# On minimally Terracini finite sets of points in projective spaces

Edoardo Ballico  Dipartimento Di Matematica, Università di Trento, Via
Sommarive 14, 38123, Povo, Trento, Italia<EMAIL_ADDRESS>and  Maria
Chiara Brambilla  Università Politecnica delle Marche, via Brecce Bianche,
I-60131 Ancona, Italia<EMAIL_ADDRESS>
###### Abstract.

We characterize the number of points for which there exist non-empty Terracini
sets of points in $\mathbb{P}^{n}$. Then we study minimally Terracini finite
sets of points in $\mathbb{P}^{n}$ and we obtain a complete description in the
case of $\mathbb{P}^{3}$, when the number of points is less than twice the
degree of the linear system.

###### Key words and phrases:

interpolation problems, minimal Terracini locus, Terracini locus, zero-
dimensional schemes

###### 2010 Mathematics Subject Classification:

Primary: 14C20; Secondary:14N07

Partially supported by GNSAGA of INdAM

##  1\. Introduction

The notion of Terracini locus in projective spaces has been recently
introduced in [3] and then extended to other projective varieties and
investigated in [2, 4, 5, 10]. This property encodes the fact that a set of
double points imposes independent conditions to a linear system, hence it
gives information for interpolation problems over double points in special
position. Moreover it can be interpreted in terms of special loci contained in
higher secant varieties to projective varieties, for general reference see
e.g. [11, 6].

The interest in this subject is also motivated by the connection with the
theory of tensors, see e.g. [12]. In particular, since symmetric tensors can
be identified with homogeneous polynomials, the development of geometric
methods in projective spaces can give contribution to the study of the rank
and decompositions of symmetric tensors.

A finite set of points $S$ of $\mathbb{P}^{n}$ is said to be Terracini with
respect to $\cal{O}_{\mathbb{P}^{n}}(d)$ if

$h^{0}(\cal{I}_{2S}(d))>0,\ h^{1}(\cal{I}_{2S}(d))>0,\mbox{ and }\langle
S\rangle=\mathbb{P}^{n}.$

We denote by $\mathbb{T}(n,d;x)$ all the sets of points
$S\subset\mathbb{P}^{n}$ of cardinality $x$ which are Terracini with respect
to $\cal{O}_{\mathbb{P}^{n}}(d)$.

The first result of this paper characterizes the triples $n,d,x$ such that the
Terracini locus is non-empty, as follows:

######  Theorem 1.1.

Fix positive integers $n$, $d$ and $x$.

(i) If either $n=1$ or $d=2$, then $\mathbb{T}(n,d;x)=\emptyset$ for any $x$.

(ii) $\mathbb{T}(2,3;x)=\emptyset$ for any $x$.

(iii) If $n\geq 2$, $d\geq 3$ and $(n,d)\neq(2,3)$, then
$\mathbb{T}(n,d;x)\neq\emptyset$ if and only if $x\geq n+\lceil d/2\rceil$.

In order to make a finer description it is very useful to study minimally
Terracini loci. The minimally Terracini property has been introduced in [2,
Definition 2.2] for any projective variety. A Terracini set of points
$S\subset\mathbb{P}^{n}$ is said to be minimally Terracini with respect to
$\cal{O}_{\mathbb{P}^{n}}(d)$ if

$h^{1}(\cal{I}_{2A}(d))=0\mbox{ for all }A\subsetneq S.$

We denote by $\mathbb{T}(n,d;x)^{\prime}$ the set of all
$S\in\mathbb{T}(n,d;x)$ which are minimally Terracini with respect to
$\cal{O}_{\mathbb{P}^{n}}(d)$.

In Theorem 3.1 we see that if $S\in S(\mathbb{P}^{n},x)$ is minimally
Terracini for some $\cal{O}_{\mathbb{P}^{n}}(d)$, then such $d$ is unique and
it is the maximal integer $t$ such that $h^{1}(\cal{I}_{2S}(t))>0$. On the
other hand, also the dimension $n$ for which a set can be minimally Terracini
is unique, by the concision result proved in Proposition 3.3.

Note that, for fixed $n,d$, we know that $\mathbb{T}(n,d;x)$ is not empty for
infinitely many $x$, by Theorem 1.1. On the other hand, if we consider the
subsets of minimally Terracini sets
$\mathbb{T}(n,d;x)^{\prime}\subseteq\mathbb{T}(n,d;x)$ we have that they are
not empty only for finitely many $x$, as proved in Proposition 3.5. In other
words the minimality property is a strong condition which allows us to prove
interesting bounds and characterizations of the triples $n,d,x$ for which
$\mathbb{T}(n,d;x)^{\prime}$ is or is not empty.

In Section 4 we investigate the sets of points on rational normal curves and
on their degenerations (reducible rational normal curves). In particular
Theorem 4.2 and Proposition 4.7 completely describe the minimal Terracini sets
related to such curves. Since rational normal curves give rise to elements of
$\mathbb{T}(n,d;1+\lceil nd/2\rceil)^{\prime}$, we may formulate the following
conjecture:

######  Conjecture 1.2.

For any $x\leq\lfloor\frac{nd+1}{2}\rfloor$, we have
$\mathbb{T}(n,d;x)^{\prime}=\emptyset$.

Here we prove the conjecture for $\mathbb{P}^{2}$, Proposition 5.2, and for
$\mathbb{P}^{3}$, Theorem 1.3.

After the easy description of the situation in the plane (see Section 5), we
focus on the case of $\mathbb{P}^{3}$, and we obtain the following three
results, which are the main results of this paper.

######  Theorem 1.3.

Fix integers $d\geq 4$ and $x$ such that $2x\leq 3d+1.$ Then
$\mathbb{T}(3,d;x)^{\prime}=\emptyset$.

######  Theorem 1.4.

Fix integers $d\geq 7$ and $x=1+\lceil 3d/2\rceil$. Then
$S\in\mathbb{T}(3,d;x)^{\prime}$ if and only if $S$ is contained in a rational
normal curve.

######  Theorem 1.5.

Fix integers $d\geq 17$ and $x$ such that $1+\lceil{3}d/2\rceil<x<2d$. Then
$\mathbb{T}(3,d;x)^{\prime}=\emptyset$.

The bound in the last theorem is sharp, as shown in Example 6.2, where the
points lie on an elliptic curve. Hence we notice that there are gaps and, once
fixed $n,d$ big enough, the first cases where $\mathbb{T}(3,d;x)^{\prime}$ is
not empty are $1+\lceil 3d/2\rceil$ and $2d$, corresponding respectively to
rational normal curves and to elliptic curves. The situation is completely
analogous in $\mathbb{P}^{2}$, see Section 5.

The paper is organized as follows: in Section 2 we present the preliminary
results and in particular we introduce the notion of critical scheme, which is
a crucial tool in our proofs. Section 3 contains the first properties of
Terracini and minimal Terracini sets and the proof of Theorem 1.1. In Section
4 we characterize the minimally Terracini sets of points on rational normal
curves and their degenerations. Section 5 is devoted to the plane, while
Section 6 to the case of $\mathbb{P}^{3}$ and to the proofs of Theorems 1.3,
1.4 and 1.5.

##  2\. Preliminaries and notation

For any $x\in\mathbb{N}$ let $S(\mathbb{P}^{n},x)$ denote the set of all the
union of $x$ points of a projective space $\mathbb{P}^{n}$. For any set
$E\subset\mathbb{P}^{n}$, let $\langle E\rangle$ denote the linear span of $E$
in $\mathbb{P}^{n}$.

We denote by $\mathbb{T}_{1}(n,d;x)$ the set of all $S\in S(\mathbb{P}^{n},x)$
such that $h^{0}(\cal{I}_{2S}(d))>0$ and $h^{1}(\cal{I}_{2S}(d))>0$.

######  Definition 2.1.

A set $S$ of points of $\mathbb{P}^{n}$ is said to be Terracini with respect
to $\cal{O}_{\mathbb{P}^{n}}(d)$ if

  * •

$h^{0}(\cal{I}_{2S}(d))>0$ and $h^{1}(\cal{I}_{2S}(d))>0$,

  * •

$\langle S\rangle=\mathbb{P}^{n}$.

We denote by $\mathbb{T}(n,d;x)$ the set of all $S\in S(\mathbb{P}^{n},x)$
which are Terracini with respect to $\cal{O}_{\mathbb{P}^{n}}(d)$.

Obviously $\mathbb{T}(n,d;x)=\emptyset$ for all $x\leq n$, since every
$S\in\mathbb{T}(n,d;x)$ spans $\mathbb{P}^{n}$.

We recall from [2, Definition 2.2] the following important definition; it
applies to any projective variety, but we write it now only the case of
$\mathbb{P}^{n}$.

######  Definition 2.2.

A set $S$ is said to be minimally Terracini with respect to
$\cal{O}_{\mathbb{P}^{n}}(d)$ if it is Terracini and moreover

  * •

$h^{1}(\cal{I}_{2A}(d))=0$ for all $A\subsetneq S$.

We denote by $\mathbb{T}(n,d;x)^{\prime}$ the set of all
$S\in\mathbb{T}(n,d;x)$ which are minimally Terracini with respect to
$\cal{O}_{\mathbb{P}^{n}}(d)$.

In the next remark we recall the exceptional cases of the Alexander-
Hirschowitz theorem, which are all the cases when any general set of points is
minimally Terracini.

######  Remark 2.3.

Assume $(n,d;x)\in\\{(2,4;5),(3,4;9),(4,4;14),(4,3;7)\\}$. Fix a general $S\in
S(\mathbb{P}^{n},x)$. Since $x\geq n+1$, we have $\langle
S\rangle=\mathbb{P}^{n}$ and hence $S\in\mathbb{T}(n,d;x)$ by the Alexander-
Hirschowitz theorem [1]. Since $\sigma_{x-1}(X_{n,d})$ is not defective, then
$S\in\mathbb{T}(n,d;x)^{\prime}$.

We collect here some preliminary results we will use in the sequel.

######  Remark 2.4.

Fix a zero-dimensional scheme $Z\subset\mathbb{P}^{n}$. Since $Z$ is zero-
dimensional, $h^{i}(\cal{O}_{Z}(t))=0$ for all $t\in\mathbb{N}$ and all $i\geq
1$. Obviously $h^{j}(\cal{O}_{\mathbb{P}^{n}}(t))=0$ for all $j>0$. Thus the
exact sequence

$0\xrightarrow{\ \ }\cal{I}_{Z}(t)\xrightarrow{\ \
}\cal{O}_{\mathbb{P}^{n}}(t)\xrightarrow{\ \ }\cal{O}_{Z}(t)\xrightarrow{\ \
}0$

gives $h^{i}(\cal{I}_{Z}(t))=0$ for all $i>1$.

We will often use the following obvious observation.

######  Remark 2.5.

Let $W\subset Z\subset\mathbb{P}^{n}$ be zero-dimensional schemes. Since $\dim
Z=0$, $h^{i}(\cal{I}_{W,Z}(d))=0$ for all $i\geq 1$. Thus

$h^{0}(\cal{I}_{Z}(d))\leq h^{0}(\cal{I}_{W}(d))\mbox{\ \ and \
}h^{1}(\cal{I}_{W}(d))\leq h^{1}(\cal{I}_{Z}(d)).$

Furtherly, it is easy to prove that

$h^{0}(\cal{I}_{Z}(d))\leq h^{0}(\cal{I}_{Z}(d+1))\mbox{\ \ and \
}h^{1}(\cal{I}_{Z}(d+1))\leq h^{1}(\cal{I}_{Z}(d)).$

######  Remark 2.6.

Fix a hyperplane $H\subset\mathbb{P}^{n}$ and any finite set $S\subset H$. The
residual exact sequence of $H$

(1) $0\xrightarrow{\ \ }\cal{I}_{S}(d-1)\xrightarrow{\ \
}\cal{I}_{2S}(d)\xrightarrow{\ \ }\cal{I}_{2S\cap H,H}(d)\xrightarrow{\ \ }0$

Remark 2.4 gives

(2) $h^{1}(H,\cal{I}_{2S\cap H,H}(d))\leq h^{1}(\cal{I}_{2S}(d))\leq
h^{1}(H,\cal{I}_{2S\cap H,H}(d))+h^{1}(\cal{I}_{S}(d-1)).$

We recall from [7] the following useful observation.

######  Remark 2.7.

Let $Z$ be a zero dimensional scheme in $\mathbb{P}^{n}$, such that
$h^{1}(\cal{I}_{Z}(d))>0$. By [7, Lemma 34], we know that if $\deg(Z)\leq
2d+1$, then there is a line $L$ such that $\deg(Z\cap L)\geq d+2$. In
particular we have that $\deg(Z)\geq d+2$.

We recall the following lemma which we learned from K. Chandler ([8, 9]).

######  Lemma 2.8.

Let $W$ be an integral projective variety, $\cal{L}$ a line bundle on $W$ with
$h^{1}(\cal{L})=0$ and $S\subset W_{\operatorname{reg}}$ a finite collection
of points. Then $h^{1}(\cal{I}_{(2S,W)}\otimes\cal{L})>0$ if and only if there
is a scheme $Z\subset 2S$ such that any connected component of $Z$ has degree
$\leq 2$ and such that $h^{1}(\cal{I}_{Z}\otimes\cal{L})>0$.

The schemes $Z$ appearing in Lemma 2.8 are curvilinear subscheme of a
collection of double points. More precisely in the following definition we
introduce the notion of critical schemes, which are the crucial tools in our
proofs.

######  Definition 2.9.

Given $S$ a collection of $x$ points in $\mathbb{P}^{n}$, we say that a zero-
dimensional scheme $Z$ is $d$-critical for $S$ if:

  * •

$Z\subseteq 2S$ and any connected component of $Z$ has degree $\leq 2$,

  * •

$h^{1}(\cal{I}_{Z}(d))>0$,

  * •

$h^{1}(\cal{I}_{Z^{\prime}}(d))=0$ for any $Z^{\prime}\subsetneq Z$.

The next lemmas describe the properties of a critical scheme.

######  Lemma 2.10.

Let $Z$ be a zero-dimensional scheme such that $h^{1}(\cal{I}_{Z}(d))>0$ and
$h^{1}(\cal{I}_{Z^{\prime}}(d))=0$ for any $Z^{\prime}\subsetneq Z$. Then
$h^{1}(\cal{I}_{Z}(d))=1$.

###### Proof.

Assume $h^{1}(\cal{I}_{Z}(d))\geq 2$ and take a subscheme $Z^{\prime}\subset
Z$ such that $\deg(Z^{\prime})=\deg(Z)-1$. We have
$h^{1}(\cal{I}_{Z^{\prime}}(d))\geq
h^{1}(\cal{I}_{Z}(d))-\deg(Z)+\deg(Z^{\prime})>0$. Thus $Z$ is not critical, a
contradiction. ∎

######  Lemma 2.11.

Fix $S\in\mathbb{T}(n,d;x)^{\prime}$ and take $Z$ critical for $S$. Then
$Z_{\operatorname{red}}=S$.

###### Proof.

Assume $S^{\prime}:=Z_{\operatorname{red}}\neq S$. Lemma 2.8 gives
$h^{1}(\cal{I}_{2S^{\prime}}(d))>0$. Thus
$S\nsubseteq\mathbb{T}(n,d;,x)^{\prime}$, a contradiction. ∎

######  Lemma 2.12.

Fix integers $n\geq 2$, $d>t\geq 1$ and $x>1$. Take
$S\in\mathbb{T}(n,d;x)^{\prime}$ and a critical scheme $Z$ for $S$. Take
$D\in|\cal{O}_{\mathbb{P}^{n}}(t)|$ with $Z\nsubseteq D$. Then
$h^{1}(\cal{I}_{\operatorname{Res}_{D}(Z)}(d-t))>0$.

###### Proof.

Since $Z\nsubseteq D$ and is critical, then Lemma 2.8 gives
$h^{1}(\cal{I}_{Z\cap D}(d))=0$. Thus the residual exact sequence of $D$ gives
$h^{1}(\cal{I}_{\operatorname{Res}_{D}(Z)}(d-t))>0$. ∎

##  3\. First results on minimally Terracini sets of points

We prove now the fact that if $S\in S(\mathbb{P}^{n},x)$ is minimally
Terracini for some $\cal{O}_{\mathbb{P}^{n}}(d)$, then such $d$ is unique and
it is the maximal integer $t$ such that $h^{1}(I_{2S}(t))>0$.

######  Theorem 3.1.

Fix $n\geq 2$ and $S\in\mathbb{T}(n,d;x)^{\prime}$. Then

(i) $h^{1}(\cal{I}_{2S}(d+1))=0$,

(ii) $S\notin\mathbb{T}(n,t;x)$ for any $t\geq d+1$,

(iii) $S\notin\mathbb{T}(n,t;x)^{\prime}$ for any $t\leq d-1$.

###### Proof.

We prove now (i). Assume $h^{1}(\cal{I}_{2S}(d+1))>0$. By Lemma 2.8, there is
a $(d+1)$-critical scheme for $S$, that is a zero-dimensional scheme $S\subset
Z\subset 2S$ with every connected component of degree $\leq 2$ and satisfying
$h^{1}(\cal{I}_{Z}(d+1))=1$ (by Lemma 2.11).

Fix $p\in Z_{\operatorname{red}}$ and call $Z(p)$ the connected component of
$Z$ supported at $p$. Set $L:=\langle Z(p)\rangle$. Hence $L$ is either a
line, or a point $L=Z(p)=\\{p\\}$.

Let $H\subset\mathbb{P}^{n}$ be a general hyperplane containing $L$. Since $Z$
is curvilinear, by generality of $H$ we can assume that the scheme $Z\cap H$
is equal to the scheme $Z\cap L$. Let us denote $\zeta=Z\cap H=Z\cap L$.

Now assume that $h^{1}(\cal{I}_{\zeta,H}(d+1))>0$. Then $L$ is a line. Since
$\zeta\subset L$, then we have the following diagram

${\cal{I}_{L,H}(d+1)}$${\cal{I}_{\zeta,H}(d+1)}$${\cal{I}_{\zeta,L}(d+1)}$${\cal{I}_{L,H}(d+1)}$${\cal{O}_{H}(d+1)}$${\cal{O}_{L}(d+1)}$${\cal{O}_{\zeta}(d+1)}$${\cal{O}_{\zeta}(d+1)}$

from which we get $h^{1}(\cal{I}_{\zeta,L}(d+1))>0$ (that implies $\deg(Z\cap
L)\geq d+3$).

On the other hand, this implies $h^{1}(\cal{I}_{\zeta}(d+1))>0$. Since
$\langle S\rangle=\mathbb{P}^{n}$ and $n\geq 2$, the set $S\cap L$ gives
$S\notin\mathbb{T}(n,d;x)^{\prime}$, a contradiction.

Now assume $h^{1}(\cal{I}_{Z\cap H,H}(d+1))=0$. The residual exact sequence of
$H$ gives $h^{1}(\cal{I}_{\operatorname{Res}_{H}(Z)}(d))>0$. Since
$\operatorname{Res}_{H}(Z)_{\operatorname{red}}\subseteq S\setminus\\{p\\}$,
by Lemma 2.8 we have that $h^{1}(\cal{I}_{2(S\setminus\\{p\\})}(d+1))=0$. This
contradicts the minimality of $S$, that is
$S\notin\mathbb{T}(n,d,x)^{\prime}$, a contradiction.

To prove (ii) it is enough to use (i) and Remark 2.5.

Claim (iii) follows easily by contradiction, using again (i) and Remark 2.5. ∎

######  Remark 3.2.

Take any $S\in\mathbb{T}(n,d;x)^{\prime}$. Since
$h^{1}(\cal{I}_{2S^{\prime}}(d))=0$ for all $S^{\prime}\subset S$ such that
$\\#S^{\prime}=x-1$, then $h^{1}(\cal{I}_{2S}(d))\leq n+1$. This property
holds for any projective variety of dimension $n$.

The following result is a kind of concision or autarky for Terracini loci of
Veronese varieties.

######  Proposition 3.3.

Take a finite set of points $S\subset\mathbb{P}^{n}$ such that $M:=\langle
S\rangle\subseteq\mathbb{P}^{n}$. Then

$h^{1}(M,\cal{I}_{2S\cap M,M})(d))>0\ \mbox{ if and only if }\
h^{1}(\cal{I}_{2S}(d))>0.$

###### Proof.

By Remark 2.5, we have $h^{1}(\cal{I}_{2S}(d))\geq h^{1}(\cal{I}_{2S\cap
M}(d))$. Since $M$ is arithmetically Cohen-Macaulay, we get
$h^{1}(M,\cal{I}_{2S\cap M,M}(d))=h^{1}(\cal{I}_{2S\cap M}(d))$. Hence the
only if part is obvious.

Now assume $h^{1}(\cal{I}_{2S}(d))>0$. Take a hyperplane
$H\subset\mathbb{P}^{n}$ such that $H\supseteq M$ and use induction on $n-\dim
M$. It is sufficient to prove that $h^{1}(H,\cal{I}_{2S\cap H,H}(d))>0$.

Take a critical scheme $Z$ for $S$. In order to conclude it is enough to find
a zero-dimensional scheme $W\subset H$ such that
$h^{1}(H,\cal{I}_{W,H}(d))>0$, $W_{\operatorname{red}}=Z_{\operatorname{red}}$
and for each $p\in Z_{\operatorname{red}}$ the connected components, $Z_{p}$
and $W_{p}$ of $Z$ and $W$ containing $p$ have the same degree. Fix a general
$o\in\mathbb{P}^{n}\setminus H$. Let
$h_{o}:\mathbb{P}^{n}\setminus\\{o\\}\xrightarrow{\ \ }H$ denote the linear
projection for $o$. Since $o$ is general, $o$ is not contained in one of the
finitely many lines spanned by the degree $2$ connected components of $Z$.
Since $Z_{\operatorname{red}}\subset H$, $o$ is not contained in a line
spanned by $2$ points of $Z_{\operatorname{red}}$. Thus $h_{|Z}$ is an
isomorphism. Set $W:=h(Z)$. By the semicontinuity theorem for cohomology to
prove that $h^{1}(H,\cal{I}_{W,H}(d))>0$ it is sufficient to prove that $W$ is
a flat limit of a flat family $\\{W_{c}\\}_{c\in\mathbb{K}\setminus\\{0\\}}$
of schemes projectively equivalent to $Z$. Fix a system $x_{0},\dots,x_{n}$ of
homogeneous coordinates of $\mathbb{P}^{n}$ such that $H=\\{x_{0}=0\\}$ and
$o=[1:0:\dots:0]$. For any $c\in\mathbb{K}\setminus\\{0\\}$ let $h_{c}$ denote
the automorphism of $\mathbb{P}^{n}$ defined by the formula
$h_{c}([x_{0}:x_{1}:\dots:x_{n}])=[cx_{0}:x_{1}:\dots:x_{n}]$. Note that
$h_{c|H}:H\xrightarrow{\ \ }H$ is the identity map. Set $W_{c}:=h_{c}(W)$.∎

We start now the classification of Terracini and minimal Terracini sets of
points in $\mathbb{P}^{n}$. Obviously $\mathbb{T}(n,d;x)=\emptyset$ for all
$x\leq n$, since every $S\in\mathbb{T}(n,d;x)$ spans $\mathbb{P}^{n}$.

######  Lemma 3.4.

$\mathbb{T}(1,d;x)=\mathbb{T}_{1}(1,d;x)=\emptyset$ for all $d>0$ and $x>0$.

###### Proof.

Assume the existence of $S\in\mathbb{T}_{1}(1,d;x)$. Thus
$h^{1}(\cal{I}_{2S}(d))>0$ and hence $2x\geq d+2$ and
$h^{0}(\cal{I}_{2S}(d))>0$ and hence $2x\leq d+1$, a contradiction. ∎

The following lemma shows a key difference between $\mathbb{T}(n,d;x)$ and its
subset $\mathbb{T}(n,d;x)^{\prime}$. In particular for fixed $n$ and $d$, we
have $\mathbb{T}(n,d;x)^{\prime}\neq\emptyset$ for only finitely many integers
$x$.

######  Proposition 3.5.

Fix integers $n\geq 2$ and $d\geq 3$. Set
$\rho:=\lceil(\binom{n+d}{n}+1)/(n+1)\rangle$. Then
$\mathbb{T}(n,d;x)^{\prime}=\emptyset$ for all $x>\rho$. Moreover, if
$\binom{n+d}{n}/(n+1)\in\mathbb{Z}$, then
$\mathbb{T}(n;d;1+\binom{n+d}{n}/(n+1))^{\prime}=\emptyset$.

###### Proof.

Assume $S\in\mathbb{T}(n,d;x)$ and take $S^{\prime}\subset S$ with
$\\#(S^{\prime})=x-1\geq\rho$. Hence $h^{1}(\cal{I}_{2S^{\prime}}(d))>0$. On
the other hand $h^{1}(\cal{I}_{2S^{\prime}}(d))>0$ by Remark 2.5. Thus $S$ is
not minimally Terracini.

Now assume $\binom{n+d}{n}\equiv 0\mod{n+1}$. In this case
$\mathbb{T}(n,d;1+\binom{n+d}{n}/(n+1))^{\prime}=\emptyset$ for the following
reason. Take $S$ with $\\#S=1+\binom{n+d}{n}/(n+1)$ and such that
$h^{0}(\cal{I}_{2S}(d))>0$ and $h^{1}(\cal{I}_{2S}(d))>0$. Let
$S^{\prime}\subset S$ be such that $\\#S^{\prime}=\binom{n+d}{n}/(n+1)$. Since
$h^{1}(\cal{I}_{2S^{\prime}}(d))=h^{0}(\cal{I}_{2S^{\prime}}(d))$ and
$h^{0}(\cal{I}_{2S^{\prime}}(d))>0$, then
$S\notin\mathbb{T}(n,d;1+\binom{n+d}{n}/(n+1))^{\prime}$.

∎

######  Lemma 3.6.

$\mathbb{T}(n,2;x)=\emptyset$ for all $x>0$ and all $n>0$.

###### Proof.

We may assume $x\geq n+1$. By the Alexander-Hirschowitz theorem any quadric
form is the sum of at most $n+1$ squares of linear forms. Hence Terracini
lemma gives $h^{0}(\cal{I}_{2S}(2))=0$ for any $S\in S(\mathbb{P}^{n},x)$
spanning $\mathbb{P}^{n}$. ∎

The following result shows that many elements of
$\mathbb{T}_{1}(n,d;x)\setminus\mathbb{T}(n,d;x)$ are easily produced and not
interesting.

######  Lemma 3.7.

Fix $n\geq 2$, $d\geq 2$ and $x\geq\lceil d/2\rceil+1$. Let $S$ be a
collection of $x$ points on a line $L\subset\mathbb{P}^{n}$. Then
$S\in\mathbb{T}_{1}(n,d;x)$.

###### Proof.

We need to prove that $h^{1}(\cal{I}_{2S}(d))>0$ and
$h^{0}(\cal{I}_{2S}(d))>0$. Fix a hyperplane $H$ containing $L$. Take $G:=2H$
if $d=2$ and call $G$ the union of $2H$ and a hypersurface of degree $d-2$ if
$d>2$. Since $S\subset\mathrm{Sing}(G)$, $h^{0}(\cal{I}_{2S}(d))>0$. Since
$\deg(2S\cap L)=2x\geq d+2$, $h^{1}(\cal{I}_{2S\cap L}(d))>0$. Thus
$h^{1}(\cal{I}_{2S}(d))>0$, by Remark 2.5. ∎

######  Lemma 3.8.

For any $x>0$, we have $\mathbb{T}(3,3;x)^{\prime}=\emptyset$.

###### Proof.

We know that $\mathbb{T}(3,3;x)^{\prime}=\emptyset$ for $x\leq 4$ by
Proposition 3.10.

We assume $x\geq 5$ and let $S\in\mathbb{T}(3,3;x)^{\prime}$. If four of the
points of $S$ are in a plane $H$, then we have that $S$ is not minimal.
Therefore the points of $S$ are in linear general position and thus
$h^{0}(\cal{I}_{2S}(3))=0$, because the set of unions of $n+2$ points of
$\mathbb{P}^{n}$ in linear general position is an open orbit for the action of
$\mathrm{Aut}(\mathbb{P}^{n})$ and the secant varieties of
$\nu_{3}(\mathbb{P}^{3})$ have the expected dimension by Alexander-Hirschowitz
theorem. ∎

###  3.1. Proof of Theorem 1.1

We are now in position to give the proof of Theorem 1.1 which classifies
Terracini loci. We start with the following lemma.

######  Lemma 3.9.

Assume $n\geq 1$ and $d\geq 2$. Let $Z\subset\mathbb{P}^{n}$ be a zero-
dimensional scheme such that $\deg(Z)\leq d+n+1$, $h^{1}(\cal{I}_{Z}(d))>0$
and $\langle Z\rangle=\mathbb{P}^{n}$. Then there is a line $L$ such that
$\deg(L\cap Z)\geq d+2$ and $\deg(Z)=d+n+1$.

###### Proof.

The lemma is trivial for $n=1$. Now assume $n=2$. Since $\deg(Z)\leq 2d+1$,
there is a line $L$ such that $\deg(Z\cap L)\geq d+2$, by Remark 2.7. Clearly,
since $\langle Z\rangle=\mathbb{P}^{2}$, we get $\deg(Z)=d+3$.

Now assume $n>2$. Take a hyperplane $H\subset\mathbb{P}^{n}$ such that
$w:=\deg(Z\cap H)$ is maximal. Since $\langle Z\rangle=\mathbb{P}^{n}$, we
have $n\leq w<z$ and $\langle Z\cap H\rangle=H$. If $h^{1}(H,\cal{I}_{Z\cap
H,H}(d))>0$, then we conclude by induction on $n$.

Now assume $h^{1}(H,\cal{I}_{Z\cap H,H}(d))=0$ and by the residual exact
sequence of $H$

(3) $0\xrightarrow{\ \ }\cal{I}_{\operatorname{Res}_{H}(Z)}(d-1)\xrightarrow{\
\ }\cal{I}_{Z}(d)\xrightarrow{\ \ }\cal{I}_{Z\cap H,H}(d)\xrightarrow{\ \ }0,$

we have $h^{1}(\cal{I}_{\operatorname{Res}_{H}(Z)}(d-1))>0$. By Remark 2.7,
since $\deg(\operatorname{Res}_{H}(Z))\leq z-w\leq d+1\leq 2d+1$, we have a
line $L$ with $\deg(L\cap\operatorname{Res}_{H}(Z))\geq d+2$, a contradiction.
∎

The following proposition proves the emptyness of the Terracini locus for
small number of points.

######  Proposition 3.10.

Assume $n,d\geq 2$ and fix an integer $x$ such that $x\leq n+\lceil
d/2\rceil-1$. Then $\mathbb{T}(n,d;x)=\emptyset$.

###### Proof.

The case $d=2$ is true by Lemma 3.6, hence we can assume $d\geq 3$.

Assume $n=2$. If $S\in\mathbb{T}(2,d;x)$ and $Z$ is a critical scheme for $S$,
then we have $\deg(Z)\leq 2x\leq d+3$. Hence by Lemma 3.9 there exists a line
$L$ such that $\deg(Z\cap L)\geq d+2$ and hence $x>\\#(S\cap L)\geq\lceil
d/2\rceil+1$, a contradiction.

Assume $n\geq 3$ and we use induction on $n$. Consider $S\in\mathbb{T}(n,d;x)$
and let $S^{\prime}\subseteq S$ be the minimal subset such that
$h^{1}(\cal{I}_{2S^{\prime}}(d))>0$. Set $y:=\\#S^{\prime}$, $M:=\langle
S^{\prime}\rangle$, and $m:=\dim M$. Proposition 3.3 gives
$h^{1}(\cal{I}_{2S^{\prime}\cap M,M}(d))>0$.

If $m<n$ then we get that either $y\geq m+\lceil d/2\rceil$ by induction (and
hence $x\geq y+(m-n)=n+\lceil d/2\rceil$, a contradiction), or
$h^{0}(M,\cal{I}_{2S^{\prime}\cap M}(d))=0$ and hence
$y(m+1)\geq\binom{m+d}{m}$, which is possible only if $M$ is a line. In this
case we have again a contradiction because, since $S$ spans $\mathbb{P}^{n}$
and $h^{1}(\cal{I}_{2S^{\prime}\cap M,M}(d))>0$, we have

$x\geq y+(n-1)\geq\frac{d+2}{2}+n-1=n+\frac{d}{2}.$

Thus we may assume $m=n$. Let $H\subset\mathbb{P}^{n}$ be any hyperplane
spanned by $S^{\prime}\cap H$. Let $S^{\prime\prime}=S^{\prime}\cap H$. Thus
$n\leq\\#(S^{\prime\prime})<y$. Since
$\operatorname{Res}_{H}(2S^{\prime\prime})=S^{\prime\prime}$, we have the
exact sequence:

$0\xrightarrow{\ \ }\cal{I}_{S^{\prime\prime}}(d-1)\xrightarrow{\ \
}\cal{I}_{2S^{\prime\prime}}(d)\xrightarrow{\ \
}\cal{I}_{2S^{\prime\prime}\cap H,H}(d)\xrightarrow{\ \ }0$

The minimality of $S^{\prime}$ and Proposition 3.3 give
$h^{1}(H,\cal{I}_{{2S^{\prime\prime}\cap H},H}(d))=0$. Now, if
$h^{1}(\cal{I}_{S^{\prime\prime}}(d-1))>0$, then either
$\\#(S^{\prime\prime})\geq n+d$, a contradiction, or
$\\#(S^{\prime\prime})\leq n+d-1$. Then Lemma 3.9 applied in $H$ gives
$\\#(S^{\prime\prime})=n+d-1$ which is again impossible. Hence
$h^{1}(\cal{I}_{S^{\prime\prime}}(d-1))=0$. Thus from the previous exact
sequence we have $h^{1}(\cal{I}_{2S^{\prime\prime}})(d))=0$.

We consider now the exact sequence with respect to the quadric hypersurface
$2H$:

$0\xrightarrow{\ \ }\cal{I}_{S^{\prime}\setminus
S^{\prime\prime}}(d-2)\xrightarrow{\ \ }\cal{I}_{2S^{\prime}}(d)\xrightarrow{\
\ }\cal{I}_{2S^{\prime\prime},2H}(d)\xrightarrow{\ \ }0$

where $\operatorname{Res}_{2H}({2}S^{\prime})=S^{\prime}\setminus
S^{\prime\prime}$.

Since the quadric hypersurface $2H$ in $\mathbb{P}^{n}$ is arithmetically
Cohen-Macaulay, we get $h^{1}(\cal{I}_{2S^{\prime\prime},2H}(d))=0$, which
implies $\cal{I}_{S^{\prime}\setminus S^{\prime\prime}}(d-2)>0$. But since
$\\#(S^{\prime}\setminus S^{\prime\prime})\leq y-n\leq\lceil d/2\rceil-1$ we
have a contradiction. ∎

We now give the proof of the main result of this section.

###### Proof of Theorem 1.1:.

Part (i) is true by Lemmas 3.4 and 3.6.

Now assume $n=2$ and $d=3$. A singular plane cubic $C$ with at least $3$
singular points is either the union of $3$ lines or the triple lines or the
union of a double line and another line. Thus if $\mathrm{Sing}(C)$ spans
$\mathbb{P}^{2}$, then $\\#\mathrm{Sing}(C)=3$ and $\mathrm{Sing}(C)$ is
projectively equivalent to $3$ non-collinear points. Hence
$\mathbb{T}(2,3;x)=\emptyset$ for all $x\geq 4$. Thus we have proved (ii)
because clearly $\mathbb{T}(2,3;3)=\emptyset$.

Assume now $n\geq 2$, $d\geq 3$ and $(n,d)\neq(2,3)$. By Proposition 3.10 we
have that if $x<n+\lceil d/2\rceil$, then $\mathbb{T}(n,d;x)=\emptyset$. Hence
it is enough to prove the other implication.

Consider first the case $n=2$ and $d\geq 4$. We have $x\geq\lceil
d/2\rceil+2$. Let $L,M,N$ three lines and $G:=(d-2)L\cup M\cup N$. Take as $S$
the point $M\cap N$ and $x-1$ points of $L\setminus(M\cup N)$. Since
$S\subset\mathrm{Sing}(G)$, $h^{0}(\cal{I}_{2S}(d))>0$. We have
$h^{1}(\cal{I}_{2S}(d))>0$. Indeed $L$ contains at least $\lceil d/2\rceil+1$
points of $L$, hence $\deg(2S\cap L)\geq d+2$ and by Remark 2.6 we have
${h^{1}(\cal{I}_{2S}(d))\geq h^{1}(\cal{I}_{2S\cap L,L}(d))>0.}$

Now assume $n\geq 3$, $d=3$ and $x\geq n+2$. Fix hyperplanes $H,K,U$ of
$\mathbb{P}^{n}$ such that $\dim H\cap K\cap U=n-3$. Since $H\cap K$ and
$H\cap U$ are $2$ different codimension $1$ subspaces of $H$, their union
spans $H$. Let $S$ be the union of $n+1$ general points in $(H\cap
K)\cup(H\cap U)$ and a point on $(K\cap U)\setminus(H\cap K\cap U)$. Hence
$\langle S\rangle=\mathbb{P}^{n}$, $h^{0}(\cal{I}_{2S}(3)\neq 0$ and it is
easy to show (by induction on $n$) that $h^{1}(\cal{I}_{2S}(3))\neq 0$. Hence
by Remark 2.5, it follows that $\mathbb{T}(n,3;x)\neq\emptyset$ for all $x\geq
n+2$ and $n\geq 3$.

Now assume $n\geq 3$ and $d\geq 4$. Recall that $x\geq n+\lceil d/2\rceil$.
Take a line $L\subset H$ and set $G:=(d-2)H\cup K\cup U$. Consider a
collection $E$ of $x-n+1$ points on the line $L$. Since $\\#E\geq\lceil
d/2\rceil+1$, by Lemma 3.7 we have $h^{1}(\cal{I}_{2E}(d))>0$. Let $A\subset
H$ be a collection of $n-2$ general points. Note that $\langle E\cup
A\rangle=H$. Take as $S$ the union of $A\cup E$ and a point of $(U\cap
K)\setminus(H\cap K\cap U)$. Obviously $S$ spans $\mathbb{P}^{n}$ and
$h^{1}(\cal{I}_{2S}(d))>0$ by Remark 2.5. ∎

Notice that the set of points $S\in\mathbb{T}(3,3;5)$ produced in the previous
proof is not minimally Terracini, because $4$ points belong to a plane. Indeed
by Lemma 3.8 we already know that $\mathbb{T}(3,3;5)^{\prime}=\emptyset$.

##  4\. Rational normal curves

We start now to analyze the set of points laying on a rational normal curve.
For each $n>1$, we denote by $\cal{C}_{n}$ the set of all rational normal
curves of $\mathbb{P}^{n}$.

######  Lemma 4.1.

Fix integers $n\geq 2$, $d\geq 4$ and $x=\lceil nd/2\rceil$. Take a rational
normal curve $C\in\cal{C}_{n}$ and let $S\subset C$ be a collection of $x$
points on $C$. Then $h^{1}(\cal{I}_{2S}(d))=0$.

###### Proof.

Assume $h^{1}(\cal{I}_{2S}(d))>0$. By Lemma 2.8, there exists a $d$-critical
scheme $Z$ for $S$. Since $C$ is scheme-theoretically cut-out by quadrics,
there is $Q\in|\cal{I}_{C}(2)|$ such that $Q\cap Z=C\cap Z:=\zeta$ and we have

$0\xrightarrow{\ \ }\cal{I}_{C,Q}(d)\xrightarrow{\ \
}\cal{I}_{\zeta,Q}(d)\xrightarrow{\ \ }\cal{I}_{\zeta,C}(d)\xrightarrow{\ \
}0$

Since $\deg(Z)\leq 2x\leq nd+1$, we have $h^{1}(\cal{I}_{\zeta,C}(d))=0$, and
since $C$ is projectively normal we get $h^{1}(\cal{I}_{\zeta,Q}(d))=0$.

Thus the residual exact sequence of $Q$ gives
$h^{1}(\cal{I}_{\operatorname{Res}_{Q}(Z)}(d-2))>0$.

Since $\operatorname{Res}_{Q}(Z)\subseteq S\subset C$, we have

$0\xrightarrow{\ \ }\cal{I}_{C}(d-2)\xrightarrow{\ \
}\cal{I}_{\operatorname{Res}_{Q}(Z)}(d-2)\xrightarrow{\ \
}\cal{I}_{\operatorname{Res}_{Q}(Z),C}(d-2)\xrightarrow{\ \ }0,$

hence, to get a contradiction, it is sufficient to prove that
$\deg(\operatorname{Res}_{Q}(Z))<n(d-2)+2$. But indeed

$\deg(\operatorname{Res}_{Q}(Z))\leq x\leq\lceil nd/2\rceil\leq n(d-2)+1,$

where the last inequality is true for $d\geq 4$. ∎

######  Theorem 4.2.

Fix integers $n\geq 2$, $d\geq 3$ and assume $(n,d)\neq(2,3)$. Given a
rational normal curve $C\in\cal{C}_{n}$ and a collection $S\subset C$ of $x$
points on the curve.

Then

(i) if $n\geq 3,d\geq 4$ and $x\geq 1+\lceil nd/2\rceil$, then
$S\in\mathbb{T}(n,d;x)$;

(ii) if $n\geq 4,d=3$ and $x=1+\lceil nd/2\rceil$, then
$S\in\mathbb{T}(n,d;x)$;

(iii) if $n\geq 2$, $d\geq 4$ and $x=1+\lceil nd/2\rceil$, then
$S\in\mathbb{T}(n,d;x)^{\prime}$.

###### Proof.

By the exact sequence

$0\xrightarrow{\ \ }\cal{I}_{C\cup 2S}(d)\xrightarrow{\ \
}\cal{I}_{2S}(d)\xrightarrow{\ \ }\cal{I}_{2S\cap C,C}(d)\xrightarrow{\ \ }0$

since $h^{1}(\cal{I}_{2S\cap
C,C}(d))=h^{1}(\cal{O}_{\mathbb{P}^{1}}(nd-2x))>0$, we have
$h^{1}(\cal{I}_{2S}(d))>0$. Since $x\geq n+1$ and $C$ is a rational normal
curve, $\langle S\rangle=\mathbb{P}^{n}$.

If $n\geq 3$ then $h^{0}(\cal{I}_{C}(2))\geq 2$, hence $C$ is contained in $2$
different quadric hypersurfaces. Thus if $d\geq 4$, we have
$h^{0}(\cal{I}_{2S}(d))>0$. Hence $S\in\mathbb{T}(n,d;x)$ and we have proved
(i).

Assume now $x=1+\lceil nd/2\rceil$. Fix any collection $A$ of $x$ general
points and note that $h^{0}(\cal{I}_{2S}(d))\geq h^{0}(\cal{I}_{2A}(d))$.

Hence, assuming $d=3$ we have

$h^{0}(\cal{I}_{2S}(3))\geq\binom{n+3}{3}-(n+1)x>0$

where the last inequality is true for any $n\geq 5$. If $n=4$ and $x=7$, we
have $h^{0}(\cal{I}_{2S}(3))\geq h^{0}(\cal{I}_{2A}(3))=1$, by the Alexander-
Hirschowitz theorem. Thus we have proved (ii).

Now assume $n=2$ and $x=d+1$. We have
$h^{0}(\cal{I}_{2S}(d))\geq\binom{d+2}{2}-3(d+1)>0$, for $d\geq 5$. If $d=4$
and $x=5$, then $h^{0}(\cal{I}_{2S}(4))\geq h^{0}(\cal{I}_{2A}(4))=1$, again
by the Alexander-Hirschowitz theorem. Hence $S\in\mathbb{T}(n,d;x)$ for $n=2$
and $d\geq 4$.

In order to complete the proof of (iii) we need to prove the minimality, and
this follows by Lemma 4.1. ∎

######  Remark 4.3.

Recall that by Theorem 1.1 we know that $\mathbb{T}(2,3;x)=\emptyset$ for all
$x>0$. Moreover in the proof of Lemma 3.8 we have seen that $x\geq 5$ in
$\mathbb{P}^{3}$ points are not Terracini if they are in linearly general
position. Hence if $S$ is a collection of $x\geq 5$ points on a rational
normal cubic we have $S\not\in\mathbb{T}(3,3;5)$.

###  4.1. Degenerations of rational normal curves.

We introduce now the notion of reducible rational normal curves.

######  Definition 4.4.

A reduced, connected and reducible curve $T\subset\mathbb{P}^{n}$, for $n\geq
2$, such that $\deg(T)=n$, $\langle T\rangle=\mathbb{P}^{n}$ is called
reducible rational normal curve.

Of course, in $\mathbb{P}^{2}$ a reducible rational normal curve is a
reducible conic.

Since $T$ is connected, there is an ordering $T_{1},\dots,T_{s}$ of the
irreducible component such that each $T[i]:=T_{1}\cup\cdots\cup T_{i}$, $1\leq
i\leq s$, is connected. We say that each such ordering of the irreducible
components of $T$ is a good ordering.

Set $n_{i}:=\deg(T_{i})$. Note that $n=n_{1}+\cdots+n_{s}$ and $\dim\langle
T_{i}\rangle\leq n_{i}$ with equality if and only if $T_{i}$ is a rational
normal curve in its linear span. For $i=1,\dots,s-1$ we have the following
Mayer-Vietoris exact sequence

(4) $0\xrightarrow{\ \ }\cal{O}_{T[i+1]}(t)\xrightarrow{\ \
}\cal{O}_{T[i]}(t)\oplus\cal{O}_{T_{i+1}}(t)\xrightarrow{\ \
}\cal{O}_{T[i]\cap T_{i+1}}(t)\xrightarrow{\ \ }0,$

in which $T[i]\cap T_{i+1}$ is the scheme-theoretic intersection. Since
$T[i+1]$ is connected, $\deg(T[i]\cap T[i+1])>0$. Thus (4) gives $\dim\langle
T[i+1]\rangle\leq\dim\langle T[i]\rangle+n_{i}$ with equality if and only if
$\deg(T[i]\cap T[i+1])=1$, $T_{i+1}$ is a rational normal curve in its linear
span and $\langle T[i]\rangle\cap\langle T_{i+1}\rangle$ is the point
$T[i]\cap T_{i+1}$.

Since $n=n_{1}+\cdots+n_{s}$, by induction on $i$ we get $p_{a}(T)=0$ and that
each $T_{i}$ is a rational normal curve in its linear span. Using (4) and
induction on $t$ we also get $h^{1}(\cal{O}_{T}(t))=0$ and
$h^{0}(\cal{O}_{T}(t))=nt+1$ for all $t\geq 0$, and that the restriction map
$H^{0}(\cal{O}_{\mathbb{P}^{n}}(t))\xrightarrow{\ \ }H^{0}(\cal{O}_{T}(t))$ is
surjective, i.e. $T$ is arithmetically Cohen-Macaulay. In the same way we see
that each $T[i]$ is arithmetically Cohen-Macaulay in its linear span.

Recall that each $T_{i}$ is smooth. For any $p\in T_{i}$ let $L_{i}(p)$ denote
the tangent line of $T_{i}$ at $(p)$. Take $p\in\mathrm{Sing}(T)$ and let
$T_{i_{1}},\dots T_{i_{k}}$, $k\geq 2$, be the irreducible components of $T$
passing through $p$. Since $n=n_{1}+\cdots+n_{s}$ and $p_{a}(T)=0$, the $k$
lines $L_{i_{1}}(p),\dots,L_{i_{k}}(p)$ through $p$ span a $k$-dimensional
linear space (such a singularity is often called a seminormal or a weakly
normal curve singularity).

An irreducible component $T_{i}$ of $T$ is said to be a final component if
$\\#(T_{i}\cap\mathrm{Sing}(T))=1$. Since $s\geq 2$, $T$ has at least $2$
final components (e.g. $T_{1}$ and $T_{s}$ for any good ordering of the
irreducible components of $T$), but it may have many final components (e.g.
for some $T$ with $s\geq 3$ we may have $\\#\mathrm{Sing}(T_{i})=1$ for all
$i\geq 2$ and there is one $T$, unique up to a projective transformation,
formed by $n$ lines through the same point).

######  Remark 4.5.

Take a (reducible) rational normal curve $T\subset\mathbb{P}^{n}$. Since
$h^{1}(\cal{O}_{T})=0$, the exact sequence

$0\xrightarrow{\ \ }\cal{I}_{T}\xrightarrow{\ \
}\cal{O}_{\mathbb{P}^{n}}\xrightarrow{\ \ }\cal{O}_{T}\xrightarrow{\ \ }0$

gives $h^{2}(\cal{I}_{T})=0$. Since $h^{1}(\cal{I}_{T}(1))=0$, the
Castelnuovo-Mumford Lemma gives that the homogeneous ideal of $T$ is generated
by quadrics. Thus $T$ is scheme-theoretically cut out by quadric.

######  Lemma 4.6.

Fix $n\geq 2$, $d\geq 4$. Let $T$ be a reducible rational normal curve in
$\mathbb{P}^{n}$ and $S\in S(\mathbb{P}^{n},x)$ such that $S\subset
T_{\operatorname{reg}}$ and $\langle S\rangle=\mathbb{P}^{n}$. If $2x\geq
dn+2$, then $S\in\mathbb{T}(n,d;x)$.

###### Proof.

Since $h^{0}(\cal{I}_{T}(2))=\binom{n}{2}$, we have that
$h^{0}(\cal{I}_{2S}(d))>0$ if $d\geq 4$.

Set $Z:=2S\cap T$. Since $S\cap\mathrm{Sing}(T)=\emptyset$, $\deg(Z)=2x$ and
$Z$ is a Cartier divisor of $T$. Since $h^{0}(\cal{O}_{T}(d))=nd+1$, then
$h^{1}(\cal{I}_{Z,T}(d))\geq 1$. Hence $h^{1}(\cal{I}_{Z}(d))\geq 1$, since
$T$ is arithmetically Cohen-Macaulay, and $S\in\mathbb{T}_{1}(n,d;x)$. ∎

######  Proposition 4.7.

Assume $n\geq 2$ and $d\geq 5$ and set $x=1+\lceil nd/2\rceil$. Fix a
reducible rational normal curve $T=T_{1}\cup\cdots\cup
T_{s}\subset\mathbb{P}^{n}$, $s\geq 2$. Assume the existence of
$S\in\mathbb{T}(n,d;x)^{\prime}$ such that $S\subset T$. Set
$n_{i}:=\deg(T_{i})$ and $x_{i}:=\\#(S\cap T_{i})$. Then:

  1. (1)

$S\subset T_{\operatorname{reg}}$;

  2. (2)

$n$ is even and $d$ is odd;

  3. (3)

every final component $T_{i}$ of $T$ has $n_{i}$ odd and $2x_{i}=n_{i}d+1$.

###### Proof.

Note that $x_{1}+\cdots+x_{s}\geq x$ and that $x_{1}+\cdots+x_{s}=x$ if and
only if $S\subset T_{\operatorname{reg}}$. We have $n=n_{1}+\cdots+n_{s}$,
$2x=nd+2$ if $nd$ is even and $2x=nd+3$ if $n$ and $d$ are odd. Set $W:=2S\cap
T$ as schemes.

Set $S_{1}:=S\cap\mathrm{Sing}(T)$ and $S_{2}:=S\setminus S_{1}$. For each
$o\in\mathrm{Sing}(T)$ let $m(o)$ denote the number of irreducible components
of $T$ passing through $o$. We saw that $T$ has Zariski tangent of dimension
$m(o)$ and hence the connected component $W(o)$ of $W$ with $o$ as its
reduction has degree $m(o)+1$. Thus $w:=\deg(W)=2\\#S_{2}+\sum_{o\in
S_{1}}(m(o)+1)$. Since $T$ has at most $s-1$ singular points,
$S_{2}\neq\emptyset$. Fix $u\in S_{2}$ and set
$S^{\prime}:=S\setminus\\{u\\}$. If $\deg(W)\geq nd+4$, then
$h^{1}(\cal{I}_{2S^{\prime}}(d))>0$ and hence
$S\notin\mathbb{T}(n,d;x)^{\prime}$, a contradiction. Now assume $\deg(W)\leq
nd+3$. Thus $nd$ is even, $\\#S_{1}=1$, say $S_{1}=\\{u\\}$, and $T$ is nodal
at $u$. Since $p_{a}(T)=0$, $T$ is connected, the irreducible components of
$T$ are smooth and $T$ is a nodal at $u$, $T\setminus\\{u\\}$ has $2$
connected components. Call $T^{\prime}$ and $T^{\prime\prime}$ the closures in
$\mathbb{P}^{n}$ of the $2$ connected components of $T\setminus\\{u\\}$. Note
that $\deg(W)=\deg(W\cap T^{\prime})+\deg(W\cap T^{\prime\prime})$ and
$n=\dim\langle T^{\prime}\rangle+\dim\langle T^{\prime\prime}\rangle$, either
$\deg(W\cap T^{\prime})\geq\dim\langle T^{\prime}\rangle+2$ or $\deg(W\cap
T^{\prime\prime})\geq d\dim\langle T^{\prime\prime}\rangle+2$. Thus
$S\notin\mathbb{T}(n,d;x)^{\prime}$. Note that $2x_{i}\leq n_{i}d+1$ is
equivalent to $2x_{i}\leq n_{i}d$ if $n_{i}d$ is even. This is sufficient to
exclude the case $d$ even.

From now on we assume $d$ odd. Recall that $2x_{i}\leq n_{i}d+1$ for all odd
$n_{i}$ and $2x_{i}\leq n_{i}d$ for all even $n_{i}$. Since $d\geq 5$ and
$2x_{i}\leq n_{i}d+1$ for all $i$, a good ordering of the irreducible
components of $T$ and $s-1$ Mayer-Vietoris exact sequences give
$h^{1}(\cal{I}_{S}(d-2))=0$. Let $Z$ be a critical scheme for $S$. Since
$h^{1}(\cal{I}_{T}(1))=0$ and $h^{2}(\cal{I}_{T})=h^{2}(\cal{O}_{T}(1))=0$,
the Castelnuovo-Mumford’s lemma gives that $\cal{I}_{T}(2)$ is globally
generated. Since $\cal{I}_{T}(2)$ is globally generated and every connected
component of $Z$ has degree $\leq 2$, $Q\cap Z=T\cap Z$ for a general
$Q\in|\cal{I}_{T}(2)|$. Since $\operatorname{Res}_{Q}(Z)\subseteq S$ and
$h^{1}(\cal{I}_{S}(d-2))=0$ and $Q$ is arithmetically Cohen-Macaulay, the
residual exact sequence of $Q$ gives $h^{1}(\cal{I}_{Z\cap Q}(d))=0$ and hence
$Z\subset T$. Thus $Z\subseteq W$. Since $T$ is arithmetically Cohen-Macaulay,
we get $h^{1}(\cal{I}_{Z,T}(d))>0$ and hence $h^{1}(\cal{I}_{W,T}(d))>0$.

(a) Assume $n$ odd. In particular $s\geq 3$ and there are at least three odd
$n_{i}$ with $2x_{i}=n_{i}d+1$. Let $T^{\prime}$ be a minimal connected
subcurve of $T$ such that $\deg(T^{\prime}\cap W)\geq 2+d\dim(\langle
T^{\prime}\rangle)$. Since $2x_{i}\leq n_{i}d+1$ for all $i$ and each subcurve
$T^{\prime\prime}$ of $T$ has at least one final component (a final component
of $T^{\prime\prime}$, not necessarily of $T$) the minimality of $T^{\prime}$
gives $\deg(T^{\prime}\cap W)=2+d\dim(\langle T^{\prime}\rangle)$. The set
$S\cap T^{\prime}$ shows that $S\notin\mathbb{T}(n,d;x)^{\prime}$, a
contradiction.

(b) Assume $n$ even and let $T_{i}$ any final component of $T$. Let $Y$ be the
union of all other components of $T$. Since $T_{i}$ is a final component, $Y$
is connected (and hence $\deg(Y)=\dim\langle Y\rangle$ and hence $Y$ is a,
possibly reducible, rational normal curve in $\langle Y\rangle$, $\langle
Y\rangle\cap\langle T_{i}\rangle$ is a point, $p$, and $\\{p\\}$ is the
scheme-theoretic intersection of $T_{i}$ and $Y$. We proved that $p\notin S$.
Since $\langle S\rangle=\mathbb{P}^{n}$ and $p\notin S$, then $\langle S\cap
T_{i}\rangle=\langle T_{i}\rangle$ and $\langle S\cap Y\rangle=\langle
Y\rangle$ and in particular $S\cap T_{i}\neq\emptyset$ and $S\cap
Y\neq\emptyset$. Since $S$ is minimal and $T$ is arithmetically Cohen-
Macaulay, $h^{1}(\cal{I}_{Z\cap T_{i}}(d))=h^{1}(\cal{I}_{Z\cap Y,T}(d))=0$.
The following Mayer-Vietoris type sequence on $T$

(5) $0\xrightarrow{\ \ }\cal{I}_{Z,T}(d)\xrightarrow{\ \ }\cal{I}_{Z\cap
T_{i},T_{i}}(d)\oplus\cal{I}_{Z\cap Y,Y}(d)\xrightarrow{\ \
}\cal{O}_{p}(d)\xrightarrow{\ \ }0$

is exact, because $p\notin S$. We proved that $h^{1}(\cal{I}_{Z\cap
T_{i},T_{i}}(d))=h^{1}(\cal{I}_{Z\cap Y,Y}(d))=0$. Assume $2x_{i}\leq n_{i}d$
(which is always the case if $n_{i}$ is even). The restriction map
$H^{0}(\cal{I}_{Z\cap T_{i},T_{i}}(d))\xrightarrow{\ \ }H^{0}(\cal{O}_{p}(d))$
is surjective, because $T_{i}\cong\mathbb{P}^{1}$ and $\deg(Z\cap
T_{i})\leq\deg(\cal{O}_{T_{i}}(d)$. Thus (5) gives
$h^{1}(T,\cal{I}_{Z,T}(d))=0$, a contradiction.∎

##  5\. Minimally Terracini finite sets in the plane

In this section we focus on the case of the plane. We deduce from [ep] the
following result, which we will need in the sequel.

######  Remark 5.1.

Fix positive integers $d,z$ such that $z\leq 3d$. Let $Z\subset\mathbb{P}^{2}$
be a zero-dimensional, $Z\neq\emptyset$. If $\deg(Z)=z$ and $d$ is the maximal
integer $t$ such that $h^{1}(I_{2S}(t))>0$, then either there is line $L$ such
that $\deg(L\cap Z)\geq d+2$ or there is a conic such that $\deg(Z\cap D)\geq
2d+2$ or $z=3d$ and $Z$ is the complete intersection of a plane cubic and a
degree $d$ plane curve (see [ep, Remarque (i) p. 116]).

######  Proposition 5.2.

Fix integers $x>0$ and $d\geq 4$.

(a) If $x\leq d$, then $\mathbb{T}(2,d;x)^{\prime}=\emptyset$.

(b) Let $S\in S(\mathbb{P}^{2},d+1)$. Then $S\in\mathbb{T}(2,d,d+1)^{\prime}$
if and only if $S$ is contained in a reduced conic $D$. Moreover, if $D=R\cup
L$ is reducible (with $L$ and $R$ lines), then $d$ is odd, $\\#(S\cap
R)=\\#(S\cap L)=(d+1)/2$ and $S\cap R\cap L=\emptyset$.

(c) Assume $d\geq 5$. Then $\mathbb{T}(2,d;x)^{\prime}=\emptyset$ for all $x$
such that $d+2\leq x<3d/2$.

###### Proof.

Fix $S\in\mathbb{T}(2,d;x)^{\prime}$ and let $Z$ be a critical scheme for $S$.
We have $\deg(Z)\leq 2x$ and $d$ is the maximal integer such that
$h^{1}(\cal{I}_{Z}(d))>0$ by Theorem 3.1.

Assume first $x\leq d$. Then $\deg(Z)\leq 2d$ and, by Remark 2.7, there is a
line $L$ such that $\deg(Z\cap L)\geq d+2$. Thus $h^{1}(\cal{I}_{Z\cap
L}(d))>0$. Since $\langle S\rangle=\mathbb{P}^{2}$, $S$ is not minimal, and
this prove (a) .

The if implication of part (b) follows from Theorem 4.2 (iii).

We prove now the other implication. Take $S\in\mathbb{T}(2,d;d+1)^{\prime}$
and let $Z$ be a critical scheme for $S$. By Lemma 2.11,
$Z_{\operatorname{red}}=S$. Assume that $S$ is not contained in a reduced
conic. Since $\langle S\rangle=\mathbb{P}^{2}$, $S$ is not contained in a
double line, therefore $S$ is not contained in a conic. Hence Remark 5.1
implies that there is a line $L\subset\mathbb{P}^{2}$ such that $\deg(L\cap
Z)\geq d+2$ and hence $h^{1}(\cal{I}_{Z\cap L}(d))>0$. Hence $S$ is not
minimal.

Finally Proposition 4.7 gives the last part of (b).

We prove now (c). Assume $d+2\leq x<3d/2$ and let $S\in\mathbb{T}(2,d;x)$ with
$Z$ critical for $S$. Since $S$ is minimal $\\#(S\cap L)\leq(d+1)/2$ for all
lines $L$ and $\\#(S\cap D)\leq 2d+1$ for each conic. Since $Z$ is critical,
$\deg(Z\cap L)\leq d+1$ for each line $L$ and $\deg(D\cap Z)\leq 2d+1$ for any
conic $D$. Thus since $\deg(Z)\leq 3d-1$, by Remark 5.1 we have
$h^{1}(\cal{I}_{Z}(d))=0$, a contradiction. ∎

Just above the range covered by Proposition 5.2 we have the following
examples.

######  Example 5.3.

Assume $d=2k$ even, $d\geq 6$ and take $x:=3k$. Let $C\subset\mathbb{P}^{2}$
be a smooth plane cubic and $T$ a smooth plane curve of degree $k$. Take as
$S$ the complete intersection $C\cap T$. Set $Z:=C\cap 2T=2S\cap C$. Since
$\deg(Z)=3d$ and $h^{0}(\cal{O}_{C}(d))=3d$, then
$h^{1}(\cal{I}_{Z,C}(d))=h^{0}(\cal{I}_{Z,C}(d))=1$. Since
$h^{0}(\cal{O}_{C}(d-3)))=3d-9\geq 3k=\\#S$, we get
$h^{1}(\cal{I}_{S,C}(d-3))=0$. Since $C$ is arithmetically normal,
$h^{1}(\cal{I}_{S}(d-3))=0$. Thus the residual exact sequence of $C$ gives
$h^{1}(\cal{I}_{2S}(d))=h^{1}(\cal{I}_{{Z},C}(d)){=1}$. We also get
$h^{1}(\cal{I}_{2S^{\prime}\cap C,C}(d))=0$ for all $S^{\prime}\subsetneq S$,
since $\deg(2S^{\prime}\cap C)\leq 3d-2$. Thus
$S\in\mathbb{T}(2,d;3d/2)^{\prime}$.

######  Example 5.4.

Take $d$ odd, $d\geq 7$, and set $x:=(3d+1)/2$. Let $C\subset\mathbb{P}^{2}$
an smooth plane cubic. Take $S\subset C$ such that $\\#S=(3d+1)/2$. By
assumption $S$ is a Cartier divisor of $C$. Since $p_{a}(C)=1$ and
$\deg(\cal{O}_{C}(d-3)))=3d-9>\\#S$, then $h^{1}(C,\cal{I}_{S,C}(d-3))=0$.
Since $C$ is arithmetically-normal, $h^{1}(\cal{I}_{S}(d-3))=0$. Thus the
residual exact sequence of $C$ gives
$h^{1}(\cal{I}_{2S}(d))=h^{1}(\cal{I}_{2S\cap C,C}(d))$. Since $p_{a}(C)=1$,
we get $h^{1}(\cal{I}_{2S\cap C,C}(d))=1$. We also have
$h^{1}(\cal{I}_{2S^{\prime}\cap C,C}(d))=0$ for all $S^{\prime}\subsetneq S$,
since $\deg(2S^{\prime}\cap C)\leq 3d-1$, hence
$S\in\mathbb{T}(2,d;(3d+1)/2)^{\prime}$.

##  6\. Minimally Terracini finite sets in $\mathbb{P}^{3}$

Now we consider the case of finite sets of points in $\mathbb{P}^{3}$. The
following proposition extends Remark 5.1 to the case of schemes of
$\mathbb{P}^{3}$.

######  Proposition 6.1.

Fix a positive integer $d$. Let $Z\subset\mathbb{P}^{3}$ be a zero-dimensional
scheme such that $\langle Z\rangle=\mathbb{P}^{3}$, its connected components
have degree $\leq 2$ and $z:=\deg(Z)\leq 3d+1$. We have
$h^{1}(\cal{I}_{Z}(d))>0$ if and only if one of the following cases occur:

(i) there is a line $L\subset\mathbb{P}^{3}$ such that $\deg(Z\cap L)\geq
d+2$;

(ii) there is a conic $D$ such that $\deg(D\cap Z)\geq 2d+2$;

(iii) there is a plane cubic $T$ such that $\deg(T\cap Z)=3d$ and $T\cap Z$ is
the complete intersection of $T$ and a degree $d$ plane curve.

###### Proof.

Since the if part is trivial, we only need to prove the only if part.

We use induction on $d$. The case $d=1$ is obvious, since conditions
$\deg(Z)\leq 4$ and $\langle Z\rangle=\mathbb{P}^{3}$ imply that $Z$ is
linearly independent.

Assume $d\geq 2$ and that the proposition is true for lower degrees. If there
is a plane $H$ such that $h^{1}(\cal{I}_{Z\cap H}(d))>0$, then we may use
Remark 5.1.

Now assume that $h^{1}(\cal{I}_{Z\cap H}(d))=0$ for any plane
$H\subset\mathbb{P}^{3}$. Take a plane $H\subset\mathbb{P}^{3}$ such that
$w:=\deg(Z\cap H)$ is maximal. Since $\langle Z\rangle=\mathbb{P}^{3}$ then
$z\geq 4$, and $w\geq 3$ and hence $\deg(\operatorname{Res}_{H}(Z))=z-w\leq
3(d-1)+1$. Since $h^{1}(\cal{I}_{Z\cap H}(d))=0$, the residual exact sequence
of $H$ gives $h^{1}(\cal{I}_{\operatorname{Res}_{H}(Z)}(d-1))>0$. The
inductive assumption applied to the scheme $\operatorname{Res}_{H}(Z)$ implies
that either there is a line $R$ such that
$\deg(R\cap\operatorname{Res}_{H}(Z))\geq d+1$, or there is a conic $D$ such
that $\deg(D\cap\operatorname{Res}_{H}(Z))\geq 2d$, or there is a plane cubic
$C$ such that $\deg(\operatorname{Res}_{H}(Z)\cap C)=3d-3$ and
$\operatorname{Res}_{H}(Z)\cap C$ is the complete intersection of $C$ and a
degree $d-1$ plane curve.

Assume for the moment $d=2$. In this case either there is a conic $D$ such
that $\deg(D\cap\operatorname{Res}_{H}(Z))\geq 4$ (and hence $w\geq 4$ and
$z\geq 8$, a contradiction), or there is a line $R$ such that
$\deg(R\cap\operatorname{Res}_{H}(Z))\geq d+1$. If $d\geq 3$ the existence of
the conic or the cubic implies $z\geq 4d$, a contradiction.

Thus for any $d\geq 2$ there is a line $R$ such that
$\deg(R\cap\operatorname{Res}_{H}(Z))\geq d+1$.

Since we already excluded case (i) (because $h^{1}(\cal{I}_{Z\cap M}(d))=0$
for every plane $M$) we have $\deg(R\cap
Z)=d+1=\deg(R\cap\operatorname{Res}_{H}(Z))$.

Take any plane $M$ containing $R$ and spanned by $M\cap Z$. Since
$h^{1}(\cal{I}_{M\cap Z}(d))=0$, the residual exact sequence of $M$ first
gives $h^{1}(\cal{I}_{\operatorname{Res}_{M}(Z)}(d-1))>0$ and then the
existence of a line $L$ such that $\deg(L\cap\operatorname{Res}_{M}(Z))\geq
d+1$. As before we have $\deg(L\cap
Z)=d+1=\deg(L\cap\operatorname{Res}_{M}(Z))$.

First assume $L=R$. Since each connected component of $Z$ has degree $\leq 2$,
we get $\\#(Z_{\operatorname{red}}\cap R)=d+1$,
$\operatorname{Res}_{M}(Z)=Z_{\operatorname{red}}$ and each connected
component of $Z$ with reduction contained in $R$ has degree $2$. Fix $p\in
Z_{\operatorname{red}}\cap R$ and let $Z(p)$ denote the connected component of
$Z$ supported at $p$.

Note that $N:=\langle R\cup Z(p)\rangle$ is a plane. Taking $N$ as $M$ we see
(since $\deg(R\cap\operatorname{Res}_{N}(Z))\leq d$) that there is at least
one line $J\neq R$ (maybe $J=L$) such that
$\deg(J\cap\operatorname{Res}_{{N}}(Z))=d+1$. If $J\cap R\neq\emptyset$, then
$\deg((J\cup R)\cap Z)\geq\deg(J\cap\operatorname{Res}_{M}(Z))+\deg(R\cap
Z)\geq 2d+2$ (since $M\supset R$). Hence we are in case (ii) with a reducible
conic, but this case is impossible because we assumed $h^{1}(\cal{I}_{N\cap
Z}(d))=0$.

Thus we may assume $J\cap R=\emptyset$ and $\deg(J\cap Z)=\deg(R\cap Z)=d+1$.
Let $Q$ be a general quadric containing $J\cup R$. Since each connected
component of $Z$ has degree $\leq 2$ and $Q$ is general, $Z\cap Q=Z\cap(J\cup
R)$. Since $R\cap J=\emptyset$, $Q$ is a smooth quadric and $J,R$ are in the
same ruling. Since $\deg(J\cap Z)=\deg(R\cap Z)=d+1$, the residual exact
sequence of $J\cup R$ in $Q$ gives $h^{1}(\cal{I}_{Z\cap Q,Q}(d))=0$. Thus the
residual exact sequence of $Q$ gives
$h^{1}(\cal{I}_{\operatorname{Res}_{Q}(Z)}(d-2))>0$. Hence
$\deg(\operatorname{Res}_{Q}(Z))\geq d$. Thus $z\geq d+2d+2$, a contradiction.
∎

Notice that if $z\leq 3d$, case (iii) of the previous theorem never occurs
since $\langle Z\rangle=\mathbb{P}^{3}$.

Thanks to Proposition 6.1, we can easily prove Theorem 1.3 which states the
emptyness of the minimal Terracini loci $\mathbb{T}(3,d;x)^{\prime}$ for
$0<2x\leq 3d+1$.

###### Proof of Theorem 1.3.

Consider $S\in\mathbb{T}(3,d;x)^{\prime}$ and let $Z$ be a critical scheme for
$S$. By Lemma 2.11 we know that $Z_{\operatorname{red}}=S$ hence $\langle
Z\rangle=\mathbb{P}^{3}$. Since $\deg(Z)\leq 2x\leq 3d+1$, we can apply
Proposition 6.1.

In any of the three cases there is a plane $H$ and a subset $S^{\prime}=S\cap
H$ which contradicts the minimality of $S$. ∎

Now we will prove Theorem 1.4, which characterizes the first non-empty minimal
Terracini loci in $\mathbb{P}^{3}$, i.e. $\mathbb{T}(3,d;1+\lceil
3d/2\rceil)^{\prime}$. Notice that one implication follows from Theorem 4.2
(iii). By Proposition 4.7, we also know that if $S$ is contained in a
reducible rational normal curve, then $S\not\in\mathbb{T}(3,d;1+\lceil
3d/2\rceil)^{\prime}$.

###### Proof of Theorem 1.4.

We only need to prove that any $S\in\mathbb{T}(3,d;1+\lceil
3d/2\rceil)^{\prime}$ is contained in a rational normal curve.

Given $d\geq 7$ and $x=1+\lceil 3d/2\rceil$, we set $\varepsilon:=1$ if $d$ is
even and $\varepsilon:=0$ if $d$ is odd. Given
$S\in\mathbb{T}(3,d;x)^{\prime}$, let $Z$ be a critical scheme for $S$ and
$z:=\deg(Z)$. Recall that $Z_{\operatorname{red}}=S$ and $z\leq
3d+3-\varepsilon$.

Take a quadric $Q\in|\cal{O}_{\mathbb{P}^{3}}(2)|$ such that $w:=\deg(Z\cap
Q)$ is maximal.

(a) Assume $Z\nsubseteq Q$. Since $\dim|\cal{O}_{\mathbb{P}^{3}}(2)|=9$, we
have $w\geq 9$. By minimality of $S$, we also have $h^{1}(\cal{I}_{Z\cap
Q}(d)=0$, hence $h^{1}(\cal{I}_{\operatorname{Res}_{Q}(Z)}(d-2))>0$. Since
$\deg(\operatorname{Res}_{Q}(Z))=z-w\leq 3d+3-9=3(d-2)$, if
$\langle\operatorname{Res}_{Q}(Z)\rangle=\mathbb{P}^{3}$, then Proposition 6.1
implies that:

(i) either there is a line $L$ such that $\deg(\operatorname{Res}_{Q}(Z)\cap
L)\geq d$,

(ii) or there is a plane conic $D$ such that
$\deg(\operatorname{Res}_{Q}(Z)\cap D)\geq 2d-2$.

If $\dim\langle\operatorname{Res}_{Q}(Z)\rangle=1$, then by Remark 2.7 we are
in case (i); if $\dim\langle\operatorname{Res}_{Q}(Z)\rangle=2$, by Remark 5.1
we are in case (i) or (ii). Indeed we can exclude that $z-w=3(d-2)$, because
in this case $w\leq 9$ and this contradicts again the assumption on $Q$.

First we exclude case (ii): assume the existence of the conic $D$. Since
$\dim|\cal{I}_{D}(2)|=4$, from the assumption on $Q$ we have that
$w\geq\max(2d-2+4,z)$. Hence $\deg(\operatorname{Res}_{Q}(Z))=z-w<2d-2$, a
contradiction.

Hence assume the existence of a line $L$ as in case (i). Note that there is a
plane $H$ such that $L\subset H$ and $\deg(H\cap Z)\geq d+1$. We have
$h^{1}(\cal{I}_{\operatorname{Res}_{H}(Z)}(d-1))>0$, by the minimality of $S$,
and $\deg(\operatorname{Res}_{H}(Z))\leq 3d+3-d-1=2d+2<3(d-1)$.

Now we consider again the following possibilities: if
$\dim\langle\operatorname{Res}_{H}(Z)\rangle=1$ then obviously there is a line
$R$ such that $\deg(\operatorname{Res}_{H}(Z)\cap R)\geq d+1$. If
$\dim\langle\operatorname{Res}_{H}(Z)\rangle=2$ then by Remark 5.1 either
there is a line $R$ as above, or there is a conic $D^{\prime}$ such that
$\deg(\operatorname{Res}_{H}(Z)\cap D^{\prime})\geq 2d$.

Finally if $\operatorname{Res}_{H}(Z)$ spans $\mathbb{P}^{3}$, then we can
apply Proposition 6.1 and we have again the same two cases: a line $R$ or a
conic $D^{\prime}$ as above.

We exclude first the existence of the conic $D^{\prime}$. As above we get
$w\geq\max(2d+3,z)$ and hence $z=3d+3$, $d$ odd,
$\deg(\operatorname{Res}_{Q}(Z))=d$ and $\operatorname{Res}_{Q}(Z)=Z\cap L$.
We also have $\deg(Z\cap D^{\prime})=2d$ and $Q\supset D^{\prime}$. There
exists a quadric $Q^{\prime}$ containing $D^{\prime}$ and a degree $3$
subscheme $W$ of $L\cap Z$. We get $\deg(\operatorname{Res}_{Q^{\prime}}(Z)=d$
and the existence of a line $L^{\prime}\neq L$ such that
$\operatorname{Res}_{Q^{\prime}}(Z)\subset R$ and
$\deg(\operatorname{Res}_{Q^{\prime}}(Z))=d$. Since $\deg(L\cap
L^{\prime})\leq 1$ and $\operatorname{Res}_{Q}(Z)$ and
$\operatorname{Res}_{Q^{\prime}}(Z)$ differ at most for a degree $3$ scheme,
we get a contradiction (since $d>4$).

Now assume the existence of the line $R$.

(a1) First assume $R=L\subset H$. Since $Z$ is critical, every connected
component of $\operatorname{Res}_{H}(Z)$ is a simple point, hence we get
$\\#(S\cap R)\geq d+1$, contradicting the minimality of $S$.

(a2) Now assume $R\neq L$ and $R\cap L\neq\emptyset$. Consider the plane
$M=\langle L\cup R\rangle$. Since $\deg(L\cap R)=1$, then $\deg(Z\cap M)\geq
2d$. Since $h^{1}(\cal{I}_{\operatorname{Res}_{M}(Z)}(d-1))>0$ and
$\deg(\operatorname{Res}_{M}(Z))\leq d+3$, there is a line $E$ such that
$\deg(E\cap\operatorname{Res}_{M}(Z))\geq d+1$. As above we get $E\neq L$ and
$E\neq R$. Take $Q^{\prime}\in|\cal{I}_{E\cup L\cup R}(2)|$. Since
$Z\nsubseteq Q$ and $w$ is maximal, $Z\nsubseteq Q^{\prime}$. Hence
$h^{1}(\cal{I}_{\operatorname{Res}_{Q^{\prime}}(Z)}(d-2))>0$, hence by Remark
2.7, we have $\deg(\operatorname{Res}_{Q^{\prime}}(Z))\geq d-1$. Hence
$z\geq(d-1)+\deg(Z\cap(L\cup R\cup E))=(d-1)+(2d+d+1-3)=4d-3$, a
contradiction.

(a3) Now assume $R\cap L=\emptyset$. Take $Q^{\prime\prime}\in|\cal{I}_{R\cup
L}(2)|$ with $\deg(Z\cap Q^{\prime\prime})$ maximal. The maximality of $w$
gives $Z\nsubseteq Q^{\prime\prime}$. Thus
$h^{1}(\cal{I}_{\operatorname{Res}_{Q^{\prime\prime}}(Z)}(d-2))>0$ and
$\deg(\operatorname{Res}_{Q^{\prime\prime}})\leq 3d+3-(d+1+d)=d+2\leq
2(d-2)+1$. Hence there is a line $F$ such that
$\deg(F\cap\operatorname{Res}_{Q^{\prime\prime}}(Z))\geq d$. We conclude as in
case (a2), using $L$, $R$ and $F$ instead of $L$, $R$ and $E$.

(b) Now we know that $Z\subset Q$. Here we will show that $Q$ is integral.

First note that $Z$ is not contained in a double plane, $2H$. Indeed since
each component of $Z$ has degree $\leq 2$, we would get $S\subset H$, a
contradiction.

Now assume that $Z$ is contained in a reducible quadric $H\cup M$.

With no loss of generality we may assume $a:=\deg(H\cap Z)\geq\deg(Z\cap M)$.
Clearly $a\geq z/2$. Since $\deg(\operatorname{Res}_{H}(Z))=z-a\leq z/2\leq
2(d-1)+1$, and $h^{1}(\operatorname{Res}_{H}(Z)(d-1))>0$, then, by Remark 2.7,
there is a line $L$ such that $\deg(L\cap\operatorname{Res}_{H}(Z))\geq d+1$.
Let $N$ be a general plane containing $L$. Since
$\deg(\operatorname{Res}_{H\cup N}(Z))\leq x-d-1$, we have
$h^{1}(\cal{I}_{\operatorname{Res}_{H\cup N}(Z)}(d-2))=0$, again by Remark
2.7. Thus $Z\subset H\cup N$ and the generality of $N$ implies $Z\subset H\cup
L$. Since $\\#(L\cap S)\geq\lceil(d+1)/2\rceil$ and
$S\in\mathbb{T}(3,d;x)^{\prime}$, we get $d$ odd, $\\#(S\cap L)=(d+1)/2$ and
$L\cap H\notin S$.

Let $N$ be again a general plane containing $L$. Since
$\deg(\operatorname{Res}_{N}(Z))\leq 3d+3-d-1$, by Proposition 6.1, either
there is a line $R$ such that $\deg(R\cap\operatorname{Res}_{N}(Z))\geq d+1$,
or there is a conic $D$ with $\deg(D\cap\operatorname{Res}_{N}(Z))\geq 2d$.

First assume the existence of $D$. There exists a cubic surface $\Sigma$ such
that $L\cup D\subset\Sigma$ and $H\cap Z\cap\Sigma=D\cap Z$. Since
$\deg(\operatorname{Res}_{\Sigma}(Z))\leq z-(d+1)-2d\leq 2$, we get
$h^{1}(\cal{I}_{\operatorname{Res}_{\Sigma}(Z)}(d-3))=0$ and $Z\subset\Sigma$,
hence we have $Z\subset D\cup L$. Since $H\cap L\notin S$, we get $\\#(S\cap
D)=d+1$. Thus $S\notin\mathbb{T}(3,d;x)^{\prime}$, a contradiction.

Now assume the existence of the line $R$. Since $R\cap L\cap S=\emptyset$ and
$S\in\mathbb{T}(3,d;x)^{\prime}$, we have $L\cap R=\emptyset$ and $\deg(R\cap
Z)=d+1$. We get $\\#(S\cap R)=(d+1)/2$. Take a general
$Q^{\prime}\in|\cal{I}_{R\cup L}(2)|$. Since $\cal{I}_{R\cup L}(2)$ is
globally generated, $Q^{\prime}$ is smooth and $Z\cap Q^{\prime}=Z\cap(R\cup
L)$. Since $h^{1}(\cal{I}_{\operatorname{Res}_{Q^{\prime}}}(d-2))>0$ and
$\deg(\operatorname{Res}_{Q^{\prime}}(Z))\leq d+1$, there is a line $E$ such
that $\deg(E\cap Z)\geq d$. Since $d$ is odd, we get $\\#(E\cap S)=(d+1)/2$.
Since $Z\subset H\cup L$ and $L\subset H$, $L\cap E\neq\emptyset$. The conic
$L\cup E$ gives $S\notin\mathbb{T}(3,d;x)^{\prime}$, a contradiction.

(c) We have proved that $Z$ in contained in an integral quadric $Q$. Since
$\dim|\cal{O}_{\mathbb{P}^{3}}(2)|=9$, there is quadric
$T\subset\mathbb{P}^{3}$ such that $\deg(T\cap Z)\geq 8-\varepsilon$ and
$T\neq Q$. We want to prove that $Z\subset T$.

Assume $Z\not\subset T$. Since $\deg(\operatorname{Res}_{T}(Z))\leq 3(d-2)+1$,
the residual exact sequence of $T$ gives
$h^{1}(\cal{I}_{\operatorname{Res}_{T}(Z)}(d-2))>0$. First assume
$\deg(\operatorname{Res}_{T}(Z))=3d-5$ and that
$\langle\operatorname{Res}_{T}(Z)\rangle$ is contained in a plane $M$. Since
$Q$ is irreducible, $Q\cap M$ is a conic containing at least
$\lceil(3d-5)/2\rceil$ points of $S$, a contradiction with the minimality of
$S$.

Since $\operatorname{Res}_{T}(Z)$ is not a scheme of degree $3d-5$ contained
in a plane, Proposition 6.1 implies that either there is a line $L_{1}$ such
that $\deg(L_{1}\cap\operatorname{Res}_{T}(Z))\geq d$ or there is a conic
$D_{1}$ such that $\deg(D_{1}\cap\operatorname{Res}_{T}(Z))\geq 2d-2$ or there
is a plane cubic $C_{1}$ such that
$\deg(C_{1}\cap\operatorname{Res}_{T}(Z))\geq 3d-6$.

The existence of the plane cubic $C_{1}$ is excluded, for the following
reason: $Z$ is contained in an integral quadric $Q$, hence $\langle
C_{1}\rangle\not\subset Q$. Then $\deg(C_{1}\cap Q)\leq 6$ and this gives a
contradiction because $6<3d-6$.

Assume the existence of the conic $D_{1}$. The scheme
$\operatorname{Res}_{\langle D_{1}\rangle}(Z)$ has degree $\leq
d+5-\varepsilon$ and $h^{1}(\cal{I}_{\operatorname{Res}_{\langle
D_{1}\rangle}(Z)}(d-1))>0$, because $Z$ is critical. Thus by Remark 2.7, there
is a line $L_{2}$ such that $\deg(Z\cap(L_{2}\cup
D_{1}))\geq{(d+1)+(2d-2)}=3d-1$. Bézout’s theorem gives $L_{2}\subset Q$ and
$D_{1}\subset Q$. Since $Q$ is an integral quadric, the conic $D_{1}$ is a
plane section of $T$. Hence $L_{2}\cap D_{1}\neq\emptyset$. Since $Z$ is
critical, $L_{2}\nsubseteq\langle D_{1}\rangle$. Thus $D_{1}\cup L_{2}$ is a
reducible rational normal curve. Note that $\cal{I}_{L_{2}\cup D_{1}}(2)$ is
globally generated. Since each connected component of $Z$ has degree $\leq 2$,
a general $Q_{1}\in|\cal{I}_{L_{2}\cup D_{1}}(2)|$ satisfies $Q_{1}\cap
Z=(L_{2}\cup D_{1})\cap Z$. Since $\deg(\operatorname{Res}_{Q_{1}}(Z))\leq
4-\varepsilon\leq d$ and $Z$ is critical, $Z\subset L_{2}\cup D_{1}$,
contradicting Proposition 4.7.

Now assume the existence of the line $L_{1}$. Bézout’s theorem gives
$L_{1}\subset Q$. Take a general plane $U\supset L_{1}$. Since each connected
component of $Z$ has degree $\leq 2$, then $L_{1}\cap Z=U\cap Z$. Since
$\deg(\operatorname{Res}_{U}(Z))\leq 2d+3-\varepsilon$ and $d\geq 6$,by
Proposition 6.1 either there is a line $L_{3}$ such that
$\deg(\operatorname{Res}_{U}(Z)\cap L_{3})\geq d+1$ or there is a conic
$D_{3}$ such that $\deg(D_{3}\cap\operatorname{Res}_{U}(Z))\geq 2d$.

The existence of $D_{3}$ is excluded again by Proposition 4.7. Thus there
exists $L_{3}$ and we have $\\#(S\cap(L_{1}\cup L_{3}))\geq\lceil
d/2\rceil+\lceil(d+1)/2\rceil=d+1$; we also get that $d$ is odd. Since $S$ is
minimal, $L_{1}\cap L_{3}=\emptyset$ and we get a contradiction as in step
(a3), using the lines $L_{1}$ and $L_{3}$ instead of $R$ and $L$.

(d) By the previous steps, $Z$ is contained in no reducible quadric and in
infinitely many integral quadrics. Moreover, every quadric containing a degree
$8-\varepsilon$ subscheme of $Z$ contains $Z$.

As in every pencil of quadrics at least one is singular, let $T$ be a quadric
cone containing $Z$. Call $o$ its vertex. Every line $L$ such that $\deg(L\cap
Z)\geq 3$ is contained in $T$ and any union of $2$ lines of $T$ is a reducible
conic, because they contain $o$. Set $E:=Q\cap T$, where $Q$ is a general
quadric containing $Z$. Since $E$ is the complete intersection of $2$ quadric
surfaces, the adjunction formula gives $\omega_{E}\cong\cal{O}_{E}$. The
Koszul complex of the equations of $Q$ and $T$ gives $h^{0}(\cal{O}_{E})=1$.
Hence by duality we have $h^{1}(\cal{O}_{E})=1$.

First assume $E$ integral. Since the rank $1$ torsion free sheaf
$\cal{I}_{Z,E}(d)$ has degree $4d-\deg(Z)>0$, then
$h^{1}(\cal{I}_{Z,E}(d))=0$. Since $E$ is arithmetically Cohen-Macaulay,
$h^{1}(\cal{I}_{Z}(d))=0$, a contradiction.

Now assume that $E$ is not integral. If $E$ is not reduced, it may have
multiple components, but no embedded point. If $E_{\operatorname{red}}\neq E$,
then $E_{\operatorname{red}}$ is a reduced curve of degree $\leq 3$ containing
$S$. Since $h^{0}(\cal{O}_{E})=1$, $E_{\operatorname{red}}$ is connected,
hence Proposition 4.7 gives a contradiction.

Thus $E=E_{\operatorname{red}}$ is reduced and each irreducible component of
$E$ is either a line or a smooth conic or a rational normal curve. First
assume $E=E_{1}\cup E_{2}$ with $E_{1}$ and $E_{2}$ reduced conics. Since $Z$
is critical, $h^{1}(\cal{I}_{\operatorname{Res}_{\langle
E_{i}\rangle}(Z)}(d-1))>0$, $i=1,2$ and hence $\deg(Z\cap E_{1})+\deg(Z\cap
E_{2})-\deg(Z\cap E_{1}\cap E_{2})\geq 4d$, a contradiction. Thus $E$ has at
most one smooth conic among its irreducible components and it is not formed by
$4$ lines through $o$. Hence there is a connected degree three curve $C\subset
E$, which is either a rational normal curve, or a reducible rational normal
curve. If $Z\subset C$, then $C$ is a rational normal curve by Proposition
4.7. In order to conclude it is enough to prove that this is the only
possibility.

Indeed assume $Z\nsubseteq C$. Since $\cal{I}_{C}(2)$ is globally generated
and every connected component of $Z$ has degree $\leq 2$, for a general
$Q^{\prime}\in|\cal{I}_{C}(2)|$ we have $Q^{\prime}\cap Z=C\cap Z$. Hence
$h^{1}(\cal{I}_{\operatorname{Res}_{Q^{\prime}}(Z)}(d-2))>0$. We write
$E=L_{4}\cup C$ with $L_{4}$ a line. We have
$\operatorname{Res}_{Q^{\prime}}(Z)\subset L_{4}$ and
$\deg(\operatorname{Res}_{Q^{\prime}}(Z))\geq d$. Take a general plane
$M\supset L_{4}$.

Since $h^{1}(\cal{I}_{\operatorname{Res}_{M}(Z)}(d-1))>0$ and
$\deg(\operatorname{Res}_{M}(Z))\leq 2d+3-\varepsilon\leq 3d$, then by
Proposition 6.1 either there is a line $L_{5}\subset C$ such that
$\deg(L_{5}\cap\operatorname{Res}_{M}(Z))\geq d+1$ (excluded because $E$ would
be a union of $2$ reduced conics) or there is a conic $D_{4}$ such that
$\deg(\operatorname{Res}_{M}(Z)\cap D_{4})\geq 2d$ (excluded because because
$E$ would be a union of $2$ reduced conics). ∎

We are going finally to prove our last main result, which is Theorem 1.5. We
point out that the bound obtained in the theorem is sharp, as shown in the
following example, which implies that
$\mathbb{T}(3,d;2d)^{\prime}\neq\emptyset$ for all $d\geq 5$.

######  Example 6.2.

Take $d\geq 5$. Let $C\subset\mathbb{P}^{3}$ be a smooth linearly normal
elliptic curve. Let $\cal{L}$ be a line bundle on $C$ such that
$\cal{L}^{\otimes 2}\cong\cal{O}_{C}(d)$. Since $\deg(\cal{L})=2d$ and $C$ has
genus 1, $\cal{L}$ is very ample.

Fix any $S\subset|\cal{L}|$ formed by $2d$ points. We will show that
$S\in\mathbb{T}(3,d;2d)^{\prime}$. Obviously $\langle
S\rangle=\mathbb{P}^{3}$. Since $2S\cap C\in|\cal{O}_{C}(d)|$, we have
$h^{i}(\cal{I}_{2S\cap C,C}(d))=1$, $i=0,1$.

The curve $C$ is the smooth complete intersection of $2$ quadric surfaces, say
$C=Q\cap Q^{\prime}$. Clearly $Q$ and $Q^{\prime}$ are smooth at each point of
$S$ and $\operatorname{Res}_{Q}(2S)=S$ and
$\operatorname{Res}_{Q^{\prime}}(2S\cap Q)=S$, hence the residual exact
sequence of $Q$ in $\mathbb{P}^{3}$ and of $C$ in $Q$ gives:

(6) $0\xrightarrow{\ \ }\cal{I}_{S}(d-2)\xrightarrow{\ \
}\cal{I}_{2S}(d)\xrightarrow{\ \ }\cal{I}_{2S\cap Q,Q}(d)\xrightarrow{\ \ }0,$
(7) $0\xrightarrow{\ \ }\cal{I}_{S,Q}(d-2)\xrightarrow{\ \ }\cal{I}_{2S\cap
Q,Q}(d)\xrightarrow{\ \ }\cal{I}_{2S\cap C,C}(d)\xrightarrow{\ \ }0.$

Since $d\geq 5$, we have $\\#S=2d<4d-8=\deg(\cal{O}_{C}(d-2))$. Thus
$h^{1}(\cal{I}_{S,C}(d-2))=0$. Since $C$ is arithmetically Cohen-Macaulay, we
have $h^{1}(\cal{I}_{S}(d-2))=0$, and hence $h^{1}(\cal{I}_{S,Q}(d-2))=0$.
Using (7) and (6), we get $h^{1}(\cal{I}_{2S}(d))=1$ and
$h^{0}(\cal{I}_{2S}(d))\geq 1$.

Take now $S^{\prime}\subsetneq S$. Since $\deg(2S^{\prime}\cap C)<4d$, we have
$h^{1}(\cal{I}_{(2S^{\prime}\cap C,C}(d))=0$. Moreover
$h^{1}(Q,\cal{I}_{S^{\prime},Q}(d-2))=0$, by Remark 2.5. Hence, using again
(7) and (6) (with $S^{\prime}$ instead of $S$), we get
$h^{1}(\cal{I}_{2S^{\prime}}(d))=0$.

Thus $S\in\mathbb{T}(3,d;2d)^{\prime}$.

From the previous example we can deduce the following remark.

######  Remark 6.3.

Fix integers $x<2d$. Let $E\subset\mathbb{P}^{3}$ be an integral complete
intersection of two quadric surfaces. Let $S$ be a collection of points on
$E$, then $h^{1}(\cal{I}_{2S}(d))=0$.

The following technical lemma generalizes the previous remark to reducible
quartic curves satisfying further suitable conditions.

######  Lemma 6.4.

Fix $d\geq 5$. Let $T\subset\mathbb{P}^{3}$ be a reduced curve with
$\deg(T)\leq 4$ and such that any irreducible component of $T$ is a line or a
conic or a rational normal cubic. Assume also that no plane contains a
subcurve of $T$ of degree $\geq 3$. Let $S\subset T$ be a collection of points
such that $\\#(S)\leq 2d-1$ and

  * •

$\\#(S\cap L)\leq\lceil d/2\rceil$ for any line $L\subseteq T$

  * •

$\\#(S\cap C)\leq d$ for any conic $C\subseteq T$

  * •

$\\#(S\cap D)\leq(3d+1)/2$ for any rational normal cubic $D\subseteq T$.

Let $Z\subset T$ be a zero-dimensional scheme such that
$Z_{\operatorname{red}}=S$, any connected component of $Z$ has degree $\leq
2$, $Z$ is contained in an integral quadric surface and $Z$ is not contained
in any reducible quadric. Then $h^{1}(\cal{I}_{Z}(d))=0$.

###### Proof.

Since $h^{1}(\cal{I}_{T}(t))=0$ for all $t\geq 5$, it is sufficient to prove
that $h^{1}(\cal{I}_{Z,T}(d))=0$. We already analized all cases with
$\deg(T)\leq 3$ and $T$ connected. Thus we may assume that $T$ is connected
and $\deg(T)=4$.

Consider a good ordering $T_{1},\dots,T_{s}$ of the irreducible components of
$T$ and set $Y=T_{1}\cup\cdots\cup T_{s-1}$. The components $T_{1}$ and
$T_{s}$ are final components and for every final component $T_{i}$ of $T$
there is a good ordering with $T_{i}$ as its first component. Thus, changing
if necessary the good ordering, we may assume $\deg(T_{1})\geq\deg(T_{s})$.
Thus $\deg(T_{s})\leq 2$ and $\deg(T_{s})=2$ if and only if $s=2$ and
$\deg(T_{1})=2$. This case is excluded, because $T$ would be contained in a
reducible quadric.

Hence $\deg(T_{1})\geq\deg(T_{s})=1$. Set $E:=T_{s}\cap Y$ (scheme-theoretic
intersection). Since $T$ contains no plane subcurves of degree $\geq 3$, then
we can assume, up to choosing a good ordering that $\deg(T_{s}\cap Y)\leq 2$.
Set $e:=\\#(S\cap E)$ and $z:=\deg(Z)\leq 2(\\#S)$. Note that $\\#S=\\#(S\cap
T_{s})+\\#(S\cap Y)-e$. We have the following Mayer-Vietoris type sequence on
$T$

(8) $0\xrightarrow{\ \ }\cal{I}_{Z,T}(d)\xrightarrow{\ \ }\cal{I}_{Z\cap
T_{s},T_{s}}(d)\oplus\cal{I}_{Z\cap Y,Y}(d)\xrightarrow{\ \ }\cal{I}_{Z\cap
E,E}(d)\xrightarrow{\ \ }0.$

(a) Assume $\\#(S\cap T_{s})\leq\lceil d/2\rceil-1$. Thus
$h^{1}(\cal{I}_{E\cup(Z\cap T_{s}),T_{s}}(d))=0$, since $\deg(E\cup(Z\cap
T_{s}))\leq 2+2(\lceil d/2\rceil-1)$. Then the restriction map
$H^{0}(\cal{I}_{Z\cap T_{s},T_{s}}(d))\xrightarrow{\ \ }H^{0}(\cal{I}_{Z\cap
E,E}(d))$ is surjective. Thus the exact sequence (8) gives
$h^{1}(\cal{I}_{Z,T}(d))=0$ and we conclude.

(b) Assume $\\#(S\cap T_{s})=\lceil d/2\rceil$. If $S\cap Y\cap
T_{s}=\emptyset$, then we have $\cal{I}_{Z\cap E,E}(d)=\cal{O}_{E}(d)$ and we
conclude as in step (a). Thus from now on we assume $S\cap T_{s}\cap
Y\neq\emptyset$. Let $M$ be a plane containing $L_{s}$ such that $\deg(Z\cap
M)$ is maximal.

(b1) Assume that $M$ contains another irreducible component, $T_{i}$, of $T$.
Since $T$ contains no planar subcurve of degree $\geq 3$, $\deg(T_{i})=1$ and
$T_{i}$ is unique in $M$. Since $T_{s}\cup T_{i}$ is a conic,
$\\#(S\cap(T_{s}\cup T_{i}))\leq d$. The closure $A$ of $T\setminus(T_{s}\cup
T_{i})$ is either a reduced conic or the union of $2$ disjoint lines. The
first case is excluded, because $T$ is not contained in a reducible quadric.
Now assume that $A$ is the union of $2$ disjoint lines, say $A=L\cup R$. The
lines $L$ and $R$ are final components of $T$. By step (a) we may assume
$\\#(S\cap L)=\\#(S\cap R)=\lceil d/2\rceil$. Thus $L\cap T_{s}=L\cap
R=\emptyset$. Let $Q$ be the unique quadric containing $L\cup R\cup T_{s}$.
Since $L\cap R=\emptyset$, $Q$ is a smooth quadric. Changing if necessary the
names of the $2$ rulings of $Q$ we may assume $L\cup R\cup
T_{s}\in|\cal{O}_{Q}(3,0)|$. Since $T_{i}$ meets each connected component of
$L\cup R\cup T_{s}$, Bézout’s theorem gives $T_{i}\subset Q$ and
$T_{i}\in|\cal{O}_{Q}(0,1)|$. Let $Z^{\prime}\subset Q$ be the residual of $Z$
with respect to the divisor $L\cup R\cup T_{s}$. It is sufficient to prove
that $h^{1}(Q,\cal{I}_{Z^{\prime}}(d-3,d))=0$. Since $T_{i}\cup T_{s}$ is a
reducible conic, $\\#(S\cap T_{i}\cup T_{s})\leq d$ and hence $\\#(S\cap
T_{i})\leq d-\lceil d/2\rceil$ with strict inequality if $S\cap T_{i}\cap
T_{s}\neq\emptyset$. Thus $\deg(Z^{\prime})\leq 4d-2-6\lceil d/2\rceil\leq
d-2$ and hence $h^{1}(\cal{I}_{Z^{\prime},Q}(d-3,d))=0$, and we conclude that
$h^{1}(\cal{I}_{Z,T}(d))=0$.

(b2) Assume that $T_{s}$ is the unique connected component of $T$ contained in
$M$. Thus $\deg((Y\cap(M\setminus T_{s}))\leq 3$. Hence $h^{1}(\cal{I}_{Z\cap
M}(d))=0$. By the residual exact sequence of $M$ it is sufficient to prove
that $h^{1}(\cal{I}_{\operatorname{Res}_{M}(Z)}(d-1))=0$. Assume by
contradiction that $h^{1}(\cal{I}_{\operatorname{Res}_{M}(Z)}(d-1))>0$. Since
$\deg(M\cap Z)>\deg(Z\cap T_{s})$, we have
$\deg(\operatorname{Res}_{M}(Z))\leq 4d-2-2\lceil d/2\rceil-1\leq 3(d-1)$.
Since $T$ contains no plane curve of degree $\geq 3$, Proposition 6.1 gives
that either there is a line $L_{1}$ such that
$\deg(L_{1}\cap\operatorname{Res}_{M}(Z))\geq d+1$ or there is a conic $D_{1}$
such that $\deg(D_{1}\cap\operatorname{Res}_{M}(Z))\geq 2d$.

(b2.1) Assume first the existence of the line $L_{1}$. Since $\\#(S\cap
L_{1})\leq\lceil d/2\rceil$, we get $d$ odd and $\deg(Z\cap L_{1})=d+1$. Since
$\\#(S\cap J)\leq d$ for all conics $J\subset T$ and $d$ is odd, $L_{1}\cap
T_{s}=\emptyset$. Let $A_{1}$ denote the closure of $T\setminus(L_{1}\cup
T_{s})$. Either $A_{1}$ is a reduced conic or it is the union of $2$ disjoint
lines. We have $\\#(S\cap(T\setminus(T_{s}\cup L_{1})))\leq d-2$. There is an
integral quadric $Q$ containing $T_{s}\cup L_{1}$ and at least one point of
$S\cap(T\setminus T_{s}\cup R_{1}))$ for each component of $A_{1}$. Thus
$h^{1}(\cal{I}_{\operatorname{Res}_{Q}(Z)}(d-2))=0$. Thus it is sufficient to
prove that $h^{1}(\cal{I}_{Z\cap Q,Q}(d))=0$. Since $L_{1}\cap
T_{s}=\emptyset$, $Q$ is a smooth quadric. We get $h^{1}(\cal{I}_{Z\cap
Q,Q}(d))=0$, unless $Q$ contains another irreducible component of $T$. First
assume $A_{1}\subset Q$. Since $Q$ is a smooth quadric, we get (for a suitable
choice of the $2$ rulings of $Q$) that either $T\in|\cal{O}_{Q}(4,0)|$
(excluded, because $T$ is reduced and connected) or $T\in|\cal{O}_{Q}(3,1)|$
or $T\in|\cal{O}_{Q}(2,2)|$, which are also excluded. Now assume that $Q$ only
contains one component, $R$, of $A_{1}$. Write $A_{1}=R\cup R_{2}$ and
$A_{2}:=L_{1}\cup T_{s}\cup R$. Either $A_{2}\in|\cal{O}_{Q}(3,0)|$ or
$A_{2}\in|\cal{O}_{Q}(2,1)|$. In both cases we get $h^{1}(Q,\cal{I}_{Z\cap
A_{2},Q}(d,d))=0$. To conclude the proof we need to consider $R_{2}\cap Z\cap
Q$. We have $\deg(R_{2}\cap Z\cap Q)\leq 4$ and hence $h^{1}(Q,\cal{I}_{Z\cap
Q,Q}(d,d))=0$.

(b2.2) Assume the existence of the conic $D_{1}$. Since $\\#(S\cap D_{1})\leq
d$, we get $\\#(S\cap D_{1})=d$ and hence $\deg(Z\cap D_{1})=2d$. By step (b1)
we may assume that if $D_{1}$ is reducible, then none of its component
contains $\lceil d/2\rceil$ points of $S$. We get $T=D_{1}\cup R\cup T_{s}$
with $R$ a line and $\\#(S\cap(T\setminus D_{1}\cup T_{s}))\leq d-1-\lceil
d/2\rceil$. If $R$ is a final component of $T$, then we use step (a) and that
$\\#(R\cap S)<\lceil d/2\rceil$. Now assume that $R$ is not a final component
of $T$. Assume for the moment $T_{s}\cap D_{1}\neq\emptyset$. Since $T$
contains no degree $3$ planar subcurve, $D_{1}\cup T_{s}$ is a reducible
rational normal curve and we may find a quadric $Q_{1}$ containing $D_{1}\cup
T_{s}$, but not $R$. To conclude in this case we need
$\deg(\operatorname{Res}_{Q_{1}}(Z))\leq d-1$. We have $\\#(S\cap R)\leq
2d-1-d-\lceil d/2\rceil$, and we can conclude. Now assume $D_{1}\cap
T_{s}=\emptyset$. Since $T$ is connected, $R$ meet $T_{s}$ and $D_{1}$ at a
different point. In this case $T$ is contained in the reducible quadric
$\langle R\cup T_{s}\rangle\cup\langle D_{1}\rangle$, a contradiction. ∎

We give now the proof of Theorem 1.5, which states that
$\mathbb{T}(3,d;x)^{\prime}$ is empty if $1+\lceil{3}d/2\rceil<x<2d$.

###### Proof of Theorem 1.5:.

Assume the existence of $S\in\mathbb{T}(3,d;x)^{\prime}$ and fix a critical
scheme $Z$ of $S$. Set $z:=\deg(Z)\leq 4d-2$.

Set $Z_{0}=Z$. For any $i>0$, let $Q_{i}$ be a quadric surface such that
$z_{i}:=\deg(Z_{i-1}\cap Q_{i})$ is maximal and set
$Z_{i}:=\operatorname{Res}_{Q_{i}}(Z_{i-1})$. The sequence $\\{z_{i}\\}_{i\geq
1}$ is weakly decreasing. Let $e$ be the maximal $i$ such that $z_{i}\neq 0$.
Then $z=z_{1}+\cdots+z_{e}$ and $Z_{e}=\emptyset$. Since
$h^{0}(\cal{O}_{\mathbb{P}^{3}}(2))=10$, $z_{i}\geq 9$ for all $i<e$, hence we
have $e\leq(4d+6)/9$, for $z\leq 4d-2$. By Lemma 2.12, since $Z$ is critical
and $S\in\mathbb{T}(3,d;x)^{\prime}$, we have
$h^{1}(\cal{I}_{Z_{e-1}}(d-2e+2))>0$.

(I) Assume first $e\geq 2$, i.e. $Z$ is not contained in any quadric surface.
Since $h^{1}(\cal{I}_{Z_{e-1}}(d-2e+2))>0$, then Proposition 6.1 implies that
either $z_{e}\geq 3(d-2e+2)+1$ or there is a line $L$ such that
$\deg(Z_{e-1}\cap L)\geq d-2e+4$ or there is a plane conic $D$ such that
$\deg(Z_{e-1}\cap D)\geq 2d-4e+6$.

(I.a) First assume $z_{e}\geq 3(d-2e+2)+1$. Since the sequence $z_{i}$ is
weakly decreasing, we get $z\geq e(3d-6e+6)$. It is easy to check that
$e(3d-6e+6)>4d-2$ for any $d\geq 13$ and $2\leq e\leq(4d+6)/9$. This
contradicts our hypotesis.

(I.b) Now assume the existence of a plane conic $D$ such that
$\deg(Z_{e-1}\cap D)\geq 2d-4e+6$. Since $h^{0}(\cal{I}_{D}(2))=5$, we get
$z_{i}\geq(2d-4e+6)+4$ for all $i<e$. Thus $z\geq e(2d-4e+10)-4$. It is easy
to check that $e(2d-4e+10)-4>4d-2$ for any $2\leq e\leq(4d+6)/9$, and this
gives again a contradiction.

(I.c) Finally assume the existence of a line $L$ such that $\deg(Z_{e-1}\cap
L)\geq d-2e+4$. Since $h^{0}(\cal{I}_{L}(2))=7$, we have $z_{i}\geq(d-2e+4)+6$
for all $i<e$. Hence $z\leq e(d-2e+10)-6$. It is easy to check that
$e(2d-4e+11)-6>4d-2$ for any $4\leq e\leq(4d+6)/9$. Hence we get
$e\in\\{2,3\\}$.

Let $H$ be a general plane containing $L$. Since each connected component of
$Z$ has degree $\leq 2$, we may assume $Z\cap L=Z\cap H$.

(I.c1) First assume $e=3$. Since $z_{1}\geq z_{2}\geq z_{3}\geq d-2$ and
$z_{1}+z_{2}\geq\lceil 2z/3\rceil$, we have
$\deg(\operatorname{Res}_{Q_{1}\cup Q_{2}\cup H}(Z))\leq z-\lceil
2z/3\rceil-(d-2)=\lfloor z/3\rfloor-d+2<d-3=(d-5)+2$, since $d\geq 7$. Since
$S$ is minimally Terracini, we get $Z\subset Q_{1}\cup Q_{2}\cup L$. Since
$e>2$ and $H$ is contained in a quadric surface, $Z\nsubseteq Q_{1}\cup H$.
Since $S$ is minimally Terracini,
$h^{1}(\cal{I}_{\operatorname{Res}_{Q_{1}\cup H}(Z)}(d-3))>0$. We have:
$\deg(\operatorname{Res}_{Q_{1}\cup H}(Z))\leq(z-z_{1})-(d-2)\leq$ $z-\lceil
z/3\rceil-d+2\leq\frac{5d+2}{3}\leq 2(d-3)+1$, for $d\geq 17$. Hence there is
a line $R$ such that $\deg(R\cap\operatorname{Res}_{Q_{1}\cup H}(Z))\geq d-1$.
Taking a general plane containing $R$ and taking again the residual, we get
$Z\subset Q_{1}\cup L\cup R$. But since $h^{0}(\cal{I}_{R\cup L}(2))>0$ and
$e\leq 2$, we have a contradiction.

(I.c2) Now assume $e=2$ and hence $z_{1}\geq\lceil z/2\rceil$. We have
$\deg(\operatorname{Res}_{H}(Z))\leq z-d$ and
$h^{1}(\cal{I}_{\operatorname{Res}_{H}(Z)}(d-1))>0$.

First assume $\langle\operatorname{Res}_{H}(Z)\rangle=\mathbb{P}^{3}$. Since
$z-d\leq 3(d-1)+1$, Proposition 6.1 implies that either there is a plane cubic
$T_{3}$ with $T_{3}\cap\operatorname{Res}_{H}(Z)$ the complete intersection of
$T_{3}$ and a degree $d-1$ plane curve or there is a conic $T_{2}$ such that
$\deg(T_{2}\cap\operatorname{Res}_{H}(Z))\geq 2d$ or there is a line $T_{1}$
such that $\deg(T_{1}\cap\operatorname{Res}_{H}(Z))\geq d+1$.

First assume the existence of $T_{3}$. Since
$\deg(\operatorname{Res}_{H\cup\langle T_{3}\rangle}(Z))\leq 1$, by minimality
of $S$ we get $Z\subset H\cup\langle T_{3}\rangle$, contradicting the
assumption $e>1$.

Assume the existence of $T_{2}$. Since $\deg(\operatorname{Res}_{H\cup\langle
T_{2}\rangle}(Z))\leq z-3d\leq d-1$, we get $Z\subset H\cup\langle
T_{2}\rangle$, again a contradiction.

Now assume the existence of $T_{1}$. Take a general quadric
$U\in|\cal{I}_{L\cup{T_{1}}}(2)|$. Since $\deg(\operatorname{Res}_{U}(Z))\leq
z-2d-1\leq 2(d-2)+1$, by Remark 2.7 there is a line $R_{1}$ such that
$\deg(R_{1}\cap\operatorname{Res}_{U}(Z))\geq d$. Take a general
$U^{\prime}\in|\cal{I}_{L\cup T_{1}\cup R_{1}}(2)|$. Since
$\deg(\operatorname{Res}_{U^{\prime}}(Z))\leq z-3d-1<d$ and $S$ is minimally
Terracini, $Z\subset U^{\prime}$, contradicting the assumption $e>1$.

Now assume $\dim\langle\operatorname{Res}_{H}(Z)\rangle\leq 2$. The only new
case is if $\deg(\operatorname{Res}_{H}(Z))=3d-2$ and
$\operatorname{Res}_{H}(Z)$ is contained in a plane cubic $C$. Since
$\deg(\operatorname{Res}_{\langle C\rangle}(Z))\leq d$, $S$ is not minimally
Terracini.

(II). Assume now $e=1$, that is $Z$ is contained in a quadric $Q$.

If $Q$ is reducible we argue as in step (b) of the proof of Theorem 1.4 and we
get a contradiction. So we can assume that $Z$ is not contained in any
reducible quadric. In particular $Q$ is irreducible and reduced.

Set $W_{0}:=Z$. Take $D_{1}\in|\cal{O}_{Q}(2)|$, such that
$w_{1}=\deg(W_{0}\cap D_{1})$ is maximal and set
$W_{1}:=\operatorname{Res}_{D_{1}}(W_{0})$. For $i\geq 2$, we iterate the
construction: choose divisors $D_{i}\in|\cal{O}_{Q}(2)|$ such that
$w_{i}:=\deg(W_{i-1}\cap D_{i})$ is maximal and set
$W_{i}:=\operatorname{Res}_{D_{i}}(W_{i-1})$. The sequence $\\{w_{i}\\}_{i\geq
1}$, is weakly decreasing. Let $c\geq 1$ be the maximal $i$ such that
$w_{i}\neq 0$, i.e. $W_{c}=\emptyset$ and and $z=w_{1}+\ldots+w_{c}$.

By Lemma 2.12, since $Z$ is critical for $S$ minimal, we have
$h^{1}(\cal{I}_{W_{c-1}}(d-2c+2))>0$. Since $\dim|\cal{O}_{Q}(2)|=8$, if
$w_{i}\leq 7$, then $w_{i+1}=0$ and $W_{i+1}=\emptyset$. Thus $w_{i}\geq 8$
for $1\leq i<c$, hence we get $c\leq\frac{4d+5}{8}$, since $z\leq 4d-2$.

(II.a) If $c=1$, then we have $Z\subset D_{1}=Q\cap Q^{\prime}$ where
$Q^{\prime}$ is an integral quadric. Hence $D_{1}$ is a complete intersection
of two quadrics. If $D_{1}$ is integral, then by Remark 6.3 we have
$h^{1}(\cal{I}_{Z}(d))=0$, a contradiction. If $D_{1}$ is reducible we have
again a contradiction by Lemma 6.4 and by the minimality of $S$.

(II.b) Now we assume $c=\lceil d/2\rceil$. Hence either $d$ is even and
$h^{1}(\cal{I}_{W_{c-1}}(2))>0$, or $d$ is odd and
$h^{1}(\cal{I}_{W_{c-1}}(1))>0$.

First assume $d$ odd and $c=\lceil d/2\rceil$. Then we have $8(\lceil
d/2\rceil-1)+\deg(W_{c-1})\leq 4d-2$, then $\deg(W_{c-1})\leq 2$, which is a
contradiction. Now assume $d$ even and $c=d/2$. Since
$8(d/2-1)+\deg(W_{c-1})\leq 4d-2$, then $\deg(W_{c-1})\leq 6$. Thus either
there is a line $L$ such that $\deg(W_{c-1}\cap L)\geq 4$ or $\deg(W_{c-1})=6$
and $W_{c-1}$ is contained in a conic $D$.

First assume the existence of the line $L$ such that $\deg((W_{c-1})\cap
L)\geq 4$. Bézout’s theorem implies $L\subset Q$. Since
$h^{0}(\cal{I}_{L,Q}(2))=6$, the maximality of the integer $w_{c-1}$ implies
$w_{c-1}\geq w_{c}+5\geq 9$. Thus $4d-2\geq(d/2-1)9+4$, a contradiction, since
$d\geq 7$.

Now assume $\deg(W_{c-1})=6$ and that $W_{c-1}$ is contained in a conic $D$.
If $D$ is reducible we may assume that no irreducible component $J$ of $D$
satisfied $\deg(J\cap W_{c-1})\geq 4$. With these assumptions Bézout’s theorem
implies $D\subset Q$. Since $h^{0}(\cal{I}_{D,Q}(2))=4$, the maximality of the
integer $w_{c-1}$ gives $w_{c-1}\geq w_{c}+3=9$, which leads again to a
contradiction.

(II.c) Now we may assume $2\leq c<d/2$.

Assume for the moment $w_{c}\geq 3(d-2c+2)$. Since the sequence $\\{w_{i}\\}$
is weakly decreasing, $4d-2\geq z\geq 3c(d-2c+2)$. Since $c<d/2$, we get $c=1$
a contradiction.

Now assume $w_{c}<3(d-2c+2)$. By applying Proposition 6.1 we know that either
there is a conic $D$ such that $\deg(D\cap W_{c-1})\geq 2(d-2c+2)+2=2d-4c+6$,
or there is a line $L$ such that $\deg(L\cap W_{c-1})\geq d-2c+4$.

(II.c1) In the first case, since $h^{0}(\cal{I}_{D,Q}(2))=4$, we have
$w_{i}\geq(2d-4c+6)+3$ for all $i<c$. Hence $z\geq c(2d-4c+9)-3$. Since $z\leq
4d-2$, then we have again a contradiction.

(II.c2) Assume now the existence of $L$. Since $h^{0}(\cal{I}_{L,Q}(2))=6$, we
get $w_{i}\geq(d-2c+4)+5$ for all $i<c$. Thus $z\geq c(d-2c+9)-5$. It is easy
to check that $2\leq c\leq 3$, hence $\deg(L\cap Z)\geq d-2$.

Take a quadric $U\in|\cal{O}_{Q}(2))|$ containing $L$ and such that
$\deg(Z\cap U)$ is maximal. Since $h^{0}(\cal{I}_{L,Q}(2))=6$, we have
$\deg(L\cap U)\geq(d-2)+5=d+3$. Thus $\deg(\operatorname{Res}_{U}(Z))\leq
4d-2-d-3=3(d-2)+1$. By Proposition 6.1 either there is a plane cubic $E$ such
that $\deg(E\cap\operatorname{Res}_{U}(Z))\geq 3(d-2)$ or there is a conic $F$
such that $\deg(\operatorname{Res}_{U}(Z)\cap F)\geq 2d-2$ or there is a line
$R$ such that $\deg(\operatorname{Res}_{U}(Z)\cap R)\geq d$. In all cases
(since $d\geq 5$) Bézout’s theorem implies that $R$, $F$ and $E$ are contained
in $Q$ (or at least all the components supporting $Z$). Since $Q$ is an
integral quadric, we exclude the plane cubic $E$.

(II.c2.1) Assume the existence of a conic $F$. Even if $Q$ is not assumed to
be smooth, $F$ is a plane section of $Q$ and $F\cup L$ is a reducible rational
normal curve.

Thus $Z\nsubseteq F\cup L$.

Since $\cal{I}_{F\cup L}(2)$ is globally generated, a general
$Q^{\prime}\in|\cal{I}_{F\cup L}(2)|$ has $Q^{\prime}\cap Z=(F\cup L)\cap Z$
and hence $\operatorname{Res}_{Q^{\prime}}(Z)\neq\emptyset$. Since
$h^{1}(\cal{I}_{\operatorname{Res}_{Q^{\prime}}(Z)}(d-2))>0$ and
$\deg(\operatorname{Res}_{Q^{\prime}}(Z))\leq 4d-2-3d+4$ and $d\geq 7$, there
is a line $R^{\prime}$ such that $\deg(\operatorname{Res}_{Q^{\prime}}(Z)\cap
R^{\prime})\geq d$. Since $\cal{I}_{F\cup L\cup R^{\prime}}(t)$ is globally
generated for, say, $t=4$, we get $Z\subset F\cup L\cup R^{\prime}$. Hence we
conclude by Lemma 6.4.

(II.c2.2) Assume finally the existence of the line $R$. Since each connected
component of $Z$ has degree $\leq 2$ and no line contains $d-2$ points of $S$,
$R\neq L$.

(II.c2.2.1) First assume $R\cap L\neq\emptyset$. Thus $H:=\langle R\cup
L\rangle$ is a plane. Since $\deg(\operatorname{Res}_{H}(Z))\leq 4d-2-2d+2$
and $h^{1}(\cal{I}_{\operatorname{Res}_{H}(Z)}(d-1))>0$ either
$\deg(\operatorname{Res}_{H}(Z))=2d$ and $\operatorname{Res}_{H}(Z)$ is
contained in a conic $F_{1}$ or there is a line $R_{1}$ such that
$\deg(R_{1}\cap\operatorname{Res}_{H}(Z))\geq 2$. In the first case we get
$Z\subset L\cup R\cup F_{1}$ and we conclude by Lemma 6.4.

(II.c2.2.2) Now assume $R\cap L=\emptyset$. Take a general
$Q_{1}\in|\cal{I}_{R\cup L}(2)|$. Thus $Q_{1}\cap Z=(R\cup L)\cap Z$. We get
$h^{1}(\cal{I}_{\operatorname{Res}_{Q_{1}}(Z)}(d-2))>0$ with
$\deg(\operatorname{Res}_{Q_{1}}(Z))\leq 2d$. We get that either there is a
conic $F_{2}$ with $\deg(F_{2}\cap\operatorname{Res}_{Q_{1}}(Z))\geq 2d-2$ or
a line $R_{2}$ such that $\deg(R_{2}\cap\operatorname{Res}_{Q_{1}}(Z))\geq d$.
If $F_{2}$ exist, we get $Z\subset R\cup L\cup F_{2}$ and we use Lemma 6.4. If
$R_{2}$ exists, we take a general $U_{1}\in|\cal{I}_{R\cup L\cup R_{2}}(3)|$
and get that $Z$ is contained in the union of $4$ lines. Hence we conclude
again by Lemma 6.4. ∎

## References

  * [1] J. Alexander and A. Hirschowitz, Polynomial interpolation in several variables, J. Alg. Geom. 4 (1995), 201–222.
  * [2] E. Ballico, A. Bernardi and P. Santarsiero, Terracini loci for $3$ points on a Segre variety, arXiv:2012.00574, to appear in Asian Journal of Mathematics.
  * [3] E. Ballico and L. Chiantini, On the Terracini locus of projective varieties. Milan J. Math. 89 (2021), No. 1, 1-17.
  * [4] E. Ballico and L. Chiantini, Terracini loci of curves, arXiv:2305.02035 to appear in Revista Matematica Complutense.
  * [5] E. Ballico and E. Ventura, A note on very ample Terracini loci, arXiv:2303.12509 to appear in Rend. Circ. Mat. Palermo.
  * [6] . A. Bernardi, E. Carlini, M. Catalisano, A. Gimigliano, and A. Oneto. The Hitchhiker Guide to: Secant Varieties and Tensor Decomposition. Mathematics, 6(12):314, 2018.
  * [7] A. Bernardi, A. Gimigliano, M. Idà, Computing symmetric rank for symmetric tensors, J. Symbolic. Comput. 46 (2011), 34–55.
  * [8] K. A. Chandler, Hilbert functions of dots in linearly general positions, in: Zero-dimensional schemes (Ravello 1992), de Gruyter, Berlin (1994), pp. 65–79.
  * [9] K. Chandler, A brief proof of a maximal rank theorem for generic 2-points in projective space. Trans. Amer. Math. Soc. 353 (2000), no. 5, 1907–1920.
  * [10] L. Chiantini, F. Gesmundo, Decompositions and Terracini loci of cubic forms of low rank, arXiv:2302.03715.
  * [11] A. Iarrobino and V. Kanev. Power Sums, Gorenstein Algebras, and Determinantal Loci. Volume 1721 of Lecture Notes in Mathematics. Springer, Berlin, New York NY, 1999.
  * [12] J. M. Landsberg, Tensors: Geometry and Applications. Graduate Studies in Mathematics, Vol. 128, Amer. Math. Soc. Providence, 2012.