On the ranks of the third secant variety of Segre-Veronese embeddings

TL;DR

This paper establishes bounds on the tensor rank and border rank for partially symmetric tensors, particularly in the Segre case, revealing the range of possible ranks under various conditions.

Contribution

It provides new upper bounds for the rank of border rank 3 tensors and characterizes the spectrum of attainable ranks in the Segre case.

Findings

All ranks from 3 to k-1 occur for border rank 3 tensors.

If all dimensions are at least 3, ranks between k and 2k-1 also occur.

The results give a comprehensive view of rank possibilities for these tensors.

Abstract

We give an upper bound for the rank of the border rank 3 partially symmetric tensors. In the special case of border rank 3 tensors $T\in V_1\otimes \cdots \otimes V_k$ (Segre case) we can show that all ranks among 3 and $k-1$ arise and if $\dim V_i \geq 3$ for all $i$'s, then also all the ranks between $k$ and $2k-1$ arise.

Equations

$$T\in S^{d_1}{V}_1\otimes \cdots \otimes S^{d_k}{V}_k$$

$$T=\sum _{i=1}^{r(T)}{\lambda }_i{v}_{1,i}^{\otimes d_1}\otimes \cdots \otimes v_{k,i}^{\otimes d_k}$$

$$r(T)\le -1+\sum _{i=1}^{k}2d_i.$$

$${\nu }_{n_1^1,\dots ,n_k^1}:{\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}\to {\mathbb{P}}^M,whereM=\left(\prod _{i=1}^k(n_i+1)\right)-1$$

$${\pi }_i:{\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}\to {\mathbb{P}}^{n_i}$$

$${\tau }_i:{\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}\to {\mathbb{P}}^{n_1}\times \cdots \times \hat{}{\mathbb{P}}^{n_i}\times \cdots \times {\mathbb{P}}^{n_k}$$

$$h^{\ast }\left({\mathcal{O}}_{{\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}}({\epsilon }_i)\right)\cong {\mathcal{O}}_{{\mathbb{P}}^1}(a_i).$$

$${\nu }_{n_1^{d_1},\dots ,n_k^{d_k}}:{\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}\to {\mathbb{P}}^N,whereN=\left(\prod _{i=1}^k\left(\genfrac{}{}{0ex}{}{d_i+n_i}{n_i}\right)\right)-1$$

$$X:={\nu }_{n_1^{d_1},\dots ,n_k^{d_k}}({\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k})$$

$$X={\sigma }_1(X)\subset \tau (X)\subset {\sigma }_2(X)\subset \cdots \subset {\sigma }_{s-1}(X)\subset {\sigma }_s(X)\subset \cdots \subset {\mathbb{P}}^N.$$

$$\mathcal{S}(p):=\left\{\{x_1,\dots ,x_s\}\subset {\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}|r_X(p)=sandp\in ⟨{\nu }_{n_1^{d_1},\dots ,n_k^{d_k}}(x_1),\dots ,{\nu }_{n_1^{d_1},\dots ,n_k^{d_k}}(x_s)⟩\right\}.$$

$$\mathrm{Cr}(p):=\min \left\{\mathrm{deg}(Z)|{\nu }_{n_1^{d_1},\dots ,n_k^{d_k}}(Z)\subset X,Zcurvilinear,p\in ⟨{\nu }_{n_1^{d_1},\dots ,n_k^{d_k}}(Z)⟩\right\}.$$

$$p\in {\nu }_{1^{d_1},\dots ,1^{d_a},2^{d_{a+1}},\dots 2^{d_k}}({\mathbb{P}}^1{)}^a\times ({\mathbb{P}}^2{)}^b.$$

$$f_i:{\mathbb{P}}^1\to {\mathbb{P}}^{n_i}with{f}_i(3u)={\pi }_i(\Gamma ).$$

$$f=(f_1,\dots ,f_k):{\mathbb{P}}^1\to {\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}.$$

$$C:={\nu }_{n_1^{d_1},\dots ,n_k^{d_k}}\left(f\left({\mathbb{P}}^1\right)\right)andZ:={\nu }_{n_1^{d_1},\dots ,n_k^{d_k}}(\Gamma ).$$

$$\alpha :=\sum _{j\in I^{\prime }}{d}_j+\sum _{j\in J^{\prime }}{d}_j+d_i$$

$$T:=L\cup \left(\bigcup _{j\in I^{\prime }}{D}_j\right)\cup \left(\bigcup _{j\in J^{\prime }}{T}_j\right)$$

$${\nu }_{1^{(k)}}:({\mathbb{P}}^1{)}^k\to {\mathbb{P}}^r,r=2^k-1$$

$${\nu }_{1^{(k-1)}}:({\mathbb{P}}^1{)}^{k-1}\to {\mathbb{P}}^{r^{\prime }},r^{\prime }=2^{k-1}-1$$

$$o:=\mathrm{Supp}(\Gamma )\in ({\mathbb{P}}^1{)}^k.$$

$$C:={\nu }_{1^{(k)}}(f({\mathbb{P}}^1))$$

$$r_X(p)\le r_C(p).$$

$$D:={\pi }_k^{-1}(v_k).$$

$$\ell :{\mathbb{P}}^{2^k-1}\setminus ⟨{\nu }_{1^{(k)}}(D)⟩\to {\mathbb{P}}^{2^{k-1}-1}$$

$$\ell \circ {\nu }_{1^{(k)}}={\nu }_{1^{(k-1)}}\circ {\tau }_k.$$

$$g:E\to Z\hookrightarrow {\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}.$$

$$h=(h_1,\dots ,h_k):{\mathbb{P}}^1\to {\mathbb{P}}^{n_1}\times \cdots \times {\mathbb{P}}^{n_k}$$

$$D:=h({\mathbb{P}}^1)$$

$$r_X(p)\le r_{{\nu }_{n_1^{d_1},\dots ,n_k^{d_k}}(D)}(p).$$

Full text

Introduction

In this paper we deal with the problem of finding a bound for the minimum integer $r(T)$ needed to write a given tensor $T$ as a linear combination of $r(T)$ decomposable tensors. Such a minimum number is now known under the name of rank of $T$. In order to be as general as possible we will consider the tensor $T$ to be partially symmetric, i.e.

[TABLE]

where the $d_{i}$’s are positive integers and $V_{i}$’s are finte dimensional vector spaces defined over an algebraically closed field $K$. The decomposition that will give us the rank of such a tensor will be of the following type:

[TABLE]

where $\lambda_{i}\in K$ and $v_{j,i}\in V_{j}$, $i=1,\ldots,r(T)$ and $j=1,\ldots,k$.

Another very interesting and useful notion of “ rank ” is the minimum $r(T)$ such that a tensor $T$ can be written as a limit of a sequence of rank $r(T)$ tensors. This last integer is called the border rank of $T$ (Definition 1.5) and clearly it can be strictly smaller than the rank of $T$ (Remark 1.6). It has become a common technique to fix a class of tensors of given border rank and then study all the possible ranks arising in that family (cf. [7, 3, 10, 14, 6]). The rank of tensors of border rank 2 is well known (cf. [7] for symmetric tensors, [2] for tensors without any symmetry, [4] for partially symmetric tensors). The first not completely classified case is the one of border rank 3 tensors. In [7, Theorem 37] the rank of any symmetric order $d$ tensor of border rank 3 has been computed and it is shown that the maximum rank reached is $2d-1$. In the present paper, Theorem 1.7, we prove that the rank of partially symmetric tensors $T$ as in (1) of border rank 3 can be at most

[TABLE]

In [10, Theorem 1.8] J. Buczyński and J.M. Landsberg described the cases in which the inequality in Theorem 1.7 is an equality: when $k=3$ and $d_{1}=d_{2}=d_{3}=1$ they show that there is an element of rank 5. All ranks for border rank 3 partially symmetric tensors are described in [9] when $k=3$, $d_{1}=d_{2}=d_{3}=1$ and $n_{i}=1$ for at least one integer $i$. Therefore our Theorem 1.7 is the natural extension of the two extreme cases (tensors without any symmetry where $d_{i}=1$ for all $i=1,\ldots,k$ and totally symmetric case where $k=1$).

In the special case of tensors without any symmetry, i.e. $T\in V_{1}\otimes\cdots\otimes V_{k}$, we will be able to show, in Theorem 1.8, that all ranks among 3 and $k-1$ arise and if $\dim V_{i}\geq 3$ for all $i$’s then also all the ranks between $k$ and $2k-1$ arise, therefore this result is sharp (cf. Remark 3.11). In the proof of this theorem we will describe the structure of our solutions: they are all obtained from $(\mathbb{P}^{2})^{k}$ taking as a border scheme a degree $3$ connected curvilinear scheme. The two critical cases are rank $k-1$ on $(\mathbb{P}^{1})^{k}$ and rank $2k-1$ on $(\mathbb{P}^{2})^{k}$ and the other cases can be deduced from one of these two.

In [5] we defined the notion of curvilinear rank for symmetric tensors to be the minimum length of a curvilinear scheme whose span contains a given symmetric tensor. We can extend some of the ideas in [5] and some of those used in our proof of Theorem 1.7 to the case of partially symmetric tensors and prove that, if a partially symmetric tensor is contained in the span of a special degree $c$ curvilinear scheme with $\alpha$ components, the rank of this tensor is bounded by $2\alpha+c\left(-1+\sum_{i=1}^{k}d_{i}\right)$ (cf. Theorem 1.11).

1. Notation, Definitions and Statements

In this section we introduce the basic geometric tools that we will use all along the paper.

Notation 1.1**.**

We indicate with

[TABLE]

the Segre embedding of the multi-projective space $\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}$, i.e. the embedding of $\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}$ by the complete linear system $|\cal{O}_{\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}}(1,\dots,1)|$.

For each $i\in\{1,\dots,k\}$ let

[TABLE]

denote the projection onto the $i$-th factor.

Let

[TABLE]

denote the projection onto all the factors different from $\mathbb{P}^{n_{i}}$.

Let $\varepsilon_{i}\in\mathbb{N}^{k}$ be the $k$-tuple of integers $\varepsilon_{i}=(0,\dots,1,\dots,0)$ with $1$ only in the $i$-th position. We say that a curve $C\subset{\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}}$ has multi-degree $(a_{1},\dots,a_{k})$ if for all $i=1,\dots,k$ the line bundle $\cal{O}_{C}(\varepsilon_{i})$ has degree $a_{i}$.

We say that a morphism $h:\mathbb{P}^{1}\to{\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}}$ has multi-degree $(a_{1},\dots,a_{k})$ if, for all $i=1,\ldots,k$:

[TABLE]

Let

[TABLE]

denote the Segre-Veronese embedding of ${\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}}$ of multi-degree $(d_{1},\dots,d_{k})$ and define

[TABLE]

to be the Segre-Veronese variety.

The name “ Segre-Veronese ” is classically due to the fact that when the $d_{i}$’s are all equal to 1, then the variety $X$ is called “ Segre variety ”; while when $k=1$ then $X$ is known to be a “ Veronese variety ”.

Remark 1.2**.**

An element of $X$ is the projective class of a decomposable partially symmetric tensor $T\in S^{d_{1}}V_{1}\otimes\cdots\otimes S^{d_{k}}V_{k}$ where $\mathbb{P}(V_{i})=\mathbb{P}^{n_{i}}$. More precisely $p\in X$ if there exists $T\in S^{d_{1}}V_{1}\otimes\cdots\otimes S^{d_{k}}V_{k}$ such that $p=[T]=[v_{1,i}^{\otimes d_{1}}\otimes\cdots\otimes v_{k,i}^{\otimes d_{k}}]$ with $[v_{j,i}]\in\mathbb{P}^{n_{i}}$.

Definition 1.3**.**

The $s$-th secant variety $\sigma_{s}(X)$ of $X$ is the Zariski closure of the union of all $s$-secant $\mathbb{P}^{s-1}$ to $X$. The tangential variety $\tau(X)$ is the Zariski closure of the union of all tangent lines to $X$.

Observe that

[TABLE]

Definition 1.4**.**

The $X$-rank $r_{X}(p)$ of an element $p\in\mathbb{P}^{N}$ is the minimum integer $s$ such that there exist a $\mathbb{P}^{s-1}\subset\mathbb{P}^{N}$ which is $s$-secant to $X$ and containing $p$.

We indicate with $\mathcal{S}(p)$ the set of sets of points of $\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}$ “ evincing ” the $X$-rank of $p\in\mathbb{P}^{N}$, i.e.

[TABLE]

Definition 1.5**.**

The $X$-border rank $br_{X}(p)$ of an element $p\in\mathbb{P}^{N}$ is the minimum integer $s$ such that $p\in\sigma_{s}(X)$.

Remark 1.6**.**

For any $p\in\mathbb{P}^{N}=\mathbb{P}(S^{d_{1}}V_{1}\otimes\cdots\otimes S^{d_{k}}V_{k})$ we obviously have that $br_{X}(p)\leq r_{X}(P)$. In fact $p\in\mathbb{P}^{N}$ of rank $r$ is such that there exist a tensor $T\in S^{d_{1}}V_{1}\otimes\cdots\otimes S^{d_{k}}V_{k}$ that can be minimally written as in (2); while an element $p\in\mathbb{P}^{N}$ has border rank $s$ if and only if there exist a sequence of rank $r$ tensors $T_{i}\in S^{d_{1}}V_{1}\otimes\cdots\otimes S^{d_{k}}V_{k}$ such that $\lim_{i\rightarrow\infty}T_{i}=T$ and $p=[T]$.

The first result that we prove in Section 2 is an upper bound for the rank of points in $\sigma_{3}(X)$.

Theorem 1.7**.**

The rank of an element $p\in\sigma_{3}(X)$ is $r_{X}({p})\leq-1+\sum_{i=1}^{k}2d_{i}$.

In the case $d_{i}=1$ for all $i=1,\ldots,k$, i.e. if $X$ is the Segre variety, we fill in all low ranks with points of $\sigma_{3}(X)\setminus\sigma_{2}(X)$. In Section 3 we prove the following result.

Theorem 1.8**.**

Assume $k\geq 3$ and let $X\subset\mathbb{P}^{M}$ be the Segre variety of $k$ factors as in Notation 1.1. Let $\alpha$ be the cardinality of the set $\{i\in\{1,\dots,k\}\mid n_{i}\geq 2$}. For each $x\in\{3,\dots,\alpha+k-1\}$ there is $p\in\sigma_{3}(X)\setminus\sigma_{2}(X)$ with $r_{X}({p})=x$.

Remark 1.9**.**

If $\alpha=k$, i.e. if $n_{i}\geq 2$ for all $i$’s, Theorems 1.7 and 1.8 give the ranks of the points of $\sigma_{3}(X)\setminus\sigma_{2}(X)$. In Remark 3.11 we discuss the reason why we do not know the rank of a specific $p\in\sigma_{3}(X)\setminus\sigma_{2}(X)$.

Moreover, in the case of Segre varieties where factors have dimension $n_{i}\geq 2$, Theorem 1.8 says that all ranks from $3$ to $2k-1$ can be attained. Therefore the above result is sharp.

As remarked in the Introduction, we can extend some of the ideas of [5] on the notion of curvilinear rank to some tools used in our proof of Theorem 1.7 to the case of partially symmetric tensors.

Definition 1.10**.**

A scheme $Z\subset{\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}}$ is curvilinear if it is a finite union of disjoint schemes of the form $\mathcal{O}_{C_{i},P_{i}}/{{m}}_{p_{i}}^{e_{i}}$ for smooth points $p_{i}\in\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}$ on reduced curves $C_{i}\subset\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}$. Equivalently $Z$ is curvilinear if the tangent space at each of its connected component supported at the $p_{i}$’s has Zariski dimension $\leq 1$. We define the curvilinear rank $\mathrm{Cr}(p)$ of a point $p\in\mathbb{P}^{N}$ as:

[TABLE]

In Section 4 we prove the following result.

Theorem 1.11**.**

If there exists a degree $c$ curvilinear scheme $Z\subset{\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}}$ such that $p\in\langle\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(Z)\rangle$ and $Z$ has $\alpha$ connected components, each of them mapped by $\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}$ into a linearly independent zero-dimensional sub-scheme of $\mathbb{P}^{N}$, then $r_{X}({p})\leq 2\alpha+c\left(-1+\sum_{i=1}^{k}d_{i}\right)$.

2. Proof of Theorem 1.7

Remark 2.1**.**

Fix a degree $3$ connected curvilinear scheme $E\subset\mathbb{P}^{2}$ not contained in a line and a point $u\in\mathbb{P}^{1}$. The scheme $E$ is contained in a smooth conic. Hence there is an embedding $f:\mathbb{P}^{1}\to\mathbb{P}^{2}$ with $f(\mathbb{P}^{1})=C$ and $f(3u)=E$.

Remark 2.2**.**

For any couple of points $u,o\in\mathbb{P}^{1}$, there is an isomorphism $f:\mathbb{P}^{1}\to\mathbb{P}^{1}$ such that $f(u)=o$. For any such $f$ we have $f(3u)=3o$.

Remark 2.3**.**

Fix two points $u,o\in\mathbb{P}^{1}$. There is a morphism $f:\mathbb{P}^{1}\to\mathbb{P}^{1}$ such that $f(u)=o$, $f$ is ramified at $u$ and $\deg(f)=2$, i.e. $f^{\ast}(\cal{O}_{\mathbb{P}^{1}}(1))\cong\cal{O}_{\mathbb{P}^{1}}(2)$. Since $\deg(f)=2$, $f$ has only order $1$ ramification at $u$. Thus $f(3u)=2o$ (as schemes).

We recall the following lemma proved in [4, Lemma 3.3].

Lemma 2.4** (Autarky).**

Let $p\in\langle X\rangle$ with $X$ being the Segre-Veronese variety of $\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}$ embedded in multidegree $(d_{1},\ldots,d_{k})$. If there exist $\mathbb{P}^{m_{i}}$, $i=1,\ldots,k$, with $m_{i}\leq n_{i}$, such that $p\in\langle\nu_{m_{1}^{d_{1}},\ldots,m_{k}^{d_{k}}}(\mathbb{P}^{m_{1}}\times\cdots\times\mathbb{P}^{m_{k}})\rangle$, then the $X$-rank of $p$ is the same as the $Y$-rank of $p$ where $Y$ is the Segre-Veronese $\nu_{m_{1}^{d_{1}},\ldots,m_{k}^{d_{k}}}(\mathbb{P}^{m_{1}}\times\cdots\times\mathbb{P}^{m_{k}})$.

Corollary 2.5**.**

Let $\Gamma\subset\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}$ be a 0-dimensional scheme of minimal degree such that $p\in\langle\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(\Gamma)\rangle$, then the $X$-rank of $p$ is equal to its $Y$-rank where $Y$ is the Segre-Veronese embedding of $\mathbb{P}^{m_{1}}\times\cdots\times\mathbb{P}^{m_{k}}$ where each $m_{i}=\dim\langle\pi_{i}(\Gamma)\rangle-1\leq\deg(\pi_{i}(\Gamma))-1$ ($\pi_{i}$ as in Notation 1.1). If there exists an index $i$ such that $\deg(\pi_{i}(\Gamma))=1$, then we can take $Y$ to be the Segre-Veronese embedding of $\mathbb{P}^{m_{1}}\times\cdots\times\hat{}\mathbb{P}^{m_{i}}\times\cdots\times\mathbb{P}^{m_{k}}$.

Proof.

Consider the projections $\pi_{i}:{\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}}$ onto the $i$-th factor $\mathbb{P}^{n_{i}}$ as in Notation 1.1. It may happen that $\deg(\pi_{i}(\Gamma))$ can be any value from $1$ to $\deg(\Gamma)$.

By the just recalled Autarky Lemma (cf. Lemma 2.4), we may assume that each $\pi_{i}(\Gamma)$ spans the whole $\mathbb{P}^{n_{i}}$. Therefore if there is an index $i\in\{1,\ldots,k\}$ such that $\deg(\pi_{i}(\Gamma))=1$ we can take $p\in\mathbb{P}^{n_{1}}\times\cdots\times\hat{}\mathbb{P}^{n_{i}}\times\cdots\times\mathbb{P}^{n_{k}}$. Moreover the autarkic fact that we can assume $\mathbb{P}^{n_{i}}$ to be $\langle\pi_{i}(\Gamma)\rangle$ implies that we can replace each $\mathbb{P}^{n_{i}}$ with $\mathbb{P}^{\dim\langle\pi_{i}(\Gamma)\rangle-1}$ and clearly $\dim\langle\pi_{i}(\Gamma)\rangle\leq\deg(\pi_{i}(\Gamma))$. ∎

Proof of Theorem 1.7:.

Because of the filtration of secants varieties (3), for a given element $p\in\sigma_{3}(X)$, it may happen that either $p\in X$, or $p\in\sigma_{2}(X)\setminus X$ or $p\in\sigma_{3}(X)\setminus\sigma_{2}(X)$. We distinguish among these cases.

1. (1)

If $p\in X$, then $r_{X}({p})=1$. 2. (2)

If $p\in\sigma_{2}(X)\setminus X$ then either $p$ lies on a honest bisecant line to $X$ (and in this case obviously $r_{X}({p})=2$) or $p$ belongs to certain tangent line to $X$. In this latter case, the minimum number $h\leq k$ of factors containing such a tangent line is the minimum integer such that $p\in\langle\nu_{n_{1}^{d_{1}},\dots,n_{h}^{d_{h}}}(\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{h}})\rangle$ (maybe reordering factors). In [4, Theorem 3.1] we proved that, if this is the case, then $r_{X}(p)=\sum_{i=1}^{h}d_{i}$. 3. (3)

From now on we assume that $p\in\sigma_{3}(X)\setminus\sigma_{2}(X)$. By [10, Theorem 1.2] there is short list of zero-dimensional schemes $\Gamma\subset{\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}}$ such that $p\in\langle\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(\Gamma)\rangle$, therefore, in order to prove Theorem 1.7, it is sufficient to bound the rank of the points in $\langle\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(\Gamma)\rangle$ for each $\Gamma$ in their list.

Since $p\in\sigma_{3}(X)\setminus\sigma_{2}(X)$, The possibilities for $\Gamma$ are only the following: either $\Gamma$ is a smooth degree 3 zero-dimensional scheme (case (3a) below), or it is the union of a degree 2 scheme supported at one point and a simple point (case (3b)), or it is a curvilinear degree $3$ scheme (case (3c)) or, finally, a very particular degree $4$ scheme with $2$ connected components of degree $2$ (case (3d)).

1. (3a)

If $\Gamma$ is a set of $3$ distinct points, then obviously $r_{X}({p})=3$ ([10, Case (i), Theorem 1.2]). 2. (3b)

If $\Gamma$ is a disjoint union of a simple point $a$ and a degree $2$ connected scheme ([10, Case (ii), Theorem 1.2]), then there is a point $q$ on a tangent line to $X$ such that $p\in\langle\{\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(a),q\}\rangle$. Hence $r_{X}({p})\leq 1+r_{X}(q)\leq 1+\sum_{i=1}^{k}d_{i}$ (for the rank on the tangential variety of $X$ see [2]). Since $d_{i}>0$ for all $i$’s and $k\geq 2$, then $1+\sum_{i=1}^{k}d_{i}\leq-1+\sum_{i=1}^{k}2d_{i}$. 3. (3c)

Assume $\deg(\Gamma)=3$ and that $\Gamma$ is connected ([10, Case (iii), Theorem 1.2]) supported at a point $\{o\}:=\Gamma_{\mathrm{red}}$. Since the case $k=1$ is true by [7, Theorem 37], we can prove the theorem by using induction on $k$, with the case $k=1$ as the starting case.

Since $\deg(\Gamma)=3$, by Corollary 2.5, we can assume that $p$ belongs to a Segre-Veronese variety of $k$ factors all of them being either $\mathbb{P}^{1}$’s or $\mathbb{P}^{2}$’s, i.e., after having reordered the factors,

[TABLE]

The $\mathbb{P}^{1}$’s correspond to the cases in which either $\deg(\pi_{i}(\Gamma))=3$ and $\dim\langle\pi_{i}(\Gamma)\rangle=1$ (i.e. $\pi_{i}(\Gamma)$ is contained in a line of the original $\mathbb{P}^{n_{i}}$), or $\deg(\pi_{i}(\Gamma))=2$ (notice that in this case $\pi_{i\mid\Gamma}$ is not an embedding). The $\mathbb{P}^{2}$’s correspond to the cases in which $\dim\langle\pi_{i}(\Gamma)\rangle=2$, $\deg(\pi_{i}(\Gamma))=3$. Finally we can exclude all the cases in which $\deg(\pi_{i}(\Gamma))=1$ because, again by Corollary 2.5, we would have that $p$ belongs to a Segre-Veronse variety of less factors and then this won’t give the highest bound for the rank of $p$.

Now fix a point $u\in\mathbb{P}^{1}$. By Remarks 2.1, 2.2 and 2.3 there is

[TABLE]

Consider the map

[TABLE]

We have $f(u)=\{o\}$ and $\pi_{i}(f(3u))=f_{i}(3u)=\pi_{i}(\Gamma)$. Since $\pi_{i}(f(3u))=\pi_{i}(\Gamma)$ for all $i$’s, the universal property of products gives $f(3u)=\Gamma$. The map $f$ has multi-degree $(a_{1},\dots,a_{k})$ where $a_{i}=1$ if $n_{i}=1$ and $\deg(\pi_{i}(\Gamma))=3$, and $a_{i}=2$ in all other cases. Notice that $f_{i}$ is an embedding if $\deg(\pi_{i}(\Gamma))\neq 2$. Since $\deg(\pi_{i}(\Gamma))=2$ if and only if $\pi_{i}^{-1}(o_{i})$ contains the line spanned by the degree $2$ sub-scheme of $\Gamma$, we have $\deg(\pi_{i}(\Gamma))=2$ for at most one index $i$. Since $k\geq 2$, $f$ is an embedding. Set

[TABLE]

The curve $C$ is smooth and rational of degree $\delta:=\sum_{i=1}^{k}a_{i}d_{i}$. Note that $\delta\leq\sum_{i=1}^{k}2d_{i}$. Hence to prove Theorem 1.7 in this case it is sufficient to show that $r_{C}({p})\leq\delta-1$ because clearly $r_{C}(p)\geq r_{X}(p)$ since $C\subset X$.

By assumption $p\in\langle Z\rangle$. Since $p\notin\sigma_{2}(X)$, $\langle Z\rangle$ is not a line of $\mathbb{P}^{N}$. Hence $\langle Z\rangle$ is a plane because $\deg(Z)=\deg(\Gamma)=3$. Since $C$ is a degree $\delta$ smooth rational curve, we have $\dim\langle C\rangle\leq\delta$. By [14, Proposition 5.1] we have $r_{C}({p})\leq\dim\langle C\rangle$. Hence it is sufficient to prove the case $\delta=\dim\langle C\rangle$, i.e. we may assume that $C$ is a rational normal curve in its linear span.

If $\delta\geq 4$, since $Z$ is connected and of degree $3$, by Sylvester’s theorem (cf. [11]) we have $p$ has $C$-border rank $3$ and $r_{C}({p})=\delta-1$, concluding the proof in this case.

If $\delta\leq 3$, we have $\sigma_{2}({C})=\langle C\rangle$ and hence $p\in\sigma_{2}(X)$, contradicting $p\in\sigma_{3}(X)\setminus\sigma_{2}(X)$. 4. (3d)

Assume that $\Gamma$ has degree $4$ ([10, Case (iv), Theorem 1.2]). J. Buczyński and J.M. Landsberg show that $p$ belongs to the span of two tangent lines to $X$ whose set theoretic intersections with $X$ span a line which is contained in $X$. This means that $\Gamma=v\sqcup w$ with $v,w$ being two degree 2 reduced zero-dimensional schemes with support contained in a line $L\subset{\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}}$ and moreover that the multi-degree of $L$ is $\varepsilon_{i}$ for some $i=1,\ldots,k$ (cfr. Notation 1.1). This case occurs only when $d_{i}=1$, i.e. when $\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(L)=\nu_{n_{1}^{1},\dots,n_{k}^{1}}(L)=\tilde{L}$ is a line.

Observe that $\tilde{v}:=\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(v)$ and $\tilde{w}:=\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(w)$ are two tangent vectors to $X$. In [2, Theorem 1] we prove that the $X$-rank of a point $p\in T_{o}(X)$ for a certain point $o=(o_{1},\ldots,o_{k})\in X$, is the minimum number $\eta_{X}(p)$ for which there exist $E\subseteq\{1,\dots,k\}$ such that $\sharp(E)=\eta_{X}(p)$ and $T_{o}(X)\subseteq\langle\cup_{i\in E}Y_{o,i}\rangle$ where $Y_{o,i}$ is the $n_{i}$-dimensional linear subspace obtained by fixing all coordinates $j\in\{1,\dots,k\}\setminus\{i\}$ equal to $o_{j}\in\mathbb{P}^{n}_{i}$. Let $I$ and $J$ be the sets playing the role of $E$ for $\langle\tilde{v}\rangle$ and $\langle\tilde{w}\rangle$ respectively and set $I^{\prime}=I\setminus\{i\}$ (meaning that $I^{\prime}=I$ if $i\notin I$ and $I^{\prime}=I\setminus\{i\}$ otherwise) and $J^{\prime}=J\setminus\{i\}$ . Now take

[TABLE]

and note that $\alpha\leq-1+\sum_{h=1}^{k}2d_{h}$, therefore if we prove that $r_{X}(p)\leq\alpha$ we are done. Let $D_{j}\subset\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}$, $j\in I^{\prime}$, be the line of multi-degree $\varepsilon_{J}$ containing $\pi_{j}(v)$, and let $T_{j}$, $j\in J^{\prime}$, be the line of $X$ of multi-degree $\varepsilon_{j}$ containing $\pi_{j}(w)$. The curve $L\cup\left(\bigcup_{j\in I^{\prime}}D_{j}\right)$ contains $v$ and the curve $L\cup\left(\bigcup_{j\in J^{\prime}}T_{j}\right)$ contains $w$. Hence the curve

[TABLE]

is a reduced and connected curve containing $\Gamma$. Since $p\in\langle\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(\Gamma)\rangle$, we have that if we call $\tilde{T}:=\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(T)$ then $p\in\langle\tilde{T}\rangle$ and $r_{X}({p})\leq r_{\tilde{T}}({p})$. The curve $\tilde{T}$ is a connected curve whose irreducible components are smooth rational curves and with $\deg(\tilde{T})=\alpha$. Hence $\dim\langle\tilde{T}\rangle\leq\alpha$. Since $\tilde{T}$ is reduced and connected, as in [14, Proposition 4.1] and in [11], we get $r_{\tilde{T}}({p})\leq\alpha$. Summing up $r_{X}(p)\leq r_{\tilde{T}}(p)\leq\alpha\leq-1+\sum_{h=1}^{k}2d_{h}$.

∎

3. Proof of Theorem 1.8

Autarky Lemma (proved in [4, Lemma 3.3] and recalled here in Lemma 2.4) is true also for the border rank ([9, Proposition 2.1]). This allows to formulate the analog of Corollary 2.5 for border rank. Therefore, in order to prove Theorem 1.8 and $x\leq k-1$, we can limit ourselves to the study of the case $n_{i}=1$ for all $i$’s. This is the reason why in the first part of this section we will always work with the Segre variety of $\mathbb{P}^{1}$’s. Let

[TABLE]

be the Segre embedding of $k$ copies of $\mathbb{P}^{1}$’s and $X:=\nu_{1^{(k)}}((\mathbb{P}^{1})^{k})$; and let

[TABLE]

be the the Segre embedding of $k-1$ copies of $\mathbb{P}^{1}$’s and $X^{\prime}:=\nu_{1^{(k-1)}}((\mathbb{P}^{1})^{k-1}).$

Proposition 3.1**.**

Assume $k\geq 3$. Let $\Gamma\subset(\mathbb{P}^{1})^{k}$ be a degree $3$ connected curvilinear scheme such that $\deg(\pi_{i}(\Gamma))=3$ for all $i$’s, and let $\beta$ be the only degree $2$ sub-scheme of $\Gamma$. For all $p\in\langle\nu_{1^{(k)}}(\Gamma)\rangle\setminus\langle\nu_{1^{(k)}}(\beta)\rangle$ we have that

1. (a)

if $k=3$, then $2\leq r_{X}({p})\leq 3$ and $r_{X}({p})=2$ if $p$ is general in $\langle\nu_{1^{(k)}}(\Gamma)\rangle$; 2. (b)

if $k\geq 4$, then $r_{X}({p})=k-1$.

Proof.

Since $\Gamma\subset(\mathbb{P}^{1})^{k}$ is connected, it has support at only one point; all along this proof we set

[TABLE]

First of all recall that in step (3a) of the proof of Theorem 1.7 we obtained an embedding $f=(f_{1},\dots,f_{k})$ with $f_{i}:\mathbb{P}^{1}\to\mathbb{P}^{1}$ an isomorphism (see (4)); moreover we can fix a point $u\in\mathbb{P}^{1}$ such that $f(u)=o$ and $\Gamma=f(3u)$. We proved that

[TABLE]

is a degree $k$ rational normal curve in its linear span. Obviously

[TABLE]

If $k\geq 4$ Sylvester’s theorem implies $r_{C}({p})=k-1$.

Now assume $k=3$. Since a degree $3$ rational plane curve has a unique singular point, for any $q\in\langle C\rangle$ there is a unique line $L\subset\langle C\rangle=\mathbb{P}^{3}$ with $\deg(L\cap C)=2$. Thus $r_{C}({p})=2$ (resp. $r_{C}({p})=3$) if and only if $p\notin\tau({C})$ (resp. $p\in\tau({C})$, cfr. Definition 1.3). Since $\tau({C})$ is a degree $4$ surface, by Riemann-Hurwitz, we see that both cases occur and that $r_{C}({p})=2$ (and hence $r_{X}({p})=2$ if $p$ is general in $\langle\nu_{1^{(k)}}(\Gamma)\rangle$).

Claim 1**.**

Let the point $o\in(\mathbb{P}^{1})^{k}$ be, as in (7), the support of $\Gamma$. Fix any $F\in|\cal{O}_{(\mathbb{P}^{1})^{k}}(\varepsilon_{k})|$ such that $o\notin F$. Then $\langle\nu_{1^{(k)}}(\Gamma)\rangle\cap\langle\nu_{1^{(k)}}(F)\rangle=\emptyset$.

Proof of Claim 1.

It is sufficient to show that $h^{0}(\cal{I}_{F\cup\Gamma}(1,\dots,1))=h^{0}(\cal{I}_{F}(1,\dots,1))-3$, i.e. $h^{0}(\cal{I}_{\Gamma}(1,\dots,1,0))=h^{0}(\cal{O}_{(\mathbb{P}^{1})^{k}}(1,\dots,1,0))-3$. This is true because $f_{1},\ldots,f_{k-1}$ (recalled at the beginning of the proof this Proposition 3.1 and introduced in (4)) are isomorphisms. ∎

1. (a)

Assume $k=3$. Since $r_{X}({p})\leq r_{C}({p})\leq 3$ and $r_{C}({p})=2$ for a general $p$ in $\langle\nu_{1^{(3)}}(\Gamma)\rangle$, we only need to prove that $r_{X}({p})>1$. The case $r_{X}({p})=1$ corresponds to a completely decomposable tensor: $p=\nu_{1^{(3)}}(q)$ for some $q\in{(\mathbb{P}^{1})^{3}}$. Clearly $r_{X}(\nu_{1^{(3)}}(o))=1$ but $o\in\langle\beta\rangle$ then, since we took $p\in\langle\nu_{1^{(3)}}(\Gamma)\rangle\setminus\langle\nu_{1^{(3)}}(\beta)\rangle$, we have $p\neq\nu_{1^{(3)}}(o)$ and in particular $q\neq o$. In this case we can add $q$ to $\Gamma$ and get that $h^{1}(\cal{I}_{q\cup\Gamma}(1,1,1))>0$ by [1, Lemma 1]. Since $\deg(f_{i}(\Gamma))=3$, for all $i$’s, every point of $\langle\beta\rangle\setminus\{o\}$ has rank $2$. Since $q:=(q_{1},q_{2},q_{3})\neq o:=(o_{1},o_{2},o_{3})$ we have $q_{i}\neq o_{i}$ for some $i$, say for $i=3$. Take $F\in|\cal{O}_{(\mathbb{P}^{1})^{3}}(\varepsilon_{3})|$ such that $q\in F$ and $o\notin F$. Hence $F\cap(\Gamma\cup\{q\})=\{q\}$. We have $h^{1}(F,\cal{I}_{q,F}(1,1,1))=0$, because $\cal{O}_{(\mathbb{P}^{1})^{3}}(1,1,1)$ is spanned. Claim 1 gives $h^{1}(\cal{I}_{\Gamma}(1,1,0))=0$. The residual exact sequence of $F$ in ${(\mathbb{P}^{1})^{3}}$ gives $h^{1}(\mathcal{I}_{\Gamma\cup\{q\}}(1,1,1))=0$, a contradiction. 2. (b)

From now on we assume $k\geq 4$ and that Proposition 3.1 is true for a smaller number of factors. Since $X\supset C$, we have $r_{X}({p})\leq k-1$ (in fact, as we already recalled above, $r_{C}({p})=k-1$ by Sylvester’s theorem). We need to prove that we actually have an equality, so we assume $r_{X}({p})\leq k-2$ and we will get a contradiction.

Take a set of points $S\in\cal{S}({p})$ of $(\mathbb{P}^{1})^{k}$ evincing the $X$-rank of $p$ (see Definition 1.4) and consider $v=(v_{1},\dots,v_{k})\in S\subset(\mathbb{P}^{1})^{k}$ to be a point appearing in a decomposition of $p$. We can always assume that, if $o=(o_{1},\dots,o_{k})$, then $v_{k}\neq o_{k}$: such a $v\in S\subset\mathcal{S}(p)$ exists because, by Autarky (here recalled in Lemma 2.4), no element of $\cal{S}({p})$ is contained in $(\mathbb{P}^{1})^{k-1}\times\{o_{k}\}$.

Consider the pre-image

[TABLE]

Clearly by construction $o\notin D$ hence for any $q\in{(\mathbb{P}^{1})^{k}}\setminus D$ we have $h^{1}(\cal{I}_{q\cup D}(1,\dots,1))=h^{1}(\cal{I}_{q}(1,\dots,1,0))=0$, because $\cal{O}_{(\mathbb{P}^{1})^{k}}(1,\dots,1,0)$ is globally generated. This implies that $\langle\nu_{1^{(k)}}(D)\rangle$ intersects $X$ only in $\nu_{1^{(k)}}(D)$.

Now consider

[TABLE]

the linear projection from $\langle\nu_{1^{(k)}}(D)\rangle$. Since $p\notin\langle\nu_{1^{(k)}}(D)\rangle$ (Claim 1), $\ell$ is defined at $p$. Moreover the map $\ell$ induces a rational map $\nu_{1^{(k)}}({(\mathbb{P}^{1})^{k}}\setminus D)\to\nu_{1^{(k-1)}}({(\mathbb{P}^{1})^{k-1}})$ which is induced by the projection $\tau_{k}:{(\mathbb{P}^{1})^{k}}\to{(\mathbb{P}^{1})^{k-1}}$ defined in Notation 1.1. We have

[TABLE]

Since $o\notin D$, we have $\ell(\langle\Gamma\rangle)=\langle\nu_{1^{(k-1)}}(\Gamma^{\prime})\rangle$, where $\Gamma^{\prime}=\tau_{k}(\Gamma)$. Hence $p^{\prime}:=\ell({p})\in\langle\nu_{1^{(k-1)}}(\Gamma^{\prime})\rangle$. By [2] every element of $\langle\nu_{1^{(k-1)}}(\beta)\rangle\setminus\nu_{1^{(k-1)}}(o^{\prime})$, with $o^{\prime}:=\tau_{k}(o)$, has $X^{\prime}$-rank $k-1$. Since $\deg(\pi_{i}(\Gamma))=3$ for all $i$’s, we have $\deg(\pi_{i}(\beta))=2$ for $i=1,\dots,k-1$. This implies that the minimal sub-scheme $\alpha$ of $\Gamma^{\prime}$ such that $p^{\prime}\in\langle\nu_{1^{(k-1)}}(\alpha)\rangle$ is such that $\alpha\neq\beta$ where $\beta$ is the degree $2$ sub-scheme of $\Gamma^{\prime}$. Now let $S^{\prime}\subset(\mathbb{P}^{1})^{k-1}$ be the projection by $\tau_{k}$ of the set of points of $S\subset\mathcal{S}(p)$ that are not in $D$, i.e. $S^{\prime}:=\tau_{k}(S\setminus S\cap D)$. Since $\sharp(S^{\prime})\leq k-2$ and $p^{\prime}\in\langle\nu_{1^{(k-1)}}(\Gamma^{\prime})\rangle$, the inductive assumption gives $\alpha\neq\Gamma^{\prime}$ (it works even when $k=4$). Hence $\alpha=\{o^{\prime}\}$. Thus $p\in\langle\nu_{1^{(k)}}(\{o\}\cup D)\rangle$. Hence $\dim(\langle\nu_{1^{(k)}}(\Gamma\cup D)\rangle)\leq\dim(\langle\nu_{1^{(k)}}(D)\rangle)+2$, contradicting Claim 1.

∎

We need the following lemma, which is the projective version of an obvious linear algebra exercise.

Lemma 3.2**.**

Fix two linear spaces $L_{1}\subsetneq L_{2}\subset\mathbb{P}^{m}$ and a finite set $E\subset L_{2}$ spanning $L_{2}$. Let $\ell:\mathbb{P}^{m}\setminus L_{1}\to\mathbb{P}^{z}$, $z:=m-1-\dim L_{1}$, be the linear projection from $L_{1}$. Then $\ell(L_{2}\setminus L_{1})$ is a linear space spanned by the set $\ell(E\setminus E\cap L_{1})$.

Notation 3.3**.**

Fix $(a,b)\in\mathbb{N}^{2}\setminus\{(0,0)\}$. Let $\Delta_{a,b}$ be the set of all pairs $(f,o)$, where $o\in\mathbb{P}^{1}$, $f:\mathbb{P}^{1}\to(\mathbb{P}^{2})^{a}\times(\mathbb{P}^{1})^{b}$, each $\pi_{i}\circ f$, $1\leq i\leq a$, is a degree $2$ embedding and, for $a+1\leq i\leq b$, $\pi_{i}\circ f$ is an isomorphism.

Lemma 3.4**.**

Set $\tilde{G}=\mathrm{Aut}(\mathbb{P}^{2})^{a}\times\mathrm{Aut}(\mathbb{P}^{1})^{b}$, $G:=\tilde{G}\times\mathrm{Aut}(\mathbb{P}^{1})$. Let $G$ acts on $\Delta_{a,b}$ via $(g,h)(f,o)=(g\circ f\circ h^{-1},h(o))$. Then this action is transitive, i.e., for $(f,o),(f^{\prime},o^{\prime})$ we have $(g,h)\in G$ such that $h(o)=o^{\prime}$ and $g\circ f\circ h^{-1}=f^{\prime}$.

Proof.

Fix any $h\in\mathrm{Aut}(\mathbb{P}^{1})$ such that $h(o)=o^{\prime}$ and write $\tilde{f}:=f\circ h^{-1}$.

Write $\tilde{f}=(\tilde{f_{1}},\ldots,\tilde{f}_{a},\tilde{f}_{a+1},\ldots,\tilde{f}_{a+b})$ and $f^{\prime}=(f^{\prime}_{1},\ldots,f^{\prime}_{a},f^{\prime}_{a+1},\ldots,f^{\prime}_{a+b})$ with $\tilde{f_{i}}:=\pi_{i}\circ\tilde{f}$ and $f^{\prime}_{i}:=\pi_{i}\circ f^{\prime}$. We need to find $g=(g_{1},\dots,g_{a},g_{a+1},\dots,g_{a+b})\in\tilde{G}$ such that $g\circ\tilde{f}=f^{\prime}$, i.e. by the universal property of maps to products, we need to find $g=(g_{1},\dots,g_{a},g_{a+1},\dots,g_{a+b})\in\tilde{G}$ such that $g_{i}\circ\tilde{f}_{i}=f^{\prime}_{i}$ for all $i$.

If $a+1\leq i\leq a+b$ take $g_{i}:=f^{\prime}_{i}\circ\tilde{f_{i}}^{-1}$.

Now we fix $i$ such that $1\leq i\leq a$. We have two degree $2$ embeddings $f^{\prime}_{i}:\mathbb{P}^{1}\to\mathbb{P}^{2}$ and $\tilde{f_{i}}:\mathbb{P}^{1}\to\mathbb{P}^{2}$. Any two such maps are equivalent, up to an automorphism of $\mathbb{P}^{2}$, because these embeddings are induced by the complete linear system of the anticanonical line bundle of $\mathbb{P}^{1}$. Thus there is $g_{i}\in\mathrm{Aut}(\mathbb{P}^{2})$ such that $g_{i}\circ\tilde{f}_{i}=f^{\prime}_{i}$. ∎

Notation 3.5**.**

Take $Y=(\mathbb{P}^{2})^{a}\times(\mathbb{P}^{1})^{b}$ and let $\nu_{2^{(a)},1^{(b)}}:Y\to\mathbb{P}^{N}$, $N:=3^{a}2^{b}-1$, be the Segre embedding of $Y$. Let $\Gamma_{a,b}$ (resp. $\Gamma^{\prime}_{a,b}$) be the set of all $p\in\mathbb{P}^{N}$, such there is $(f,o)\in\Delta_{a,b}$ with $p\in\langle\nu_{2^{(a)},1^{(b)}}(f(3o))\rangle$ (resp. and $p\notin\langle\nu_{2^{(a)},1^{(b)}}(f(2o))\rangle$).

Since the image of an algebraic set by a morphism is constructible, $\Gamma_{a,b}$ and $\Gamma^{\prime}_{a,b}$ of Notation 3.5 are constructible sets. The closure of $\Gamma_{a,b}$ in $\mathbb{P}^{N}$ is irreducible. Therefore we are allowed to inquire about the rank of a general element of $\Gamma_{a,b}$. If either $a>0$ or $b\geq 2$, then $\Gamma^{\prime}_{a,b}\neq\emptyset$ and the closures in $\mathbb{P}^{N}$ of $\Gamma_{a,b}$ and $\Gamma^{\prime}_{a,b}$ are the same.

Lemma 3.6**.**

For all $k\geq 3$ we have $r_{X}({p})=2k-1$ for a general $p\in\Gamma_{k,0}$ as in Notation 3.5.

Proof.

We use induction on $k$, the case $k=3$ being true by [10, Theorem 1.8].

Now assume $k\geq 4$. Call $\nu_{2^{(k)}}:(\mathbb{P}^{2})^{k}\to\mathbb{P}^{r}$, $r:=3^{k}-1$, the Segre embedding. Fix $a\in\mathbb{P}^{1}$. For each $1\leq i\leq k$ let $f_{i}:\mathbb{P}^{1}\to\mathbb{P}^{2}$ be a degree $2$ embedding. Let $f=(f_{1},\dots,f_{k}):\mathbb{P}^{1}\to(\mathbb{P}^{2})^{k}$ be the embedding with $f_{i}=\pi_{i}\circ f$ for all $i$. As in step (3c) of the proof of Theorem 1.7 we see that the curve $C:=\nu_{2^{(k)}}(f(\mathbb{P}^{1}))$ is a rational normal curve of degree $2k$ in its linear span. Fix $a\in\mathbb{P}^{1}$ and set $o:=(o_{1},\dots,o_{k}):=f(a)$ and $A:=f(3a)$. The scheme $\nu_{2^{(k)}}(A)$ has degree $3$ and it is curvilinear. Fix a general $p\in\langle\nu_{2^{(k)}}(A)\rangle\setminus\langle\nu_{2^{(k)}}(2o)\rangle$. Since $p$ has border rank $3$ with respect to the rational normal curve $C$, Sylvester’s theorem gives $r_{C}({p})=2k-1$. Hence $r_{X}({p})\leq 2k-1$. To prove the lemma for the integer $k$ it is sufficient to prove that $r_{X}({p})\geq 2k-1$.

Assume $r_{X}({p})\leq 2k-2$ and fix $B\in\cal{S}({p})$.

1. (a)

In this step we assume the existence of a line $L\subset\mathbb{P}^{n_{k}}$ such that $o_{k}\notin L$ and $\sharp(Y^{\prime}\cap B)\geq 2$, where $Y^{\prime}:=\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k-1}}\times L$. We have $Y^{\prime}\in|\cal{O}_{Y}(\varepsilon_{k})|$. Since $o_{k}\notin L$, we have $o\notin Y^{\prime}$ and hence $A\cap Y^{\prime}=\emptyset$. Set $B^{\prime}:=B\setminus B\cap Y^{\prime}$. Set $A^{\prime}:=\tau_{k}(A)$ where $\tau_{k}$ is defined in Notation 1.1. Since $k\geq 3$ and $(f_{1},f_{2}):\mathbb{P}^{1}\to\mathbb{P}^{2}\times\mathbb{P}^{2}$ is an embedding, we have $\deg(A^{\prime})=3$. Let $\nu_{2^{(k-1)}}:(\mathbb{P}^{2})^{k-1}\to\mathbb{P}^{s}$, $s=3^{k-1}-1$, be the Segre embedding of $(\mathbb{P}^{2})^{k-1}$. Note that the linear projection from $L$ of $\mathbb{P}^{2}\setminus L$ sends $\mathbb{P}^{2}\setminus L$ onto a point. Set $E:=\langle\nu_{2^{(k)}}(Y^{\prime})\rangle$. We have $\dim E=2\cdot 3^{k-1}-1$. Let $\ell:\mathbb{P}^{M}\setminus E\to\mathbb{P}^{s}$ denote the linear projection from $E$. Since $A\cap Y^{\prime}=\emptyset$, $\ell(\nu_{2^{(k)}}(A))$ is a well-defined zero-dimensional scheme. Note that $\nu_{2^{1},2^{1}}(f_{1},f_{2})(\mathbb{P}^{1})$ is not a line of the Segre embedding of $\mathbb{P}^{2}\times\mathbb{P}^{2}$. Since $k\geq 3$, we get that $\nu_{2^{(k-1)}}(A^{\prime})$ spans a plane. Hence $\ell(\nu_{2^{(k)}}(A))=A^{\prime}$ is linearly independent, i.e. $\langle\nu_{2^{(k)}}(A)\rangle\cap E=\emptyset$. Hence $p^{\prime}:=\ell({p})$ is well-defined and in particular it is well-defined its rank with respect to the Segre variety $X^{\prime}:=\nu_{2^{(k-1)}}((\mathbb{P}^{2})^{k-1})$. Since $\dim\langle\nu_{2^{(k)}}(A)\rangle=\dim\langle\nu_{2^{(k)}}(A^{\prime})\rangle$ and $p$ is general in $\langle\nu_{2^{(k)}}(A)\rangle$, $p^{\prime}$ is general in $\langle\nu_{2^{(k-1)}}(A^{\prime})\rangle$. By the inductive assumption (case $k\geq 5$) or by [10, Theorem 1.8] (case $k=4$), we have $r_{X^{\prime}}(p^{\prime})=2k-3$. Since $p\in\langle\nu_{2^{(k)}}(B)\rangle$, Lemma 3.2 applied to $E:=\nu_{2^{(k)}}(B)$, $m=3^{k}-1$ and $L_{1}=E$, gives $p^{\prime}\in\langle\nu_{2^{(k-1)}}(B^{\prime})\rangle$. Since $\sharp(B^{\prime})\leq\sharp(B)-2<3k-3$, we get a contradiction. 2. (b)

Assume the non-existence of a line $L\subset\mathbb{P}^{n_{k}}$ such that $o_{k}\notin L$ and $\sharp(Y^{\prime}\cap B)\geq 2$. By Autarky we have $B\nsubseteq(\mathbb{P}^{2})^{k-1}\times\{o_{k}\}$. Hence the assumption of this step is equivalent to assuming the existence of $b\in B$ such that $\pi_{k}(b)\neq o_{k}$, but $\pi_{k}(B)$ is contained in the line $R\subset\mathbb{P}^{n_{2}}$ spanned by $o_{k}$ and $\pi_{k}(b)$. Hence $B\subset(\mathbb{P}^{2})^{k-1}\times R$, contradicting Autarky, because $n_{k}=2$ and $f_{k}(3a)$ spans $\mathbb{P}^{2}$.

∎

Lemma 3.7**.**

Let $\nu_{2^{(1)},1^{(1)}}(Y)$ be the Segre embedding of $\mathbb{P}^{2}\times\mathbb{P}^{1}$. We have $\Gamma_{1,1}\nsubseteq\nu_{2^{(1)},1^{(1)}}(Y)$.

Proof.

We have $\tau(\nu_{2^{(1)},1^{(1)}}(Y))\supsetneq\nu_{2^{(1)},1^{(1)}}(Y)$. Since a general tangent vector of $Y$ is of the form $f(2o)$ with $(f,o)\in\Delta_{1,1}$, we get $\Gamma_{1,1}\nsubseteq\nu_{2^{(1)},1^{(1)}}(Y)$. ∎

Definition 3.8**.**

Let $X\subset\mathbb{P}^{N}$ be any variety, $Z$ a zero-dimensional scheme and $H$ an effective Cartier divisor. We define the scheme $Res_{H}(Z)\subset\mathbb{P}^{N}$ to be the residue scheme of $Z$ with respect to $H$, namely the subscheme of $\mathbb{P}^{N}$ whose ideal sheaf is $\mathcal{I}_{Z}:\mathcal{I}_{H}$.

Lemma 3.9**.**

Take $Y=(\mathbb{P}^{2})^{2}$. For every $p\in\Gamma^{\prime}_{2,0}$ we have $r_{X}({p})>2$ (cf. Notation 3.5).

Proof.

Assume the existence of a set $B\subset Y$ such that $\sharp(B)\leq 2$ and $p\in\langle\nu_{2^{(2)}}(B)\rangle$. Since $B\in\cal{S}({p})$, we have $p\notin\langle\nu_{2^{(2)}}(B^{\prime})\rangle$ for any $B^{\prime}\subsetneq B$. Take $(f,o)\in\Delta_{2,0}$ such that $p\in\langle\nu_{2^{(2)}}(f(3o))\rangle$ and $p\notin\langle f(2o)\rangle$. Set $A:=f(3o)$. By assumption we have $p\notin\langle\nu_{2^{(2)}}(A^{\prime})\rangle$ for any $A^{\prime}\subsetneq A$. In particular $B\neq\{o\}$. By [1, Lemma 1] we have $h^{1}(\cal{I}_{A\cup B}(1,1))>0$. Since $\sharp(B)\leq 2$, there is a line $R\subset\mathbb{P}^{2}$ such that $\pi_{1}(B)\subset R$. Set $H:=R\times\mathbb{P}^{2}\in|\cal{O}_{Y}(1,0)|$ and call $\nu^{\prime}:H\to\mathbb{P}^{5}$ the Segre embedding of $H$. We have $\mathrm{Res}_{H}(A\cup B)\subseteq A$. Since $\pi_{2}(A)$ spans $\mathbb{P}^{2}$ by the definition of $\Gamma_{2,0}$, $\pi_{2|\mathrm{Res}_{H}(A\cup B)}$ is an embedding and $\pi_{2}(A\cup B)$ is linearly independent. The residual exact sequence of $H$ in $Y$ gives $h^{1}(H,\cal{I}_{H\cap(A\cup B),H}(1,1))>0$. Hence $\langle\nu^{\prime}(H\cap A)\rangle\cap\langle\nu^{\prime}(H\cap B)\rangle\neq\emptyset$. Since $\pi_{1}(A)$ spans $\mathbb{P}^{2}$, we have $A\nsubseteq H$. Thus $H\cap A\subsetneq A$. By the definition of $\Gamma^{\prime}_{2,0}$ we have $p\notin\langle\nu_{2^{(2)}}(H\cap A)\rangle$. Set $J:=\langle\nu_{2^{(2)}}(A)\rangle\cap\nu_{2^{(2)}}(Y)$. Since the only linear subspaces of $\nu_{2^{(2)}}(Y)$ are the ones contained in a ruling of $Y$ and $(f,o)\in\Delta_{2,0}$, the plane $\langle\nu_{2^{(2)}}(A)\rangle$ is not contained in $\nu_{2^{(2)}}(Y)$. Hence $J\nsubseteq\langle\nu_{2^{(2)}}(A)\rangle$. Since $\nu_{2^{(2)}}(Y)$ is scheme-theoretically cut out by quadrics, $J$ is cut out by plane conics. Write $J=\nu_{2^{(2)}}(I)$ with $I\subset Y$. $J$ is not a reducible conic or a double line or a line, because $\pi_{i}(A)$ spans $\mathbb{P}^{2}$, $i=1,2$, while all linear subspaces of $\nu_{2^{(2)}}(Y)$ are contained in a ruling of $Y$. If $J$ were a smooth conic we would have that either $\pi_{1}(I)$ spans $\mathbb{P}^{2}$ and $\pi_{2}(I)$ is a point, or $\pi_{2}(I)$ spans $\mathbb{P}^{2}$ and $\pi_{1}(I)$ is a point or $\pi_{1}(I)$ and $\pi_{2}(I)$ are lines, contradicting the assumption that each $\pi_{i}(A)$ spans $\mathbb{P}^{2}$. Thus $J$ is a zero-dimensional scheme of degree $\leq 4$. Since $A\cup B\subseteq I$, we get that either $B=\{o\}$ (and we excluded this case) or $B=\{o,q\}$ for some $q\in A$ with $q\neq o$. Thus $\deg(A\cup B)=4$. We have $h^{1}(\cal{I}_{A\cup B}(1,1))\neq 0$ ([1, Lemma 1]). Since $p$ has not rank $2$ with respect to $\nu_{2^{(2)}}(C)$, we have $q\notin C$. Thus there is $M\in|\cal{O}_{Y}(1,1)|$ with $M\supset C$ and $q\notin M$. Thus $M\cap(A\cup B)=A$ and $\mathrm{Res}_{M}(A\cup B)=\{q\}$. Thus $h^{1}(\cal{I}_{Q})=0$. Since $h^{1}(\cal{I}_{A}(1,1))=0$, the residual exact sequence of $M$ in $Y$ gives a contradiction. ∎

Lemma 3.10**.**

Fix integers $a\geq 0$ and $b\geq 0$ with $a+b\geq 3$. We have $r_{X}({p})=2a+b-1$ for a general $p\in\Gamma_{a,b}$ (cf. Notation 3.5).

Proof.

The case $a=0$ is true by Proposition 3.1. The case $b=0$ is true by Lemma 3.6. Thus we may assume that $a>0$ and $b>0$. Set $k:=a+b$. Take $(f,o)\in\Delta_{a,b}$ such that $p$ is a general element of $\langle\nu_{2^{(a)},1^{(b)}}(A)\rangle$ with $A:=3o$. Set $C:=f(\mathbb{P}^{1})$, $f_{i}:=\pi_{i}\circ f$ and $o_{i}:=\pi_{i}(f(o))$. Since $\nu_{2^{(a)},1^{(b)}}({C})$ is a degree $2a+b$ rational normal curve in its linear span and $2a+b\geq 4$, Sylvester’s theorem gives $r_{\nu_{2^{(a)},1^{(b)}}({C})}=2a+b-1$. Thus $r_{X}({p})\leq 2a+b-1$. Assume $r_{X}({p})\leq 2a+b-2$ and take $B\in\cal{S}({p})$. By Autarky we have $B\nsubseteq(\mathbb{P}^{2})^{a}\times(\mathbb{P}^{1})^{b-1}\times\{o_{k}\}$. Take $z\in B$ such that $b_{k}:=\pi_{k}(z)\neq o_{k}$. Set $Y^{\prime}:=(\mathbb{P}^{2})^{a}\times(\mathbb{P}^{1})^{b-1}\times\{b_{k}\}$. Let $\nu_{2^{(a)},1^{(b-1)}}:=(\mathbb{P}^{2})^{a}\times(\mathbb{P}^{1})^{b-1}\to\mathbb{P}^{s}$, $s:=-1+3^{a}2^{b-1}$, be the Segre embedding of $(\mathbb{P}^{2})^{a}\times(\mathbb{P}^{1})^{b-1}$. Set $E:=\langle\nu_{2^{(a)},1^{(b)}}(Y^{\prime})\rangle$. We have $\dim E+1=2\cdot 3^{2}$. Let $\ell:\mathbb{P}^{M}\setminus E\to\mathbb{P}^{s}$ the linear projection from $E$. Set $A^{\prime}:=\tau_{k}(A)$ (as in Notation 1.1). As in the proof of Lemma 3.6 we get $E\cap\langle\nu_{2^{(a)},1^{(b)}}(A)\rangle=\emptyset$, $\nu_{2^{(a)},1^{(b-1)}}(A^{\prime})=\ell(A)$, $p^{\prime}:=\ell({p})$ is a general element of $\langle\nu_{2^{(a)},1^{(b-1)}}(A^{\prime})\rangle$.

1. (a)

Assume $(a,b)=(1,2)$. Since $\nu_{2^{1},1^{2}}(Y)\nsubseteq\Gamma_{1,2}$, $p$ is general in $\Gamma_{1,2}$ and $\sharp(B)\leq 2$, we have $\sharp(B)=2$. Thus $\sharp(A^{\prime})=1$ and so $p^{\prime}\in\nu_{2^{(1)},1^{(1)}}(\mathbb{P}^{2}\times\mathbb{P}^{1})$. Hence a general element of $\Gamma_{1,1}$ has rank $1$, contradicting Lemma 3.7. 2. (b)

Assume $(a,b)=(2,1)$. We use Lemma 3.9. 3. (c)

By the previous steps we may assume $a+b\geq 4$, $a>0$, $b>0$ and use induction on the integer $a+b$. (and hence by the inductive assumption applied to $(a,b-1)$ it has rank $2a+b-2$), while $p^{\prime}\in\langle\nu_{2^{(a)},1^{(b-1)}}(B\setminus B\cap Y^{\prime})\rangle$ with $\sharp(B\setminus B\cap Y^{\prime})\leq x-2$ (because $b_{k}\in\pi_{k}(Y^{\prime})\cap\pi_{k}(B)$), a contradiction.

∎

Proof of Theorem 1.8:.

First assume $x\leq k-1$. If $x=3$, then we may take as $p$ a general point of $\sigma_{3}(X)$. Now assume $x\geq 4$ and hence $k\geq 5$. Apply Proposition 3.1 to $(\mathbb{P}^{1})^{x+1}$ and then use Autarky (Lemma 2.4). Now assume $k\leq x\leq 2k-1$. For $x=2k-1$ use Lemma 3.6 and Autarky. For each $x\in\{4,\dots,2k-2\}$ use the case $a=x+1-k$ and $b=k-a$ of Lemma 3.10 and then apply Autarky.∎

Remark 3.11**.**

Take the set-up of Theorem 1.8. If $n_{i}\geq 2$ for all $i$, then Theorem 1.8 gives all ranks of points of $\sigma_{3}(X)\setminus\sigma_{2}(X)$, but it does not say the rank of each point of $\sigma_{3}(X)\setminus\sigma_{2}(X)$. One problem is that in Lemma 3.6 we do not check all ranks of points of $\Gamma^{\prime}_{1,1}$. A bigger problem is that the inductive proof should be adapted and the induction must start. These problems may be not deal-breakers, but there is a class of points of $\sigma_{3}(X)\setminus\sigma_{2}(X)$ (occurring even if $n_{i}=1$ for some $i$) for which we do not have a good upper bound for the rank (except that $r_{X}({p})\leq 2k-1$). These are the points $p\in\langle\nu_{n_{1}^{1},\ldots,n_{k}^{1}}(A)\rangle$ with $A\subset Y$ a connected curvilinear scheme of degree $3$ and $\deg(\pi_{i}(A))=2$ for some $i$, because in this case $A\nsubseteq C$ with $C\subset Y$ and $\nu_{n_{1}^{1},\ldots,n_{k}^{1}}({C})$ a rational normal curve in its linear span. We have no idea about the rank of these points.

4. Proof of Theorem 1.11

Lemma 4.1**.**

Fix an integer $c>0$ and $u\in\mathbb{P}^{1}$. Let $E=cu\subset\mathbb{P}^{1}$ be the degree $c$ effective divisor of $\mathbb{P}^{1}$ with $u$ as its support. Let $g:E\to\mathbb{P}^{n}$ be any morphism. Then there is a non-negative integer $e\leq c$ and a morphism $h:\mathbb{P}^{1}\to\mathbb{P}^{n}$ such that $h^{\ast}\left(\cal{O}_{\mathbb{P}^{n}}(1)\right)\cong\cal{O}_{\mathbb{P}^{1}}(e)$ and $h_{|E}=g$.

Proof.

Every line bundle on $E$ is trivial. We fix an isomorphism between $g^{\ast}\left(\cal{O}_{\mathbb{P}^{n}}(1)\right)$ and $\cal{O}_{E}({c})$. After this identification, $g$ is induced by $n+1$ sections $u_{0},\dots,u_{n}$ of $\cal{O}_{E}({c})$ such that at least one of them has a non-zero restriction at $\{u\}$. The map $H^{0}\left(\cal{O}_{\mathbb{P}^{1}}({c})\right)\to H^{0}\left(\cal{O}_{E}({c})\right)$ is surjective and its kernel is the section associated to the divisor $cu$. Hence there are $v_{0},\dots,v_{n}\in H^{0}\left(\cal{O}_{\mathbb{P}^{1}}({c})\right)$ with $v_{i|E}=u_{i}$ for all $i$. Not all sections $v_{0},\dots,v_{n}$ vanish at [math]. If they have no common zero, then they define a morphism $\mathbb{P}^{1}\to\mathbb{P}^{n}$ extending $g$ and we may take $e=c$. Now assume that they have a base locus and call $F$ the scheme-theoretic base locus of the linear system associated to $v_{0},\dots,v_{m}$. We have $\deg(F)\leq c$. Set $e:=c-\deg(F)$ and $S:=F_{\mathrm{red}}$. The sections $v_{0},\dots,v_{n}$ induce a morphism $f:\mathbb{P}^{1}\setminus S\to\mathbb{P}^{n}$ with $f_{|E}=g$. See $v_{0},\dots,v_{n}$ as elements of $|\cal{O}_{\mathbb{P}^{1}}({c})|$ and set $u_{i}:=u-F\in|\cal{O}_{\mathbb{P}^{1}}(e)|$. By construction the linear system spanned by $u_{0},\dots,u_{n}$ has no base points, hence it induces a morphism $h:\mathbb{P}^{1}\to\mathbb{P}^{n}$ such that $h^{\ast}\left(\cal{O}_{\mathbb{P}^{n}}(1)\right)\cong\cal{O}_{\mathbb{P}^{1}}(e)$. We have $h_{|\mathbb{P}^{1}\setminus S}=f$ and hence $h_{|E}=g$. ∎

Proof of Theorem 1.11:.

Let $Z\subset{\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}}$ such that $p\in\langle\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(Z)\rangle$ and $Z$ has $Z_{1}$, $\ldots$ , $Z_{\alpha}$ connected components, By assumption there is $p_{i}\in\langle\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(Z_{i})\rangle$ such that $p\in\langle\{p_{1},\dots,p_{\alpha}\}\rangle$. Note that if Theorem 1.11 is true for each $(Z_{i},p_{i})$, then it is true for $Z$. Hence it is sufficient to prove Theorem 1.11 under the additional assumption that $Z$ is connected, so from now on we assume

• •

$Z$ connected.

Moreover, since $r_{X}({p})=1\leq 2-1+\sum_{i}d_{i}$ if $c=1$, we may also assume that

• •

$\deg Z=c\geq 2$.

Finally, since the real-valued function $x\mapsto x\left(-1+\sum_{i=1}^{k}d_{i}\right)$ is increasing for $x\geq 1$, with no loss of generality we may assume that, for any $G\subsetneq Z$,

• •

$p\notin\langle\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(G)\rangle$.

Fix $u\in\mathbb{P}^{1}$ and let $E=cu\subset\mathbb{P}^{1}$ be the degree $c$ effective divisor of $\mathbb{P}^{1}$ with $u$ as its support. Since $Z$ is curvilinear and $\deg(Z)=c$, we have $Z\cong E$ as abstract zero-dimensional schemes. Let $g:E\to{\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}}$ be the composition of an isomorphism $E\to Z$ with the inclusion $Z\hookrightarrow{\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}}$:

[TABLE]

Set $g_{i}:=\pi_{i}\circ g$. If we apply Lemma 4.1 to each $g_{i}$, we get the existence of an integer $c_{i}\in\{0,\dots,c\}$ and of a morphism $h_{i}:\mathbb{P}^{1}\to\mathbb{P}^{n_{i}}$ such that $h_{i|Z}=g_{i}$ and ${h_{i}}^{\ast}\left(\cal{O}_{\mathbb{P}^{n_{i}}}(1)\right)\cong\cal{O}_{\mathbb{P}^{1}}(c_{i})$. The map

[TABLE]

has multi-degree $(c_{1},\dots,c_{k})$. The curve

[TABLE]

is an integral rational curve containing $Z$. Since $p\in\langle\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(Z)\rangle$, we have

[TABLE]

Thus it is sufficient to prove that, if we call $\tilde{D}:=\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(D)$, then $r_{\tilde{D}}({p})\leq 2+c\left(-1+\sum_{i=1}^{k}d_{i}\right)$. Since $c_{i}\leq c$ for all $i$, it is sufficient to prove that $r_{\tilde{D}}({p})\leq 2-c+\sum_{i=1}^{k}c_{i}d_{i}$.

Set ${\tilde{Z}}:=\nu_{n_{1}^{d_{1}},\dots,n_{k}^{d_{k}}}(Z)$, $m:=\dim(\langle\tilde{D}\rangle)$ and

[TABLE]

By assumption ${\tilde{Z}}$ is linearly independent in $\langle\tilde{D}\rangle\cong\mathbb{P}^{m}$ and in particular $c\leq m+1$.

1. (a)

Assume that the map $h$ defined in (8) is birational onto its image. The curve $\tilde{D}\subset\mathbb{P}^{N}$ just defined is a rational curve of degree $a:=\sum_{i=1}^{k}c_{i}d_{i}$ contained in the projective space $\mathbb{P}^{m}:=\langle\tilde{D}\rangle$ and non-degenerate in $\mathbb{P}^{m}$. Note that $a\geq m$ and that $p\in\langle{\tilde{Z}}\rangle$.

1. (1)

First assume that $a=m$. In this case $\tilde{D}$ is a rational normal curve of $\mathbb{P}^{m}$. If $c\leq\left\lceil(a+1)/2\right\rceil$, then Sylvester’s theorem implies that $r_{\tilde{D}}({p})=a+2-c=2-c+\sum_{i=1}^{k}c_{i}d_{i}$. Now assume $c>\left\lceil(a+1)/2\right\rceil$. Since ${\tilde{Z}}$ is connected and curvilinear and $p\notin\langle G\rangle$ for any $G\subsetneq{\tilde{Z}}$, Sylvester’s theorem implies $r_{\tilde{D}}({p})\leq c$. 2. (2)

Now assume $m<a$. There is a rational normal curve $C\subset\mathbb{P}^{a}$ and a linear subspace $W\subset\mathbb{P}^{a}$ such that $\dim(W)=a-m-1$, $C\cap W=\emptyset$ and $h$ is the composition of the degree $a$ complete embedding $j:=\mathbb{P}^{1}\hookrightarrow\mathbb{P}^{a}$ and the linear projection $\ell:\mathbb{P}^{a}\setminus W\to\mathbb{P}^{m}$ from $W$. The scheme $E^{\prime}:=j(E)$ is a degree $c$ curvilinear scheme and $\ell$ maps $E^{\prime}$ isomorphically onto ${\tilde{Z}}$. Since ${\tilde{Z}}$ is linearly independent, then $\langle E^{\prime}\rangle\cap W=\emptyset$ and $\ell$ maps isomorphically $\langle E^{\prime}\rangle$ onto $\langle{\tilde{Z}}\rangle$. Thus there is a unique $q\in\langle E^{\prime}\rangle$ such that $\ell(q)=p$. Take any finite set $S\subset j(\mathbb{P}^{1})$ with $q\in\langle S\rangle$. Since $C\cap W=\emptyset$, $\ell(S)$ is a well-defined subset of ${\tilde{D}}$ with cardinality $\leq\sharp(S)$. Hence $r_{\tilde{D}}({p})\leq r_{C}(q)$. As in step (a1) we see that either $r_{C}(q)=a+2-c$ (case $c\leq\left\lceil(a+1)/2\right\rceil$) or $r_{C}(q)\leq c$ (case $c>\left\lceil(a+1)/2\right\rceil$). 2. (b)

Now assume that $h$ is not birational onto its image, but it has degree $k\geq 2$. Note that $k$ divides $c_{i}$ for all $i$. In this case we will prove that $r_{\tilde{D}}({p})\leq 2-c+\sum_{i=1}^{k}c_{i}d_{i}/k$. Let $h^{\prime}:\mathbb{P}^{1}\to h(\mathbb{P}^{1})$ denote the normalization map. There is a degree $k$ map $h^{\prime\prime}:\mathbb{P}^{1}\rightarrow\mathbb{P}^{1}$ such that $h$ is the composition of $h^{\prime}\circ h^{\prime\prime}$ and the inclusion $h(\mathbb{P}^{1})\subset{\mathbb{P}^{n_{1}}\times\cdots\times\mathbb{P}^{n_{k}}}$. We have $Z=h^{\prime}(E^{\prime})$, where $E^{\prime}=cu^{\prime}$ and $u^{\prime}=h^{\prime\prime}(u)$. We use $E^{\prime}$ and $h^{\prime}$ instead of $E$ and $h$ and repeat verbatim step (a).

∎

Acknowledgements

We want to thank the anonymous referee and the Handling Editor Jan Draisma for their careful jobs that improved the presentation of this paper. A special thank to the referee for her/his very interesting questions that encouraged us in giving a better version of Theorem 1.8.