we may assume $a\ge 2$. By Lemma 7, it is sufficient to use the case $A\subset L$ with L, a curve of multidegree ${ϵ}_1$. Note that ${\pi }_k\left(L\right)$ is a point, o. Set $D:={\left({\mathbb{P}}^1\right)}^{k-1}\times \left\{o\right\}\subset Y$. Note that $A\subset L\subset Y$ and that $D\in |{\mathcal{O}}_Y\left({ϵ}_k\right)|$.

Set $\alpha :=h^0\left({\mathcal{O}}_D(t_1,\dots ,t_k)\right)$. We have $\alpha ={\prod }_{i=1}^{k-1}(t_i+1)$. Thus, for all $t\in \mathbb{N}$, we have $h^0\left({\mathcal{O}}_Y(\underline{t},t)\right)=\alpha (t+1)$. We have ${\mathcal{O}}_D(\underline{t},t)\cong {\mathcal{O}}_D(\underline{t},0)$ for all $t\in \mathbb{Z}$. Remember that we will take $t_i:=d_i$ in the proofs of Theorems 6 and 7, but we want to state and prove Theorems 8 and 9 in a more general way.

Let e, f be integers satisfying

$$a+ke+f=\alpha ,$$

(12)

$e+kf\le \alpha$ and with $\alpha -e-kf$ minimum. Since $kx+y=k(x-1)+(y+k)$ and

$(x-1)+k(y+k)=x+ky+k^2-1$, the integers e and f satisfy the following inequalities:

$$\alpha -k^2+1\le e+kf\le \alpha$$

(13)

If $\alpha \ge a+k^2-1$, then $(e,f)\in {\mathbb{N}}^2$, since $a>0$, $f\ge e$.

Remark 18.

Since $t_i\ge 2$ for all $2\le i\le k-1$ and $t_1\ge \max \{3,a-1\}$, we have $\alpha \ge 2^{k-2}(t_1+1)$ and $\alpha \ge a+k^2$.

Theorem 8.

Assume $\alpha \ge a+k^2$, $t_1\ge \max \{3,a-1\}$ and that $(D,\underline{t})$ is not secant defective. Assume that for all $b\in \mathbb{N}$ and a general $B\in S(D,b)$, either $h^0(D,{\mathcal{I}}_{A\cup 2B,D}\left(\underline{t}\right))=0$ or $h^1(D,{\mathcal{I}}_{A\cup 2B,D}\left(\underline{t}\right))=0$. Take a general $S\in S(Y,e+f)$ and a general $S^{\prime }\in S(Y,e+f+k)$. Then, $h^1\left({\mathcal{I}}_{A\cup 2S}(\underline{t},1)\right)=0$, $h^0\left({\mathcal{I}}_{A\cup 2S}(\underline{t},1)\right)=\alpha -kf-e\le k^2-1$ and $h^0\left({\mathcal{I}}_{A\cup 2S^{\prime }}(\underline{t},1)\right)=0$.

Proof. 

Since $A\subset L$, $t_1\ge \max \{3,a-1\}$ and $t_i>0$ for all $i<k$, we have $h^1(D,{\mathcal{I}}_{A,D}\left(\underline{t}\right))=0$. Hence, the assumptions on $A\cup 2B$ are not satisfied for the trivial reason that we have $h^0(D,{\mathcal{I}}_{A,D}\left(\underline{t}\right))=0$.

By (12) and (13), to prove the result for S, it is sufficient to prove the equality $h^1\left({\mathcal{I}}_{A\cup 2S}(\underline{t},1)\right)=0$. Take a general $(S_1,S_2)\in S(D,e)\times S(D,f)$. By the assumption on ${\mathcal{I}}_{A,D}\left(\underline{t}\right)$ and (12), we have the equalities $h^i(D,{\mathcal{I}}_{A\cup 2S_1\cup S_2}(\underline{t},1))=0$, $i=0,1$, $h^1(D,{\mathcal{I}}_{S_1\cup 2S_2}(\underline{d},0))=0$ and $h^0(D,{\mathcal{I}}_{S_1\cup 2S_2}(\underline{d},1))=\alpha -e-kf$. We degenerate e of the points of S to different points of $S_1$ and apply the Differential Horace Lemma to each point of $S_2$. We obtain $h^1\left({\mathcal{I}}_{A\cup 2S}(\underline{t},1)\right)\le h^1\left({\mathcal{I}}_{S_1\cup 2S_2}(\underline{t},0)\right)=0$.

We degenerate $S^{\prime }$ to $(S_1,S_2^{\prime })$ with $S_2^{\prime }$ general in $S(D,f+k)$ and use the assumption that $(D,\underline{t})$ is not secant defective to obtain $h^0(D,{\mathcal{I}}_{S_1\cup (2S_2^{\prime },D),D}(\underline{t},0))=0$.   □

Remark 19.

Take the assumptions of Theorem 8 and add the conditions that $\alpha \equiv 0modk+1$ and $a\equiv 0(mod{k}^2-1)$. Set $\tilde{e}:=\frac{a}{k+1}-\frac{ka}{k^2-1}$ and $\tilde{f}:=\frac{\alpha }{k+1}+\frac{a}{k^2-1}$. In the proof of Theorem 11, take a general $({\tilde{S}}_1,{\tilde{S}}_2)\in S(D,\tilde{e})\times S(D,\tilde{f})$ instead of $(S_1,S_2)$. The same proof gives $h^i\left({\mathcal{I}}_S(\underline{t},1)\right)=0$, $i=0,1$, for a general $S\in S(Y,(2\alpha -a)/(k+1\left)\right)$.

Lemma 10.

Take the assumptions of Theorem 8 with $\alpha \ge a+2{(k+1)}^2$. We have $e+f\ge k$. Fix an integer z such that $0\le z\le k$. Let $Z\subset Y$ be a general union of $e+f-z$ double points of Y and z double points of D. Then, $h^1\left({\mathcal{I}}_{A\cup Z}(\underline{t},1)\right)=0$.

Proof. 

We have $(k+1)(e+f)\ge 2\alpha -a-k^2+1$. We follow the proof of Theorem 8 taking instead of $2S_1\cup 2S_2$, the same reduction, $S_1\cup S_2$, but with z of its connected component general double points of D, not general double points of Y with reduction general in D.   □

Proof of Theorem 7.

Since the case $k=2$ is true by Proposition 3, we assume $k>2$ and use induction on the integer k. Fix $D\in |{\mathcal{O}}_Y\left({ϵ}_k\right)|$. We take $t_i=d_i$ for $i<k$ in the statement of Theorem 8. Hence, $\alpha ={\prod }_{i=1}^{k-1}(d_i+1)$. By [7], $(D,\underline{d})$ is not secant defective. Since $a\ge 2$, $d_1\ge a-1$ and $d_2\ge 3$, we have $\alpha \ge a+2{(k+1)}^2$, and hence, we may apply Theorem 8 and Lemma 10.

(a) In this step, we use the case $d_k=2$. Let w be the maximal integer such that $h^1\left({\mathcal{I}}_{A\cup 2S}(\underline{d},1)\right)=0$ and set $\beta :=2\alpha -a-(k+1)w$. By Theorem 8, we have $w\ge e+f$ and $\beta \le k^2-1$. Set $y:=\lfloor 3\alpha /(k+1)\rfloor$ and $x_1:=\lceil 3\alpha /(k+1)\rceil$. By Remark 2, it is sufficient to prove that $h^1\left({\mathcal{I}}_{A\cup 2S}(\underline{d},2)\right)=0$ for a general $S\in S(Y,y)$ and $h^0\left({\mathcal{I}}_{A\cup 2S^{\prime }}(\underline{d},2)\right)=0$ for a general $S^{\prime }\in S(Y,x_1)$.

(a1) In this step, we prove that $h^1\left({\mathcal{I}}_{A\cup 2S}(\underline{d},2)\right)=0$ for a general $S\in S(Y,y)$. Take a general $S\in S(Y,y)$. Set $g:=\lfloor (\alpha -a)/k\rfloor$ and $h:=\alpha -a-kg$. We have $0\le h\le k-1$.

(a1.1) Assume $y\ge g+h$. Take a general

$$(S_1,S_2,S_3)\in S(Y,y-g-h)\times S(D,g)\times S(D,h).$$

By the assumptions on D and $\underline{d}$, we have $h^i(D,{\mathcal{I}}_{(2S_2,D)\cup S_3\cup A,D}\left(\underline{d}\right))=0$, $i=0,1$. We degenerate g of the points of S to different points of $S_2$ and apply the Differential Horace Lemma to each point of $S_3$. By Lemma 2, to prove that $h^1\left({\mathcal{I}}_{A\cup 2S}(\underline{d},2)\right)=0$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{2S_1\cup S_2\cup (2S_3,D)}(\underline{d},1)\right)=0$.

We prove that $h^1\left({\mathcal{I}}_{2S_1\cup (2S_3,D)}(\underline{d},1)\right)=0$. Since $\#S_3\le k$, to apply Lemma 10, it is sufficient to prove that $y-g\le e+f$. To handle the case with $x=x_1$, we prove the following stronger numerical result.   □

Claim 5.

We have $x_1\le e+f+g$.

Proof of Claim 5.

Assume $x_1\ge g+e+f+1$. Since $(k+1)x_1\le 3\alpha +k$, we find $3k\alpha \ge k(k+1)g+k(k+1)e+k(k+1)f+1$. Since $ke+f=\alpha -a$ and $kg+h=\alpha$, we find $3k\alpha \ge (2k+2)\alpha -2(k+1)a+(k^2-1)f-(k+1)h$. Since $h\le k$, we find a contradiction.   □

Claim 1 and Lemma 10 with $z:=h$ give $h^1\left({\mathcal{I}}_{2S_1\cup (2S_3,D)}(\underline{d},1)\right)=0$, and hence, $h^0\left({\mathcal{I}}_{2S_1\cup (2S_3,D)}(\underline{d},1)\right)=2\alpha -(k+1)$. By Lemma 1, to prove that $h^1\left({\mathcal{I}}_{2S_1\cup (2S_3,D)}(\underline{d},1)\right)=0$, it is sufficient to prove that $h^0({\mathcal{I}}_{2S_1}(\underline{d},0)\le 2\alpha -(k+1)-g$. Since $S_1$ is general in Y, its image by the projection of Y onto its first factors is general. Since $({({\mathbb{P}}^1)}^{k-1},\underline{d})$ is not secant defective, $h^0\left({\mathcal{I}}_{2S_1}(\underline{d},0)\right)=\max \{0,\alpha -k(y-k-g)\}$. By Lemma 1, to conclude for y, it is sufficient to prove that $2\alpha \ge (k+1)(y-g-h)$. Since $(k+1)y\le 3\alpha$, it is sufficient to prove that $(k+1)(g+h)\ge \alpha$. This inequality is true because $kg+h=\alpha -a$.

(a1.1) Assume $y<g+h$. Instead of h, take $h^{\prime }:=\max \{0,g+h-y\}$, $g^{\prime }:=y-h^{\prime }$ and $S_1=\varnothing$.

(a2) The case $x=x_1$ is easier, because we proved Claim 1 for it and we only need to check that $h^0\left({\mathcal{I}}_{2S_1}(\underline{d},0)\right)\le \max \{0,\alpha -k(y-k-g)\}$.

(b) In this step, we take $d_k>2$ and prove this case by induction on the integer $d_k$. To prove the inductive step, we use the case $d_k-1$ of Lemma 10. Set $y^{\prime }:=\lfloor d_k\alpha /(k+1)\rfloor$. For all integers z such that $0\le z\le k$, we check that $h^1\left({\mathcal{I}}_{A\cup Z}(\underline{t},d_k-1)\right)=0$ with Z a general union of $y^{\prime }-z$ double points of Y and of z double points of D.

Let $e_1$, $f_1$ be integers satisfying

$$ka+k{e}_1+f_1=\alpha ,$$

(14)

$a+e_1+k{f}_1\le \alpha$ such that $\alpha -a-e_1-k{f}_1$ is minimal. We saw that

$$\alpha -k^2+1\le a+e_1+k{f}_1\le \alpha$$

(15)

We have $f_1\ge e_1$. If $\alpha \ge ka+k^2-1$, then $e_1\ge 0$.

Theorem 9.

Assume $\alpha \ge ka+k^2$, $t_1\ge \max \{5,2a-1\}$ and that $(D,\underline{t})$ is not secant defective. Assume that for all $b\in \mathbb{N}$ and a general $B\in S(D,b)$, either $h^0\left({\mathcal{I}}_{2A\cup 2B,D}\left(\underline{t}\right)\right)=0$ or $h^0\left({\mathcal{I}}_{2A\cup 2B,D}\left(\underline{t}\right)\right)=0$. Take a general $S\in S(Y,e_1+f_1)$ and a general $S^{\prime }\in S(Y,e_1+f_1+k)$. Then, $h^1\left({\mathcal{I}}_{2A\cup 2S}(\underline{t},1)\right)=0$, $h^0\left({\mathcal{I}}_{2A\cup 2S}(\underline{t},1)\right)=\alpha -kf-e\le k^2-1$ and $h^0\left({\mathcal{I}}_{2A\cup 2S^{\prime }}(\underline{t},1)\right)=0$.

Proof. 

Since $A\subset L$, $t_1\ge 2a-1$ and $t_i>0$ for all $i<k$, we have $h^1(D,{\mathcal{I}}_{2A,D}\left(\underline{t}\right)=0$. Hence, the assumptions on $2A\cup 2B$ are not satisfied for the trivial reason that $h^0\left({\mathcal{I}}_{2A}\left(\underline{t}\right)\right)=0$.

By (14) and 15, to prove the result for S, it is sufficient to prove that $h^1\left({\mathcal{I}}_{2A\cup 2S}(\underline{t},1)\right)=0$. Take a general $(S_1,S_2)\in S(D,e_1)\times S{(D,f)}_1$. By the inductive assumption on k and (14) and (15), we have the equalities $h^i(D,{\mathcal{I}}_{2A\cup 2S_1\cup S_2}(\underline{t},1))=0$, $i=0,1$, $h^1(D,{\mathcal{I}}_{A\cup S_1\cup 2S_2}(\underline{t},0))=0$ and $h^0(D,{\mathcal{I}}_{S_1\cup 2S_2}(\underline{t},0))=\alpha -e_1-kf$. We degenerate $e_1$ of the points of S to different points of $S_1$ and apply the Differential Horace Lemma to each point of $S_2$.

We obtain $h^1\left({\mathcal{I}}_{2A\cup 2S}(\underline{t},1)\right)\le h^1\left({\mathcal{I}}_{A\cup S_1\cup 2S_2}(\underline{t},0)\right)=0$.

We degenerate $S^{\prime }$ to $(S_1,S_2^{\prime })$ with $S_2^{\prime }$ general in $S(D,f+k)$ and use the inductive assumption to obtain $h^0(D,{\mathcal{I}}_{S_1\cup \left(2S_2^{\prime }\right),D}(\underline{t},0))=0$.   □

Remark 20.

In the set-up of Theorem 9, assume $\alpha \equiv 0(modk+1)$ and $\alpha \equiv 0(modk-1)$.

Set ${\tilde{e}}_1:=\frac{\alpha }{k+1}-\frac{ka}{k-1}$ and ${\tilde{f}}_1:=\frac{\alpha }{k+1}+\frac{a}{k-1}$. The proof of Theorem 9 gives the equalities $h^i\left({\mathcal{I}}_{2A\cup 2S}(\underline{t},1)\right)=0$, $i=0,1$, for a general $S\in S(Y,{\tilde{e}}_1+{\tilde{f}}_1)$.

Proof of Theorem 6.

It is sufficient to mimic the proof of Theorem 7 using Theorem 9 and lemma similar to lemma 19 instead of Theorem 8.   □

7. Examples

In this section we study a few cases with $\#A\le 4$ in ${\mathbb{P}}^n$ and ${\mathbb{P}}^1\times {\mathbb{P}}^1$. For ${\mathbb{P}}^n$, we use the group $G:=\mathrm{Aut}\left({\mathbb{P}}^n\right)$. For ${\mathbb{P}}^1\times {\mathbb{P}}^1$, we use $G:=\mathrm{Aut}\left({\mathbb{P}}^1\right)\times \mathrm{Aut}\left({\mathbb{P}}^1\right)$. For Y, either ${\mathbb{P}}^n$ or ${\mathbb{P}}^1\times {\mathbb{P}}^1$, we describe the G-orbits of the set $S(Y,a)$ for all $a\le 4$.

At the end of the section, we recall the definition of multiple points and give an easy example to see why our tools are useful.

Remark 21.

Take $Y={\mathbb{P}}^n$, $n\ge 2$. Since ${\mathbb{P}}^n$ is 2-transitive, for $a\le 2$, the action of G on $S(Y,a)$ is transitive. Hence, $2A\cup 2S$ has the Hilbert function of a general $S^{\prime }\in S({\mathbb{P}}^n,x+a)$ for a general $S\in S({\mathbb{P}}^n,x)$. For $a=3,4$, the set $S({\mathbb{P}}^n,a)$ has an open orbit (for $n=4$, we need that $n\ne 1$). If A is in the open orbit, then $2A\cup 2S$ has the Hilbert function of a general $S^{\prime }\in S({\mathbb{P}}^n,x+a)$ for a general $S\in S({\mathbb{P}}^n,x)$.

(a) Take $a=3$. The action has 2 orbits: the set of collinear points and the set of non-collinear ones. The open orbit is the set of all non-collinear ones.

(b) Take $a=4$. Here, the set of collinear points has infinitely many orbits (they are parametrized by the cross ratio of the 4 points). The set of collinear points is a closed and irreducible subset of ${\mathbb{P}}^n$ of dimension $2n+4$. There are only a finite number of other orbits, and they are described below in (b1) and (b2).

(b1) Assume $n=2$. The open orbit is formed by sets such that no 3 of the points are collinear. The other orbit is formed by the sets A such that $\#(L\cap A)=3$ for some line L.

(b2) Assume $n>2$. The open orbit is formed by 4 linearly independent points. Another orbit is formed by the sets $A\in S({\mathbb{P}}^n,4)$ contained in a plane but such that no 3 of the points are collinear. The other orbit is formed by the sets A such that $\#(L\cap A)=3$ for some line L.

Remark 22.

Take $d=2$, $a\ge 2$, $x\ge 0$, $A\in S({\mathbb{P}}^n,a)$ and a general $S\in S({\mathbb{P}}^n,x)$.

Set $m:=\dim ⟨A⟩$. The singular locus of a quadric hypersurface is a linear space. Thus $h^0\left({\mathcal{I}}_{2A\cup 2S}\left(2\right)\right)=\max \{0,n-m-x\}$.

Example 1.

Take $d=3$, $a\in \{3,4\}$, $x\ge 0$, $A\in S({\mathbb{P}}^n,a)$ and a general $S\in S({\mathbb{P}}^n,x)$ with the convention $S=\varnothing$ if $x=0$.

(a) Assume $A\subset L$ for some line L. We have $\mathrm{deg}(2A\cap L)=2a$, and hence, $h^1(L,{\mathcal{I}}_{2A\cap L}\left(3\right))=2a-4$. Remark 2 gives $h^1\left({\mathcal{I}}_{2A}\left(3\right)\right)\ge 2a-4$. We will check in steps (a1) and (a2) that $h^1\left({\mathcal{I}}_{2A}\left(3\right)\right)=2$ if $a=3$ and $h^1\left({\mathcal{I}}_{2A}\left(3\right)\right)=n+3$ if $a=4$. We will also see that $h^{1}1\left({\mathcal{I}}_{2B}\left(3\right)\right)=2+(n+1)(b-3)$ for any line L and any $B\in S(L,b)$ with $b\ge 3$.

(a1) Assume $n=2$. Since $|{\mathcal{I}}_{2A}|$ is the set of all plane cubics $2L\cup R$ with R a line, we have $h^0\left({\mathcal{I}}_{2A\cup 2S}\left(3\right)\right)=3$ (resp. 0) if $x=0$ (resp. $x\ge 1$). Since $h^0\left({\mathcal{I}}_{2A}\left(3\right)\right)=3$, we have $h^1\left({\mathcal{I}}_{2A}\left(3\right)\right)=3a-7$.

(a2) Assume $n>2$ and take a general hyperplane $H\supset L$. The case $n=2$ and Remark 2 give $h^1\left({\mathcal{I}}_{2A}\left(d\right)\right)\ge 3a-7$. Consider the residual exact sequence of H:

$$0\to {\mathcal{I}}_A\left(2\right)\to {\mathcal{I}}_{2A}\left(3\right)\to {\mathcal{I}}_{H\cap 2A,H}\left(3\right)\to 0$$

(16)

Since $h^1\left({\mathcal{I}}_A\left(2\right)\right)=0$ if $a=3$ and $h^1\left({\mathcal{I}}_A\left(2\right)\right)=1$ is $a=4$, the long cohomology exact sequence of (16) gives $h^1\left({\mathcal{I}}_{2A}\left(3\right)\right)=2$ if $a=3$. Note that for $a=3$, L is contained in the base locus of ${\mathcal{I}}_A\left(2\right)$. We find that for $a=3$, all cubic hypersurfaces containing $2A$ are singular at all points of L. Hence, for all $p\in L\setminus A$, we have $h^0\left({\mathcal{I}}_{2A\cup 2p}\left(3\right)\right)=h^0\left({\mathcal{I}}_{2A}\left(3\right)\right)$. Hence, for $a=4$, we have $h^1\left({\mathcal{I}}_{2A}\left(3\right)\right)=n+3$.

(b) Assume $a=4$ and the existence of a line L such that $\#(A\cap L)=3$.

(b1) Assume $n=2$. Case (a1) gives $h^0\left({\mathcal{I}}_{2A}\left(3\right)\right)=0$. Hence, $h^1\left({\mathcal{I}}_{2A}\left(3\right)\right)=2$.

(b2) Assume $n>2$. We want to prove by induction on n that $h^1\left({\mathcal{I}}_{2A}\left(3\right)\right)=2$. Let H be a general hyperplane containing the plane $⟨A⟩$. We have $h^1\left({\mathcal{I}}_A\left(3\right)\right)=0$. Hence, it is sufficient to use the long cohomology exact sequence of (16).

(c) Assume $n>2$, the non-existence of a line L such that $\#(A\cap L)=3$ and that $M:=⟨A⟩$ is a plane. We have $h^0(M,{\mathcal{I}}_{2A\cap M,M}\left(3\right))=0$, and hence, $h^1(M,{\mathcal{I}}_{2A\cap M,M}\left(3\right))=2$. Then, as in step (b2), by induction on n, we obtain $h^1\left({\mathcal{I}}_{2A}\left(3\right)\right)=2$.

Proposition 4.

Take $A\in S({\mathbb{P}}^2,4)$ and a general $S\in S({\mathbb{P}}^2,x)$.

(a) Assume that A is contained in a line.

We have $h^1\left({\mathcal{I}}_{2A}\left(6\right)\right)=1$, $h^1\left({\mathcal{I}}_{2A}\left(5\right)\right)=h^1\left({\mathcal{I}}_{2A\cup 2S}\left(5\right)\right)=2$ for $x\le 4$, $h^0\left({\mathcal{I}}_{2A\cup 2S}\left(6\right)\right)=1$ and $h^1\left({\mathcal{I}}_{2A\cup 2S}\left(6\right)\right)=2$ for $x=5$, $h^0\left({\mathcal{I}}_{2A\cup 2S}\left(6\right)\right)=0$ if and only if $x\ge 6$. Moreover, we have $h^0\left({\mathcal{I}}_{2A\cup 2S}\left(5\right)\right)=0$ if and only if $x\ge 4$.

(b) Assume that A is not contained in a line. We have $h^0\left({\mathcal{I}}_{2A\cup 2S}\left(6\right)\right)=0$ if $x\ge 6$ and $h^1\left({\mathcal{I}}_{2A\cup 2S}\left(6\right)\right)=0$ if $x\le 5$. We have $h^1\left({\mathcal{I}}_{2A\cup 2S}\left(5\right)\right)=h^0\left({\mathcal{I}}_{2A\cup 2S}\left(5\right)\right)=0$ for $x=3$.

Proof. 

Take A contained in a line L. We have $\mathrm{deg}(L\cap 2A)=8$, and hence, we obtain $h^1(L,{\mathcal{I}}_{2A\cap L,L}\left(6\right))=1$ and $h^1(L,{\mathcal{I}}_{2A\cap L,L}\left(5\right))=2$. Remark 2 gives $h^1\left({\mathcal{I}}_{2A}\left(6\right)\right)>0$ and $h^1\left({\mathcal{I}}_{2A}\left(5\right)\right)\ge 2$. Since ${\mathrm{Res}}_L\left(2A\right)=A$ and $h^1\left({\mathcal{I}}_A\left(x\right)\right)=0$ for all $x\ge 3$, the residual exact sequence of L gives $h^1\left({\mathcal{I}}_{2A}\left(6\right)\right)=1$, $h^1\left({\mathcal{I}}_{2A}\left(5\right)\right)=2$ and $h^i\left({\mathcal{I}}_{2A\cup 2S}\left(6\right)\right)=h^i\left({\mathcal{I}}_{A\cup 2S}\left(5\right)\right)$, $i=0,1$. If $x\le 4$, we have $h^1\left({\mathcal{I}}_{2S}\left(4\right)\right)=0$ by the Alexander–Hirschowitz theorem. Hence, the residual exact sequence of L gives $h^1\left({\mathcal{I}}_{2A\cup 2S}\left(6\right)\right)=h^1\left({\mathcal{I}}_{2A}\left(6\right)\right)=2$ for $x\ge 4$. Take $x=5$ and let D be the only conic containing S. Using the residual exact sequence of $2L$, we obtain $|{\mathcal{I}}_{2A\cup 2S}\left(6\right)|=\{2L\cup 2D\}$. Hence, $h^1\left({\mathcal{I}}_{2A\cup 2S}\left(6\right)\right)=2$. Adding a general point, we obtain $h^0\left({\mathcal{I}}_{2A\cup 2S}\left(6\right)\right)=0$ if $x\ge 6$. The residual sequence of $2L$ gives $h^0\left({\mathcal{I}}_{2A\cup 2S}\left(5\right)\right)=0$ if and only if $x\ge 4$.

Now, assume that A is not contained in a line. By the semicontinuity theorem for cohomology (ref. [16], Th. III.12.8) and Lemma 4, it is sufficient to use the case in which there is a line L such that $\#(L\cap A)=3$. Set $E:=L\cap A$ and $\left\{o\right\}:=A\setminus E$. Since $h^0\left({\mathcal{O}}_{{\mathbb{P}}^2}\left(6\right)\right)=28$, to prove the part on ${\mathcal{O}}_{{\mathbb{P}}^2}\left(6\right)$, it is sufficient to prove that $h^0\left({\mathcal{I}}_{2A\cup 2S}\left(6\right)\right)=1$ for a general $S\in S({\mathbb{P}}^2,5)$. Take $p\in L\setminus E$ and a general $S_1\in S({\mathbb{P}}^2,4)$. We have $h^i(L,{\mathcal{I}}_{2A\cup \left\{p\right\},L}\left(6\right))=0$, $i=0,1$. By the Differential Horace Lemma applied to p to prove that $h^1\left({\mathcal{I}}_{2A\cup 2S}\left(6\right)\right)=0$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{A\cup (2p,L)\cup 2S_1\cup 2o}\left(5\right)\right)=0$. Take a general $q\in L$ and a general $S_2\in S({\mathbb{P}}^2,3))$. Note that $h^i(L,{\mathcal{I}}_{A\cup (2p,L)\cup \left\{q\right\},L}\left(5\right))=0$, $i=0,1$. Applying the Differential Horace Lemma to q, we find that to prove that $h^1\left({\mathcal{I}}_{A\cup (2p,L)\cup 2S_1\cup 2o}\left(5\right)\right)=0$, it is sufficient to prove that $h^0\left({\mathcal{I}}_{2o\cup 2S_2\cup (2q,L)}\left(4\right)\right)=1$. Since $S_2$ is general and q is general in q, there is a unique conic D containing $S_2\cup \{o,q\}$ and it is not tangent to L at q. Thus ${\mathrm{Res}}_D(2o\cup 2S_2\cup (2q,L))=\{o,q\}\cup S_2$, and hence, we obtain that $|{\mathcal{I}}_{2o\cup 2S_2\cup (2q,L)}\left(4\right)|=\left\{2D\right\}$.

Let U denote the set of all $A\in S({\mathbb{P}}^2,4)$ such that there is a line L with $\#(A\cap L)=3$. The set U is a G-orbit. Hence, $h^i\left({\mathcal{I}}_{2S\cup 2A}\left(5\right)\right)$, $i=0,1$, may be computed in the following way. Set $\left\{o\right\}:=A\setminus L\cap A$. By the residual exact sequence of L for $i=0,1$, we have the equality $h^i\left({\mathcal{I}}_{2A\cup 2}\left(5\right)\right)=h^i\left({\mathcal{I}}_{2o\cup 2S\cup (S\cap L)}\left(4\right)\right)$ with $\left\{o\right\}\cup S$, a general element of $S({\mathbb{P}}^2,4))$ (and hence $h^1\left({\mathcal{I}}_{2A\cup 2o}\left(4\right)\right)=0$ and $h^0\left({\mathcal{I}}_{2o\cup 2S}\left(4\right)\right)=3$ by the Alexander–Hirschowitz theorem) and 3 general points of a general line of ${\mathbb{P}}^2$. Thus, $h^i\left({\mathcal{I}}_{2o\cup 2S\cup (S\cap L)}\left(4\right)\right)=0$, $i=0,1$.   □

Remark 23.

Take $Y={\mathbb{P}}^1\times {\mathbb{P}}^1$. For any $A\in S(Y,a)$, $a\ge 2$, we have the pair $(\#{\pi }_1\left(A\right),\#{\pi }_2\left(A\right))$ of positive integers and $(\#{\pi }_1\left(A\right),\#{\pi }_2\left(A\right))=(\#{\pi }_1\left(B\right),\#{\pi }_2\left(B\right))$ if A and B are contained in the same G-orbit. Not all pairs of integers $(e,f)$ with $1\le e\le a$ and $1\le f\le a$ occur as $(\#{\pi }_1\left(A\right),\#{\pi }_2\left(A\right))$ for some $S(Y,a)$. For instance, if $\#{\pi }_i\left(A\right)=1$, then $\#{\pi }_{3-i}\left(A\right)=a$.

Note that $h^0\left({\mathcal{I}}_{2A}(1,1)\right)=0$ for all $a\ge 2$ and all $A\in S(Y,a)$.

(a) The variety $S(Y,2)$ has 3 G-orbits, the open one, the one with points with the same first projection and the one with the same second projection.

(b) The variety $S(Y,3)$ has 5 G-orbits, distinguished by the pair $(\#{\pi }_1\left(A\right),\#{\pi }_2\left(A\right))$.

(c) The set $S(Y,4)$ has infinitely many orbits, even among the ones contained in a curve of multidegree ${ϵ}_1$ or a curve of multidegree ${ϵ}_2$. Since $\mathrm{Aut}\left({\mathbb{P}}^1\right)$ is 3-transitive, there is a unique G-orbit associated to $(3,2)$ and a unique G-orbit associated to $(2,3)$. There are 2 different G-orbits associated to $(2,2)$. One of them is formed by the elements of $S(Y,4)$, which are the complete intersection of a reduced curve of multidegree $(2,0)$ and a reduced curve of multidegree $(0,2)$. The other one is formed by the set of all $A\in S(Y,4)$ contained in the smooth locus of a reducible element of $|{\mathcal{O}}_Y(1,1)|$. The classical notion of cross-ratio shows the existence of a 2-dimensional family of orbits with pair $(4,4)$. We have $h^0\left({\mathcal{I}}_A(1,1)\right)=2$ if and only if there is $i\in \{1,2\}$ and A is contained in a curve of multidegree ${ϵ}_i$. Note that 4 different orbits are formed by sets A with $h^0\left({\mathcal{I}}_A(1,1)\right)=1$.

Notation 1.

For any smooth manifold M, any $p\in M$ and any positive integer m, let $(mp,M)$ denote the closed subscheme of M with ${\left({\mathcal{I}}_p\right)}^m$ as its ideal sheaf. If $n:=\dim M$, then $\mathrm{deg}\left((mp,M)\right)=\left(\genfrac{}{}{0pt}{}{n+m-1}{n}\right)$. Set $(0p,M):=\varnothing$. For all finite subsets of M, set $(mS,M)={\cup }_{p\in S}(mp,M)$. Write $mp$ and $mS$ instead of $(mp,M)$ and $(mS,M)$, respectively, if $M=Y$.

The zero-dimensional scheme $(mp,M)$ is called a multiple point of M of multiplicity m. L. Chiantini and Th. Mawking considered the postulation of the union of a triple point and finitely many general double points [23,24]. The next result considers the same problem for 2 multiple points of ${\mathbb{P}}^1\times {\mathbb{P}}^1$, the ones corresponding to one of the smaller orbits in part (a) of Remark 23.

Proposition 5.

Take $Y={\mathbb{P}}^1\times {\mathbb{P}}^1$ and $A\in S(Y,2)$ such that $\#{\pi }_1\left(A\right))=1$. Let L be the only element of $|{\mathcal{O}}_Y(0,1|$ containing A. For all positive integers m, d and all zero-dimensional schemes Z such that $Z\cap L=\varnothing$, we have $h^0\left({\mathcal{I}}_{mA\cup Z}(1,d)\right)=h^0\left({\mathcal{I}}_Z(1,d-m)\right)$.

Proof. 

Note that $\mathrm{deg}(xA\cap L)=2x$ for all integers $x\ge 0$ and that $\mathrm{deg}\left({\mathcal{O}}_L(1,t)\right)=1$ for all $t\in \mathbb{Z}$. Thus, it is sufficient to use m times the residual exact sequence of L.   □

For another case of multiple points with a fixed support, see [25].

8. Discussion

Motivated by its relation with the computation of the dimensions of the secant varieties of an embedded variety X, we consider an interpolation problem for unions of $a+x$ double points, a of them prescribed, while the other x are general in X. Our results are in cases of Veronese embeddings of projective spaces and of Segre–Veronese embeddings of multiprojective spaces. The multiprojective spaces we consider are the ones with all factors of dimension 1 and the ones with 2 factors, one of them one-dimensional. We use the known fundamental results for general unions of double points in these cases [4,7,8,11]. We suggest that other places to be explored are Grassmannians and the Schur factors of tensors [26,27,28]. We proposed a related problem when a of the double points are allowed to vary in a prescribed subvariety. This related problem often comes in some inductive proofs on the interpolation of general double points (the so-called Horace Method). We suggest to explore here even when instead of a subvariety, we have a flag of subvarieties and prescribe the number of double points whose reduction is contained in each stratum. In the last section, we gave several examples on how to use the tools of the paper for related problems, even with points of higher multiplicities.