Multiple Points with Increasing Multiplicities on a Fixed Projective Set

Abstract

Take a finite subset S of an n-dimensional projective space. We study the Hilbert function of the multiples $mS$ of S, mainly when S is general or at least very general. We recall several classical conjectures on this problem, raise new open questions, and prove some particular cases. An open question is if all $mS$ have the expected Hilbert function. We find cases in which there are Zariski open subsets of sets S with maximal rank for all m and pairs $(n,\#S)$ for which no such open set exists. We start the study of the m-Terracini sets proving when the first one is nonempty for Veronese embeddings.

1. Introduction

The aim of this paper is to study the Hilbert function of the zero-dimensional schemes $mS$ (finite unions of points with equal multiplicity m) in a projective space.

Let X be an integral projective variety of dimension $n>0$. Let $X_{\mathrm{reg}}$ denote the set of all smooth points of X. For each $P\in X_{\mathrm{reg}}$ and each positive integer m the zero-dimensional scheme $mP$ is the closed subscheme of X with ${\left({\mathcal{I}}_P\right)}^m$ as its ideal sheaf. We have $\mathrm{deg}\left(mP\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)$ and $m{P}_{\mathrm{red}}=\left\{P\right\}$.

For all irreducible quasi-projective varieties W and all positive integers x, let $S(W,x)$ denote the set of all subsets of X of cardinality x. The set $S(W,x)$ is an irreducible quasi-projective variety of dimension $x\dim W$. Often we take $W=X_{\mathrm{reg}}$, because we only consider multiples of smooth points of X. The set $S(X_{\mathrm{reg}},x)$ is irreducible and of dimension $nx$. For each finite set $S\subset X_{\mathrm{reg}}$ set $mS:={\cup }_{P\in S}mP$. Our main objectives of study are the sets $mS$, $S\in S(X_{\mathrm{reg}},x)$, and we investigate their Hilbert functions.

Let $Z\subset X$ be a zero-dimensional scheme and let L be a line bundle on X. We recall that Z is said to have maximal rank with respect to the line bundle L if the restriction map ${\rho }_{L,Z}:H^0\left(L\right)\to H^0(Z,L_{|Z})$ has maximal rank as a linear map between two finite-dimensional vector spaces; i.e., it is either injective or surjective. Assume that we know the integers $h^0\left(L\right)$ and $\mathrm{deg}\left(Z\right)$. Since $h^0(Z,L_{|Z})=\mathrm{deg}\left(Z\right)$, the linear map ${\rho }_{L,Z}$ has maximal rank if and only if $h^0({\mathcal{I}}_Z\otimes L)=\max \{0,h^0\left(L\right)-\mathrm{deg}\left(Z\right)\}$. Hence if Z has maximal rank with respect to L we know the integer $h^0({\mathcal{I}}_Z\otimes L)$ in terms of the known integers $h^0\left(L\right)$ and $\mathrm{deg}\left(Z\right)$. The integer $h^0\left(L\right)-h^0({\mathcal{I}}_Z\otimes L)$ is the number of independent conditions that Z imposes to the vector space $H^0\left(L\right)$ of the global sections of L. This is an interpolation problem and quite often a very important one. Let F be a family of line bundles on X. We say that Z has maximal rank for F if it has maximal rank with respect to all $L\in F$. We say that Z has maximal rank with respect to all powers of L if Z has maximal rank with respect to all $L^{\otimes t}$; t a positive integer.

In Section 8 we consider the following problem. Fix X, a very ample line bundle L on X and integers $x>0$ and $m\ge 2$. Describe the set of all $S\in S(X_{\mathrm{reg}},x)$ such that $mS$ has not maximal rank with respect to L. The case $x=1$ is related to the set of osculating spaces of X with respect to the complete linear system $\left|L\right|$. In the case $m=2$ this is related to the Terracini sets of X with respect to $\left|L\right|$ [1]. We call them m-Terracini sets. In characteristic 0 being $2S$ not of maximal rank all $S\in S(X_{\mathrm{reg}},x)$ are equivalent to the defectivity of the x-secant variety of X as an embedded variety. The paper [1] describes all cases in which the 2-Terracini loci for the d-Veronese embedding of ${\mathbb{P}}^2$ have codimension 1 in $S\in S({\mathbb{P}}^2,x)$. In Section 8 we give four open questions and give some basic results of these m-Terracini sets. Our main result is the computation of the first integer x for which they are nonempty for the Veronese embeddings (Theorem 11). If $m=2$ a more precise result is known ([2], Theorem 6.6). We encourage the reader to go further. Galuppi, Santarsiero, Torrance, and Turatti in [2], §5 and §7, gave the first integer x with a nonempty 2-Terracini set for the Del Pezzo surfaces and for the Segre–Veronese varieties. For the Hirzebruch surfaces $F_e$, $e\ge 0$, A. Laface proved for $m=2$ and $m=3$ when they are m-defective [3]. Hence it seems to be easy to describe for low m all m-Terracini sets with a very low codimension of $S(F_e,x)$.

We raise the following conjecture.

Conjecture 1: Let X be an integral projective variety and L a very ample line bundle on X. Is there an integer $x_0$ (depending on L and X) such that for all $x\ge x_0$ and all positive integers m the scheme $mS$ has maximal rank with respect to all powers of L for a very general $S\in S(X_{\mathrm{reg}},x)$?

We discuss this conjecture in Remark 12. In the introduction we only want to point out the case $X={\mathbb{P}}^n$, raised by B. Harbourne in 1998 [4], but apparently never studied outside the case ${\mathbb{P}}^2$.

The words ‘very general’ are often used in algebraic geometry, and we explain them here, because they are sometimes confused with the notion of ‘general’. Let W be an irreducible quasi-projective variety defined over an algebraically closed field $\mathbb{K}$. Consider a property, say being blue, that elements of W may or may not satisfy. Let E be the set of all blue elements of W. The words ‘a general $S\in W$ is blue’ mean that E contains a nonempty Zariski open subset of W. By definition a very general element of W is blue if and only if $W\setminus E$ is contained in a countable union of irreducible subvarieties of W of dimension $<\dim W$. We refer to the second category subset of W as synonymous of ‘very general’. Thus, a general element of W is very general. If $\mathbb{K}$ is uncountable, then proving that a very general element of W is blue implies that $E\ne \varnothing$. For a countable $\mathbb{K}$ the subset E may have a second category and be empty. In our study we get some cases in which $E\ne \varnothing$ over $\overline{\mathbb{Q}}$, but $E=\varnothing$ over ${\overline{\mathbb{F}}}_p$ and hence no solution is defined over a finite field (Remarks 9, 15 and 20). If $\mathbb{K}=\mathbb{C}$ in some cases we get density for the Euclidean topology (Remarks 16 and 20). For a case in which we may get a generality and not only very generality, see Proposition 3. For cases with very generality, but not generality, see Theorems 4 and 8. If we are silent about generality in a statement proving the second category it means that we do not know if it is true and we think that it is very difficult to get ‘generality’.

Conjecture 2: Fix an integer $n\ge 2$. Is there an integer $x_0\left(n\right)>0$ such that for all $x\ge x_0\left(n\right)$ the set of all $S\in S({\mathbb{P}}^n,x)$ such that $mS$ has maximal rank for all $m>0$ is a second category subset of $S({\mathbb{P}}^n,x)$?

We prove that if $x<2^n$ no $S\in S({\mathbb{P}}^n,x)$ has the property that all $mS$, $m\gg 0$ have maximal rank (Theorem 1). For $n=3$ this bound is sharp (see Theorem 8 on $S({\mathbb{P}}^3,8)$). For $n=2$ in characteristic zero the set of all $S\in S({\mathbb{P}}^2,x)$, x a square, such that $mS$ has maximal rank for all $m>0$ is a second category subset of $S({\mathbb{P}}^2,x)$ (Corollary 3). We discuss this result and a similar conjectural one for all $x\ge 10$ in Section 4.

In Section 5, Section 6 and Section 7 we study the geometry of $mS$ in ${\mathbb{P}}^2$ and ${\mathbb{P}}^3$ for very low $\#S$. There is a detailed description of the cases $(n,x)=(2,9)$ and $(n,x)=(3,8)$ with the aim to describe the Hilbert functions of the multiple $mS$ for all $S\in S({\mathrm{P}}^n,x)\setminus F$ with $\dim F\le nx-2$ (Theorems 4 and 8).

Macaulay more than a century ago described all possible Hilbert functions of finite sets and of zero-dimensional schemes.

Question 1: Given positive integers n, x and m describe the Hilbert functions of all $mS$, $S\in S({\mathbb{P}}^n,x)$. Take two different integers m and $m_1$. Given the Hilbert function of $mS$, what are the restrictions on the Hilbert function of $m_{1}S$?

Of course, in general only assuming Conjecture 2 we know the Hilbert function of all $mS$, $m\in \mathbb{N}$ for a general $S\in S({\mathbb{P}}^n,x)$, $x\gg 0$. Before the Conclutions Section (Section 10) there is a short section with five open questions and some guesses on them.

We thank the referees for several useful suggestions.

2. Preliminary Results

We work over an algebraically closed field $\mathbb{K}$. In a few cases we point out when our solutions are defined over a smaller field, e.g., $\mathbb{R}$ or $\mathbb{Q}$ or a finite field.

For any quasi-projective variety W set $S(W,0):=\{\varnothing \}$. For any finite set S set $0S:=\varnothing$. Several times we have projective varieties $Y\subset X$ and finite sets $S\subset Y_{\mathrm{reg}}\cap X_{\mathrm{reg}}$. In this case $(mS,Y)$ denotes the union of the m-points $mP$, $P\in S$, as multiple points of Y. Since $S\subset Y_{\mathrm{reg}}\cap X_{\mathrm{reg}}$, we have $(mS,Y)=(mS,X)\cap Y$ (scheme–theoretic intersection). Often we write $mS$ instead of $(mS,X)$.

Let X be a projective variety. Take positive integers m, x, e, $i\in \mathbb{N}$ and a line bundle L on X. By the semicontinuity theorem for cohomology the set of all $S\in S(X_{\mathrm{reg}},x)$ such that $h^i({\mathcal{I}}_{mS}\otimes L)\ge e$ is closed in $S(X_{\mathrm{reg}},x)$ ([5], III.12.8).

Remark 1.

Fix integers $n\ge 2$ and $d\ge 0$. Let A and B be zero-dimensional subschemes of ${\mathbb{P}}^n$ such that $A\subset B$. Then

$$h^1\left({\mathcal{I}}_A\left(d\right)\right)\le h^1\left({\mathcal{I}}_B\left(d\right)\right)\le h^1\left({\mathcal{I}}_A\left(d\right)\right)+\mathrm{deg}\left(B\right)-\mathrm{deg}\left(A\right).$$

Remark 2.

Let $Z\subset {\mathbb{P}}^n$ be a zero-dimensional scheme. Assume the inequality $h^1\left({\mathcal{I}}_Z\left(d\right)\right)>0$. Then we have $h^1\left({\mathcal{I}}_Z(d+1)\right)<h^1\left({\mathcal{I}}_Z\left(d\right)\right)$.

Remark 3.

The connected algebraic group $\mathrm{Aut}\left({\mathbb{P}}^n\right)$ acts on $S({\mathbb{P}}^n,x)$. If $S,S_1\in S({\mathbb{P}}^n,x)$ and there is $g\in \mathrm{Aut}\left({\mathbb{P}}^n\right)$ such that $g\left(S\right)=S_1$, then $g\left(mS\right)=m{S}_1$ for all positive integers m. Thus the Hilbert functions of $mS$ and $m{S}_1$ are the same.

Remark 4.

Assume $x\le n+2$. It is well-known that there is a nonempty open subset U of ${\left({\mathbb{P}}^n\right)}^x$ such that U is an orbit for the action of $\mathrm{Aut}\left({\mathbb{P}}^n\right)$. Indeed, for $x\le n+1$ it is sufficient to take the set of all $(p_1,\dots ,p_x)$ such that $p_i\ne p_j$ for all $i\ne j$ and these x points are linearly independent. Now assume $x=n+2$. The set U is formed by $n+2$ distinct points such that any $n+1$ of them span ${\mathbb{P}}^n$. These sets are usually said to be in linear general position. In each of these cases we may assign different multiplicities to the points of U and still get the same Hilbert function for all $S\in U$ by Remark 3.

Let X be an integral projective variety, $Z\subset X$ a zero-dimensional subscheme and $D\subset W$ and effective Cartier divisor of X. The residual scheme ${\mathrm{Res}}_D\left(Z\right)$ of Z with respect to D is the closed subscheme of X with ${\mathcal{I}}_Z:{\mathcal{I}}_D$ as its ideal sheaf. For all line bundles L we have an exact sequence

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

(1)

often called the residual exact sequence of D. We have ${\mathrm{Res}}_D\left(Z\right)\subseteq Z$ and

$$\mathrm{deg}\left(Z\right)=\mathrm{deg}(Z\cap D)+\mathrm{deg}\left({\mathrm{Res}}_D\left(Z\right)\right).$$

Remark 5.

Let $Z\subset {\mathbb{P}}^1$ be a zero-dimensional scheme. Set $z:=\mathrm{deg}\left(Z\right)$. Fix $d\in \mathbb{N}$. The cohomology of line bundles on ${\mathbb{P}}^1$ gives $h^0({\mathbb{P}}^1,{\mathcal{O}}_{{\mathbb{P}}^1}\left(d\right)(-Z))=\max \{d+1-z,0\}$ and $h^1({\mathbb{P}}^1,{\mathcal{O}}_{{\mathbb{P}}^1}\left(d\right)(-Z))=\max \{z-d-1,0\}$.

Remark 6.

Let X be an integral projective variety of dimension $n\ge 2$. Fix an integer $x\ge 2$. Let $X_0^x$ be the complement of the big diagonals of $X^x$, i.e., the set of all $(p_1,\dots ,p_x)\in X^x$ such that $p_i\ne p_j$ for all $i\ne j$. Since $X^x\setminus X_0^x$ has codimension n in $X^x$ and $n\ge 2$, a general $(p_1,\dots ,p_x)$ is contained in an integral projective curve $D\subset X_0^x$. Taking the quotient by the symmetric group $S_x$ we get that a general $S\in S(X,x)$ is contained in an integral projective curve contained in $S(X,x)$.

Lemma 1.

Fix integers $n\ge 2$, $m\ge 1$ and $d>0$ and a zero-dimensional scheme $Z\subset {\mathbb{P}}^n$. Assume $h^1\left({\mathcal{I}}_{Z\cup mp}\left(d\right)\right)>h^1\left({\mathcal{I}}_Z\left(d\right)\right)$ for a general $p\in {\mathbb{P}}^m$. Then

$$h^1\left({\mathcal{I}}_{Z\cup mE}\left(d\right)\right)\ge h^1\left({\mathcal{I}}_Z\left(d\right)\right)+2[h^1\left({\mathcal{I}}_{Z\cup mp}\left(d\right)\right)-h^1\left({\mathcal{I}}_Z\left(d\right)\right)]$$

for a general $E=\{p,q\}\in S({\mathbb{P}}^n,2)$.

Proof. 

Set $e:=h^1\left({\mathcal{I}}_{Z\cup mp\}}\left(d\right)\right)-h^1\left({\mathcal{I}}_Z\left(d\right)\right)$ and $z:=\left({\displaystyle \genfrac{}{}{0pt}{}{m+n-1}{n}}\right)=\mathrm{deg}\left(mp\right)$. Since E is general, $E\cap Z=\varnothing$. Since q is general, we have $h^1\left({\mathcal{I}}_{Z\cup mp}\left(d\right)\right)=h^1\left({\mathcal{I}}_{Z\cup mq}\left(d\right)\right)$. By Remark 1 we have $e\le z+1$. The definition of e gives that $W_p:=H^0\left({\mathcal{I}}_{Z\cup mp}\left(d\right)\right)$ is a linear subspace of $H^0\left({\mathcal{I}}_Z\left(d\right)\right)$. Since q is general, $W_q:=H^0\left({\mathcal{I}}_{Z\cup mq}\left(d\right)\right)$ is a linear subspace of $H^0\left({\mathcal{I}}_Z\left(d\right)\right)$ of codimension $z+1-e$. Since $H^0\left({\mathcal{I}}_{Z\cup mE}\left(d\right)\right)=W_p\cap W_q$, $H^0\left({\mathcal{I}}_{Z\cup mE}\left(d\right)\right)$ has at most codimension $\mathrm{deg}\left(mE\right)-2e$ in $H^0\left({\mathcal{I}}_{Z\cup mE}\left(d\right)\right)$ by Grassmann’s formula, concluding the proof. □

Remark 7.

Let $Z\subset {\mathbb{P}}^n$, $n\ge 2$, be a zero-dimensional scheme. Since $\dim Z=0$, we have $h^i(Z,L)=0$ for all line bundles on Z for all $i\ge 1$. Hence the long cohomology exact sequence of the exact sequence

$$0\to {\mathcal{I}}_Z\left(t\right)\to {\mathcal{O}}_{{\mathbb{P}}^2}\left(t\right)\to {\mathcal{O}}_Z\left(t\right)\to 0$$

gives $h^i\left({\mathcal{I}}_Z\left(t\right)\right)=0$ for all $i\ge n+1$, $h^i\left({\mathcal{I}}_Z\left(t\right)\right)=0$ for all $2\le n<i$ and all $t\in \mathbb{Z}$ and $h^n\left({\mathcal{I}}_Z\left(t\right)\right)=0$ for all $t\ge -n$.

Remark 8.

Let $C\subset {\mathbb{P}}^2$ be an effective divisor and $Z\subset C$ a zero-dimensional scheme. Let x be a positive integer. Since C is arithmetically normal, ${\mathrm{Res}}_C\left(Z\right)=\varnothing$ and $h^2\left({\mathcal{I}}_Z\left(x\right)\right)=0$ (Remark 7), the long cohomology exact sequence of the residual sequence (1) of ${\mathcal{I}}_Z\left(x\right)$ with respect to C gives $h^1\left({\mathcal{I}}_Z\left(x\right)\right)=h^1(C,{\mathcal{I}}_{Z,C}\left(x\right))$ and $h^0\left({\mathcal{I}}_Z\left(x\right)\right)=h^0(C,{\mathcal{I}}_{C,Z}\left(x\right))+h^0({\mathcal{O}}_{{\mathbb{P}}^2}(x-\mathrm{deg}\left(C\right))$.

Remark 9.

Let C be an integral projective curve of arithmetic genus 1, L a degree x line bundle on C and $Z\subset C$ a zero-dimensional scheme contained in the smooth locus of C. Thus Z is a Cartier divisor of C. Set $z:=\mathrm{deg}\left(Z\right)$. The cohomology of line bundles on C and Riemann–Roch give that $h^0(C,L(-Z))=0$ and $h^1(C,L(-Z))=z-x$ if $z>x$, and $h^0(C,L(-Z))=x-z$ and $h^1(C,L(-Z))=0$ if $z<x$. Now assume $x=z$. If $L(-Z)\ne {\mathcal{O}}_C$, then Riemann–Roch gives $h^0(C,L(-Z))=h^1(C,L(-Z))=0$. Since C has genus 1, if $L(-Z)\cong {\mathcal{O}}_C$, then $h^0(C,L(-Z))=h^1(C,L(-Z))=1$. Now assume $z=x>0$. We have $h^0\left(L\right)=x$ and $L(-Z)\cong {\mathcal{O}}_C$ if and only if $Z\in \left|L\right|$.

Remark 10.

Let C be an elliptic curve. Fix a finite set $S\subset C$. For each $L\in {\mathrm{Pic}}^3\left(C\right)$ the linear system $\left|L\right|$ embeds C in ${\mathbb{P}}^2$ as a smooth plane cubic. For each $L\in {\mathrm{Pic}}^4\left(C\right)$ the linear system $\left|L\right|$ embeds C in ${\mathbb{P}}^3$ as a smooth complete intersection of two quadric surfaces. Fix an integer $k\ge 2$. The set of all $R\in {\mathrm{Pic}}^0\left(C\right)$ such that $R^{\otimes km}\ne {\mathcal{O}}_C$ for all m is a second category subset of ${\mathrm{Pic}}^0\left(C\right)$.

3. Negative Results on the Maximal Rank for All mS

Proposition 1.

Take integers $2\le x\le n$ and $m\ge 2$. There is no $S\in S({\mathbb{P}}^n,x)$ such that $mS$ has maximal rank with respect to ${\mathcal{O}}_{{\mathbb{P}}^n}(2m-2)$.

Proof. 

Fix $S\in S({\mathbb{P}}^n,x)$. Since $x\le n$, there is a hyperplane $H\supset S$. The hypersurface $mH$ gives $h^0\left({\mathcal{I}}_{mS}\left(m\right)\right)>0$. Since $m\ge 2$, we get $h^0\left({\mathcal{I}}_{mS}(2m-2)\right)>0$. Fix $S^{\prime }\subseteq S$ such $\#S^{\prime }=2$ and let $L\subset {\mathbb{P}}^n$ be the line spanned by $S^{\prime }$. Since $\#(S\cap L)\ge 2$, we have $\mathrm{deg}(L\cap mS)\ge 2m$. Hence $h^1(L,{\mathcal{I}}_{L\cap mS,L}(2m-2))>0$. Remark 1 gives $h^1\left({\mathcal{I}}_{mS}(2m-2)\right)>0$. Thus $mS$ has not maximal rank. □

Theorem 1.

Fix integers n and x such that $n\ge 2$ and $2\le x<2^n$. Take $S\in S({\mathbb{P}}^n,x)$. Then for all $m\gg 0$ the scheme $mS$ has not maximal rank with respect to ${\mathcal{O}}_{{\mathbb{P}}^n}(2m-2)$.

Proof. 

Fix $S\in S({\mathbb{P}}^n,x)$. Fix $S^{\prime }\subseteq S$ such that $\#S^{\prime }:=2$ and let $L\subset {\mathbb{P}}^n$ be the line spanned by $S^{\prime }$. Since $\#(S\cap L)\ge 2$, we have $\mathrm{deg}(L\cap mS)\ge 2m$. Hence $h^1(L,{\mathcal{I}}_{L\cap mS,L}(2m-2))>0$. Thus $h^1(L,{\mathcal{I}}_{L\cap mS}(2m-2))>0$.

The restriction map $H^0\left({\mathcal{O}}_{{\mathbb{P}}^n}(2m-2)\right)\to H^0(L,{\mathcal{O}}_L(2m-2))$ gives the inequality $h^1\left({\mathcal{I}}_{L\cap mS}(2m-2)\right)>0$. Remark 1 gives $h^1\left({\mathcal{I}}_{mS}(2m-2)\right)>0$. Thus $mS$ has not maximal rank in degree $2m-2$ if and only if $h^0\left({\mathcal{I}}_{mS}(2m-2)\right)>0$.

We have $h^0\left({\mathcal{O}}_{{\mathbb{P}}^n}(2m-2)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n+2m-2}{n}}\right)$ and $\mathrm{deg}\left(mS\right)=x\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)$. The polynomial $\left({\displaystyle \genfrac{}{}{0pt}{}{n+2m-2}{n}}\right)\in \mathbb{K}\left[m\right]$ has degree n with leading term $2^n{m}^n/n!$. The polynomial $x\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)$ has degree n with leading term $x{m}^n/n!$. Thus if $x<2^n$ for $m\gg 0$ the zero-dimensional scheme $mS$ has not maximal rank in degree $2m-2$ for all large integers m. □

Remark 11.

Take x, $S\in S({\mathbb{P}}^n,x)$ and m as in Theorem 1 with $h^1\left({\mathcal{I}}_{mS}(2m-2)\right)>0$. By Remark 2 we have $h^1\left({\mathcal{I}}_{mS}\left(d\right)\right)\ge 2m-1-d$ for all $d<2m-2$.

The following result covers a smaller range of $(n,x)$ than the one covered by Theorem 1, but it is easy to check if it may be applied to a triple $(n,m,x)$.

Proposition 2.

Fix integers $n\ge 2$, $m\ge 2$ and x such that $2\le x\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)<\left({\displaystyle \genfrac{}{}{0pt}{}{2m-2+n}{n}}\right)$. Then there is no $S\in S({\mathbb{P}}^n,x)$ such that $mS$ has maximal rank.

Proof. 

Take $S\in S({\mathbb{P}}^n,x)$. We have $\mathrm{deg}\left(mS\right)=x\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)$. Hence $h^0\left({\mathcal{I}}_{mS}(2m-2)\right)>0$. Fix $S^{\prime }\subseteq S$ such $\#S^{\prime }:=2$ and let $L\subset {\mathbb{P}}^n$ be the line spanned by $S^{\prime }$. Since $\#(S\cap L)\ge 2$, we have $\mathrm{deg}(L\cap mS)\ge 2m$. Hence $h^1(L,{\mathcal{I}}_{L\cap mS,L}(2m-2))>0$. Thus $h^1\left({\mathcal{I}}_{L\cap mS}(2m-2)\right)>0$. Remark 1 gives $h^1\left({\mathcal{I}}_{mS}(2m-2)\right)>0$. Thus $mS$ has not maximal rank. □

Proposition 3.

For all integers $n\ge 2$ and $1\le x\le n+3$ there is a nonempty Zariski open subset U of $S({\mathbb{P}}^n,x)$ such that $h^0\left({\mathcal{I}}_{mA}\left(d\right)\right)=h^0\left({\mathcal{I}}_{mB}\left(d\right)\right)$ for all $A,B\in U$ and all positive integers d and m.

Proof. 

Since the case $x\le n+2$ is true by [6], Theorem 4.1, we may assume $x=n+3$. Let $S_1({\mathbb{P}}^n,n+3)$ denote the set of all $A\in S({\mathbb{P}}^n,n+3)$ such that each subset of A of cardinality $n+1$ spans ${\mathbb{P}}^n$. The set $S_1({\mathbb{P}}^n,n+3)$ is a nonempty Zariski open subset of $S({\mathbb{P}}^n,n+3)$. Every $A\in S_1({\mathbb{P}}^n,n+3)$ is contained in a unique rational normal curve $D\subset {\mathbb{P}}^n$ ([7], Example 1.22). We may take $U=S_1({\mathbb{P}}^n,n+3)$ by [8], Theorem 2.1 and Remark 2.2. □

Lemma 2.

Take $S\in S({\mathbb{P}}^2,5)$. Then $h^1\left({\mathcal{I}}_{mS}\left(2m\right)\right)>0$ for all $m\ge 2$.

Proof. 

The case $m=2$ is true by the Alexander–Hirschowitz theorem [9,10,11,12]. For $m\ge 3$ the lemma is true, because $5(m+1)m/2>\left({\displaystyle \genfrac{}{}{0pt}{}{2m+2}{2}}\right)$ for all $m\ge 3$. □

Lemma 3.

Fix $A\in S({\mathbb{P}}^2,4)$ and $P\in {\mathbb{P}}^2\setminus A$. Then $h^1\left({\mathcal{I}}_{mA\cup (m-1)P}\left(2m\right)\right)>0$ for all $m\ge 4$.

Proof. 

We have $\mathrm{deg}\left(mA\right)+\mathrm{deg}\left(\right(m-1\left)P\right)=4(m+1)m/2+m(m-1)/2$. Note that we have $4(m+1)m/2+m(m-1)/2>\left({\displaystyle \genfrac{}{}{0pt}{}{2m+2}{2}}\right)$ if and only if $m\ge 4$. □

4. Conjecture 1 and the Consequences of Previous Works in the Plane

First, we discuss the differences between Conjecture 1 and the following theorem proved by J. Alexander and A. Hirschowitz [13].

Theorem 2

(J. Alexander and A. Hirschowitz [13]). Fix an integral projective variety X, a line bundle L and a positive integer m. Then there is an integer $t_0$ such that for all $t\ge t_0$ a general multiple points scheme $Z\subset X_{\mathrm{reg}}$ with each point of multiplicity at most m has maximal rank with respect to all $L^{\otimes t}$.

Remark 12.

Theorem 2 is due to J. Alexander and A. Hirschowitz and it is very strong. It is a maximal rank for all line bundles $L^{\otimes t}$, $t\ge t_0$, and all numbers of points, not just a large number of points. The main difference with respect to Conjecture 1 is that the multiplicity is upper bounded, while in Conjecture 1 we require all positive multiplicities. The paper [13] also contains some extensions of Theorem 2, i.e., [13], Th. 1.1 and Cor. 1.2. The extension of Conjecture 1 (the twist by ${\mathcal{I}}_Z\otimes M$, where M is a line bundle and Z is a zero-dimensional scheme) is as far as we know open in the plane. Now assume that Z is a sum of multiple points, $m_i{P}_i$. If $m\ge m_i$ for all i and $m_i=m$ for at least 10 indices i, then this case fits in the set-up of quasi-uniform multiple points studied in [14].

From now on in this section we take $X={\mathbb{P}}^2$ and discuss Theorem 3, Corollary 1 and related references.

The story started with M. Nagata [15] and his results (and conjectures) of the Hilbert functions of $mS$ for S general in $S({\mathbb{P}}^2,x)$. Later, several mathematicians refined his conjectures and B. Harbourne was/is a main source of these new results. One can see in [4], Fact. 1.1, a summary of the description of the Hilbert functions of general $mS$ for $x\le 9$. Some of these papers also give conjectures for the minimal free resolution of a general $mS$ [4,14,16,17,18,19]. In all cases even assuming the conjecture for each m there would be a Zariski open subset $U_m$ of $S({\mathbb{P}}^2,x)$ with $mS$ of the expected Hilbert function. Thus ${\cap }_n{U}_m$ would be a second category set such that for all $S\in {\cap }_n{U}_m$ all $mS$ have the expected Hilbert function. We do not think there is a Zariski open set with the same property (see Proposition 4 for $x=9$).

Many years of works on the uniform Segre–Harbourne–Gimigliano–Hirschowitz (SHGH) conjecture gave the following maximal rank results for all $mS$ in the plane (Theorem 3 and Corollary 1).

Theorem 3.

Assume $\mathrm{char}\left(\mathbb{K}\right)=0$. Fix an integer $x\ge 10$. If the uniform SHGH conjecture is true for the integer x, then the set of all $S\in S({\mathbb{P}}^2,x)$ such that $mS$ has maximal rank for all $m>0$ is a second category subset of $S({\mathbb{P}}^2,x)$.

Proof. 

The case $x=9$ is true by Remark 13. Hence we may assume $x\ge 10$. The uniform part of the SHGH conjecture was explained in the introduction of [19] which explained the content of the more general [14] (for our case [4] would be enough). □

Corollary 1.

Assume $\mathrm{char}\left(\mathbb{K}\right)=0$. If x is a square $\ge 9$, then the set of all $S\in S({\mathbb{P}}^2,x)$ such that $mS$ has maximal rank for all $m>0$ is a second category subset of $S({\mathbb{P}}^2,x)$.

Proof. 

The case $x=9$ is true by Remark 13. Hence we may assume $x\ge 10$. This is the statement (with $x=k^2$) of [20], Theorem at page 1950. See also [21]. □

Note that Corollary 1 implies the following result not depending on the uniform SHGH conjecture.

Corollary 2.

Assume $\mathrm{char}\left(\mathbb{K}\right)=0$. For all integers y let $a(m,y)$ be a minimal integer such that $y\left({\displaystyle \genfrac{}{}{0pt}{}{m+1}{2}}\right)\le \left({\displaystyle \genfrac{}{}{0pt}{}{a(m,y)+2}{2}}\right)$ and $b(m,y)$ the maximal integer such that $y\left({\displaystyle \genfrac{}{}{0pt}{}{m+1}{2}}\right)\ge \left({\displaystyle \genfrac{}{}{0pt}{}{b(m,y)+2}{2}}\right)$. Fix integers $x\ge 9$, $m\ge 2$ and take a general $S\in S({\mathbb{P}}^2,x)$. Then $h^1\left({\mathcal{I}}_{mS}\left(a(m,{\lceil \sqrt{x}\rceil }^2)\right)\right)=0$ and $h^0\left({\mathcal{I}}_{mS}\left(b(m,{\lfloor \sqrt{x}\rfloor }^2)\right)\right)=0$.

5. 9 Points in ${\mathbb{P}}^{\mathbf{2}}$

We take $X={\mathbb{P}}^2$ and $x=9$. We need a bit more than [6], §5. One motivation is to discuss the cases $\mathbb{K}={\overline{\mathbb{F}}}_p$ and $\mathbb{K}=\overline{\mathbb{Q}}$ and see what we are able to prove for a finite field or for $S\subset {\mathbb{P}}^2\left(\mathbb{Q}\right)$. Another motivation is that we may find a nonempty open subset U of $S({\mathbb{P}}^2,9)$ such that for each $S\in S({\mathbb{P}}^2,9)$ all schemes $mS$ have almost maximal rank in a very precise way (Theorems 4 and 5). We use it in Example 5.

Lemma 4.

Take an integral $C\in |{\mathcal{O}}_{{\mathbb{P}}^2}\left(3\right)|$, $S\in S(C_{\mathrm{reg}},9)$ and an integer $m\ge 1$.

(a) 

We have $h^0\left({\mathcal{I}}_{mS}\left(3m\right)\right)=h^1\left({\mathcal{I}}_{mS}\left(3m\right)\right)+1$. Moreover, $mS\subset mC$.

(b) 

We have $h^1\left({\mathcal{I}}_{mS}\left(3m\right)\right)>0$ if and only if there it $t\in \{1,\dots ,m\}$ such that

$$\left(tS\right)\cap C\in |{\mathcal{O}}_C\left(3t\right)|.$$

(c) 

The integer $h^1\left({\mathcal{I}}_{mS}\left(3m\right)\right)$ is the number of integers $t\in \{1,\dots ,m\}$ such that

$$\left(tS\right)\cap C\in |{\mathcal{O}}_C\left(3t\right)|.$$

Proof. 

Since $S\subset C_{\mathrm{reg}}$, we have $h^0\left({\mathcal{O}}_{{\mathbb{P}}^2}\left(3m\right)\right)=(3m+2)(3m+1)/2=\mathrm{deg}\left(mS\right)+1$. Hence part (a) is true.

Since (b) is a particular case of (c), it is sufficient to prove part (c). By Remark 7 when we use a long cohomology exact sequences we will only need to check the vector spaces $H^0$ and $H^1$.

Since C is arithmetically normal, the case $m=1$ is true. Hence we may assume that $m\ge 2$ and that part (c) is true for all positive integers $<m$. Since S is contained in the smooth locus of C, ${\mathrm{Res}}_C\left(tS\right)=(t-1)S$, with the convention $0S=\varnothing$. By (1) the residual exact sequence of $mS$ with respect to C is the following exact sequence

$$0\to {\mathcal{I}}_{(m-1)S}(3m-3)\to {\mathcal{I}}_{mS}\left(3m\right)\to {\mathcal{I}}_{C\cap \left(mS\right),C}\left(3m\right)\to 0$$

(2)

Since $S\subset C_{\mathrm{reg}}$, we have $\mathrm{deg}(C\cap mS)=m\#S=9m$ and $C\cap mS$ is a Cartier divisor of C. Since $\mathrm{deg}\left(C\right)=3$, we have $\mathrm{deg}\left({\mathcal{O}}_C\left(3m\right)\right)=9m$. Hence Remark 9 gives the inequality

$$h^0(C,{\mathcal{I}}_{C\cap \left(mS\right),C}\left(3m\right))=h^1(C,{\mathcal{I}}_{C\cap \left(mS\right),C}\left(3m\right))\le 1.$$

It also gives that we have $h^0(C,{\mathcal{I}}_{C\cap \left(mS\right),C}\left(3m\right))=1$ if and only if $C\cap mS\in |{\mathcal{O}}_C\left(3m\right)|$. If $h^0(C,{\mathcal{I}}_{C\cap \left(mS\right),C}\left(3m\right))=0$, then part (c) follows by the inductive assumption on m and the long cohomology exact sequence of (2).

Now assume $h^0(C,{\mathcal{I}}_{C\cap \left(mS\right),C}\left(3m\right))\ne 0$. This inequality implies the equality $h^0(C,{\mathcal{I}}_{C\cap \left(mS\right),C}\left(3m\right))=h^1(C,{\mathcal{I}}_{C\cap \left(mS\right),C}\left(3m\right))=1$. The long cohomology exact sequence of (2) gives $h^0\left({\mathcal{I}}_{mS}\left(3m\right)\right)\le h^0\left({\mathcal{I}}_{(m-1)S}(3m-3)\right)+1$. Hence to conclude the proof of the lemma it is sufficient to prove that $h^0\left({\mathcal{I}}_{mS}\left(3m\right)\right)>h^0\left({\mathcal{I}}_{(m-1)S}(3m-3)\right)$. The map induced in $H^0$ from (2) is the map $f\mapsto f_{|C}$. Since $h^0(C,{\mathcal{I}}_{C\cap \left(mS\right),C}\left(3m\right))\ne 0$, Remark 8 gives the existence of $f\in H^0\left({\mathcal{I}}_{mS}\left(3m\right)\right)$ such that $f_{|C}\ne 0$. Since the injective map $H^0\left({\mathcal{I}}_{(m-1)S}(3m-3)\right)\to H^0\left({\mathcal{I}}_{mS}\left(3m\right)\right)$ is induced by the multiplication by an equation of C, we get $h^0\left({\mathcal{I}}_{mS}\left(3m\right)\right)>h^0\left({\mathcal{I}}_{(m-1)S}(3m-3)\right)$. □

Remark 13.

Let G be the set of all $S\in S({\mathbb{P}}^2,9)$ such that $h^0\left({\mathcal{I}}_S\left(3\right)\right)=1$ and the unique plane cubic $C_S$ containing S is smooth. Fix $S\in G$ and set $C:=C_S$. Remark 9 gives $h^1\left({\mathcal{I}}_{mS}\left(x\right)\right)=0$ for all $d>3m$, $h^0\left({\mathcal{I}}_{mS}\left(d\right)\right)=0$ for all $d<3m$, $h^0\left({\mathcal{I}}_{mS}\left(3m\right)\right)=h^1\left({\mathcal{I}}_{mS}\left(3m\right)\right)\le 1$ and $h^0\left({\mathcal{I}}_{mS}\left(2m\right)\right)=1$ if and only if ${\mathcal{O}}_C\left(mS\right)\cong {\mathcal{O}}_C\left(3m\right)$. The torsion of ${\mathrm{Pic}}^0\left(C\right)$ is countable and dense in ${\mathrm{Pic}}^0\left(C\right)$.

Theorem 4.

Take $S\in S({\mathbb{P}}^2,9)$ contained in the smooth locus of an integral degree 3 plane curve T. Then:

(a) 

$h^1\left({\mathcal{I}}_{mS}(3m+1)\right)=0$ for all $m\ge 1$.

(b) 

$h^0\left({\mathcal{I}}_{mS}(3m-1)\right)=0$ for all $m\ge 1$.

(c) 

Let E be the set of all $A\in S({\mathbb{P}}^2,9)$ such that $h^i\left({\mathcal{I}}_{mA}\left(3m\right)\right)=0$ for all $m\ge 1$. Then E is a second category subset of $S({\mathbb{P}}^2,9)$, but it contains no nonempty open subset of $S({\mathbb{P}}^2,9)$.

Proof. 

Since $S\subset T_{\mathrm{reg}}$, $mS$ is a Cartier divisor of T, $\mathrm{deg}(T\cap mS)=m$ and ${\mathrm{Res}}_T\left(mS\right)=(m-1)S$.

Since T is arithmetically Cohen–Macaulay, to prove part (a) it is sufficient to use that $h^1(T,{\mathcal{I}}_{mS}(3m+1))=0$ by Remark 9.

Now we prove part (b). First, assume $m=1$. Since $\mathrm{deg}\left(S\right)=9$ and T is irreducible, the theorem of Bezout gives that S is not contained in a conic.

Now assume $m>1$ and that $h^0\left({\mathcal{I}}_{(m-1)S}(3m-4)\right)=0$. By Remark 9 we have the equality $h^0(T,{\mathcal{I}}_{mS}(3m-1))=0$. Hence the long cohomology exact sequence of the residual sequence of T gives $h^0\left({\mathcal{I}}_{mS}(3m-1)\right)=0$.

Part (c) follows from Lemma 4 and Remark 13 using induction on m. □

Theorem 5.

Fix a real number ϵ such that $0<ϵ<1$. For each positive integer m set $m_1(ϵ,m):=\lfloor (3+ϵ)m\rfloor$ and $m_1(ϵ,m):=\lceil (3-ϵ)m\rceil$. Take $S\in S({\mathbb{P}}^2,9)$ in the smooth locus of an integral plane curve. Then

$$\underset{m\to +\infty }{\lim }{h}^0({\mathcal{I}}_{mS}\left(m_1(ϵ,m)\right)=0,\underset{m\to +\infty }{\lim }{h}^1({\mathcal{I}}_{mS}\left(m_2(ϵ,m)\right)=0.$$

Proof. 

Obviously $m_1(ϵ,m)$ and $m_2(ϵ,m)$ are integers. For large m we have $m_1(ϵ,m)<3m$ and $m_2\left({ϵ}_m\right)>3m$. Thus the theorem is a Corollary of Theorem 4 and Remark 2. □

Remark 14.

Let $T\subset {\mathbb{P}}^2$ be an integral and singular plane curve. If T has a cusp, then ${\mathrm{Pic}}^0\left(T\right)$ is isomorphic to the additive group and hence it has no torsion in characteristic zero. If we have $\mathrm{char}\left(\mathbb{K}\right)=p>0$, then the additive group $\mathbb{K}$ as only p-torsion, the p-torsion being an increasing sequence of the groups $\mathbb{Z}/p{\mathbb{Z}}^e$, $e\ge 1$. Now assume T nodal. The algebraic group ${\mathrm{Pic}}^0\left(T\right)$ is isomorphic to the multiplicative group, which has torsions of all degrees (the roots of 1) coprime to the characteristic if $\mathbb{K}$ has positive characteristics.

Remark 15.

A more subtle problem is if (assuming the base field $\mathbb{K}$ is algebraically closed) there is $S\in S({\mathbb{P}}^2,9)$ such that $h^1\left({\mathcal{I}}_{mS}\left(3m\right)\right)=0$ for all $m>0$. A necessary condition is that S is contained in a unique plane cubic. Let C be the unique cubic containing C. If S contains a singular point of C, then $\mathrm{deg}(C\cap 2S)>18$ and hence $h^1(C,{\mathcal{I}}_{C\cap 2S,C}\left(6\right))>0$. Thus $h^1\left({\mathcal{I}}_{2S}\left(6\right)\right)>0$. Now assume that S is contained in a smooth plane cubic C. By Remark 10 $h^1\left({\mathcal{I}}_{mS}\left(3m\right)\right)=0$ for all $m>0$ if and only if there is $L\in {\mathrm{Pic}}^0\left(C\right)$ which is not torsion. This is false if $\mathbb{K}={\overline{\mathbb{F}}}_p$, but it is true in characteristic 0, since it is true for $\overline{\mathbb{Q}}$. We may even take C defined over $\mathbb{Q}$, because there are elliptic curves over $\mathbb{Q}$ with positive rank ([22], p. 234).

Proposition 4.

There is no nonempty Zariski open subset U of $S({\mathbb{P}}^2,9)$ such that we have $h^0\left({\mathcal{I}}_{mA}\left(d\right)\right)=h^0\left({\mathcal{I}}_{mB}\left(d\right)\right)$ for all d, all $m\ge 2$ and all $A,B\in U$.

Proof. 

By the semicontinuity theorem for cohomology and the openness of smoothness there is a nonempty open subset E of $S({\mathbb{P}}^2,9)$ such that for all $S\in E$ we have $h^0\left({\mathcal{I}}_S\left(3\right)\right)=1$ and the unique plane cubic containing S is smooth. Let V denote the set of all smooth plane cubics. Take $S\in E$ and let C be the unique plane cubic containing S. Fix an integer $m\ge 2$ we have $\mathrm{deg}\left(mS\right)=9(m+1)m/2=h^0\left({\mathcal{O}}_{{\mathbb{P}}^2}\left(3m\right)\right)-1$. Obviously $mS$ is contained in the multiple curve $mC$ and hence $h^1\left({\mathcal{I}}_{mS}\left(3m\right)\right)=0$ if and only if $|{\mathcal{I}}_{mS}\left(3m\right)|=\left\{mC\right\}$. Let $\eta :U\to V$ be the map $S\to \left\{\right|{\mathcal{I}}_S\left(3\right)\left|\right\}$. For each $C\in V$ a general $S\in S(C,9)$ is an element of U. Let $U_t$ be the set of all $S\in U$ such that $h^0\left({\mathcal{I}}_{tS}\left(3t\right)\right)\ne 0$. For any $C\in V$ let $C_t$ be the set of all $S\in S(C,9)$ such that $h^0\left({\mathcal{I}}_{tS}\left(3t\right)\right)>0$, i.e., $tS\cong {\mathcal{O}}_C\left(3t\right)$. Each $U_t$ is proper and closed in U. By Remarks 9, 10 and 15 applied to all sets $C_t$ the set ${\cup }_{t\ge 2}{U}_t$ is Zariski dense in $S({\mathbb{P}}^2,9)$. □

Remark 16.

Now assume that $\mathbb{K}=\mathbb{C}$. Let $C\subset {\mathbb{P}}^2$ be a smooth cubic curve. We claim that set of all $S\in S(C,9)$ such that $\left(mS\right)\cap C\in |{\mathcal{O}}_C\left(m\right)|$ for some m is dense in $C\left(\mathbb{C}\right)$ for the Euclidean topology. It is sufficient to give the identification as topological groups ${\mathrm{Pic}}^0\left(C\left(\mathbb{C}\right)\right)\cong S^1\times S^1$ with $S^1$ the unit circle of ${\mathbb{R}}^2$ and that the roots of unity are dense in $S^1$ for the Euclidean topology. As in the proof of Proposition 4 we get that there is no Euclidean open subset U of $S({\mathbb{P}}^2\left(\mathbb{C}\right),9)$ such that all $S\in U$ have maximal rank. This means that each $E\in S({\mathbb{P}}^2\left(\mathbb{C}\right),9)$ may be approximated as much as you want for any distance giving the Euclidean topology by elements of $S({\mathbb{P}}^2\left(\mathbb{C}\right),9)$ which have not maximal rank.

6. Subsets of the Plane

In this section we always take $X={\mathbb{P}}^2$. Recall that $\dim |{\mathcal{O}}_{{\mathbb{P}}^2}\left(t\right)|=(t^2+3t)/2$ and that every $C\in |{\mathcal{O}}_{{\mathbb{P}}^2}\left(t\right)|$ has arithmetic genus $(t-1)(t-2)/2$. If $C\in |{\mathcal{O}}_{{\mathbb{P}}^2}\left(c\right)|$, then ${\omega }_C\cong {\mathcal{O}}_C(c-3)$. If $t\ge c-2$, then $h^0(C,{\mathcal{O}}_C\left(t\right))=tc+1-(c-1)(c-2)/2$. If $t<c$, then $h^0(C,{\mathcal{O}}_C\left(t\right))=\left({\displaystyle \genfrac{}{}{0pt}{}{t+2}{2}}\right)$. For each integer $x\ge 4$ let $c\left(x\right)$ be the only integer c such that $({(c-1)}^2+3(c-1))/2<x\le (c^2+3c)/2$. For every integer $x\ge 4$ let $S_1({\mathbb{P}}^2,x)$ denote the set of all $S\in S({\mathbb{P}}^2,x)$ such that $h^1({\mathcal{I}}_S\left(c\left(x\right)\right)=0$ and S is contained in a smooth degree $c\left(x\right)$ curve.

Fix an integer $c\ge 4$. Let $\Delta \left(c\right)$ be the set of all $S\in S({\mathbb{P}}^2,c^2)$ such that $h^0\left({\mathcal{I}}_S\left(c\right)\right)=1$, the only element, $C_S$, of $|{\mathcal{I}}_S\left(c\right)|$ is irreducible and S is contained in the smooth locus of $C_S$. The set $\Delta \left(c\right)$ is a locally closed irreducible subset of $S({\mathbb{P}}^2,c^2)$ of dimension $c^2+(c^2+3c)/2$. Fix $S\in \Delta \left(c\right)$ and set $C:=C_S$. We have ${\omega }_C\cong {\mathcal{O}}_C(c-3)$ and C has arithmetic genus $(c^2-3c)/2+1$. Since $S\subset C_{\mathrm{reg}}$, $\mathrm{deg}(C\cap mS)=m\#S=m{c}^2$ for all $m\ge 1$. Note that for all $m\ge 1$ the divisor $C\cap mS$ is non-special.

Remark 17.

Assume $\mathrm{char}\left(\mathbb{K}\right)=0$. Take $S\in \Delta \left(c\right)$ and set $C:=C_S$. Since $h^0\left({\mathcal{I}}_S\left(c\right)\right)=1$, we have $h^0\left({\mathcal{I}}_S\left(t\right)\right)=0$ for all $t<c$ and hence $h^0\left({\mathcal{I}}_{mS}\left(t\right)\right)=0$ for all $m\ge 1$ and all $t<c$. Since $S\subset C$, $mS\subset mC$ and hence $h^0\left({\mathcal{I}}_{mS}\left(mc\right)\right)>0$. We have $h^0(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=0$ if $tc<m{c}^2$, i.e., if $t<mc$. We have $h^1(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=0$ if $ct\ge 1+m{c}^2+c(c-3)$, i.e., $t\ge mc+c-2$. Now we fix m and t and assume that for a fixed C the set $S\in S(C,c^2)$ is general. By [23] we have the equality $h^0(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=\max \{0,h^0(C,{\mathcal{O}}_C\left(t\right))-m{c}^2\}$. Hence we have the equality $h^1(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=\max \{0,m{c}^2-h^0(C,{\mathcal{O}}_C\left(t\right))\}$. Thus under these restricted assumptions on m, t and S we have $h^0(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=0$ if and only if $h^0(C,{\mathcal{O}}_C\left(t\right))\le m{c}^2$ and $h^1(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=0$ if and only if $h^0(C,{\mathcal{O}}_C\left(t\right))\ge m{c}^2$. Taking all positive integers m and t we see that the subset of $S(C,c^2)$ formed by the sets with these properties is a second category subset of $S(C,c^2)$. Varying C we get that the subset of $\Delta \left(c\right)$ with the same properties is a second category subset of $\Delta \left(c\right)$. Now we restrict to the integers $t\ge c-2$. With this new restriction we have $h^0(C,{\mathcal{O}}_C\left(t\right))=ct-c(c-3)/2$ and hence $h^0(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=0$ if and only if $t-(c-3)/2\le mc$ and $h^1(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=0$ if and only if $t\ge (c-3)/2+mc$. For $m=1$ we get $h^1(C,{\mathcal{I}}_{C\cap S,C}\left(t\right))=0$ if $t\ge c-2$ for all $S\in \Delta \left(c\right)$, while for a general $S\in S(C,c^2)$ we get $h^0(C,{\mathcal{I}}_{C\cap S,S}\left(t\right))>0$ if and only if $t\ge c+(c-3)/2$ and $h^1(C,{\mathcal{I}}_{C\cap S,C}\left(t\right))=0$ if and only if $t\ge c+(c-3)/2$.

For all $t\in \mathbb{Z}$ we have the residual exact sequence

$$0\to {\mathcal{I}}_{(m-1)S}(t-c)\to {\mathcal{I}}_{mS}\left(t\right)\to {\mathcal{I}}_{C\cap mS,C}\left(t\right)\to 0$$

(3)

Note that $\mathrm{deg}\left((m-1)S\right)=\mathrm{deg}\left(mS\right)-c^2$.

Theorem 6.

For all $S\in \Delta \left(c\right)$ we have $h^0\left({\mathcal{I}}_{mS}\left(t\right)\right)>0$ if $t\ge mc$ and $h^1\left({\mathcal{I}}_{mS}\left(t\right)\right)=0$ if $t\ge mc+c-2$. There is a second category subset G of $\Delta \left(c\right)$ such that for all $S\in G$ we have $h^0\left({\mathcal{I}}_{mS}(cm-1)\right)=0$ and $h^1\left({\mathcal{I}}_{mS}\left(t\right)\right)=0$ if and only if $t\ge (c-3)/2+mc$.

Proof. 

We saw (Remark 17) that $h^0\left({\mathcal{I}}_{mS}\left(t\right)\right)>0$. For the other assertions of the theorem we use induction on m.

First assume $m=1$. Since C is arithmetically normal, $h^1\left({\mathcal{I}}_S\left(t\right)\right)=h^1(C,{\mathcal{I}}_{S,C}\left(t\right))$. Hence the case $m=1$ of Remark 17 concludes this case.

Now assume $m>1$ and that the theorem is true for smaller multiplicities. The first part and the $h^1$-part of the second part are sufficient to use Remark 17 for $mS$, the inductive assumption for $(m-1)S$ and the long cohomology exact sequence of (3). Now we check the $h^0$-part of the second part. Remark 17 gives the case $m=1$. Assume $m\ge 2$ and take $t=mc-1$. Assume $H^0\left({\mathcal{I}}_{mS}(mc-1)\right)\ne 0$ and take $G\in |{\mathcal{I}}_{mS}(mc-1)|$. Since $h^0(C,{\mathcal{I}}_{C\cap mS}(mc-1))=0$, G has C as a component. Use that $h^0\left({\mathcal{I}}_{(m-1)S}((m-1)c-1)\right)=0$ and the long cohomology exact sequence of the exact sequence (3) with $t=mc-1$. □

The following example shows that in the definition of $\Delta \left(c\right)$ the assumption that $C_S$ is integral is essential.

Example 1.

Fix an integer $c\ge 4$. Take a smooth degree $c-1$ curve $D\subset {\mathbb{P}}^2$ and a line $L\subset {\mathbb{P}}^2$ transversal to D. Fix $A\in S(L\setminus L\cap D,c+1)$ and a general $B\in S(D\setminus D\cap L,c^2-c-1)$. Set $S:=A\cup B$. The set S is contained in the smooth locus of $D\cup L$ and $\#S=c^2$. Since $\#A=c+1$ and $A\subset L$, we have $\mathrm{deg}(L\cap mS)=mc+m$. Hence $h^1(L,{\mathcal{I}}_{L\cap mS,L}\left(t\right))>0$ for all $t\le mc+m-2$. Note that $mc+m-2>mc+c-2$ for all $m>c$. The theorem of Bezout gives that each element of $|{\mathcal{I}}_S\left(c\right)|$ contains L. Remark 17 for the integer $c-1$ and the generality of B gives $h^0(D,{\mathcal{I}}_{B,D}(c-1))=0$. Thus $h^0\left({\mathcal{I}}_S\left(c\right)\right)=1$.

For all integers $c\ge 2$ and $z>c^2$ let $\Delta (c,z)$ denote the set of all $S\in S({\mathbb{P}}^2,z)$ contained in the smooth locus of some irreducible $C_S\in \left|{\mathcal{O}}_{{\mathbb{P}}^2}\left(c\right)\right|$.

Remark 18.

Assume $\mathrm{char}\left(\mathbb{K}\right)=0$. Fix an integer $c\ge 2$ and take $S\in \Delta (c,z)$ with associated curve $C_S$. Set $C:=C_S$. Since C is irreducible and $z>c^2$, the theorem of Bezout gives the equality $h^0\left({\mathcal{I}}_S\left(c\right)\right)=1$. Thus $h^0\left({\mathcal{I}}_S\left(t\right)\right)=0$ for all $t<c$. Since $S\subset C$, $mS\subset mC$ and hence $h^0\left({\mathcal{I}}_{mS}\left(mc\right)\right)>0$. If $tc<mz$, then $h^0(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=0$. We have $h^1(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=0$ if $\mathrm{deg}\left({\mathcal{I}}_{C\cap mS,C}\left(t\right)\right)\ge 1+c(c-3)$, i.e., if $ct\ge 1+mz+c(c-3)$. Now we fix m and t and assume that for a fixed C the set $S\in S(C,z)$ is general. Since C is an integral curve, ref. [23] gives $h^0(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=\max \{0,h^0(C,{\mathcal{O}}_C\left(t\right))-mz\}$ and $h^1(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=\max \{0,mz-h^0(C,{\mathcal{O}}_C\left(t\right))\}$. Take $S^{\prime }\subset S$ such that $\#S^{\prime }=c^2$. Since $S^{\prime }$ is general in C, Remark 6 gives $h^0\left({\mathcal{I}}_{mS}(mc-1)\right)=0$ if $c\ge 4$. Hence $h^0\left({\mathcal{I}}_{mS}(mc-1)\right)=0$ if $c\ge 4$. For $c=2,3$ it is easy to check that $h^0\left({\mathcal{I}}_{mS}(mc-1)\right)=0$. Thus under these assumptions on m, t and S we have $h^1(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=0$ if and only if $h^0(C,{\mathcal{O}}_C\left(t\right))\ge mz$. Taking all positive integers m and t we see that the subset of $S(C,z)$ formed by the sets with these properties is of the second category. Now we restrict to the integers $t\ge c-2$. With this new restriction we have $h^0(C,{\mathcal{O}}_C\left(t\right))=ct-c(c-3)/2$ and hence $h^0(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=0$ if and only if $ct-c(c-3)/2\le mz$ and $h^1(C,{\mathcal{I}}_{C\cap mS,C}\left(t\right))=0$ if and only if $ct\ge c(c-3)/2+mz$. We used the assumption $\mathrm{char}\left(\mathbb{K}\right)=0$ to quote the very short, but very useful, note [23] by C. Ciliberto and R. Miranda. This note requires this assumption and uses it in an essential way in its proof.

Theorem 7.

Assume $\mathrm{char}\left(\mathbb{K}\right)=0$. For all $S\in \Delta (c,z)$ we have $h^0\left({\mathcal{I}}_{mS}\left(t\right)\right)>0$ if $t\ge mc$ and $h^1\left({\mathcal{I}}_{mS}\left(t\right)\right)=0$ if $ct\ge 1+mz+c(c-3)$. There is a subset G of second category in $\Delta (c,z)$ such that for all $S\in G$ we have $h^0\left({\mathcal{I}}_{mS}(cm-1)\right)=0$ and $h^1\left({\mathcal{I}}_{mS}\left(t\right)\right)=0$ if and only if $ct\ge c(c-3)/2+mz$.

Proof. 

Mimic the proof of Theorem 6 using Remark 18 and the long cohomology exact sequence of (3). □

7. $\mathit{X}={\mathbb{P}}^{\mathbf{3}}$

In this section we take $X={\mathbb{P}}^3$.

The cases with $S({\mathbb{P}}^3,x)$, $x\le 5$, are solved (Zariski open subset with constant Hilbert function) by Remark 4.

First, we describe the case $x=6$ (Propositions 5 and 6 and Corollary 3).

Proposition 5.

Let U be the Zariski open subset of $S({\mathbb{P}}^3,6)$ formed by points in linear general position. Then for all $m\ge 1$ and all $S\in U$ we have $h^1\left({\mathcal{I}}_{mS}\left(2m\right)\right)=0$ and $h^1\left({\mathcal{I}}_{mS}(2m-1)\right)\ge 2$. We have $h^0\left({\mathcal{I}}_{mS}(2m-1)\right)>0$ for all $m\ge 4$.

Proof. 

Fix $S\in U$. Let $C\subset {\mathbb{P}}^3$ be the rational normal curve containing S. We have $\mathrm{deg}\left({\mathcal{O}}_C\left(t\right)\right)=3t$ and $\mathrm{deg}(mS\cap C)=6m$. Thus $h^1(C,{\mathcal{I}}_{mS\cap C,C}(2m-1))=2$ and $h^1(C,{\mathcal{I}}_{mS\cap C,C}\left(2m\right))=0$. Remark 1 gives $h^1\left({\mathcal{I}}_{mS}(2m-1)\right)\ge 2$. We have $h^0\left({\mathcal{I}}_{mS}(2m-1)\right)=h^1\left({\mathcal{I}}_{mS}(2m-1)\right)+(m+1)(2m+1)2m/3-(m+2)(m+1)m$. Hence $h^0\left({\mathcal{I}}_{mS}(2m-1)\right)\ge 2+(m+1)(2m+1)2m/3-(m+2)(m+1)m$, which is positive if $m\ge 4$.

Now we prove that $h^1\left({\mathcal{I}}_{mS}\left(2m\right)\right)=0$.

Since C is arithmetically Cohen–Macaulay, Remark 5 gives $h^1\left({\mathcal{I}}_S\left(2\right)\right)=0$, proving the case $m=1$. Thus we may assume $m>1$ and use induction on m.

The homogeneous ideal of C is generated by quadrics and the general $Q\in |{\mathcal{I}}_C\left(2\right)|$ is smooth. Since Q is smooth, ${\mathrm{Res}}_Q\left(mS\right)=(m-1)S$. The residual exact sequence of Q is the following exact sequence

$$0\to {\mathcal{I}}_{(m-1)S}(2m-2)\to {\mathcal{I}}_{mS}\left(2m\right)\to {\mathcal{I}}_{Q\cap mS,Q}\left(2m\right)\to 0$$

(4)

The inductive assumption gives $h^1\left({\mathcal{I}}_{(m-1)S}(2m-2)\right)=0$. Thus the long cohomology exact sequence of (4) shows that to conclude the proof on the proposition it is sufficient to prove that $h^1(Q,{\mathcal{I}}_{Q\cap mS,Q}\left(2m\right))=0$. Since Q is a smooth quadric surface, $Q\cong {\mathbb{P}}^1\times {\mathbb{P}}^1$ and ${\mathcal{O}}_Q\left(1\right)\cong {\mathcal{O}}_{{\mathbb{P}}^1\times {\mathbb{P}}^1}(1,1)$. Since C is an irreducible degree 3 curve contained in Q, up to renaming the two rulings of Q we may assume that $C\in |{\mathcal{O}}_Q(2,1)|$. Consider the residual exact sequence of C in ${\mathbb{P}}^1\times {\mathbb{P}}^1$:

$$\begin{array}{c}0\to {\mathcal{I}}_{(m-1)S\cap Q,Q}(2m-2,2m-1)\to {\mathcal{I}}_{mS\cap Q,Q}(2m,2m)\to \hfill \\ {\mathcal{I}}_{C\cap mS,C}(2m,2m)\to 0\hfill \end{array}$$

(5)

Since $\mathrm{deg}\left({\mathcal{O}}_C(2m,2m)\right)=6m$, Remark 3 gives $h^1(C,{\mathcal{I}}_{C\cap mS,C}(2m,2m))=0$. By induction on m we have $h^1(Q,{\mathcal{I}}_{(m-1)S\cap Q,Q}(2m-2,2m-2))=0$. Hence we obtain that $h^1(Q,{\mathcal{I}}_{(m-1)S\cap Q,Q}(2m-2,2m-1))=0$. Thus it is sufficient to use the long cohomology exact sequence of (5). □

Proposition 6.

Fix $S\in S({\mathbb{P}}^3,6)$ which is not in linear general position and a positive integer $m\ge 4$. Then $h^1\left({\mathcal{I}}_{mS}\left(2m\right)\right)>0$ and $h^0\left({\mathcal{I}}_{mS}\left(2m\right)\right)>0$.

Proof. 

Since S is not in linear general position, there is $A\subset S$ such that A is contained in a plane and $\#A\ge 4$. If $\#A\ge 5$ we use Remark 1 and Lemma 2. Now assume $\#A=4$. Write $E:=S\setminus A$. By the semicontinuity theorem for cohomology we may assume that A spans a plane, H, and that the line L spanned by E intersects H in a point $p\notin A$. The intersection of H with the base locus of $|{\mathcal{I}}_{mS}\left(2m\right)|$ contains $(mA,H)\cup \left(\right(m-1)p,H)$. Use Lemma 3. □

Corollary 3.

Fix an integer $m\ge 3$. Take a general $S\in S({\mathbb{P}}^3,6)$. There is no complete curve $T\subset S({\mathbb{P}}^3,6)$ such that all $A\in T$ satisfy the equality $h^0\left({\mathcal{I}}_{mA}\left(2m\right)\right)=h^0\left({\mathcal{I}}_{mS}\left(2m\right)\right)$.

Proof. 

Let E be the set of all $B\in S({\mathbb{P}}^3,6)$ which are in linear general position. By Propositions 6 and 5 it is sufficient to prove that E contains no complete curve. Assume that it has. Let $\tilde{S}({\mathbb{P}}^3,6)$ denote the set of all ordered sets of six points. Let $\tilde{E}$ denote the set of all pairs $(B,<)$, where $B\in E$ and < is an ordering of the six points of B. It is sufficient to prove that $\tilde{E}$ contains no complete curve. Assume that it contains a complete curve, D. Fix any $A\in S({\mathbb{P}}^3,5)$ in linear general position and order its points, $p_1,\dots ,p_5$. Up to the use of a family of elements of $PGL\left(4\right)$ we may assume that the points $p_1,\dots ,p_5$ are the first five points of all elements of all $(B,<)\in D$. Let $\mathcal{H}$ be the union of the $\left({\displaystyle \genfrac{}{}{0pt}{}{5}{2}}\right)$ planes spanned by three of the points of A. The sixth point of all $B\in D$ is a point of ${\mathbb{P}}^3\setminus \mathcal{H}$. The contradiction comes from the fact that ${\mathbb{P}}^3\setminus \mathcal{H}$ is an affine variety and hence it contains no complete curve. If all $B\in D$ have at least four non-coplanar points as $p_1,p_2,p_3,$ and $p_4$, then we use them and Lemma 3. The remaining case (all points collinear) is easier. □

Proposition 7.

$S({\mathbb{P}}^3,7)$ contains an irreducible set G of codimension 2 such that for all $S\in G$ and all positive integers m and d we have $h^1\left({\mathcal{I}}_{mS}\left(d\right)\right)\ge \max \{0,7m-3d-1\}$.

Proof. 

Let G be the set of all $S\in S({\mathbb{P}}^3,7)$ which are contained in a rational normal curve $C_S$. Each $S\in G$ is in linear position. Each $A\in S({\mathbb{P}}^3,6)$ in linear general position is contained in a unique rational normal curve. Thus each $S\in G$ is contained in a unique rational normal curve, G is irreducible and $\dim G=\dim S({\mathbb{P}}^3,6)+1=19$. Hence G has codimension two. Fix $S\in G$ and set $C:=C_S$. For all positive integers m and d we have $\mathrm{deg}(C\cap mS)=7m$ and $h^0(C,{\mathcal{O}}_C\left(d\right))=3d+1$. The cohomology of line bundles on $C\cong {\mathbb{P}}^1$ also gives $h^1(C,{\mathcal{I}}_{C\cap mS,C}\left(d\right))=\max \{0,7m-3d-1\}$. Remarks 1 and 3 give $h^1\left({\mathcal{I}}_{mS}\left(d\right)\right)\ge \max \{0,7m-3d-1\}$. Since $h^0\left({\mathcal{I}}_C\left(2\right)\right)=3$ and C is arithmetically Cohen–Macaulay, we have $h^1\left({\mathcal{I}}_S\left(2\right)\right)=0$. Obviously, $h^0\left({\mathcal{I}}_{2S}\left(2\right)\right)=0$. □

Remark 19.

Take a general $S\in S({\mathbb{P}}^3,x)$ and an integer $m\ge 2$. By Theorem 1 for all $2\le x\le 7$ we have $h^1\left({\mathcal{I}}_{mS}(2m-2)\right)>0$ and $h^0\left({\mathcal{I}}_{mS}(2m-2)\right)>0$ for all $m\gg 0$. In each of these cases it is easy to check the minimum integer m such that $h^0\left({\mathcal{I}}_{mS}(2m-2)\right)>0$. The case $x=6$ of Proposition 5 and Lemma 1 give that $h^1\left({\mathcal{I}}_{mS}(2m-1)\right)\ge 4$ for a general $S\in S({\mathbb{P}}^3,7)$. For $x=9$ the Alexander–Hirschowitz theorem gives $h^1\left({\mathcal{I}}_{2S}\left(4\right)\right)=2$ and $h^0\left({\mathcal{I}}_{2S}\left(4\right)\right)=1$ for a general $S\in S({\mathbb{P}}^3,9)$ [9,10,11,12].

Lemma 5.

Let $Q\subset {\mathbb{P}}^3$ be a smooth quadric surface. Let $C\subset Q$ be an integral curve which is the complete intersection of Q and another quadric surface. Take $S\in S(C_{\mathrm{reg}},8)$ which is not the complete intersection of C and another quadric surface. We have $h^0(Q,{\mathcal{I}}_{S,Q}\left(2\right))=1$. For all integers $m\ge 2$ we have $h^1(Q,{\mathcal{I}}_{mS,Q}(2m+1))=h^0(Q,{\mathcal{I}}_{mS,Q}(2m-1))=0$ and

$$h^0(Q,{\mathcal{I}}_{mS,Q}\left(2m\right))=1+h^1(Q,{\mathcal{I}}_{mS,Q}\left(2m\right)).$$

We have $h^1(Q,{\mathcal{I}}_{mS,Q}\left(2m\right))>0$ for some $m\ge 2$ if and only if ${\mathcal{O}}_C\left(2\right)(-S)$ is a torsion element of ${\mathrm{Pic}}^0\left(C\right)$. Moreover, if $h^1(Q,{\mathcal{I}}_{(m-1)S,Q}(2m-2))>0$, then $h^1\left({\mathcal{I}}_{mS,Q}\left(2m\right)\right)>0$. Assume C is smooth. The set of all $S\in S(C,8)$ such that $h^1(Q,{\mathcal{I}}_{mS,Q}\left(2m\right))=0$ for all $m\ge 2$ is a second category subset of $S(C,8)$, but it contains no open subset of it.

Proof. 

Since $\mathrm{deg}\left(\right(mS,Q\left)\right)=4(m+1)m$ and $h^0(Q,{\mathcal{O}}_Q\left(2m\right))={(2m+1)}^2$, we have $h^0(Q,{\mathcal{I}}_{mS,Q}\left(2m\right))=1+h^1(Q,{\mathcal{I}}_{mS,Q}\left(2m\right))$ for all m. Since S is not the complete intersection of C and another quadric surface, $h^0(Q,{\mathcal{I}}_{S,Q}\left(2\right))=1$. Hence $h^1(Q,{\mathcal{I}}_{S,Q}\left(2\right))=0$, $h^0(Q,{\mathcal{I}}_{S,Q}\left(1\right))=0$ and $h^1(Q,{\mathcal{I}}_{S,Q}\left(3\right))=0$. Since $S\subset C_{\mathrm{reg}}$, we have the following residual exact sequence of C in Q:

$$0\to {\mathcal{I}}_{(m-1)S,Q}(t-2)\to {\mathcal{I}}_{mS,Q}\left(t\right)\to {\mathcal{I}}_{mS\cap C,C}\left(t\right)\to 0$$

(6)

Note that $\mathrm{deg}\left({\mathcal{O}}_C\left(t\right)\right)=4$ and $\mathrm{deg}\left(\right(mS,C\left)\right)=8m$. We have $h^1(C,{\mathcal{I}}_{mS\cap C,C}\left(t\right))=0$ if and only if either $4t>8m$ or $4t=8m$ and ${\mathcal{O}}_C\left(t\right)(-mS)\ne {\mathcal{O}}_C$. We have $h^0(C,{\mathcal{I}}_{mS\cap C,C}\left(t\right))=0$ if and only if either $4t<8m$ or $4t=8m$ and ${\mathcal{O}}_C\left(t\right)(-mS)\ne {\mathcal{O}}_C$. Thus it is sufficient to use the long cohomology exact sequence of (6) and induction on the integer m. To prove the lemma, except the “Moreover” part, we only need to check that ${\mathcal{O}}_C\left(2\right)\ne {\mathcal{O}}_C\left(S\right)$. Assume ${\mathcal{O}}_C\left(2\right)\cong {\mathcal{O}}_C\left(S\right)$, i.e., assume $S\in |{\mathcal{O}}_C\left(2\right)|$. Since C is arithmetically Cohen–Macaulay, S is the complete intersection of C and another quadric surface, a contradiction.

Now we prove the “Moreover” part. Assume $h^1(Q,{\mathcal{I}}_{(m-1)S,Q}(2m-2))>0$ and $h^1\left({\mathcal{I}}_{mS,Q}\left(2m\right)\right)=0$. The long cohomology exact sequence of (6) gives the inequality $h^1(C,{\mathcal{I}}_{(mS,C}\left(2m\right))>0$. Thus $h^1(Q,{\mathcal{I}}_{(mS,Q}\left(2m\right))>0$ (Remark 1), a contradiction.

Now we assume C is smooth. A set $S\in S(C,8)$ satisfies $h^1(Q,{\mathcal{I}}_{mS,Q}\left(2m\right))>0$ if and only if ${\mathcal{O}}_C\left(2\right)(-mS)\cong {\mathcal{O}}_C$. Hence being a second category subset and the non-existence of the open set U follow from Remarks 9, 10 and 15. □

Notation 1.

Let $\mathcal{A}\left(8\right)$ be the set of all $S\in S({\mathbb{P}}^3,8)$ such that $h^0\left({\mathcal{I}}_S\left(2\right)\right)=2$ and S is contained in the smooth locus of a degree 4 integral curve $C_S\subset {\mathbb{P}}^3$.

Lemma 6.

Take $S\in \mathcal{A}\left(8\right)$. The curve $C_S$ is uniquely determined by S and it is contained in a smooth quadric surface. Let U be the set of all $S\in \mathcal{A}\left(8\right)$ such that $C_S$ is smooth. U is a nonempty open subset of $S({\mathbb{P}}^3,8)$. The set $S({\mathbb{P}}^3,8)\setminus \mathcal{A}\left(8\right)$ has codimension 2 in $S({\mathbb{P}}^3,8)$, while the set $S({\mathbb{P}}^3,8)\setminus U$ has codimension 1 in $S({\mathbb{P}}^3,8)$.

Proof. 

Let G be the set of all integral degree 4 space curves which are the complete intersection of two quadric surfaces. The set G is a smooth and irreducible subset of the Hilbert scheme of ${\mathbb{P}}^3$. Take $C\in G$. The adjunction formula gives ${\omega }_C\cong {\mathcal{O}}_C$. Since C is integral, it has arithmetic genus 1. Thus either C is smooth and hence an elliptic curve or it has a unique singular point which is either an ordinary node or an ordinary cusp. Take a general $Q\in |{\mathcal{I}}_C\left(2\right)|$. Since C is integral and non-degenerate, Q is irreducible. Assume that Q is singular. Hence it has a unique singular point, o, and Q is a cone with vertex o. Take another general $Q^{\prime }\in \left|{\mathcal{I}}_C\left(2\right)\right|$ and call $o^{\prime }$ its singular point. If $o\ne o^{\prime }$, then $C=Q\cap Q^{\prime }$ contains the line spanned by $\{o,o^{\prime }\}$, a contradiction. If $o=o^{\prime }$, then $Q\cap Q^{\prime }$ is a union of lines containing o, a contradiction. Take any $C\in E$. We have $S(C_{\mathrm{reg}},8)\subset \mathcal{A}\left(8\right)$. Each $S(C_{\mathrm{reg}},8)$ is irreducible of dimension 8.

Take $S\in \mathcal{A}\left(8\right)$. Since $h^0\left({\mathcal{I}}_S\left(2\right)\right)=2$, $C_S\in E$. Hence $C_S$ spans ${\mathbb{P}}^3$ and it has arithmetic genus 1. Thus either $C_S$ is smooth or it has a unique singular point which is either an ordinary cusp or an ordinary node. Conversely, every integral and non-degenerate degree 4 curve $C\subset {\mathbb{P}}^3$ is the complete intersection of two quadric surfaces. The singular ones occur in codimension 1. Thus $S({\mathbb{P}}^3,8)\setminus U$ has codimension 1 in $S({\mathbb{P}}^3,8)$.

To conclude the proof of the lemma it is sufficient to prove that $S({\mathbb{P}}^3,8)\setminus \mathcal{A}\left(8\right)$ has codimension at least two in $S({\mathbb{P}}^3,8)$. We prove it by proving that it is a finite union of algebraic sets, each of them of dimension of at most 22. Note that $h^0\left({\mathcal{I}}_S\left(2\right)\right)\ge 2$ for all $S\in S({\mathbb{P}}^3,8)$. Let E be the set of all $S\in S({\mathbb{P}}^3,8)$ contained in a reducible quadric surface. A reducible quadric surface is the union of two planes and the set of all planes has dimension 3. Since $\dim S(Q_1,8)=16$ for all quadrics $Q_1$, we have $\dim E=22$. Let $E_1$ be the set of all $S\in (S({\mathbb{P}}^3,8)\setminus E)$ such that $h^0\left({\mathcal{I}}_S\left(2\right)\right)>2$. Fix $S\in E_1$. Since each quadric containing S is irreducible, $E_1$ is formed by the complete intersections of three quadric surfaces. Since $\dim |{\mathcal{O}}_{{\mathbb{P}}^3}\left(2\right)|=9$, the Grassmannian of all two dimensional linear subspaces of $|{\mathcal{O}}_{{\mathbb{P}}^3}\left(2\right)|$ has dimension $3(10-7)=21$ and hence $\dim E_1=21$. The set $E_2:=S({\mathbb{P}}^3,8)\setminus (\mathcal{A}\left(8\right)\cup E\cup E_1)$ is formed by the sets S with $C_S$ singular and with S containing the singular point of $C_S$. Since the set of all singular $C\in G$ has dimension 15, we get $\dim E_2=15+7=22$. □

Theorem 8.

Take $S\in \mathcal{A}\left(8\right)$ and a positive integer m.

(a) 

We have $h^0\left({\mathcal{I}}_{mA}\left(2m\right)\right)=m+1+h^1\left({\mathcal{I}}_{mA}\left(2m\right)\right)$ and $h^1\left({\mathcal{I}}_{mA}(2m-1)\right)>0$ for all $A\in S({\mathbb{P}}^3,8)$.

(b) 

We have $h^1\left({\mathcal{I}}_{mS}(2m+1)\right)=h^0\left({\mathcal{I}}_{mS}(2m-1)\right)=0$.

(c) 

We have $h^1\left({\mathcal{I}}_{kS}\left(2k\right)\right)>0$ for some positive integer k if and only if ${\mathcal{O}}_C\left(2\right)(-S)$ is a torsion element of ${\mathrm{Pic}}^0\left(C\right)$.

(d) 

The set of all $S\in \mathcal{A}\left(8\right)$ such that $h^1\left({\mathcal{I}}_{kS}\left(2k\right)\right)=0$ for all $k\ge 2$ is a second category subset of $\mathcal{A}\left(8\right)$, but it contains no Zariski open subset of $S({\mathbb{P}}^3,8)$.

(e) 

Take $S\in \mathcal{A}\left(8\right)$. If $h^1\left({\mathcal{I}}_{mS}\left(2m\right)\right)>0$, then $h^1\left({\mathcal{I}}_{zS}\left(2z\right)\right)>0$ for all $z>m$.

Proof. 

We have $\left({\displaystyle \genfrac{}{}{0pt}{}{2m+3}{3}}\right)=(2m+3)(2m+2)(2m+1)/6=8\left({\displaystyle \genfrac{}{}{0pt}{}{m+2}{3}}\right)+m+1$. Part (a) is true because $\left({\displaystyle \genfrac{}{}{0pt}{}{2m+2}{3}}\right)<\mathrm{deg}\left(mA\right)$.

Fix $S\in \mathcal{A}\left(8\right)$ and let $C\subset {\mathbb{P}}^3$ be the integral degree 4 curve containing S in its smooth locus. For all positive integers m we have $\mathrm{deg}(mS\cap C)=8m$. Remark 9 gives $h^1(C,{\mathcal{I}}_{mS\cap C,C}\left(t\right))=0$ if $t\ge 2m+1$, $h^1(C,{\mathcal{I}}_{mS\cap C,C}\left(t\right))>0$ if $t\le 2m-1$, $h^0(C,{\mathcal{I}}_{mS\cap C}\left(2m\right))=h^1(C,{\mathcal{I}}_{2S\cap C,C}\left(2m\right))\le 1$ and $h^0(C,{\mathcal{I}}_{mS\cap C,C}\left(2m\right))>0$ if and only if $mS\cap C\in |{\mathcal{O}}_C\left(2m\right)|$.

By Lemma 6 C is contained in a smooth quadric surface, Q. Since Q is smooth and $S\subset Q$, ${\mathrm{Res}}_Q\left(mS\right)=(m-1)S$. Thus the residual exact sequence of Q is the following exact sequence:

$$0\to {\mathcal{I}}_{(m-2)S}(x-2)\to {\mathcal{I}}_{mS}\left(x\right)\to {\mathcal{I}}_{mS,Q}\left(x\right)\to 0$$

(7)

Note that Theorem 8 is true also for $m=0$. Since C is arithmetically Cohen–Macaulay, the assertion that S is not the complete intersection of C and another quadric surface is exactly the condition assumed in Lemma 5, which gives parts (b) and (c).

Part (d) follows from Lemma 5 varying the curve C. Part (e) follows from (7), the similar statement of Lemma 5 and Remark 1. □

Remark 20.

Remarks 15 and 16 apply to part (d) of Theorem 8 with no modification.

Notation 2.

For all integers $z\ge 9$ set $a(z,1):=\lceil (z+1)/4\rceil$ and $a_1(z,1):=\lceil z/4\rceil$. Let $E\left(z\right)$ denote the set of all $S\in S({\mathbb{P}}^3,z)$ contained in the smooth locus of a linearly normal degree 4 curve.

Lemma 7.

Fix an integer $z\ge 9$ and $S\in E\left(z\right)$. Let C be linearly normal integral degree 4 curve containing S and Q any smooth quadric surface containing C. We have $h^0(Q,{\mathcal{I}}_{mS,Q}(2m-1))=0$ and $h^0(Q,{\mathcal{I}}_{mS,Q}\left(2m\right))=2m+1$. We have $h^1(Q,{\mathcal{I}}_{mS,Q}\left(t\right))=0$ for all $t\ge a\left(z\right)$. There is a second category subset of $S(C,z)$ such that $h^1(Q,{\mathcal{I}}_{mS,Q}\left(a_1\left(z\right)\right))=0$ for all $t\ge a_1\left(z\right)$.

Proof. 

Lemma 6 gives the existence of the smooth quadric Q. Note that either $a\left(z\right)=a_1\left(z\right)$ or $a\left(z\right)=1+a_1\left(z\right)$. Use Remark 9. □

Theorem 9.

We have $h^0\left({\mathcal{I}}_{mS}\left(2m\right)\right)=m+1$ and $h^0\left({\mathcal{I}}_{mS}(2m-1)\right)=0$ for all $z\ge 9$ and all $S\in E\left(z\right)$. Define recursively for all $m\ge 2$ the integers $a(z,m)$ and $a_1(z,m)$ in the following way. Let $a(z,m)$ be the maximum between $2+a(z,m-1)$ and the maximal integer x such that $4x>mz$. Let $a_1(z,m)$ be the maximum between $2+a_1(z,m-1)$ and the maximal integer x such that $4x\ge mz$. We have $h^1\left({\mathcal{I}}_{mS}\left(a\left(z\right)\right)\right)=0$. The set of all $S\in E\left(z\right)$ such that $h^1\left({\mathcal{I}}_{mS}\left(a_1\left(z\right)\right)\right)=0$ for all m is a second category subset of $E\left(z\right)$.

Proof. 

Fix $S\in E\left(z\right)$ and let $C\subset {\mathbb{P}}^3$ be the degree 4 curve containing S in its smooth locus. Since $z>8$ and C is irreducible, the theorem of Bezout gives $h^0\left({\mathcal{I}}_S\left(2\right)\right)=h^0\left({\mathcal{I}}_C\left(2\right)\right)$. Since $h^0\left({\mathcal{I}}_C\left(2\right)\right)=2$ and the symmetric product m times of a two-dimensional vector space has dimension $m+1$, $h^0\left({\mathcal{I}}_{mS}\left(2m\right)\right)\ge m+1$. Lemma 7 and induction on m gives $h^0\left({\mathcal{I}}_{mS}\left(2m\right)\right)=m+1$ and $h^0\left({\mathcal{I}}_{mS}(2m-1)\right)=0$ for all m. For the other statements use induction on m. □

Example 2.

Fix $A,B\in S({\mathbb{P}}^n,n+1)$ such that A is linearly independent, but B is not linearly independent. We want to prove that for all $m\ge 2$, the schemes $mA$ and $mB$ have different Hilbert functions. Let H be a hyperplane containing B. Note that $mH$ contains $mB$. We claim that $h^0\left({\mathcal{I}}_{mA}\left(m\right)\right)=0$. Assume $h^0\left({\mathcal{I}}_{mA}\left(m\right)\right)\ne 0$ and take $W\in |{\mathcal{I}}_{mA}\left(m\right)|$. We use induction on the integer n. First assume $n=2$. The theorem of Bezout implies that W contains each line contained two of the three points of A and with multiplicity m. Hence $\mathrm{deg}\left(W\right)\ge 2m$, a contradiction. Now assume $n\ge 3$. By the inductive assumption W contains each hyperplane M spanned by n of the points of A. Using the residual exact sequence of M we get that W contains M with multiplicity at least m. Thus $\mathrm{deg}\left(W\right)>m$, a contradiction.

Example 3.

Take $x\le n+1$. Let $U\left(x\right)$ be the set of all $S\in S({\mathbb{P}}^n,x)$ which are linearly independent. Ordering the x-points and taking homogeneous coordinates we see that U is represented by the set of all $x\times (n+1)$ matrices with rank x. Thus $S({\mathbb{P}}^n,x)\setminus U\left(x\right)$ has a codimension at least two in $S({\mathbb{P}}^n,x)$ if $x\le n$, while it is a hypersurface (the zero of the determinant) if $x=n+1$. We also see that $U(n+1,n+1)$ is an affine variety. To see that even the part for a fixed (but large) m is false it is sufficient to use Example 2.

Corollary 4.

There is $G\subset S({\mathbb{P}}^3,7)$ of codimension 2 in G such that $h^1\left({\mathcal{I}}_{mS}\left(2m\right)\right)\ge m-1$ and $h^0\left({\mathcal{I}}_{mS}\left(2m\right)\right)-h^1\left({\mathcal{I}}_{mS}\left(2m\right)\right)\ge \left({\displaystyle \genfrac{}{}{0pt}{}{m+2}{3}}\right)+2m$ for all $m\ge 2$.

Proof. 

Use the case $d=2m$ of Proposition 7. □

For more details on the next result and its extension to the case $3\le c\le 6$, see [24,25]. For a similar range of integers $(c,g)$ in higher dimensional projective spaces, see [26,27].

Theorem 10.

Assume $\mathrm{char}\left(\mathbb{K}\right)=0$. Fix an integer $c\ge 7$ and set $x:=2c+1$. Let $g_c$ be the maximal integer $g\ge c-3$ such that $4g\ge 3c+12$. There is an irreducible set $G\subset S({\mathbb{P}}^3,x)$ of codimension 2 in $S({\mathbb{P}}^3,x)$ such that for all positive integer m and d we have $h^1\left({\mathcal{I}}_{mS}\left(c\right)\right)\ge mx+g_c-1-cd$.

Proof. 

Let $W(c,g_c)$ denote the irreducible component of the Hilbert scheme of ${\mathbb{P}}^3$ containing the degree c embeddings of the genus $g_c$ curves with general moduli [24,25]. It is known that $\dim W(c,g_c)=4c$ and that a general $S\in S({\mathbb{P}}^3,2c)$ is contained in some $C_S\in W(c,g)$ [24]. Set $C:=C_S$. As in Proposition 7 the set G is the set of all $S\in S({\mathbb{P}}^3,x)$ contained in a curve $C\in W(c,g_c)$. We use that $\mathrm{deg}(mS\cap C)=mx$, ref. [23] and that for a fixed $C\in W(c,g_c)$ and a general $S\in S({\mathbb{P}}^3,x)$ the scheme $C\cap mS$ is a general union of x m-points of C. □

8. m-Terracini Sets

In this section we fix an n-dimensional integral projective variety X, a very ample line bundle L on X and a linear subspace $V\subseteq H^0\left(L\right)$ which embeds $X_{\mathrm{reg}}$. For all zero-dimensional schemes $Z\subset X$ set $V(-Z):=V\cap H^0({\mathcal{I}}_Z\otimes L)$. We also fix an integer $m\ge 2$ and study the m-Terracini sets of $(X,V)$, i.e., the integers x and the sets $S\in S(X_{\mathrm{reg}},x)$ such that $V(-mS)\ne \varnothing$ and $\dim V(-mS)>v-x\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)$. We say that $(X,V,m)$ is not defective or that $(X,V)$ is not m-defective if for all positive integers x the zero-dimensional scheme $mS$ has maximal rank with respect to V for a general $S\in S(X_{\mathrm{reg}},x)$. The case $m=2$ is classical, but usually done in characteristic 0, where 2-defectivity is equivalent to the usual defectivity of the secant varieties of X. In [1] and references therein there is a characteristic 0 assumption, which was used in several proofs and in many references used on this topic. See [1] for a study of the codimension 1 2-Terracini loci and other general results on when 2-Terracini loci occur in codimension 1, i.e., the set $T(V,2,x)\subset S(X_{\mathrm{reg}},x)$ is nonempty, $T(V,2,x)\ne S(X_{\mathrm{reg}},x)$ and $\dim T(V,2,x)=nx-1$. In the proofs contained in many of these papers the note [23] by Ciliberto and Miranda was used several times and it cannot be used in the positive characteristics as we saw at the end of Remark 18. The results proved in this section require no assumption on the characteristic of the base field and the proofs given here are characteristic-free.

The cohomology of line bundles on ${\mathbb{P}}^1$ shows that no Veronese embedding of ${\mathbb{P}}^1$ has m-Terracini sets. Thus for the Veronese embeddings we only consider the ones of projective spaces of dimension $n>1$.

Set $\sigma :=\lfloor v/\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{m}}\right)\rfloor$ and $\rho :=\lceil v/\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)\rceil$.

Remark 21.

$(X,V,m)$ is not defective if and only if $V(-mA)=0$ for a general $A\in S(X_{\mathrm{reg}},\rho )$ and $\dim V(-mB)=v-\sigma \left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)$ for a general $S\in S(X_{\mathrm{reg}},\sigma )$.

Remark 22.

Take a zero-dimensional scheme $Z\subset X$ and $P\in X_{\mathrm{reg}}\setminus Z_{\mathrm{red}}$. By Remark 1 if $\dim V(-Z)>v-\mathrm{deg}\left(Z\right)$, then $\dim V(-Z-mP)>\dim V-\mathrm{deg}\left(Z\right)-\mathrm{deg}\left(mP\right)$.

Proposition 8.

Assume $n\ge 2$ and that $\dim T(V,m,y)\ge ny-1$ for an integer $y<\sigma$. Then $\dim T(V,m,x)\ge nx-1$ for all $y\le x\le \sigma$. If $(X,V)$ is not m-secant defective, then $\dim T(V,m,x)=nx-1$ for all $y\le x\le \sigma$.

Proof. 

Take an irreducible component K of $T(V,m,y)$ of maximal dimension. By assumption either $\dim K=ny$ of $\dim K=ny-1$. For all $y<x\le \sigma$ let $K_x$ be the set of all unions of $A\in K$ and a set $E\subset X_{\mathrm{reg}}\setminus A$ such that $\#E=x-y$. Note that $K_x$ is irreducible and $\dim K_x=\dim K+n(x-y)$. Take any $S\in K_x$. By Remark 22 we have $\dim V(-mS)>v-x\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{m}}\right)$. Since $x\le \sigma$ and $\dim V(-mS)>v-x\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{m}}\right)$, we get $V(-mS)\ne \varnothing$. Hence $S\in T(V,m,x)$. Thus $K_x\subseteq T(V,m,x)$ and hence $\dim T(V,m,x)\ge nx-1$. If $(X,V)$ is not m-defective, then $\dim T(V,m,x)\le nx-1$, concluding the proof. □

Lemma 8.

Fix a positive integer z such that $z\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)\ge v$ and $T(V,m,z)\ne \varnothing$. For all integers $x>z$ we have $\dim T(V,m,x)\le n(x-z)+\dim T(V,m,z)$.

Proof. 

For all $w\ge z$ we have $w\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)\ge v$. Hence the set $T(V,m,w)$ is the set of all $B\in S(X_{\mathrm{reg}},w)$ such that $V(-mB)\ne 0$. Fix $B\in T(V,m,x)$ and take any $A\subset B$ such that $\#A=z$. Since $V(-mB)\ne 0$, $V(-mA)\ne 0$ and hence $A\in T(V,m,z)$. Since we have $\dim S(X_{\mathrm{reg}},x-y)=n(x-y)$, we get the lemma. □

Proposition 9.

Fix a positive integer z such that $\dim V(-mS)=0$ for a general $S\in S(X_{\mathrm{reg}},z)$ and take integers $c\ge 1$ and $x\ge cz$. Then $\dim T(V,x)\le nx-c$.

Proof. 

The case $c=1$ is true, because $\dim T(V,m,z)<nz$ by assumption. By Lemma 8 it is sufficient to do the case $x=cz$. Fix an irreducible component K of $T(V,x)$. Fix a general $S\in K$ and call $p_{i,j}$, $1\le i\le z$, $1\le j\le c$, its points. Set $S_j:=\{p_{1,j},\dots ,p_{z,j}\}$, $1\le j\le c$. Remark 22 gives $S_j\in T(V,m,z)$ for all j. Since the set of all $S_j$ for some $S\in K$ has dimension at most $nz-1$, we get that K has at least codimension c in $S(X_{\mathrm{reg}},x)$. □

Question 2: Take X, V, m such that $x:=\dim V/\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)\in \mathbb{N}$. When is $\dim T(V,m,x)\ge nx-1$, if $T(V,m,x)\ne S(X_{\mathrm{reg}},x)$ is $T(V,m,x)$ equidimensional of codimension 1?

Question 3: Describe the algebraic set $T(V,m,x)$ when $\dim T(V,m,x)=nx-1$.

In the case $X={\mathbb{P}}^n$, $S({\mathbb{P}}^n,x)$ and $L={\mathcal{O}}_{{\mathbb{P}}^n}\left(d\right)$ a quadruple $(n,x,m,d)$ of positive integers is said to be perfect if $\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)=x\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{m}}\right)$. Note that x is uniquely defined by n, m and d and a triple $(n,m,d)$ is a part of a perfect quadruple if and only if $\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)/\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)$ is an integer. A quadruple $(n,x,m,d)$ is perfect if and only if $h^1\left({\mathcal{I}}_{mS}\left(d\right)\right)=h^0\left({\mathcal{I}}_{mS}\left(d\right)\right)$ for some $S\in S({\mathbb{P}}^n,x)$. If $(n,x,m,d)$ is perfect, then $h^1\left({\mathcal{I}}_{mS}\left(d\right)\right)=h^0\left({\mathcal{I}}_{mS}\left(d\right)\right)$ for all $S\in S({\mathbb{P}}^n,x)$.

Question 4: Fix integers $n\ge 2$, $m\ge 2$ and $x\ge 2$. Assume that $(n,x,m,d)$ is a perfect quadruple and that $mS$ has maximal rank in degree d for a general $S\in S({\mathbb{P}}^n,x)$. Is the set of all $S\in S({\mathbb{P}}^n,x)$ such that $h^1\left({\mathcal{I}}_{mS}\left(d\right)\right)>0$ either empty or of codimension 1, except in a few cases? Is it nonempty except in a few cases which can be listed? Take an integer $e\ge 2$. Give conditions such that the set of all $S\in S({\mathbb{P}}^n,x)$ such that $h^1\left({\mathcal{I}}_{mS}\left(d\right)\right)\ge e$ has codimension at least e in $S({\mathbb{P}}^n,x)$.

For $n=m=2$ the perfect quadruples arise for $d\equiv 1,2\left(\mathrm{mod}3\right)$ and they are all described in [1]. For any $P\in {\mathbb{P}}^2$ we have $\mathrm{deg}\left(3P\right)=6$. The integers d such that $\left({\displaystyle \genfrac{}{}{0pt}{}{d+2}{2}}\right)\equiv 0\left(\mathrm{mod}6\right)$ are the integers $d\equiv 2,7,10,11\left(\mathrm{mod}12\right)$. For $n=2$ and $m=3$ all defective cases are known. So it should be possible to extend the part of [1] to the case $m=3$. Instead of the known theory of Severi on plane nodal curves one should heavily use the Differential Horace Lemma. For all n and m there are infinitely many perfect quadruples $(n,m,x,d)$. For instance, one can take $d=-n+{\prod }_{i=1}^n(n+m-1)$ and hence $x=\left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)/\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{m}}\right)\in \mathbb{Z}$.

Question 5: Fix integers $n\ge 2$, $m\ge 2$, $x\ge 2$ and $e\ge 2$. Give conditions such that the set of all $S\in S({\mathbb{P}}^n,x)$ such that

$$\min \{h^1\left({\mathcal{I}}_{mS}\left(d\right)\right),h^0\left({\mathcal{I}}_{mS}\left(d\right)\right)\}\ge e$$

has codimension at least e in $S({\mathbb{P}}^n,x)$.

Remark 23.

The most important case of the last part of Question 5 is the case $e=2$. The case $(n,x,m,3)=(2,6,3,7)$ is an example in which $h^1\left({\mathcal{I}}_{mS}\left(d\right)\right)\ge 2$ occurs in codimension 1 (see Example 4). This case occurs because $(n,x,m,3)=(2,6,1,2)$ fails in codimension 1 and this failure propagates to a perfect quadruple. Is it a way to get long lists of exceptional cases coming from “lower” quadruples? Even non-perfect quadruples, but not with maximal rank?

Example 4.

Take $n=2$, $m=3$ and $d=7$. The quadruple $(2,6,3,7)$ is perfect. Take $S\in S({\mathbb{P}}^2,6)$ such that $|{\mathcal{I}}_S\left(2\right)|\ne \varnothing$ and take $D\in |{\mathcal{I}}_S\left(2\right)|$. Since $\mathrm{deg}(D\cap 3S)\ge 18$ and $h^0(D,{\mathcal{O}}_D\left(7\right))=15$, $h^1\left({\mathcal{I}}_{3S}\left(7\right)\right)\ge 3$. The set J of all $S\in S({\mathbb{P}}^2,6)$ such that $|{\mathcal{I}}_S\left(2\right)|\ne \varnothing$ has dimension 11 and hence it has codimension 1 in $S({\mathbb{P}}^2,6)$.

Example 5.

Let U be the open subset of all $S\in S({\mathbb{P}}^2,9)$ contained in a plane cubic, $C_S$, with $C_S$ integral and S contained in the smooth locus of $C_S$. We do not assume that $C_S$ is unique. For all $m\ge 2$ set $W:=H^0\left({\mathcal{O}}_{{\mathbb{P}}^2}(3m-1)\right)$, $V:=H^0\left({\mathcal{O}}_{{\mathbb{P}}^2}(3m-1)\right)$ and $W_1:=H^0\left({\mathcal{O}}_{{\mathbb{P}}^2}(3m+1)\right)$. By Theorem 4 we have $T(W,m,9)\cap U=\varnothing$ and $T(W_1,m,9)=\varnothing$. It is important to allow the case of the complete intersections of two cubics, because they occur in codimension 1 in $S({\mathbb{P}}^2,9)$. Theorem 4 gives that the complement of ${\cup }_{m\ge 2}T(V,m,9)$ is a second category subset of $S(({\mathbb{P}}^2,9)$, but that ${\cup }_{m\ge 2}T(V,m,9)$ is dense in $S({\mathbb{P}}^2,9)$

Example 6.

Theorem 8 describes the intersection with $\mathcal{A}\left(8\right)$ of the Terracini sets of $S({\mathbb{P}}^3,8)$. We proved that $S({\mathbb{P}}^3,8)\setminus \mathcal{A}\left(8\right)$ has a codimension at least two in $S({\mathbb{P}}^3,8)$ (Lemma 6). As in Example 5 the union of all m-Terracini sets of $S({\mathbb{P}}^3,8)$ is Zariski dense in $S({\mathbb{P}}^3,8)$.

We think that the study of Hilbert functions of multiple sets for different multiplicities has many gold nuggets. In this section we describe the “worst” Hilbert functions, while very general elements of $S({\mathbb{P}}^n,x)$ have the best ones.

Proposition 10.

Fix a positive integer m and $S\in S({\mathbb{P}}^n,x)$ such that $h^1\left({\mathcal{I}}_{mS}(mx-2)\right)>0$. Then there is a line $L\subset {\mathbb{P}}^n$ such that $S\subset L$ and $h^1\left({\mathcal{I}}_{aS}\left(d\right)\right)=\max \{0,ax-d-1\}$ for all positive integers a and d.

Proof. 

By the cohomology of line bundles on ${\mathbb{P}}^1$ (Remark 5), it is sufficient to prove that S is contained in a line. Since this is obviously true for $n=1$, we may assume $n>1$ and use induction on the integer n. By the inductive assumption on n we may assume that S spans ${\mathbb{P}}^n$. Hence $x\ge n+1$. Since $n\ge 2$, we proved the cases $x\le 2$. Thus we may also use induction on the integer x and, by contradiction, take as x the first integer such that the proposition fails for the integer x. Fix a hyperplane H such that $e:=\#(S\cap H)$ is maximal. Set $S_1:=S\setminus S\cap H$. Since S spans H, $e\ge n$. The inductive assumption on x gives that either $S_1$ is contained in a line (and hence $h^1\left({\mathcal{I}}_{m{S}_1}\left(t\right)\right)=0$ for all $t\ge m(x-e)-1$) or $h^1\left({\mathcal{I}}_{m{S}_1}(m(x-e)-2)\right)=0$. In both cases we have $h^1\left({\mathcal{I}}_{m{S}_1}(m(x-e)-1)\right)=0$. Consider the residual exact sequence of the multiple divisor $mH$:

$$0\to {\mathcal{I}}_{m{S}_1}(mx-2-m)\to {\mathcal{I}}_{mS}(mx-2)\to {\mathcal{I}}_{m(S\cap H),H}(mx-2)\to 0$$

(8)

Since $e<x$, the inductive assumption gives $h^1(H,{\mathcal{I}}_{m(S\cap H,H}(mx-2))=0$. Since $mx-2-m\ge m(x-e)-1$, we have $h^1\left({\mathcal{I}}_{m{S}_1}(mx-2-m)\right)=0$. Hence the long cohomology exact sequence of (8) gives $h^1\left({\mathcal{I}}_{mS}(mx-2)\right)=0$, a contradiction. □

Proposition 11.

Fix a positive integer m and $S\in S({\mathbb{P}}^n,x)$ such that $h^1\left({\mathcal{I}}_{mS}(mx-2-m)\right)>0$. Then there is a line $L\subset {\mathbb{P}}^n$ such that $\#(S\cap L)\ge x-1$.

Proof. 

If S is contained in a line, then the proposition is true (Remark 5). Thus we may assume that $n\ge 2$ and that the proposition is true for a lower dimensional projective space. Thus we may assume that S spans ${\mathbb{P}}^n$. Hence $x\ge n+1$. Since $n\ge 2$, we proved the cases $x\le 2$. Thus we may also use induction on the integer x. By contradiction we may also assume that x is the first integer for which the proposition fails. Fix a hyperplane H such that $e:=\#(S\cap H)$ is maximal. Set $S_1:=S\setminus S\cap H$. Since S spans H, $e\ge n$. The inductive assumption on x gives that either $S_1$ is contained in a line (and hence $h^1\left({\mathcal{I}}_{m{S}_1}\left(t\right)\right)=0$ for all $t\ge m(x-e)-1$) or $h^1\left({\mathcal{I}}_{m{S}_1}(m(x-e)-2)\right)=0$. In both cases we have $h^1\left({\mathcal{I}}_{m{S}_1}(m(x-e)-1)\right)=0$. Since $e<x$, Proposition 10 gives $h^1(H,{\mathcal{I}}_{S\cap H,H}(me-2))\le 1$ with equality only if $S\cap H$ is contained in a line. Since S spans ${\mathbb{P}}^n$, the maximality of the integer e gives that $S\cap H$ is contained in a line if and only if $n=2$. Note that $e\ge n$.

First assume $n=2$. Hence H is a line. Thus $h^1(H,{\mathcal{I}}_{m(S\cap H,H}(me-1))=0$. Since $e\ge 2$, we have $h^1\left({\mathcal{I}}_{m{S}_1}(m(x-3)-1)\right)=0$. The long cohomology exact sequence of (8) gives $e=x-1$.

Now assume $n\ge 3$. Thus $h^1(H,{\mathcal{I}}_{S\cap H,H}(me-2))=0$. Since $e\ge 3$, we have $h^1(H,{\mathcal{I}}_{m(S\cap H),H}(mx-2-m))=0$. The long cohomology exact sequence of (8) gives $h^1\left({\mathcal{I}}_{m{S}_1}(mx-2-2m)\right)>0$. Since $x-e\le x-3$, we get a contradiction. □

Theorem 11.

Fix integers $n\ge 2$, $m\ge 2$, $c\ge 3$ and $d>0$ such that $d\le cm-2$ and $c\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)$. Take $X:={\mathbb{P}}^n$, $n\ge 2$, $L:={\mathcal{O}}_{{\mathbb{P}}^n}\left(d\right)$ and $V:=H^0\left(L\right)$.

(a) 

We have $T(V,m,c)\ne \varnothing$.

(b) 

Assume also $d\ge (c-1)m-1$, i.e., assume $c=\lceil (d+2)/m\rceil$. Then $T(V,m,x)=\varnothing$ for all $x<c$.

Proof. 

Note that $\mathrm{deg}\left(mA\right)=c\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+d}{n}}\right)$ for all $A\in S({\mathbb{P}}^n,c)$. Hence if $A\in S({\mathbb{P}}^n,c)$. and $h^1\left({\mathcal{I}}_{mA}\left(d\right)\right)>0$, then $A\in T(V,m,c)$. Fix a line $L\subset {\mathbb{P}}^n$ and take $S\in S(L,c)$. Since $\mathrm{deg}(L\cap mS)=mc\ge d+2$, we have $h^1(L,{\mathcal{I}}_{mS\cap L,L}\left(d\right))>0$ (Remark 5). Remark 1 gives $h^1\left({\mathcal{I}}_{mS}\left(d\right)\right)>0$. Hence $S\in T(V,m,c)$.

Now assume $d\ge (c-1)m-1$. By [28], Lemma 34 (which is characteristic-free, since Remark 5, i.e., the cohomology of line bundles on ${\mathbb{P}}^1$, is characteristic-free and using residual exact sequences is characteristic-free), part (b) is true even for $m=1$. Thus we may use induction on the integer m. We assume that part (b) fails and that n is the minimum integer $\ge 2$ for which part (b) fails. Fix an integer $x<c$ and $A\in S({\mathbb{P}}^n,x)$ such that $h^1\left({\mathcal{I}}_{mA}\left(d\right)\right)>0$. Since $d\ge m$, $x>1$.

Assume for the moment that A is contained in a line L. Since $x<c$, we have $\mathrm{deg}(L\cap mA)=xm\le d-1$. Thus $h^1(L,{\mathcal{I}}_{L\cap mA,L}\left(d\right))=0$. Take a general hyperplane $H\supseteq L$. Either because $H=L$ (case $n=2$) or the minimality assumption on n (case $n>2$), we have $h^1(H,{\mathcal{I}}_{H\cap mA,H}\left(d\right))=0$. Thus the residual exact sequence of H gives $h^1\left({\mathcal{I}}_{(m-1)A}(d-1)\right)>0$. Since $d-1\ge (c-1)(m-1)-1$, the inductive assumption on m gives $h^1\left({\mathcal{I}}_{(m-1)A}(d-1)\right)=0$, a contradiction.

Now we drop the assumption that A is contained in a line. Fix a general codimension 2 linear subspace V. Since the base field is infinite and V is general, we have $V\cap A=\varnothing$. Hence for each $p\in A$ the linear span $V_p$ of $V\cup \left\{p\right\}$ is a hyperplane. Since the base field is infinite and V is general, $V_p\ne V_q$ for all $p,q\in A$ such that $p\ne q$. Fix a general line $R\subset {\mathbb{P}}^n$. Thus $R\cap V=\varnothing$ and $R\cup V$ spans ${\mathbb{P}}^n$. Choose a system of homogeneous coordinates $x_0,\dots ,x_n$ such that $R=\{x_2=\cdots =x_n=0\}$ and $V=\{x_0=x_1=0\}$. For each $t\in \mathbb{K}\setminus \left\{0\right\}$ let $h_t:{\mathbb{P}}^n\to {\mathbb{P}}^n$ be the automorphism defined by the formula $[x_0:\cdots :x_n]\mapsto [x_0:x_1:t{x}_2:\cdots :t{x}_n]$. Let $h_0:{\mathbb{P}}^n\setminus V\to {\mathbb{P}}^n$ be the rational map defined by the formula $[x_0:\cdots :x_n]\mapsto [x_0:x_1:0:\cdots :0]$. The rational map $h_0$ is the composition of the linear projection from V onto R and the inclusion of R in ${\mathbb{P}}^n$. Since $V\cap A=\varnothing$ and $V_p\ne V_q$ for all $p,q\in A$ such that $p\ne q$, $h_0$ is defined at each point of A, $\#\left(h_0\left(A\right)\right)=x$. Hence for t going to 0 the flat family ${\left\{h_t\left(A\right)\right\}}_{t\in \mathbb{K}\setminus \left\{0\right\}}$ has $E:=h_0\left(A\right)$ as a flat limit. Thus the family ${\left\{h_t\left(mA\right)\right\}}_{t\in \mathbb{K}\setminus \left\{0\right\}}$ has $mE$ as a flat limit. Since $h_t$ is an automorphism, $h^1\left({\mathcal{I}}_{h_t\left(mA\right)}\left(d\right)\right)=h^1\left({\mathcal{I}}_{mA}\left(d\right)\right)>0$ for all $t\ne 0$. The semicontinuity theorem for cohomology gives $h^1\left({\mathcal{I}}_{mE}\left(d\right)\right)>0$. Since E is contained in a line, we obtained a contradiction. □

Fix $n\ge 2$ and $m\ge 3$. The integer $\left({\displaystyle \genfrac{}{}{0pt}{}{n+cm-2}{n}}\right)$ is degree n polynomial in c and hence for large c we get many integers d satisfying the assumptions of Theorem 11. For the case $m=2$ Galuppi, Santarsiero, Torrance and Turatti proved a stronger result ([2], Theorem 6.6). For $m=2$ the paper [2] also contains very useful results for the Del Pezzo surface (Section 5) and the Segre–Veronese varieties (Section 7).

9. Other Open Questions

We add to our long list five open questions and discuss the relations between them and the other sections of this paper.

Question 6: Fix integers $n\ge 2$ and $x>0$. Is there an integer $e_{x,n}$ such that there is a nonempty Zariski open subset U of $S({\mathbb{P}}^n,x)$ with the following property: for all integers $m\ge 1$ there is an integer $t_{m,x,n}$ such that $h^1\left({\mathcal{I}}_{mS}(t_{m,x,n}+e_{x,n})\right)=0$ and $h^0\left({\mathcal{I}}_{mS}(t_{m,x,n}-e_{x,n})\right)=0$ for all $S\in U$? Is it true at least for increasing sequence of positive integers $m_k$, $k\ge 1$?

We do not think that Question 6 has a positive answer for most $(n,x)$. Hence we propose the following question.

Question 7: Fix integers $n\ge 2$ and $x>0$. Is there a nonempty Zariski open subset U of $S({\mathbb{P}}^n,x)$ and positive integers $t_1(m,x,n)$ and $t_2(m,x,n)$ such that ${\lim }_m{t}_1(m,x,n)/t_2(m,x,n)=1$ and $h^0({\mathcal{I}}_{mS}\left(t_1(m,x,n)\right)=h^1({\mathcal{I}}_{mS}\left(t_2(m,x,n)\right)=0$ for all $m\ge 1$ and all $S\in U$? Or at least is it true at least for increasing sequence of positive integers $m_k$, $k\ge 1$? For both questions (all $m\ge 1$ or for an increasing sequence $m_k$ of multiplicities) is $\mathrm{lim\; sup}{t}_1(m,x,n)/t_2(m,x,n)>0$?

The only cases we know ($(n,x)=(2,9)$ and $(n,x)=(3,8)$) come from more precise results (Theorems 4 and 8, respectively). We have no guess for the general case.

For $n\ge 2$ the algebraic variety $S({\mathbb{P}}^n,x)$ contains complete curves (Remark 6).

Question 8: For which integers $n\ge 2$, $x>0$, $m\ge 2$ is there a complete curve $D_m\subset S({\mathbb{P}}^n,x)$ such that all $S\in D$ have $h^1\left({\mathcal{I}}_{mS}\left(t\right)\right)=h^1\left({\mathcal{I}}_{m{S}_1}\left(t\right)\right)$ for all $t>0$, where $S_1$ is a general element of $S({\mathbb{P}}^n,x)$? When we may take the same complete curve for all $m\ge 1$?

This curve (the same for all m) exists if $2\le x\le n$, but not for $x=n+1$ (Example 3). No such curve exists for $m\ge 3$ and $(n,x)=(3,6)$ (Corollary 3). We think that the projective curve $D_m$ exists only for very few triples $(n,m,x)$.

Question 9: Fix integers $n\ge 2$, $m\ge 2$ and x. Give lower and/or upper bounds on the dimension of the complete irreducible varieties $T\subset S({\mathbb{P}}^n,x)$ such that $mA$ and $mB$ have the same Hilbert function for all $A,B\in T$.

In Question 9 we do not require that T contains a general element of $S({\mathbb{P}}^n,x)$. We think that often for $n\ge 3$ there are such positive dimensional T. We think that for $n\gg 0$ there should be T with $\dim T\gg 0$.

Question 10: For which $(n,x,\mathbb{K})$ with $n\ge 2$ there is $S\in S({\mathbb{P}}^n,x)$ such that all schemes $mS$, $m\ge 1$ have maximal rank? Is the list the same for all uncountable $\mathbb{K}$?

We recall that to appear in this list we need to have $x\ge 2^n$ (Theorem 1). For uncountable $\mathbb{K}$ or for algebraically closed fields of characteristic zero we proved that $(2,9)$ and $(3,8)$ are in this list, but that if $\mathbb{K}={\overline{\mathbb{F}}}_p$ they are not in this list (Lemma 4 and Theorem 8).

10. Conclutions

For all n-dimensional projective variety X, usually ${\mathbb{P}}^n$, and all positive integers x let $S(X_{\mathrm{reg}},x)$ denote the sets of all subsets of X formed by x smooth points of S. For each $S\in S(X_{\mathrm{reg}},x)$ and each positive integer m the zero-dimensional scheme $mS$ is the union of the multiple of the points of S with multiplicity m. The scheme $mS$ has degree $\left({\displaystyle \genfrac{}{}{0pt}{}{n+m-1}{n}}\right)$. Maximal rank for an interpolation problem associated with a zero-dimensional scheme $mS$ with respect to a finite-dimensional vector space of homogeneous polynomials fits in this framework. We started the paper with an open conjecture by B. Harbourne about the Hilbert function of $mS$ when S is general or very general in $S({\mathbb{P}}^n,x)$. The conjecture ask if $x\gg 0$ all sets $mS$ have the best possible postulation, i.e., maximal rank, with respect to all line bundles. We discuss cases in which this is true for very general, but not for general and give existence for some field, even $\mathbb{Q}$, or non-existence (all finite fields). We introduce the m-Terracini sets and raise four open questions on them. Our main result is its minimal nonempty set for the Veronese embeddings of all projective spaces (Theorem 11). We believe that these m-Terracini sets have many interesting geometric properties and that they capture many features of the variety X. We sometimes (for some n and x) prove the non-existence of a positive dimensional compact family of sets with the very general Hilbert function.