On the Secant Non-Defectivity of Integral Hypersurfaces of Projective Spaces of at Most Five Dimensions

Abstract

Let $X\subset {\mathbb{P}}^n$, where $3\le n\le 5$, be an irreducible hypersurface of degree $d\ge 2$. Fix an integer $t\ge 3$. If $n=5$, assume $t\ge 4$ and $(t,d)\ne (4,2)$. Using the Differential Horace Lemma, we prove that ${\mathcal{O}}_X\left(t\right)$ is not secant defective. For a fixed X, our proof uses induction on the integer t. The key points of our method are that for a fixed X, we only need the case of general linear hyperplane sections of X in lower-dimension projective spaces and that we do not use induction on d, allowing an interested reader to tackle a specific X for $n>5$. We discuss (as open questions) possible extensions of some weaker forms of the theorem to the case where $n>5$.

1. Introduction

Let $Y\subset {\mathbb{P}}^r$ be an integral and non-degenerate variety. We recall that, as usual in Algebraic Geometry, “integral” means “reduced and irreducible”. For any integer $s\ge 1$, the s-secant variety ${\sigma }_s\left(Y\right)$ of Y is the closure of the union of all linear spaces $\langle S\rangle$, where $\langle \rangle$ denotes the linear span, and $S\subset Y$ has cardinality s. The set ${\sigma }_s\left(Y\right)$ is an integral projective subvariety of dimensions of at most $min\{r,s(dimY+1)-1\}$. If $dim{\sigma }_s\left(Y\right)=min\{r,s(dimY+1)-1\}$, we say that the s-secant variety of Y is not defective. If $dim{\sigma }_s\left(Y\right)=min\{r,s(dimY+1)-1\}$ for all $s>0$, then Y is said to be non-secant defective or non-defective.

If Y is the order t Veronese embedding of ${\mathbb{P}}^n$, the integer $dim{\sigma }_s\left(Y\right)$ is the dimension of the set of all degree t forms in $n+1$ variables with an additive decomposition with at most s addenda [1,2]. If Y is the Segre embedding of a multiprojective space (resp., a Segre–Veronese embedding of a multiprojective space), the integer $dim{\sigma }_s\left(Y\right)$ is the dimension of the set of all tensors (resp., partially symmetric tensors) with a tensor rank (resp., partially symmetric tensor rank) of at most s and the format determined by Y. Antisymmetric tensors with a given rank are related to the secant varieties of the Plücker embeddings of the Grassmannians.

The important case of the Veronese embeddings of projective spaces was solved in complete generality by J. Alexander and A. Hirschowitz, who proved the following fundamental result.

Theorem 1

(Alexander–Hirschowitz). Fix positive integers n, t, and s. Let $Y\subset {\mathbb{P}}^r$, $r=\left({\displaystyle \genfrac{}{}{0pt}{}{n+t}{n}}\right)-1$, denote the order t Veronese embedding of ${\mathbb{P}}^n$. Then, $dim{\sigma }_s\left(Y\right)=min\{r,(n+1)s-1\}$, except in the following cases:

• Either $(n,t,s)=(n,2,s)$ with $2\le s\le n$;
• Or $(n,t,s)=(2,4,5),(3,4,9),(4,3,7),(4,4,14)$.

The starting point of their proof is the known translation of the computation of the integer $dim{\sigma }_s\left(Y\right)$ in terms of the cohomology groups of certain zero-dimensional subschemes of ${\mathbb{P}}^n$, the so-called Terracini lemma ([1,2], Corollary 1.10 in [3], Lemma 1). Then, they discovered several tools for solving the translated problem. They introduced the Differential Horace Method [4,5,6,7]. Later, several mathematicians proved Theorem 1 using some of the ideas used by J. Alexander and A. Hirschowitz [8,9,10,11]. We discuss their tools in a subsection of Section 1. In Section 6, we raise two open questions.

We use the Differential Horace Lemma (Lemma 2) to prove the following result.

Theorem 2.

Let $X\subset {\mathbb{P}}^n$, where $3\le n\le 5$, be an integral hypersurface of degree $d\ge 2$. Fix a positive integer $t\ge 3$. If $n=5$, assume $t\ge 4$ and $(t,d)\ne (4,2)$. Then, ${\mathcal{O}}_X\left(t\right)$ is not secant defective.

Let $X\subset {\mathbb{P}}^n$, where $n\ge 2$, be an integral hypersurface of degree $d\ge 2$. The line bundle ${\mathcal{O}}_X\left(1\right)$ is not secant defective (Remark 2). For all $n\ge 3$, the case $(d,t)=(2,2)$ is defective (Proposition 1, Theorem 3, and Lemma 5). We do not know whether cases $(n,t)=(5,3)$ and $(n,t,d)=(5,4,2)$ are defective.

The case $d=1$ would just be the Alexander–Hirschowitz theorem for ${\mathbb{P}}^{n-1}$. The translation given by the Terracini lemma stimulated the introduction of other notions of ranks, e.g., the cactus rank [12,13,14,15,16,17,18], and our tools also apply to these generalized notions of ranks. Knowing that certain secant varieties of an embedded variety have the expected dimension has a very important consequence for the identifiability problem, e.g., the uniqueness of the tensor decomposition of generic tensors of a prescribed subgeneric rank ([19], Theorem 1.5) with examples given in [19], as well as some obtained using later papers on the non-defectivity problem [20,21,22].

We divide the proof of Theorem 2 into three sections, one for each integer n (Proposition 1 and Theorems 3 and 4). In the statement of Theorem 2, we assume $n\ne 2$ because the case where $n=2$ would be well known and too easy: no secant variety of a non-degenerate curve is defective ([3], Corollary 1.10). Fix an integral hypersurface $X\subset {\mathbb{P}}^n$, with $n\ge 3$, of degree $d\ge 2$ and an integer $t\ge 3$. To prove Theorem 2 for hypersurface X using the Differential Horace Lemma, we use the principle that a general hyperplane section C of X has non-defective secant varieties with respect to ${\mathcal{O}}_C\left(t\right)$ and some weaker assumptions on ${\mathcal{O}}_C(t-1)$ and even weaker assumptions on ${\mathcal{O}}_C(t-2)$. For a fixed X, we also use something on ${\mathcal{O}}_X(t-1)$ and ${\mathcal{O}}_X(t-2)$ but weaker than non-defectivity, so that for $n=3,4$, we are able to handle a case where $t=3$, while for $n=5$, we are able to handle the case where $t=4$ (with one exception) and then obtain all $t>4$. Then (as in all quoted papers), the proofs require several numerical inequalities (we call them Claims in the proofs for Proposition 1 and Theorems 3 and 4). Roughly speaking, for d-degree hypersurfaces in ${\mathbb{P}}^n$, we need to prove that a few bivariate degree $n-1$ polynomials $\alpha (x,y)$ are non-negative when evaluated at $x=d$ and $y=t$. There are at least two such bivariate polynomials, one for $t<d$ and one for $t\ge d$. For $n=3$, these numerical problems are solved using double differences; for $n=4$ and much more for $n=5$, we reduce the most difficult inequalities into simple checks using the principle that ${\mathcal{O}}_C\left(t\right)$ is not defective. Regardless, for all n, we use something related to the non-defectivity of the general hyperplane section of X, and hence, the use of ${\mathcal{O}}_C\left(t\right)$ is intrinsic to the Horace Method. We are proud that our tools allow us to obtain certain values of t without knowing that ${\mathcal{O}}_X(t-1)$ or ${\mathcal{O}}_C(t-1)$ is not defective. We suggest that Theorem 2 is provable also for $n>5$, at least for some higher values of n, with our framework. Another possibility for a fixed $n>5$ and d is only to prove non-defectivity for a general hypersurface of a prescribed degree. For this weaker problem, it is very useful to use Computer Algebra computations as in [21,23,24].

Tools Used Here and Elsewhere Related to Horace

The original very difficult and long proof of Theorem 1 by J. Alexander and A. Hirschotwitz used the Differential Horace Lemma (later improved and extended to more general zero-dimensional schemes in [7]) and the Blow-up Horace Lemma (a kind of Horace Lemma that uses subvarieties of a lower codimension). K. Chandler used the Differential Horace Lemma to give a shorter proof [8,9]. M. C. Brambilla and G. Ottaviani gave an even shorter proof using the Horace Lemma with respect to codimension 3 linear subspaces [10]. E. Postinghel gave a different proof degenerating the projective space [11], and her method was used in [25] to compute the dimensions of all secant varieties of Segre–Veronese embeddings of ${\left({\mathbb{P}}^1\right)}^n$. These methods try to control the Hilbert functions of certain zero-dimensional schemes, and in particular, they compute or bound the dimensions of the secant varieties of the Segre–Veronese varieties. Another method used for these varieties was the collision of fat points [24]. Moreover, these methods were refined in a spectacular way: compare [26,27,28,29,30] with [21] and compare [31] with [20]. We have great hope that they may be used for $n>5$ (see Section 6).

There is an old and open conjecture of Segre, Gimigliano, Harbourne, and Hirschowitz concerning zero-dimensional schemes in ${\mathbb{P}}^2$. Degenerations of linear systems on ${\mathbb{P}}^2$ were used to handle unions of multiple points [32] and prove this conjecture in some cases. For the case of surfaces, there is another approach to the defectivity of secant varieties [33], and we describe it in one case (Lemma 4).

We thank the referees for useful suggestions and the staff of the journal for Latex help.

2. Preliminaries and Notation

We work over an algebraically closed field with characteristic 0.

For any irreducible quasi-projective variety T and any $x\in \mathbb{N}$, let $S(T,x)$ denote the set of all subsets of T with cardinality x, with the convention $S(T,0)=\varnothing$. Let $T_{\mathrm{reg}}$ denote the set of all smooth points of T. For any closed scheme $Z\subset T$, let ${\mathcal{I}}_{Z,T}$ denote the ideal sheaf of Z in T. Often, if $T=X$, we write ${\mathcal{I}}_Z$ instead of ${\mathcal{I}}_{Z,T}$. For any sheaf $\mathcal{F}$ on T and any $i\in \mathbb{N}$, let $H^i(T,\mathcal{F})$ denote its i-th cohomology group of $\mathcal{F}$. Set $h^i(T,\mathcal{F}):=dim{H}^i(T,\mathcal{F})$. If $T=X$, we write $H^i\left(\mathcal{F}\right)$ and $h^i\left(\mathcal{F}\right)$ instead of $H^i(X,\mathcal{F})$ and $h^i(X,\mathcal{F})$, respectively. For any $p\in T_{\mathrm{reg}}$, let $(2p,T)$ denote the closed subscheme of T with ${\mathcal{I}}_p^2$ as its ideal sheaf. The scheme $(2p,T)$ is called a two-point of T. The scheme $(2p,T)$ is zero-dimensional, $deg(2p,T)=dimT+1$, and ${(2p,T)}_{\mathrm{red}}=\left\{p\right\}$. For any finite set $S\subset T_{\mathrm{reg}}$ set $(2S,T):={\cup }_{p\in S}(2p,T)$ with the convention $(2S,T):=\varnothing$ if $S=\varnothing$. If $T=X$, we often write $2p$ and $2S$ instead of $(2p,X)$ and $(2S,X)$.

For any $s\in \mathbb{N}$, let $A\left(s\right)\subset X$ be a general union of s double points, with the convention $A\left(0\right)=\varnothing$.

Remark 1.

Let T be an integral projective variety of dimension $n\ge 1$. Let $\mathcal{L}$ be a line bundle on T, and $V\subseteq H^0(T,\mathcal{L})$, a linear subspace. Fix zero-dimensional schemes $Z\subset W\subset T$. If $V(-Z)=0$, then $V(-W)=0$. If $dimV(-W)=dimV-deg\left(W\right)$, then $dimV(-Z)=dimV-deg\left(Z\right)$.

Remark 2.

Since $d\ge 2$, we have $h^0\left({\mathcal{O}}_X\left(1\right)\right)=n+1$. Obviously, $h^1\left({\mathcal{I}}_{A\left(1\right)}\left(1\right)\right)=0$. Since two different hyperplanes of ${\mathbb{P}}^n$ span ${\mathbb{P}}^n$, $h^0\left({\mathcal{I}}_{A\left(2\right)}\left(1\right)\right)=0$. Remark 1 gives that ${\mathcal{O}}_X\left(1\right)$ is not defective.

Remark 3.

Let T be an integral curve. For any $p\in T_{\mathrm{reg}}$ and any positive integer t, let $C(T,p,t)$ denote the only connected degree t subscheme of T with p as its reduction. For all positive integers s, and $t_i$, $1\le i\le s$, let $C(T,t_1,\dots ,t_s)$ be the set of all zero-dimensional schemes $A\subset X_{\mathrm{reg}}$ with s connected components, say $A=A_1\cup \cdots \cup A_s$, with $A_i=C(T,p_i,t_i)$ with $(p_1,\dots ,p_s)$ varying in $T_{\mathrm{reg}}^s$. Note that we assume $\#A_{\mathrm{red}}=s$, i.e., in the union, we take $(p_1,\dots ,p_s)$ with $p_i\ne p_j$ for all $i\ne j$. Since $T_{\mathrm{reg}}$ is irreducible, $C(T,t_1,\dots ,t_s)$ is a non-empty and irreducible quasi-projective variety of dimension s. A particular case (the case of $dimT=1$) of short note [34] may be stated in the following way. Let $\mathcal{L}$ be a line bundle on T, and $V\subseteq H^0(T,\mathcal{L})$, a linear subspace. Then, $dimV(-Z)=max\{0,dimV-t_1-\cdots -t_s\}$ for a general $Z\in C(T,t_1,\dots ,t_s)$.

Let M be an integral projective variety; $Z\subset M$, a zero-dimensional scheme; and $D\subset M$, an effective Cartier divisor of M. The residual scheme ${\mathrm{Res}}_D\left(Z\right)$ of Z with respect to D is the closed subscheme of M with ${\mathcal{I}}_Z:{\mathcal{I}}_D$ as its ideal sheaf. We have ${\mathrm{Res}}_D\left(Z\right)\subseteq Z$ and $deg\left(Z\right)=deg(Z\cap D)+deg\left({\mathrm{Res}}_D\left(Z\right)\right)$. If p is a smooth point of D and hence of M, we have $(2p,M)\cap D=(2p,D)$ and ${\mathrm{Res}}_D\left((2p,M)\right)=\left\{p\right\}$.

For all line bundles $\mathcal{L}$ on M, we have an exact sequence, often called the residual exact sequence of D:

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

(1)

The following lemma is called the Terracini lemma ([2], Corollary 1.10 in [3]).

Lemma 1.

Let $Y\subset {\mathbb{P}}^r$ be an integral projective variety. For any positive integer s, we have $dim{\sigma }_s\left(Y\right)=dim\langle (2S,Y)\rangle$, where S is a general element of $S(Y_{\mathrm{reg}},s)$.

Remark 4.

Let $X\subset {\mathbb{P}}^r$ be an integral and non-degenerate variety. Fix an integer $x\ge 2$. If ${\sigma }_x\left(X\right)={\mathbb{P}}^r$, then obviously ${\sigma }_{x+1}\left(X\right)={\mathbb{P}}^r$, and hence, ${\sigma }_{x+1}\left(X\right)$ is not defective. Now assume that ${\sigma }_x\left(X\right)⊊{\mathbb{P}}^r$ and that ${\sigma }_x\left(X\right)$ is not defective, i.e., $dim{\sigma }_x\left(X\right)=x(dimX+1)-1$. It is well known that ${\sigma }_{x-1}\left(X\right)$ is not defective. This old and elementary result is an immediate consequence of the Terracini lemma and of Remark 1.

The following lemma is called the Differential Horace Lemma (case $y=0$, and hence, $S^{\prime }=\varnothing$ is called the Horace Lemma).

Lemma 2.

Let M be an irreducible variety of at least two dimensions, $D\subset M$ be an irreducible effective Cartier divisor of X, and $A\subset M$ be a zero-dimensional scheme. Take a line bundle $\mathcal{L}$ on M and a vector space $V\subseteq H^0\left(\mathcal{L}\right)$ and $(x,y)\in {\mathbb{N}}^2$. Take $(x,y)\in {\mathbb{N}}^2$ and a general $(S,S^{\prime })\in S(D_{\mathrm{reg}},x)\times S(D_{\mathrm{reg}},y)$. Set $\alpha :=dim{V}_{|D}(-A\cap D)-dim{V}_{|D}(-A\cap D-(2S,D)-S^{\prime })$ and $\beta :=dimV(-D-{\mathrm{Res}}_D\left(A\right)-S-(2S^{\prime },D))$. Take a general $E\in S(X_{\mathrm{reg}},x+y)$. Then, $dimV(-A)-dimV(-A-2E)\ge \alpha +\beta$.

Observe that in the statement of Lemma 2, the assumption that D is a reduced Cartier divisor of M implies that $D⊈\mathrm{Sing}\left(M\right)$ because it implies that $D_{\mathrm{reg}}\subset M_{\mathrm{reg}}$.

Lemma 3.

Let M be an integral projective variety of dimension $>1$; $\mathcal{L}$, a line bundle of M; $D\subset M$, an integral Cartier divisor of M; $Z\subset M$, a zero-dimensional scheme; and x, a positive integer. Take a general $S\subset D$ such that $\#S=x$:

(a) Assume that $h^0(M,{\mathcal{I}}_{{\mathrm{Res}}_D\left(Z\right)}\otimes \mathcal{L}(-D))\le h^0(M,{\mathcal{I}}_Z\otimes \mathcal{L})-x$. Then, $h^0(M,{\mathcal{I}}_{Z\cup S}\otimes \mathcal{L})=h^0(M,{\mathcal{I}}_Z\otimes \mathcal{L})-x$.

(b) Assume that $h^0(M,{\mathcal{I}}_{{\mathrm{Res}}_D\left(Z\right)}\otimes \mathcal{L}(-D))=0$. Then, $h^0(M,{\mathcal{I}}_{Z\cup S}\otimes \mathcal{L})=max\{0,h^0(M,{\mathcal{I}}_Z$$\otimes \mathcal{L})-x\}$.

Proof. 

Since $dimM>1$, $dimD=dimM-1\ge 1$. Since we fixed the general $S\subset D$ after fixing D, $S\cap Z=\varnothing$. Obviously, $h^0(M,{\mathcal{I}}_{Z\cup S}\otimes \mathcal{L})\ge max\{0,h^0(M,{\mathcal{I}}_Z\otimes \mathcal{L})-x\}$.

First, assume $x=1$. Parts (a) and (b) are true because $h^0(M,{\mathcal{I}}_{Z\cup S}\otimes \mathcal{L})=h^0(M,{\mathcal{I}}_Z\otimes \mathcal{L})$ if and only if D is contained in the base locus of ${\mathcal{I}}_Z\otimes \mathcal{L}$.

Now assume that $x>1$ and that the result is true for lower values of x. Take a general $S^{\prime }\subset D$ such that $\#S^{\prime }=x-1$. Since ${\mathrm{Res}}_D(Z\cup S^{\prime })={\mathrm{Res}}_D\left(Z\right)$, the inductive assumption gives $h^0(M,{\mathcal{I}}_{Z\cup S^{\prime }}\otimes \mathcal{L})=max\{0,h^0(M,{\mathcal{I}}_Z\otimes \mathcal{L})-x+1\}$. Take as S the union of $S^{\prime }$ and a general point of D. Apply the case $x=1$ to the zero-dimensional scheme $Z\cup S^{\prime }$. □

In the next 3 sections, we use the following notation.

Remark 5.

Let $X\subset {\mathbb{P}}^n$, $n\ge 3$, be an integral degree d hypersurface. Let $H\subset {\mathbb{P}}^n$ be a general hyperplane. Set $C:=H\cap H$. The theorem of Bertini gives that C is a degree d integral hypersurface of H and that $\mathrm{Sing}\left(C\right)=H\cap \mathrm{Sing}\left(X\right)$. Since H is general, C is smooth if X has at most isolated singularities.

3. Proof of Theorem 2 for $\mathit{n}=\mathbf{3}$

Although Theorem 1.3 from [33] does not cover the case of X singular of Proposition 1, the notion of weak defectivity introduced in [33] gives a very strong tool with which to show that a surface $X\subset {\mathbb{P}}^r$ is not secant defective, unless under some very strong geometric and numerical conditions. As an example, we obtain a second proof of the case $d=2$, $t\ge 3$, and X singular, i.e., the case of a quadric cone (see Lemma 4).

Proposition 1.

Let $X\subset {\mathbb{P}}^3$ be an integral degree $d>1$ surface. Take positive integers t and s. Then, either $h^0\left({\mathcal{I}}_{A\left(s\right)}\left(t\right)\right)=0$ or $h^1\left({\mathcal{I}}_{A\left(s\right)}\left(t\right)\right)=0$, unless $d=2$, $t=2$, or $s=3$. We have $h^0\left({\mathcal{I}}_{A\left(3\right)}\left(2\right)\right)=h^1\left({\mathcal{I}}_{A\left(3\right)}\left(2\right)\right)=1$.

Proof. 

We have $h^1\left({\mathcal{O}}_X\left(z\right)\right)=0$ for all $z\in \mathbb{Z}$, $h^0\left({\mathcal{O}}_X\left(t\right)\right)=0$, if $t<0$; $h^0\left({\mathcal{O}}_X\left(t\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{t+3}{3}}\right)$ if $0\le t<d$; and $h^0\left({\mathcal{O}}_X\left(t\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{t+3}{3}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{t-d+3}{3}}\right)$ if $t\ge d$. The adjunction formula gives ${\omega }_X\cong {\mathcal{O}}_X(d-4)$. Duality gives $h^2\left({\mathcal{O}}_X\left(z\right)\right)=0$ for all $z>d-4$ and $h^2\left({\mathcal{O}}_X(d-4)\right)=1$. Obviously, $h^i\left({\mathcal{O}}_X\left(z\right)\right)=0$ for all $i>2$ and all $z\in \mathbb{Z}$. Take a general plane $H\subset {\mathbb{P}}^3$ and set $C:=X\cap H$. The theorem of Bertini gives that C is an integral curve (Remark 5). Since $h^1\left({\mathcal{O}}_X(t-1)\right)=0$, a standard exact sequence gives $h^0\left({\mathcal{O}}_C\left(t\right)\right)=h^0\left({\mathcal{O}}_X\left(t\right)\right)-h^0\left({\mathcal{O}}_X(t-1)\right)$ if $t\ge 0$. For any integer $t>0$, set $u_t:=\lfloor h^0({\mathcal{O}}_X\left(t\right)/3\rfloor$ and $u_t^{\prime }:=\lceil h^0\left({\mathcal{O}}_X\left(t\right)\right)/3\rceil$. By Remark 1, line bundle ${\mathcal{O}}_X\left(t\right)$ is not secant defective if and only if $h^1\left({\mathcal{I}}_{A\left(u_t\right)}\left(t\right)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(u_t^{\prime }\right)}\left(t\right)\right)=0$.

Recall that ${\mathcal{O}}_X\left(1\right)$ is not defective (Remark 2). Hence, from now on, we assume $t\ge 2$.

All smooth quadric surfaces are projectively equivalent, and all singular integral quadric surfaces are projectively equivalent. The case $d=2$ is known ([32], Propositions 4.1 and 5.2, covers the case of a smooth quadric ($n=0$ and ${\mathcal{O}}_X\left(1\right)=H+F$ in his notation) and the case of a quadric cone ($n=2$ and H as the pull-back of ${\mathcal{O}}_X\left(1\right)$ by the minimal desingularization $F_2\to X$ in his notation)). See Lemma 4 for an alternative proof for the quadric cone.

Hence, from now on, we assume $d\ge 3$.

For any positive integer t, set $a_t:=\lfloor h^0(C,{\mathcal{O}}_C\left(t\right))/2\rfloor$ and $b_t:=h^0\left({\mathcal{O}}_C\left(t\right)\right)-2a_t$. Note that $b_t\in \{0,1\}$ and $h^0(C,{\mathcal{O}}_C\left(t\right))=2a_t+b_t$.

In steps (a) and (b), we assume $t\ge 5$ and that Proposition 1 is true for integers $t-1$ and $t-2$. We address cases $t=2,3,4$ using the same ideas, but addressing them before the general case would not only make longer the proof, but obscure why a low t is difficult and why (for higher t) we can prove non-defectivity even if $t-1$ or $t-2$ is defective in a controlled way.

Claim 1: If $t\ge 5$, we have $a_t+2b_t\le h^0(C,{\mathcal{O}}_C(t-1))$.

Proof of Claim 1: Assume $h^0(C,{\mathcal{O}}_C(t-1))\le a_t+2b_t-1$. Recall that $h^0(C,{\mathcal{O}}_C\left(t\right))=2a_t+b_t$. Hence, $a_t-b_t\le h^0(C,{\mathcal{O}}_C\left(t\right))-h^0(C,{\mathcal{O}}_C(t-1))-1$. First, assume $t\ge d+2$. We have $h^0(C,{\mathcal{O}}_C\left(t\right))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+2}{2}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{t-d+2}{2}}\right)$ and $h^0(C,{\mathcal{O}}_C(t-1))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+1}{2}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{t-d+1}{2}}\right)$, and hence, $h^0(C,{\mathcal{O}}_C\left(t\right))-h^0(C,{\mathcal{O}}_C(t-1))=t+1-(t-d+1)=d$. Since $b_t\le 1$, we obtain $a_t\le d+1$, and hence, $2a_t+1<\left({\displaystyle \genfrac{}{}{0pt}{}{t+2}{2}}\right)$, a contradiction for $t\ge d+2$. Now, assume $t=d+1$. We obtaib $h^0(C,{\mathcal{O}}_C\left(t\right))-h^0(C,{\mathcal{O}}_C(t-1))=d+1$, concluding this case in the same way. Now, assume $t=d$. We obtain $h^0(C,{\mathcal{O}}_C\left(t\right))-h^0(C,{\mathcal{O}}_C(t-1))=d+2$, concluding this case. Now, assume $t\le d-1$. We have $h^0(C,{\mathcal{O}}_C\left(t\right))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+2}{2}}\right)$ and $h^0(C,{\mathcal{O}}_C(t-1))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+1}{2}}\right)$. Hence, $2\left({\displaystyle \genfrac{}{}{0pt}{}{t+1}{2}}\right)\le \left({\displaystyle \genfrac{}{}{0pt}{}{t+2}{2}}\right)+3b_3-2$. Since $b_3\le 1$, we obtain $(t+1)t\le (t+2)(t+1)/2+1$, which is false for $t\ge 5$. □

Suppose you want to prove that the zero-dimensional scheme $A\left(s\right)$ is not defective in degree t.

(a)

Assume $t\ge 5$ and $s\ge a+b$.

Fix a general $(S^{\prime },S^{\prime \prime })\in S(C,u_t)\times S(C,b_t)$. If $b_t=1$, we also apply the Differential Horace Lemma to the point $S^{\prime \prime }$. Since C is an irreducible curve, it is not secant defective ([3], Corollary 1.10) and hence, $h^i(C,{\mathcal{I}}_{(2S,C)\cup S^{\prime }}\left(t\right))=0$, $i=0,1$. Set $E:=S\cup (2S^{\prime },C)$. In using the Differential Horace Lemma to prove that $h^0\left({\mathcal{I}}_{A\left(s\right)}\left(t\right)\right)=0$ (resp. $h^1\left({\mathcal{I}}_{A\left(s\right)}\left(t\right)\right)=0$), it is sufficient to prove that $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)\cup E}(t-1)\right)=0$ (resp. $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-1)\right)=0$).

Since we are assuming that ${\mathcal{O}}_X(t-1)$ and ${\mathcal{O}}_X(t-2)$ are not defective, either $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-1)\right)=0$ or $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-1)\right)=0$ and either $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$ or $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$.

If $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-1)\right)=0$, then $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)\cup E}(t-1)\right)=0$, and hence, $h^0\left({\mathcal{I}}_{A\left(s\right)}\left(t\right)\right)=0$ in this case. Thus, we may assume $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-1)\right)=0$.

Assume for the moment that $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$. Since $S\cup (2S^{\prime },C)$ is a general union of a finite set and a tangent vector, it has the expected postulation (Remark 3). Claim 1 gives $h^1(C,{\mathcal{I}}_{E,C}(t-1))=0$. Thus, the residual exact sequence of C gives $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-1)\right)=0$, concluding the proof in this case. Thus, we have $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$. Hence, the restriction map $\rho :h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-1)\right)\to H^0(C,{\mathcal{O}}_C(t-1))$ is injective. Thus, to prove that either $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)\cup E}(t-1)\right)=0$ or $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-1)\right)=0$, it is sufficient to use that $dim\mathrm{Im}\left(\rho \right)(-E)=max\{0,dim\mathrm{Im}\left(\rho \right)-a_t-2b_t\}$ by [34] (Remark 1).

(b)

Assume $t\ge 5$ and $s\le a+b-1$.

Since $2a+b=h^0(C,{\mathcal{O}}_C\left(t\right))$ and $b\in \{0,1\}$, we obtain $2s\le h^0(C,{\mathcal{O}}_C\left(t\right))-2+b$. We have $3s\ge h^0\left({\mathcal{O}}_X\left(t\right)\right)-2$. Hence, $2h^0\left({\mathcal{O}}_X\left(t\right)\right)-4\le 3h^0(C,{\mathcal{O}}_C\left(t\right))-6+3b$. First, assume $t\ge d+1$. We obtain $2\left({\displaystyle \genfrac{}{}{0pt}{}{t+3}{3}}\right)-2\left({\displaystyle \genfrac{}{}{0pt}{}{t-d+3}{3}}\right)\le 3\left({\displaystyle \genfrac{}{}{0pt}{}{t+2}{2}}\right)-3\left({\displaystyle \genfrac{}{}{0pt}{}{t+2-d}{2}}\right)-2+3b$. Set $\psi \left(y\right)=2(y+3)(y+2)(y+1)-9(y+2)(y+1)-2(y-d+3)(y-d+2)(y-d+1)+9(y-d+2)(y-d+1)+12-18b$. We have $\psi (d+1)=2(d+4)(d+3)(d+2)-9(d+3)(d+2)+12-18b>0$, because $d\ge 3$ and $b\le 1$. For $y\ge d+2$, we have $\psi (y+1)-\psi \left(y\right)=12(y+3)(y+2)-18(y+2)>0$, because $d\ge 3$. Hence, $2\left({\displaystyle \genfrac{}{}{0pt}{}{t+3}{3}}\right)-2\left({\displaystyle \genfrac{}{}{0pt}{}{t-d+3}{3}}\right)\le 3\left({\displaystyle \genfrac{}{}{0pt}{}{t+2}{2}}\right)-3\left({\displaystyle \genfrac{}{}{0pt}{}{t+2-d}{2}}\right)-2+3b$, a contradiction. Now, assume $t<d$. We obtain $2\left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)\ge 3\left({\displaystyle \genfrac{}{}{0pt}{}{d+2}{2}}\right)-2+3b$, contradicting the assumptions that $d>t\ge 5$ and $b\le 1$. Now, assume $t=d$. We obtain $h^0\left({\mathcal{O}}_X\left(d\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)-1$ and $h^0(C,{\mathcal{O}}_C\left(d\right))=\left({\displaystyle \genfrac{}{}{0pt}{}{d+2}{2}}\right)-1$, concluding this step by contradiction.

(c)

Assume $t=2$ and $d>2$.

We have $h^0\left({\mathcal{O}}_X\left(2\right)\right)=10$ and $h^0(C,{\mathcal{O}}_C\left(2\right))=6$. Hence, $u_2=3$, $u_2^{\prime }=4$, $a_2=3$, and $b_2=0$. The proof of step (a) gives $h^i(C,{\mathcal{I}}_{(2S,C}\left(2\right))=0$, $i=0,1$. Obviously, $h^1\left({\mathcal{I}}_S\left(1\right)\right)=0$. Hence, $h^1\left({\mathcal{I}}_{2S}\left(2\right)\right)=0$ and $h^0\left({\mathcal{I}}_{2S}\left(2\right)\right)=1$. Hence, $h^0\left({\mathcal{I}}_{A\left(z\right)}\left(2\right)\right)=0$ for all $z\ge 4$. Hence, this case is not defective.

(d)

Assume $t=3$ and $d>2$.

We have $h^0\left({\mathcal{O}}_X\left(3\right)\right)=19$ if $d=3$ and $h^0\left({\mathcal{O}}_X\left(3\right)\right)=20$ if $d>3$. In both cases, $u_3=6$ and $u_3^{\prime }=7$. Take $s\in \{6,7\}$. We have $a_3=4$ and $b_3=1$ if $d=3$ and $a_4=5$ and $b_4=0$ if $d>3$. We have $h^0(C,{\mathcal{O}}_C\left(2\right))=6$, and hence, the inequality in Claim 1 is true in this case. Step (c) gives $h^1\left({\mathcal{I}}_{A(s-5)}\left(2\right)\right)=0$. If $s=6$ (resp. $s=7$), we have $h^1\left({\mathcal{I}}_{A(s-5)}\left(1\right)\right)=0$ (resp. $h^0\left({\mathcal{I}}_{A(s-5}\left(1\right)\right)=0$). Hence, we may repeat the proof of step (a).

(e)

Assume $t=4$ and $d>2$.

To conclude the proof of Proposition 1, it is sufficient to prove that ${\mathcal{O}}_X\left(4\right)$ is not secant defective. We have $h^0\left({\mathcal{O}}_X\left(4\right)\right)=35$ if $d\ge 5$, $h^0\left({\mathcal{O}}_X\left(4\right)\right)=34$ if $d=4$, and $h^0\left({\mathcal{O}}_X\left(4\right)\right)=31$ if $d=3$. Hence, $u_4=11$ and $u_4^{\prime }=12$ if $d\ge 4$ and $u_4=10$ and $u_4^{\prime }=11$ if $d=3$. Take $s\in \{u_4,u_4^{\prime }\}$. First, assume $d\ge 5$. We have $a_4=7$ and $b_4=1$, and hence, $a_4+2b_4\le 10=h^0(C,{\mathcal{O}}_C\left(3\right))$. Step (d) gives $h^1\left({\mathcal{I}}_{A(s-8)}\left(3\right)\right)=0$. Step (c) gives $h^1\left({\mathcal{I}}_{A(u_3-8)}\left(2\right)\right)=0$ and $h^0\left({\mathcal{I}}_{A(u_3^{\prime }-8)}\left(2\right)\right)=0$. Hence, the proof of step (a) works in this case. Now, assume $d=4$. We have $a_4=7$ and $b_4=0$. Hence, $a_4+2b_4\le 10=h^0(C,{\mathcal{O}}_C\left(3\right))$. Steps (c) and (d) give $h^1\left({\mathcal{I}}_{A(s-7)}\left(3\right)\right)=0$ and $h^0\left({\mathcal{I}}_{A(s-7)}\left(2\right)\right)=0$. Hence, the proof of step (a) works in this case. Now, assume $d=3$. We have $a_4=6$ and $b_4=0$. Hence, $a_4+2b_4\le 9=h^0(C,{\mathcal{O}}_C\left(3\right))=0$. Steps (c) and (d) give $h^1\left({\mathcal{I}}_{A(s-6)}\left(3\right)\right)=0$ and $h^0\left({\mathcal{I}}_{A(s-6)}\left(2\right)\right)=0$. Hence, the proof of step (a) works in this case, concluding the proof of the proposition. □

Lemma 4.

Let $X\subset {\mathbb{P}}^3$ be a quadric cone. We have $h^0\left({\mathcal{I}}_{2S}\left(t\right)\right)>0$ and $h^1\left({\mathcal{I}}_{2S}\left(t\right)\right)>0$ for a general $S\in S(X,s)$ if and only if $t=2$ and $s=3$. If $(t,s)=(2,3)$, then $h^0\left({\mathcal{I}}_{2S}\left(t\right)\right)=h^1\left({\mathcal{I}}_{2S}\left(t\right)\right)=1$.

Proof. 

Since the union of two different planes spans ${\mathbb{P}}^3$, $h^0\left({\mathcal{I}}_{A\left(2\right)}\left(1\right)\right)=0$. Since $h^1\left({\mathcal{I}}_{A\left(1\right)}\left(1\right)\right)=0$, ${\mathcal{O}}_X\left(1\right)$ is not defective. Hence, from now on, we assume $t\ge 2$.

We have $h^0\left({\mathcal{O}}_X\left(t\right)\right)={(t+1)}^2$ for all $t\ge 1$. Set $e:=\lfloor {(t+1)}^2)/3\rfloor$ and $g:=\lceil {(t+1)}^2/3\rceil$. It is sufficient to prove that the e-th and g-th secant varieties of the embedding $Y\subset {\mathbb{P}}^{{(t+1)}^2-1}$ of X induced by $|{\mathcal{O}}_X\left(t\right)|$ have the expected dimension, unless $t=2$. By the Terracini lemma, this is equivalent to proving that $h^1\left({\mathcal{I}}_{A\left(e\right)}\left(t\right)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(g\right)}\left(t\right)\right)=0$ if $t\ne 2$, while $e=g=3$ for $t=2$ and $h^1\left({\mathcal{I}}_{A\left(3\right)}\left(2\right)\right)=h^0\left({\mathcal{I}}_{A\left(3\right)}\left(2\right)\right)=1$. Take $x\in \{e,g\}$ and assume that ${\sigma }_x\left(Y\right)$ does not have the expected dimension. Take a general $M\in |{\mathcal{I}}_{A\left(x\right)}\left(t\right)|$ and set $S:=A{\left(x\right)}_{\mathrm{red}}$. Let $\Sigma$ be the union of the irreducible components of $\mathrm{Sing}\left(M\right)$ containing at least one of the elements of S. By [33] Theorem 1.2, we have $S\subset \Sigma$, and each irreducible component of $\Sigma$ has a positive dimension. Since $dimM=1$, each irreducible component $\Sigma$ occurs with multiplicity of at least 2 in M. Moreover, as noted several times in [33], the generality of S implies that either $\Sigma$ is irreducible or each point of S is contained in a different irreducible component of $\Sigma$. Since X is singular, to show that this is false, it is easier to work (as in [32]) on the minimal desingularization $\pi :F_2\to X$ of the quadric cone. We take as a basis of $\mathrm{Pic}\left(F_2\right)$ the irreducible curve h with negative self-intersection and a fiber f of the ruling of $F_2$. In [32] §2, we have ${\Gamma }_2=h$, $F=f$ and $H=h+2f$. The linear system $|{\mathcal{O}}_X\left(t\right)|$ is induced by $|{\mathcal{O}}_{F_2}(th+2tf)|$. The linear systems with integral elements are $|{\mathcal{O}}_{F_2}\left(h\right)|$ (with $|{\mathcal{O}}_{F_2}\left(h\right)|=\left\{h\right\}$), $|{\mathcal{O}}_{F_2}\left(f\right)|$ (with $h^0(F_2,{\mathcal{O}}_{F_2}\left(f\right))=2$ and the elements of $|{\mathcal{O}}_{F_2}\left(f\right)|$ mapped to the lines of the cone X), and the linear system $|uh+bf|$, where $u\ge 1$ and $b\ge 2u$ ([32], Proposition 2.2). In the latter case, we have $h^0(F_2,uh+bf)=(u+1)(2b+2-2u)/2=(u+1)(b-u+1)$ ([32], Proposition 2.3). Let $B\left(x\right)\subset F_2$ be a general union of x double points of $F_2$. Set $S^{\prime }:=B{\left(x\right)}_{\mathrm{red}}$. The generality of $S^{\prime }$ gives $S^{\prime }\cap h=\varnothing$ and that each element of $|{\mathcal{O}}_{F_2}\left(f\right)|$ contains at most one element of $S^{\prime }$. Let $M^{\prime }\in \left|{\mathcal{I}}_{B\left(x\right)}(th+2tf)\right|$ be the element corresponding to M and ${\Sigma }^{\prime }\subset F_2$ be the associated contact locus. Either $M^{\prime }=2{\Sigma }^{\prime }$ or $|M^{\prime }-2{\Sigma }^{\prime }|\ne \varnothing$. Take integers u and v such that ${\Sigma }^{\prime }\in \left|{\mathcal{O}}_{F_2}(uh+vf)\right|$:

(a) First assume that ${\Sigma }^{\prime }$ is irreducible. Since $S^{\prime }\cap h=\varnothing$ and each element of $|{\mathcal{O}}_{F_2}\left(f\right)|$ contains at most one element of $S^{\prime }$, $u>0$, and hence, $v\ge 2u$. We obtain $2u\le t$ and $2v\le 2t$. Since $S^{\prime }\subset {\Sigma }^{\prime }$ and $S^{\prime }$ is general in $F_2$, we obtain $h^0(F_2,{\mathcal{O}}_{F_2}(uh+vf))\ge x+1$, and hence, $h^0(F_2,{\mathcal{O}}_{F_2}\left(\lfloor t/2\rfloor \right)h+tf))\ge x+1$.

First, assume t is even. Hence, $h^0(F_2,{\mathcal{O}}_{F_2}\left(\lfloor t/2\rfloor \right)h+tf){)=(t/2+1)}^2$.

Assume $t=2$, and hence, $e=g=3$. We see that $|{\mathcal{I}}_{2S^{\prime }}(2h+4f)|=\left\{2D\right\}$ with $\left\{D\right\}=|{\mathcal{I}}_{S^{\prime }}(h+2f)|$. The set $S^{\prime }$ and the Terracini lemma show that this case is defective with $h^0\left({\mathcal{I}}_{2S}\left(t\right)\right)=h^1\left({\mathcal{I}}_{2S}\left(t\right)\right)=1$.

Assume t even and $t\ge 4$. We obtain $\lfloor {(t+1)}^2/3\rfloor \le -1+{(t/2+1)}^2$, a contradiction.

Now, assume $t\ge 3$ and odd. We have $h^0(F_2,{\mathcal{O}}_{F_2}\left(\lfloor t/2\rfloor \right)h+tf))={\displaystyle \frac{t+1}{2}}(t-{\displaystyle \frac{t-1}{2}}+1)=(t+1)(t+3)/4<e\le x\le t$, a contradiction.

(b) Now, assume that each point of $S^{\prime }$ is contained in a different irreducible component of ${\Sigma }^{\prime }$. We obtain $v\ge x$, and hence, $x\le t$. Thus, $\lfloor {(t+1)}^2/3\rfloor \le x\le t$, contradicting the assumption that $t\ge 3$. □

4. Proof of Theorem 2 for $\mathit{n}=\mathbf{4}$

Theorem 3.

Fix integers $d\ge 2$, $t\ge 1$, and $s\ge 1$. Let $X\subset {\mathbb{P}}^4$ be an integral degree d hypersurface. Let $A\left(s\right)\subset X$ be a general union of s double points. Then, either $h^0\left({\mathcal{I}}_{A\left(s\right)}\left(t\right)\right)=0$ or $h^1\left({\mathcal{I}}_{A\left(s\right)}\left(t\right)\right)=0$, unless $t=2$ and $s\in \{3,4\}$.

Proof. 

We have $h^0\left({\mathcal{O}}_X\left(t\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{t+4}{4}}\right)$ if $t<d$ and $h^0\left({\mathcal{O}}_X\left(t\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{t+4}{4}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{t-d+4}{4}}\right)$ if $t\ge d$. Set $u_t:=\lfloor h^0\left({\mathcal{O}}_X\left(t\right)\right)/4\rfloor$, ${ϵ}_t:=h^0\left({\mathcal{O}}_X\left(t\right)\right)-4u_t$, and $u_t^{\prime }:=\lceil h^0\left({\mathcal{O}}_X\left(t\right)\right)/4\rceil$. By Remark 1, to prove that ${\mathcal{O}}_X\left(t\right)$ is not defective, it is sufficient to prove that $h^1\left({\mathcal{I}}_{A\left(u_t\right)}\left(t\right)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(u_t^{\prime }\right)}\left(t\right)\right)=0$. Take a general $(S,S^{\prime })\in S(C,a_t)\times S(C,b_t)$. In particular, $S\cap S^{\prime }=\varnothing$. If $t\ge 3$, Proposition 1 gives $h^i(C,{\mathcal{I}}_{(2S,C)\cup S^{\prime }}\left(t\right))=0$.

Take a general hyperplane H and set $C:=H\cap X$. The theorem of Bertini gives that $C\subset {\mathbb{P}}^3$ is an irreducible degree d surface. Set $a_t:=\lfloor h^0(C,{\mathcal{O}}_C\left(t\right))/3\rfloor$ and $b_t=h^0(C,{\mathcal{O}}_C\left(t\right))-3a_t$. We have $b_t\in \{0,1,2\}$.

(a) Assume $d=2$. If X is a smooth quadric, it is sufficient to quote [31], but we allow the singular case. For all integers $t\ge 3$, we have $h^0\left({\mathcal{O}}_X\left(t\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{t+4}{4}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{t+2}{4}}\right)=(t+2)(t+1)(2t+3)/6$. For all $t\ge 0$, we have $h^0(C,{\mathcal{O}}_C\left(t\right))={(t+1)}^2$.

(a1) Assume $t=2$. Since any four points of X are contained in a hyperplane, $h^0\left({\mathcal{I}}_{A\left(4\right)}\left(2\right)\right)>0$, and hence, $h^1\left({\mathcal{I}}_{A\left(4\right)}\left(2\right)\right)\ge 0$. The residual exact sequence of hyperplane section $C_1$ gives that $\left\{2C_1\right\}=\left|{\mathcal{I}}_{A\left(4\right)}\left(2\right)\right|$. Hence, $h^1\left({\mathcal{I}}_{A\left(4\right)}\left(2\right)\right)=3$ and $h^0\left({\mathcal{I}}_{A\left(z\right)}\left(2\right)\right)=0$ for all $z>4$. Take a general $E\in S(X,3)$ and call $C_1$ a general hyperplane section of X containing E. Since $h^1(C_1,{\mathcal{I}}_{2E,C_1}\left(2\right))=1$, Remark 1 gives $h^1\left({\mathcal{I}}_{A\left(3\right)}\left(2\right)\right)>0$. The residual exact sequence of $C_1$ gives $h^1\left({\mathcal{I}}_{A\left(3\right)}\left(2\right)\right)=1$.

Since $h^1(C,{\mathcal{I}}_{2A,C}\left(2\right))=0$ for a general $A\subset C$ with $\#A=2$ (Proposition 1), the residual exact sequence of C gives $h^1\left({\mathcal{I}}_{A\left(2\right)}\left(2\right)\right)=0$.

(a2) Assume $t\ge 5$. We also assume for the moment that ${\mathcal{O}}_X(t-1)$ and ${\mathcal{O}}_X(t-2)$ are not defective. Fix $s\in \{u_t,u_t^{\prime }\}$.

Claim 1: We have $s\ge a_t+b_t$.

Proof of Claim 1: Assume $s\le a_t+b_t-1$. Recall that $b_t\in \{0,1,2\}$. Since $3a_t+b_t={(t+1)}^2$ and $4s\ge (t+2)(t+1)(2t+3)/6-3$, we obtain $(t+2)(t+1)(2t+3)-18\le 8{(t+1)}^2-16b_t-24$. This inequality is false for all $t\ge 5$. □

Proposition 1 gives $h^i(C,{\mathcal{I}}_{(2S,C)\cup S^{\prime }}\left(t\right))=0$, $i=0,1$. With the Differential Horace Lemma applied to all points of $S^{\prime }$ to prove that $h^0\left({\mathcal{I}}_{A\left(s\right)}\left(t\right)\right)=0$ (resp. $h^1\left({\mathcal{I}}_{A\left(s\right)}\left(t\right)\right)=0$), it is sufficient to prove that $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)\cup S\cup (2S^{\prime },C)}(t-1)\right)=0$ (resp. $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)\cup S\cup (2S^{\prime },C)}(t-1)\right)=0$). By assumption, either $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-1)\right)=0$ or $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-1)\right)=0$ and either $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$ or $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$. If $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$, then Proposition 1 and the residual exact sequence of C give $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)\cup S\cup (2S^{\prime },C)}(t-1)\right)=0$. Hence, we may assume $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$. □

Claim 2: We have $s-a_t\le u_{t-1}$.

Proof of Claim 2: Assume $s-a_t\ge u_{t-1}+1$. Since $4s\le (t+2)(t+1)(2t+3)/6$, $4u_{t-1}\le (t+1)t(2t+1)/6-3$, and $3a_t\ge {(t+1)}^2-2$, we obtain $(t+2)(t+1)(2t+3)-8{(t+1)}^2+16\ge (t+1)t(2t+1)-18+24$. This inequality is false for all $t\ge 5$ because $(t+2)(2t+3)-8(t+1)-t(2t+1)=-2t-2\le -2$ for all $t\ge 5$. □

Since $b_t=\#S^{\prime }\le 2$, any two points of X are contained in a hyperplane section of X, and C is general, $A(s-a_t-b_t)\cup 2S^{\prime }$ has the Hilbert function of a general union of the $s-a_t$ 2-points of X. Since ${\mathcal{O}}_X(t-1)$ is not defective, we obtain $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)\cup 2S^{\prime }}(t-1)\right)=0$. Remark 1 gives $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)\cup (2S^{\prime },C)}(t-1)\right)=0$. Since $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$ and S is general in C, Lemma 3 gives that either $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)\cup S\cup (2S^{\prime },C)}(t-1)\right)=0$ or $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)\cup S\cup (2S^{\prime },C)}(t-1)\right)=0$.

(a3) Assume $t=3$. We have $u_3=7$, ${ϵ}_3=2$, $u_3^{\prime }=8$, $a_3=5$, and $b_5=1$. Take $s\in \{7,8\}$. We have $h^1\left({\mathcal{I}}_{A(s-6)}\left(2\right)\right)=0$. If $s=7$ (resp. $s=8$), we have $h^1\left({\mathcal{I}}_{A(s-6)}\left(1\right)\right)=0$ (resp. $h^0\left({\mathcal{I}}_{A(s-6)}\left(1\right)\right)=0$). Hence, we may use the proof of step (a2).

(a4) Assume $t=4$. We have $u_4=13$, ${ϵ}_4=3$, $u_4^{\prime }=14$, $a_4=6$, and $b_4=1$. Take $s\in \{13,14\}$. By step (a3), we have $h^1\left({\mathcal{I}}_{A(s-7)}\left(3\right)\right)=0$. By step (a1), we have $h^0\left({\mathcal{I}}_{A(s-7)}\left(2\right)\right)=0$. Hence, we may repeat the proof of step (a2).

(b) Assume $d\ge 3$. Take $s\in \{u_t,u_t^{\prime }\}$.

Observation 1: Take $(d,t)$ such that ${\mathcal{O}}_C\left(t\right)$ is not defective. We have $h^i(C,{\mathcal{I}}_{(2S,C)\cup S^{\prime }}\left(t\right))=0$, $i=0,1$.

(b1) Assume $t=2$. Since $h^0\left({\mathcal{O}}_X\left(2\right)\right)=15$, $u_2=3$, ${ϵ}_2=3$, and $u_2^{\prime }=4$. Since any four points of X are contained in a hyperplane, $A\left(4\right)$ is contained in a general hyperplane section, $C_1$. Proposition 1 and the residual exact sequence of $C_1$ give $|{\mathcal{I}}_{A\left(4\right)}\left(2\right)|=\left\{2C_1\right\}$. Hence, $h^1\left({\mathcal{I}}_{A\left(4\right)}\left(2\right)\right)=2$ and $h^0\left({\mathcal{I}}_{A\left(5\right)}\left(2\right)\right)=0$. Taking three general 2-points with reduction in C and using Proposition 1 and the residual exact sequence of C, we obtain $h^1\left({\mathcal{I}}_{A\left(3\right)}\left(2\right)\right)=0$.

(b2) Assume $t=3$. We have $h^0\left({\mathcal{O}}_X\left(3\right)\right)=34$ if $d=3$ and $h^0\left({\mathcal{O}}_X\left(3\right)\right)=35$ if $d>3$. Hence, $u_3=8$, $u_3^{\prime }=9$, ${ϵ}_3=2$ if $d=3$, and ${ϵ}_2=3$ if $d>3$. We have $a_3=6$. If $d=3$ (resp. $d>3$), then $b_3=1$ (resp. $b_3=2$).

First, assume $s=8$. Since $h^1(C,{\mathcal{I}}_{(2S,C)}\left(3\right))=0$, to prove that $h^1\left({\mathcal{I}}_{A\left(8\right)}\left(2\right)\right)=0$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{A\left(2\right)\cup S}\left(2\right)\right)=0$. Step (a) gives $h^1\left({\mathcal{I}}_{A\left(2\right)}\left(2\right)\right)=0$. Obviously, $h^0\left({\mathcal{I}}_{A\left(2\right)}\left(1\right)\right)=0$. We have $\#S+deg\left(A\left(2\right)\right)=14\le h^0\left({\mathcal{O}}_X\left(2\right)\right)$. Hence, to obtain $h^1\left({\mathcal{I}}_{A\left(2\right)\cup S}\left(2\right)\right)=0$, it is sufficient to use Lemma 3.

Now, assume $s=9$. Take a general $S_1\subset C$ such that $\#S_1=7$. Proposition 1 gives $h^0(C,{\mathcal{I}}_{(2S_1,C)}\left(3\right))=0$. Hence, the residual exact sequence of C shows that to prove that $h^0\left({\mathcal{I}}_{A\left(9\right)}\left(3\right)\right)=0$, it is sufficient to prove that $h^0\left({\mathcal{I}}_{A\left(2\right)\cup S_1}\left(2\right)\right)=0$. Step (b1) gives $h^1\left({\mathcal{I}}_{A\left(2\right)}\left(2\right)\right)=0$, and hence, $h^0\left({\mathcal{I}}_{A\left(2\right)}\left(2\right)\right)=7=\#S_1$. Obviously, $h^0\left({\mathcal{I}}_{A\left(2\right)}\left(1\right)\right)=0$. Thus, it is sufficient to use Lemma 3.

(b3) Assume $t=4$. We have $h^0\left({\mathcal{O}}_X\left(4\right)\right)=70$ if $d>4$, $h^0\left({\mathcal{O}}_X\left(4\right)\right)=69$ if $d=4$, and $h^0\left({\mathcal{O}}_X\left(4\right)\right)=65$ if $d=3$. Hence, $u_4=17$, ${ϵ}_4=2$, $u_4^{\prime }=18$ if $d>4$, $u_4=17$, ${ϵ}_4=1$, $u_4^{\prime }=18$ if $d=4$, and $u_4=16$, ${ϵ}_4=1$, and $u_4^{\prime }=17$ if $d=3$. We have $h^0(C,{\mathcal{O}}_C\left(4\right))=35$ if $d>4$, $h^0(C,{\mathcal{O}}_C\left(4\right))=34$ if $d=4$, and $h^0(C,{\mathcal{O}}_C\left(4\right))=31$ if $d=3$. Hence, $a_4=11$ and $b_2=2$ if $d>4$, $a_4=11$ and $b_2=1$ if $d=4$, and $(a_4,b_4)=(10,1)$ if $d=3$.

First, assume $d\ge 4$ and $s=17$. We take a general $S_1\subset C$ such that $\#S_1=11$. Proposition 1 gives $h^1(C,{\mathcal{I}}_{(2S_1,C)}\left(4\right))=0$. Hence, the residual exact sequence of C shows that to prove that $h^1\left({\mathcal{I}}_{A\left(17\right)}\left(4\right)\right)=0$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{A\left(6\right)\cup S_1}\left(3\right)\right)=0$. Steps (b1) and (b2) give $h^0\left({\mathcal{I}}_{A\left(6\right)}\left(2\right)\right)=0$ and $h^1\left({\mathcal{I}}_{A\left(6\right)}\left(3\right)\right)=0$. Hence, $h^0\left({\mathcal{I}}_{A\left(6\right)}\left(3\right)\right)=h^0\left({\mathcal{O}}_X\left(3\right)\right)-24=11=\#S_1$. Apply Lemma 1.

Now, assume $d\ge 4$ and $s=18$. We take a general $S_1\subset C$ such that $\#S_2=12$. Proposition 1 gives $h^0(C,{\mathcal{I}}_{(2S_2,C)}\left(4\right))=0$. Hence, the residual exact sequence of C shows that to prove that $h^0\left({\mathcal{I}}_{A\left(18\right)}\left(4\right)\right)=0$, it is sufficient to prove that $h^0\left({\mathcal{I}}_{A\left(6\right)\cup S_2}\left(3\right)\right)=0$. Steps (b1) and (b2) give $h^0\left({\mathcal{I}}_{A\left(6\right)}\left(2\right)\right)=0$ and $h^1\left({\mathcal{I}}_{A\left(6\right)}\left(3\right)\right)=0$. Hence, $h^0\left({\mathcal{I}}_{A\left(6\right)}\left(3\right)\right)=h^0\left({\mathcal{O}}_X\left(3\right)\right)-24\le \#S_2$. Apply Lemma 3.

Now, assume $d=3$ and $s=16$. We take a general $S_3\subset C$ such that $\#S_3=10$. As above, we obtain $h^1\left({\mathcal{I}}_{A\left(16\right)}\left(4\right)\right)=0$ because ${ϵ}_4=b_4$. Since $h^0\left({\mathcal{I}}_{A\left(16\right)}\left(4\right)\right)=1$, we have $h^0\left({\mathcal{I}}_{A\left(17\right)}\left(4\right)\right)=10$.

(b4) Assume $t\ge 5$ and that ${\mathcal{O}}_X(t-1)$ and ${\mathcal{O}}_X(t-2)$ are not defective.

Claim 3: We have $a_t+3b_t\le h^0(C,{\mathcal{O}}_C(t-1))$.

Proof of Claim 3: Since $3a_t+b_t=h^0(C,{\mathcal{O}}_C\left(t\right))$, it is sufficient to prove that $3h^0(C,{\mathcal{O}}_C(t-1))\ge h^0(C,{\mathcal{O}}_C\left(t\right))+8b_t$. Recall that $b_t\le 2$. Let M be a general hyperplane section of C. The set M is a degree d plane curve. Since $h^1\left({\mathcal{O}}_C(t-1)\right)=0$, a standard exact sequence gives $h^0(C,{\mathcal{O}}_C\left(t\right))=h^0\left({\mathcal{O}}_C(t-1)\right)+h^0(M,{\mathcal{O}}_M\left(t\right))$. Hence, it is sufficient to prove that $2h^0\left({\mathcal{O}}_C(t-1)\right)\ge h^0(M,{\mathcal{O}}_M\left(t\right))+16$.

First, assume $t\le d$. We have $h^0(C,{\mathcal{O}}_C(t-1))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+2}{3}}\right)$ and $h^0(M,{\mathcal{O}}_M\left(t\right))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+2}{2}}\right)-ϵ$ with $ϵ=1$ if $t=d$ and $ϵ=0$ if $t<d$. The inequality $2(t+2)(t+1)t/6\ge (t+2)(t+1)/2+16$ is equivalent to $2(t+2)(t+1)t\ge 3(t+2)(t+1)+96$, i.e., to $(t+2)(t+1)(2t-3)\ge 96$, which is true for all $t\ge 5$.

Now, assume $t\ge d+1$. We obtain $h^0(C,{\mathcal{O}}_C(t-1))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+2}{3}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{t-d+2}{3}}\right)$ and $h^0(M,{\mathcal{O}}_M\left(t\right))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+2}{2}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{t-d+2}{2}}\right)$. Set $\varphi (t,d):=2h^0(C,{\mathcal{O}}_C(t-1))-h^0(M,{\mathcal{O}}_M\left(t\right))-16$. We have $\varphi (d+1,d)=2\left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)-(\left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{2}}\right)-1)-16=2(d+3)(d+2)(d+1)/6-(d+3)(d+2)/2-15=(d+3)(d+2)(2d-1)/6-15$. Since $d\ge 3$, we have $\varphi (d+1,d)\ge 0$, concluding this case. Now, assume $t\ge d+2$. To conclude the proof for t, it is sufficient to prove that $\varphi (t,d)\ge \varphi (t-1,d)$. We have $h^0(C,{\mathcal{O}}_C(t-1))-h^0(C,{\mathcal{O}}_C(t-2))=h^0(M,{\mathcal{O}}_M(t-1))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+1}{2}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{t-d+1}{2}}\right)$ and $h^0(M,{\mathcal{O}}_M\left(t\right))-h^0(M,{\mathcal{O}}_M(t-1))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+2}{2}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{t-d+2}{2}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{t+1}{2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{t-d+1}{2}}\right)=t+2-(t-d+2)=d$. Hence, it is sufficient to prove that $(t+1)t-(t-d+1)(t-d)\ge d$. We have $(t+1)t-(t-d+1)(t-d)=t^2+t-t^2+td+td-d^2-t+d=2td-d^2+d$, which is obviously at least d since $t\ge d$. □

Claim 4: The theorem is true for the triple $(d,t,s)$ if $s\le a_t+b_t-1$.

Proof of Claim 4: Assume $s\le a_t+b_t-1$. Set $x:=min\{s,a_d\}$ and $y:=s-x$. Take a general $(S_1,S_2)\in S(C,x)\times S(C,y)$ and set $F:=2S_1$. Proposition 1 gives $h^1(C,{\mathcal{I}}_{(2S_1,C)\cup S_2}\left(t\right))=0$. Note that $x+y=s$. By the Differential Horace Lemma applied to the point of $S_2$ to prove that $h^1\left({\mathcal{I}}_{A\left(s\right)}\left(t\right)\right)=0$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{S_1\cup (2S_2,C)}(t-1)\right))=0$. Hence, it is sufficient to prove that $h^1(C,{\mathcal{I}}_{S_1\cup (2S_2,C)}(t-1))=0$. Since $y\le max\{0,b_t-1\}\le 1$ and ${\mathcal{O}}_C(t-1)$ is very ample, $h^1(C,{\mathcal{I}}_{S_1\cup (2S_2,C)}(t-1))=0$ if and only if $x+3y\le h^0(C,{\mathcal{O}}_C(t-1))$, which is true by Claim 3 and the definitions of x and y. □

By Claim 4, we may assume $s\ge a_t+b_t$. By Observation 1 and Proposition 1, we have $h^i(C,{\mathcal{I}}_{(2S,C)\cup S^{\prime }}\left(t\right))=0$, $i=0,1$. Hence, applying the Horace Differential Lemma to the points of $S^{\prime }$, we see that it is sufficient to prove that either $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)\cup S\cup (2S^{\prime },C)}(t-1)\right)=0$ or $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)\cup S\cup (2S^{\prime },C)}(t-1)\right)=0$.

Claim 5: We have $s-a_t\le u_{t-1}$.

Proof of Claim 5: Assume $s-a_t\ge u_{t-1}+1$. Since $b_t\le 2$ and ${\mathcal{O}}_X(t-1)$ is not defective, we obtain $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-1)\right)\le 7$ and $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$. Using Lemma 3 to obtain $h^0\left({\mathcal{I}}_{A\left(s\right)}\left(t\right)\right)=0$, it is sufficient to prove that $a_t\ge 7$. Assume $a_t\le 6$. We obtain $h^0\left({\mathcal{O}}_X\left(t\right)\right)\le 30$. Note that the integer $h^0\left({\mathcal{O}}_X\left(5\right)\right)$ is an increasing function of d. Since $h^0\left({\mathcal{O}}_X\left(t\right)\right)\ge h^0\left({\mathcal{O}}_X\left(5\right)\right)$, and for $d=1$, we have $h^0\left({\mathcal{O}}_X\left(5\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{9}{4}}\right)>30$, which is a contradiction. □

Claim 6: We have $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)\cup (2S^{\prime },C)}(t-1)\right)=0$.

Proof of Claim 6: It is sufficient to prove that $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)\cup 2S^{\prime }}(t-1)\right)=0$ (Remark 1). Since $b_t\le 2$, any two points of X are contained in a hyperplane section of X, and C is general, $A(s-a_t-b_t)\cup 2S^{\prime }$ and $A(s-a_t)$ have the same Hilbert function. Hence, Claim 5 and the assumption that ${\mathcal{O}}_X(t-1)$ is not defective prove Claim 6. □

Claim 6 and Proposition 1 give $h^1(C,{\mathcal{I}}_{S\cup (2S^{\prime },C)}(t-1))=0$.

Since ${\mathcal{O}}_X(t-2)$ is assumed to be non-defective, either $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$ or $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$. If $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$, then we apply Lemma 3. If $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$, we use the residual exact sequence of C and that $h^1(C,{\mathcal{I}}_{S\cup (2S^{\prime },C)}(t-1))=0$. □

5. Proof of Theorem 2 for $\mathit{n}=\mathbf{5}$

Lemma 5.

Fix an integral hypersurface $X\subset {\mathbb{P}}^n$, $n\ge 5$, of degree $d\ge 2$. We have $h^1\left({\mathcal{I}}_{A\left(s\right)}\left(2\right)\right)=0$ if and only if $s\le 2$, $h^1\left({\mathcal{I}}_{A\left(3\right)}\left(2\right)\right)\le 3$, $h^0\left({\mathcal{I}}_{A\left(n\right)}\left(2\right)\right)=1$ and $h^0\left({\mathcal{I}}_{A\left(s\right)}\left(2\right)\right)=0$ if and only if $s\ge n+1$.

Proof. 

By taking general hyperplane sections, Remark 1, an induction on n, Theorem 4, and some residual exact sequences, we obtain $h^1\left({\mathcal{I}}_{A\left(2\right)}\left(2\right)\right)=0$ and $h^1\left({\mathcal{I}}_{A\left(3\right)}\left(2\right)\right)>0$. To obtain $h^1\left({\mathcal{I}}_{A\left(3\right)}\left(2\right)\right)\le 3$, it is sufficient to use the case $n=4$ and induction on n. Since any n points of X are contained in a hyperplane section, $h^0\left({\mathcal{I}}_{A\left(n\right)}\left(2\right)\right)>0$. Using induction on n and some residual exact sequences, we obtain $h^0\left({\mathcal{I}}_{A\left(n\right)}\left(2\right)\right)=1$, and hence, $h^0\left({\mathcal{I}}_{A(n+1)}\left(2\right)\right)=0$. □

Lemma 6.

Let $X\subset {\mathbb{P}}^5$ be an integral hypersurface of degree $d\ge 3$. Then, $h^1\left({\mathcal{I}}_{A\left(10\right)}\left(3\right)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(12\right)}\left(3\right)\right)=0$.

Proof. 

Take a general hyperplane H and set $C:=H\cap X$. If $d\ge 4$, then $h^0\left({\mathcal{O}}_X\left(3\right)\right)=56$ and $h^0(C,{\mathcal{O}}_C\left(3\right))=36$. If $d=3$, then $h^0\left({\mathcal{O}}_X\left(3\right)\right)=55$ and $h^0(C,{\mathcal{O}}_C\left(3\right))=35$. Take a general $S\subset C$ such that $\#S=8$. We have $h^1(C,{\mathcal{I}}_{(2S,C),C}\left(3\right))=0$ (Theorem 3). By the Horace Lemma, to prove that $h^1\left({\mathcal{I}}_{A\left(10\right)}\left(4\right)\right)=0$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{A\left(2\right)\cup S}\left(2\right)\right)=0$. Obviously, $h^0\left({\mathcal{I}}_{A\left(2\right)}\left(1\right)\right)=0$. Lemma 5 gives $h^1\left({\mathcal{I}}_{A\left(2\right)}\left(2\right)\right)=0$, and hence, $h^0\left({\mathcal{I}}_{A\left(2\right)}\left(2\right)\right)=11$. Apply Lemma 3. Take a general $B\subset C$ such that $\#B=9$. We have $h^0(C,{\mathcal{I}}_{(2B,C)}\left(3\right))=0$ (Theorem 3). By the Horace Lemma, to prove that $h^0\left({\mathcal{I}}_{A\left(12\right)}\left(3\right)\right)=0$, it is sufficient to prove that $h^0\left({\mathcal{I}}_{A\left(3\right)\cup B}\left(2\right)\right)=0$. Obviously, $h^0\left({\mathcal{I}}_{A\left(3\right)}\left(1\right)\right)=0$. Lemma 6 gives $h^1\left({\mathcal{I}}_{A\left(3\right)}\left(2\right)\right)\le 3$, and hence, $h^0\left({\mathcal{I}}_{A\left(3\right)}\left(2\right)\right))=21-15+3\le \#B$. Apply Lemma 3. □

Lemma 7.

Let $X\subset {\mathbb{P}}^5$ be an integral quadric hypersurface. Then, $h^1\left({\mathcal{I}}_{A\left(9\right)}\left(3\right)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(10\right)}\left(3\right)\right)\le 1$.

Proof. 

We have $h^0\left({\mathcal{O}}_X\left(3\right)\right)=49$ and $h^0(C,{\mathcal{O}}_C\left(3\right))=30$. Hence, $u_3=9$, ${ϵ}_3=4$, $u_3^{\prime }=10$, $a_3=7$, and $b_3=2$. Take a general $(S,S^{\prime })\in S(7,C)\times S(2,C)$. We specialize $A\left(9\right)$ to $A\left(2\right)\cup 2S$. Since $h^1(C,{\mathcal{I}}_{(2S,C),C}\left(3\right))=0$ (Theorem 3), the semicontinuity theorem for cohomology and the Horace Lemma show that to prove that $h^1\left({\mathcal{I}}_{A\left(9\right)}\left(3\right)\right)=0$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{A\left(2\right)\cup S}\left(2\right)\right)=0$. Obviously, $h^0\left({\mathcal{I}}_{A\left(2\right)}\left(1\right)\right)=0$. Lemma 5 gives $h^1\left({\mathcal{I}}_{A\left(2\right)}\left(2\right)\right)=0$, and hence, $h^0\left({\mathcal{I}}_{A\left(2\right)}\left(2\right)\right)=20-10=10\ge \#S$. Apply Lemma 3.

We have $h^i(C,{\mathcal{I}}_{(S,C)\cup S^{\prime },C}\left(3\right))=0$, $i=0,1$ (Theorem 3). By the Differential Horace Lemma applied to $S^{\prime }$, to prove that $h^0\left({\mathcal{I}}_{A(10}\left(3\right)\right)\le 1$, it is sufficient to prove that $h^0\left({\mathcal{I}}_{A\left(1\right)\cup S\cup (2S^{\prime },C)}\left(2\right)\right)\le 1$. We have $h^0\left({\mathcal{I}}_{A\left(1\right)}\left(1\right)\right)=1$. Since $\#S+deg\left((2S^{\prime },C)\right)=15\ge 14=h^0\left({\mathcal{O}}_C\left(2\right)\right)=0$, the generality of S and Lemma 5 gives $h^0(C,{\mathcal{I}}_{S\cup (2S^{\prime },C)}\left(2\right))=0$. Apply the Horace Lemma. □

Lemma 8.

Let $X\subset {\mathbb{P}}^5$ be an integral quadric hypersurface. Then, $h^1\left({\mathcal{I}}_{A\left(21\right)}\left(4\right)\right)=h^0\left({\mathcal{I}}_{A\left(21\right)}\left(4\right)\right)\le 1$.

Proof. 

We have $h^0\left({\mathcal{O}}_X\left(4\right)\right)=105$, $h^0(C,{\mathcal{O}}_C\left(4\right))=55$, and $(u_4,{ϵ}_4,u_4^{\prime },a_4,b_4)=(21,0,21,13,3)$. We have $h^0\left({\mathcal{I}}_{A\left(21\right)}\left(4\right)\right)=h^0\left({\mathcal{I}}_{A\left(21\right)}\left(4\right)\right)$. It is sufficient to prove that $h^0\left({\mathcal{I}}_{A\left(21\right)}\left(4\right)\right)\le 1$. Take a general $(S_1,S_2)\in S(13,C)\times S(3,C)$. Theorem 3 gives $h^i(C,{\mathcal{I}}_{(2S_1,C)\cup S_2,C}\left(4\right))=0$, $i=0,1$. By the Differential Horace Lemma, it is sufficient to prove that $h^0\left({\mathcal{I}}_{A\left(5\right)\cup S_1\cup (2S_2,C)}\left(3\right)\right)=0$. We have $h^1\left({\mathcal{I}}_{A\left(8\right)}\left(3\right)\right)=0$ (Lemma 7), and hence, $h^1\left({\mathcal{I}}_{A\left(5\right)\cup (2S_2,C)}\left(3\right)\right)=0$. Hence, $h^0\left({\mathcal{I}}_{A\left(5\right)\cup (2S_2,C)}\left(3\right)\right)=49-25-12=8\le \#S_1$. Lemma 5 gives $h^0\left({\mathcal{I}}_{A\left(5\right)}\left(2\right)\right)=1$. Since $S_1$ is general in C, Lemma 3 concludes the proof. □

Theorem 4.

Let $X\subset {\mathbb{P}}^5$ be an integral hypersurface of degree $d\ge 2$. Assume $t\ge 4$ and $(t,d)\ne (4,2)$. Then, ${\mathcal{O}}_X\left(t\right)$ is not defective.

Proof. 

We have $h^0\left({\mathcal{O}}_X\left(t\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{t+5}{5}}\right)$ if $t<d$ and $h^0\left({\mathcal{O}}_X\left(t\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{t+5}{5}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{t-d+5}{5}}\right)$ if $t\ge d$. Set $u_t:=\lfloor h^0\left({\mathcal{O}}_X\left(t\right)\right)/5\rfloor$, ${ϵ}_t:=h^0\left({\mathcal{O}}_X\left(t\right)\right)-5u_t$, ${ϵ}_t:=h^0\left({\mathcal{O}}_X\left(t\right)\right)-5u_t$, and $u_t^{\prime }:=\lceil h^0\left({\mathcal{O}}_X\left(t\right)\right)/5\rceil$. We have $0\le {ϵ}_t\le 4$. By Remark 1, to prove that ${\mathcal{O}}_X\left(t\right)$ is not defective, it is sufficient to prove that $h^1\left({\mathcal{I}}_{A\left(u_t\right)}\left(t\right)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(u_t^{\prime }\right)}\left(t\right)\right)=0$. Take a general $(S,S^{\prime })\in S(C,a_t)\times S(C,b_t)$. In particular, $S\cap S^{\prime }=\varnothing$. If $t\ge 3$, Theorem 4 gives $h^i(C,{\mathcal{I}}_{(2S,C)\cup S^{\prime }}\left(t\right))=0$.

Take a general hyperplane H and set $C:=H\cap X$. The theorem of Bertini gives that $C\subset {\mathbb{P}}^4$ is an irreducible degree d hypersurface. Set $a_t:=\lfloor h^0(C,{\mathcal{O}}_C\left(t\right))/4\rfloor$ and $b_t=h^0(C,{\mathcal{O}}_C\left(t\right))-4a_t$. We have $b_t\in \{0,1,2,3\}$.

Observation 1: Take $(d,t)$ such that ${\mathcal{O}}_C\left(t\right)$ is not defective. We have $h^i(C,{\mathcal{I}}_{(2S,C)\cup S^{\prime }}\left(t\right))=0$, $i=0,1$.

(a) Assume $t=4$. If $d\ge 5$, then $h^0\left({\mathcal{O}}_X\left(4\right)\right)=126$, $h^0(C,{\mathcal{O}}_C\left(4\right))=70$ and $(u_4,{ϵ}_4,u_4^{\prime },a_4,b_4)=(25,1,26,17,2)$. If $d=4$, then $h^0\left({\mathcal{O}}_X\left(4\right)\right)=125$, $h^0(C,{\mathcal{O}}_C\left(4\right))=69$ and $(u_4,{ϵ}_4,u_4^{\prime },a_4,b_4)=(25,0,25,17,1)$. If $d=3$, then $h^0\left({\mathcal{O}}_X\left(4\right)\right)=119$, $h^0(C,{\mathcal{O}}_C\left(4\right))=65$ and $(u_4,{ϵ}_4,u_4^{\prime },a_4,b_4)=(24,4,25,16,1)$.

(a1) Assume $d\ge 4$. Take a general $(S,S^{\prime })\in S(17,C)\times S(1,C)$. Theorem 3 gives $h^1(C,{\mathcal{I}}_{(2S,C)\cup S^{\prime }}\left(4\right))=0$. By the Differential Horace Lemma, to prove that $h^1\left({\mathcal{I}}_{A\left(25\right)}\left(4\right)\right)=0$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{A\left(7\right)\cup S\cup (2S^{\prime },C)}\left(3\right)\right)=0$. Lemma 5 gives $h^0\left({\mathcal{I}}_{A\left(7\right)}\left(2\right)\right)=0$. Since $\#S^{\prime }=1$ and C is general, Lemma 6 gives $h^1\left({\mathcal{I}}_{A\left(7\right)\cup 2S^{\prime }}\left(3\right)\right)=0$, and hence, $h^1\left({\mathcal{I}}_{A\left(7\right)\cup (2S^{\prime },C)}\left(3\right)\right)=0$ (Remark 1). Use Lemma 3 and that $\#S+deg\left((2S^{\prime },C)\right)\le h^0(C,{\mathcal{O}}_C\left(3\right))$. Hence, $h^0\left({\mathcal{I}}_{A\left(25\right)}\left(4\right)\right)={ϵ}_1\le 1$. Thus, $h^0\left({\mathcal{I}}_{A\left(26\right)}\left(4\right)\right)=0$.

(a2) Assume $d=3$. Take a general $(S,S^{\prime })\in S(C,16)\times S(C,1)$. Theorem 3 gives $h^i(C,{\mathcal{I}}_{(2S,C)\cup S^{\prime }}\left(4\right))=0$. By the Differential Horace Lemma, to prove that $h^0\left({\mathcal{I}}_{A\left(25\right)}\left(4\right)\right)=0$, it is sufficient to prove that $h^0\left({\mathcal{I}}_{A\left(8\right)\cup S\cup (2S^{\prime },C)}\left(3\right)\right)=0$. Lemma 6 gives $h^1\left({\mathcal{I}}_{A\left(8\right)}\left(3\right)\right)=0$ and $h^1\left({\mathcal{I}}_{A\left(8\right)\cup (2S^{\prime },C)}\left(3\right)\right)\le 1$.

Since $h^0\left({\mathcal{I}}_{A\left(8\right)}\left(2\right)\right)=0$ (Lemma 5), it is sufficient to apply Lemma 3. To prove that $h^1\left({\mathcal{I}}_{A\left(24\right)}\left(4\right)\right)=0$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{A\left(7\right)\cup S\cup (2S^{\prime },C)}\left(3\right)\right)=0$. Lemma 5 gives $h^0\left({\mathcal{I}}_{A\left(7\right)}\left(2\right)\right)=0$. Lemma 6 gives $h^1\left({\mathcal{I}}_{A\left(7\right)\cup (2S^{\prime },C)}\left(3\right)\right)=0$, and hence, $h^0\left({\mathcal{I}}_{A\left(7\right)\cup (2S^{\prime },C)}\left(3\right)\right)=\#S$. Apply Lemma 3.

(c) Assume $t\ge 5$. By step (a) and Lemma 6, we may assume that ${\mathcal{O}}_X(t-1)$ is not defective and that ${\mathcal{O}}_X(t-2)$ is not defective, if $t\ge 6$. Lemma 6 gives $h^1\left({\mathcal{I}}_{A\left(10\right)}\left(3\right)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(12\right)}\left(3\right)\right)=0$.

Claim 1: We have $a_t+4b_t\le h^0(C,{\mathcal{O}}_C(t-1))$.

Proof of Claim 1: Since $4a_t+b_t=h^0(C,{\mathcal{O}}_C\left(t\right))$, it is sufficient to prove that $4h^0(C,{\mathcal{O}}_C(t-1))\ge h^0(C,{\mathcal{O}}_C\left(t\right))+15b_t$. Recall that $b_t\le 3$. Let M be a general hyperplane section of C. The set M is a degree d space surface. Since $h^1(C,{\mathcal{O}}_C(t-1))=0$, a standard exact sequence gives $h^0(C,{\mathcal{O}}_C\left(t\right))=h^0\left({\mathcal{O}}_C(t-1)\right)+h^0(M,{\mathcal{O}}_M\left(t\right))$. Hence, it is sufficient to prove that $3h^0\left({\mathcal{O}}_C(t-1)\right)\ge h^0(M,{\mathcal{O}}_M\left(t\right))+45$. First, assume $t\le d$. We have $h^0(C,{\mathcal{O}}_C(t-1))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+3}{4}}\right)$ and $h^0(M,{\mathcal{O}}_M\left(t\right))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+3}{3}}\right)$, concluding this case, because $t\ge 5$. Now, assume $t\ge d+1$. We obtain $h^0(C,{\mathcal{O}}_C(t-1))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+3}{4}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{t-d+3}{4}}\right)$ and $h^0(M,{\mathcal{O}}_M\left(t\right))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+3}{3}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{t-d+3}{3}}\right)$. Set $\varphi (t,d):=3h^0(C,{\mathcal{O}}_C(t-1))-h^0(M,{\mathcal{O}}_M\left(t\right))-45$. We have $\varphi (d,d)\ge 0$, concluding this case. Now, assume $t>d$. To conclude the proof for t, it is sufficient to prove that $\varphi (t,d)\ge \varphi (t-1,d)$. This is true for $t=d+1$. Now, assume $t\ge d+2$ and set $\psi (t,d)=\varphi (t,d)-\varphi (t-1,d)$. We obtain $\psi (d+2,d)\ge 0$ and then see that $\psi (t,d)\ge 0$ for all $t>d+2$ (now the second term in $\psi (t,d)$ is linear). □

Claim 2: The theorem is true for the triple $(d,t,s)$ if $s\le a_t+b_t-1$.

Proof of Claim 2: Assume $s\le a_t+b_t-1$. Set $x:=min\{s,a_d\}$ and $y:=s-x$. Take a general $(S_1,S_2)\in S(C,x)\times S(C,y)$. and set $F:=2S_1$. Theorem 3 gives $h^1(C,{\mathcal{I}}_{(2S_1,C)\cup S_2}\left(t\right))=0$. Note that $x+y=s$. By the Differential Horace Lemma applied to the point of $S_2$, to prove that $h^1\left({\mathcal{I}}_{A\left(s\right)}\left(t\right)\right)=0$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{S_1\cup (2S_2,C)}(t-1)\right))=0$. Hence, it is sufficient to prove that $h^1(C,{\mathcal{I}}_{S_1\cup (2S_2,C)}(t-1))=0$. Since $y\le max\{0,b_t-1\}\le 1$ and ${\mathcal{O}}_C(t-1)$ is very ample, $h^1(C,{\mathcal{I}}_{S_1\cup (2S_2,C)}(t-1))=0$ if and only if $x+4y\le h^0(C,{\mathcal{O}}_C(t-1))$, which is true by Claim 1 and the definitions of x and y. □

By Claim 2, we may assume $s\ge a_t+b_t$. By Observation 1 and Theorem 3, we have $h^i(C,{\mathcal{I}}_{(2S,C)\cup S^{\prime }}\left(t\right))=0$, $i=0,1$. Hence, by applying the Horace Differential Lemma to the points of $S^{\prime }$, we see that it is sufficient to prove that either $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)\cup S\cup (2S^{\prime },C)}(t-1)\right)=0$ or $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)\cup S\cup (2S^{\prime },C)}(t-1)\right)=0$

Claim 3: We have $s-a_t\le u_{t-1}$.

Proof of Claim 3: Assume $s-a_t\ge u_{t-1}+1$. Since $b_t\le 3$ and ${\mathcal{O}}_X(t-1)$ is not defective, we obtain $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-1)\right)\le 14$. We also obtain $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$ even for $t=5$, because $s-a_5-b_5\ge 12$ (Lemma 6). By Lemma 3, to obtain $h^0\left({\mathcal{I}}_{A\left(s\right)}\left(t\right)\right)=0$, it is sufficient to prove that $a_t\ge 7$. Assume $a_t\le 6$, i.e., assume $h^0(C,{\mathcal{O}}_C\left(t\right))\le 24$, and hence, $h^0(C,{\mathcal{O}}_C\left(5\right))\le 24$. The integer $h^0(C,{\mathcal{O}}_C\left(5\right))$ is an increasing function of d. Since $d\ge 3$, it is sufficient to use that for $d=3$ we have $h^0(C,{\mathcal{O}}_C\left(5\right))=\left({\displaystyle \genfrac{}{}{0pt}{}{9}{4}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{6}{4}}\right)=189-15>25$. □

Claim 4: We have $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)\cup (2S^{\prime },C)}(t-1)\right)=0$.

Proof of Claim 4: It is sufficient to prove that $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)\cup 2S^{\prime }}(t-1)\right)=0$ (Remark 1). Since $b_t\le 3$, any three points of X are contained in a hyperplane section of X and C is general, $A(s-a_t-b_t)\cup 2S^{\prime }$ and $A(s-a_t)$ have the same Hilbert function. Hence, Claim 10 and the assumption that ${\mathcal{O}}_X(t-1)$ is not defective prove Claim 4. □

Claim 4 and Theorem 3 give $h^1(C,{\mathcal{I}}_{S\cup (2S^{\prime },C)}(t-1))=0$.

Assume for the moment $t\ge 6$. Since ${\mathcal{O}}_X(t-2)$ is assumed to be non-defective, either $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$ or $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$. If $h^0\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$, then we apply Lemma 3. If $h^1\left({\mathcal{I}}_{A(s-a_t-b_t)}(t-2)\right)=0$, we use the residual exact sequence of C and that $h^1(C,{\mathcal{I}}_{S\cup (2S^{\prime },C)}(t-1))=0$.

Now, assume $t=5$. Since $s\ge u_5$, Lemma 6 shows that it is sufficient to prove that $u_5-a_5-b_5\ge 12$. First, assume $d>5$. We have $h^0\left({\mathcal{O}}_X\left(5\right)\right)=262$, $u_5=52$, $h^0(C,{\mathcal{O}}_C\left(5\right))=126$, $a_5=31$, and $b_5=2$. Now, assume $d=5$, and hence, $h^0\left({\mathcal{O}}_X\left(5\right)\right)=261$, $u_5=52$, $h^0(C,{\mathcal{O}}_C\left(5\right))=125$, $a_5=31$, and $b_5=1$. Now, assume $d=4$. We have $h^0\left({\mathcal{O}}_X\left(5\right)\right)=256$, $u_5=51$, $h^0(C,{\mathcal{O}}_C\left(5\right))=120$, $a_5=30$, and $b_5=0$. Now, assume $d=3$. We have $h^0\left({\mathcal{O}}_X\left(5\right)\right)=241$, $u_5=48$, $h^0(C,{\mathcal{O}}_C\left(5\right))=111$, $a_5=27$, and $b_5=3$.

(d) Assume $d=2$ and $t=5$. We have $h^0\left({\mathcal{O}}_X\left(5\right)\right)=206$ and $h^0(C,{\mathcal{O}}_C\left(5\right))=91$. Hence, $u_5=41$, ${ϵ}_2=1$, $u_5^{\prime }=42$, $a_5=22$, and $b_5=3$. Take $s\in \{u_5,u_5^{\prime }\}$ and a general $(S,S^{\prime })\in S(22,C)\times S(3,C)$. Theorem 3 gives $h^i(C,{\mathcal{I}}_{(2S,C)\cup S^{\prime },C}\left(5\right))=0$. Thus, to prove that either $h^0\left({\mathcal{I}}_{A\left(s\right)}\left(5\right)\right)=0$ or $h^1\left({\mathcal{I}}_{A\left(s\right)}\left(5\right)\right)=0$, it is sufficient to prove that either $h^0\left({\mathcal{I}}_{A(s-25)\cup S\cup (2S^{\prime },C)}\left(4\right)\right)=0$ or $h^0\left({\mathcal{I}}_{A(s-33)\cup S\cup (2S^{\prime },C)}\left(4\right)\right)=0$. Lemma 8 gives $h^1\left({\mathcal{I}}_{A\left(20\right)}\left(4\right)\right)=0$, and hence, $h^1\left({\mathcal{I}}_{A(s-25)\cup (2S^{\prime },C)}\left(4\right)\right)=0$. Lemma 7 gives $h^0\left({\mathcal{I}}_{A(s-25)}\left(3\right)\right)=0$. Hence, it is sufficient to use Lemma 7.

(e) If $d=2$ and $t\ge 6$, it is sufficient to use step (d), the proof of step (c), and Lemma 8. □

Remark 6.

We do not know cases $t=3,4$ of Theorem 4, i.e., we do not know if they are defective. For a fixed hypersurface X and $t\in \{3,4\}$, we only do not know the precise value of one secant variety, and the possible values of its dimension only differ by 1.

6. Discussion

We computed the dimensions of the secant varieties of the Veronese embeddings of hypersurfaces, even the singular ones, of projective spaces of at most five dimensions. We use the Differential Horace Lemma, introduced by J. Alexander and A. Hirschowitz as one of their main tools to prove their fundamental theorem, for the computation of the dimensions of the secant varieties of the Veronese varieties. In the Alexander–Hirschowitz, theorem there is a list of defective secant varieties. At the end of Section 5, we discuss two cases for which the interested reader may go further. A very strong extension would be the extension to all higher projective spaces. In the introduction, we give the following open question:

Question 1: Let $X\subset {\mathbb{P}}^n$ be an irreducible hypersurface of degree $d\ge 2$. By [7], there is a positive integer $t_0\left(X\right)$ (depending on X and an unknown) such that ${\mathcal{O}}_X\left(t\right)$ is not defective for all $t\ge t_0\left(X\right)$.

The following open questions aim to obtain an explicit and low $t_1$ integer such that ${\mathcal{O}}_X\left(t\right)$ is not defective for all $t\ge t_1$.

Question 2: Let $X\subset {\mathbb{P}}^n$, $n\ge 6$, be an irreducible hypersurface of degree $d\ge n$. Is ${\mathcal{O}}_X\left(t\right)$ not defective for all $t\ge n$?

Question 3: Let $X\subset {\mathbb{P}}^n$, $n\ge 6$, be a general degree d hypersurface. Is ${\mathcal{O}}_X\left(t\right)$ not defective for some small t?

By the semicontinuity theorem for cohomology, to give a positive answer to Question 3 for a fixed triple $(n,d,t)$, it is sufficient to obtain a single solution X. For a specific X, one can use computers to obtain the Hilbert functions of the general unions of a prescribed number of 2-points. This has been carried out heavily, at least in the last 10 years (see, for instance, [23]).

In the last part of the Introduction, we discuss the different methods used to prove these types of results and their use in similar questions in which a geometric theorem is translated into a problem on the Hilbert function. We expect that the Differential Horace Lemma (or some improved form of it),the Blow-up Horace Lemma, and collisions of fat points [24] method will be very useful. Our dream is to be able to combine (without weakening either) the Differential and Blow-up Horace Lemmas.