Reconstruction of a homogeneous polynomial from its additive decompositions when identifiability fails

Abstract

Let $X\subset {\mathrm{\mathbb{P}}}^r$ be an integral and non-degenerate complex variety. For any $q\in {\mathrm{\mathbb{P}}}^r$ let $r_X(q)$ be its $X$-rank and $\mathcal{S}(X,q)$ the set of all finite subsets of $X$ such that $|S|=r_X(q)$ and $q\in 〈S〉$, where $〈〉$ denotes the linear span. We consider the case $\left|\mathcal{S}(X,q)\right|>1$ (i.e. when $q$ is not $X$ -identifiable) and study the set $W{(X)}_q:={{{\displaystyle \cap }}^{}}_{S\in \mathcal{S}(X,q)}〈S〉$, which we call the non-uniqueness set of $q$. We study the case $\dim X=1$ and the case $X$ a Veronese embedding of ${\mathrm{\mathbb{P}}}^n$. We conclude the paper with a few remarks concerning this problem over the reals.

Citation

Ballico, E. (2021), "Reconstruction of a homogeneous polynomial from its additive decompositions when identifiability fails", Arab Journal of Mathematical Sciences, Vol. 27 No. 1, pp. 41-52. https://doi.org/10.1016/j.ajmsc.2019.09.001

1. Introduction

Let $X\subset {\mathrm{\mathbb{P}}}^r$ be an integral and non-degenerate variety defined over an algebraically closed field $\mathbb{K}$ with characteristic $0$. For any set $A\subset {\mathrm{\mathbb{P}}}^r$ let $〈A〉$ denote its linear span. Fix any $q\in {\mathrm{\mathbb{P}}}^r$. The $X$ -rank $r_X(q)$ of $X$ is the minimal cardinality of a finite set $S\subset X$ such that $q\in 〈S〉$. The notion of $X$-rank includes the notion of tensor rank of a tensor (take $X$ a multi projective space and $X\subset {\mathrm{\mathbb{P}}}^r$ its Segre embedding) and the notion of additive decomposition of a homogeneous polynomial or its symmetric tensor rank (take as $X$ a projective space and as $X\subset {\mathrm{\mathbb{P}}}^r$ one of its Veronese embeddings). See [3,13,18,19] for a long list of applications of these notions.

Notation 1.

Let $\mathcal{S}(X,q)$ denote the set of all $S\subset X$ such that $|S|=r_X(q)$ and $q\in 〈S〉$. Set $W{(X)}_q:={{{\displaystyle \cap }}^{\text{}}}_{S\in \mathcal{S}(X,q)}〈S〉$.

The set $W{(X)}_q$ is the main actor of this paper. We often write $W_q$ if $X$ is clear from the context.

Remark 1.

Note that $W_q$ is a linear subspace of ${\mathrm{\mathbb{P}}}^r$ containing $q$ and that if $W_q=\{q\}$, and $\mathcal{S}(X,q)=\mathcal{S}(X,q^{\prime })$ for some $q^{\prime }\in {\mathrm{\mathbb{P}}}^r$, then $q^{\prime }=q$. We will call $W_q$ the non-uniqueness set of $q$. We have $\dim W_q=r_X(q)-1$ if and only if $〈S〉=〈S^{\prime }〉$ for all $S,S^{\prime }\in \mathcal{S}(X,q)$. In particular $Wq=\{q\}\text{and}q\notin X$ imply $\left|\mathcal{S}\left(X,q\right)\right|>1.$

In this paper we prove one result on the Veronese variety (i.e. on the additive decomposition of homogeneous polynomials) (Theorem 3) and three results for the case $\dim X=1$ (Theorems 1 and 2 and Proposition 1). The proof of the result on the Veronese variety uses one of the results for curves.

We first prove the following two cases (with $X$ a curve) in which $W_q=\{q\}$.

Theorem 1.

Fix an even integer $r\ge 2$ . Let $X\subset {\mathrm{\mathbb{P}}}^r$ be an integral and non-degenerate curve. There is a non-empty Zariski open subset $\mathcal{U}\subset {\mathrm{\mathbb{P}}}^r$ such that $r_X(q)=r/2+1$ for all $q\in \mathcal{U}$ and the following properties hold:

1. (a)

   We have $\{q\}={\cap }_{S\in \mathcal{S}(X,q)}〈S〉$ for all $q\in \mathcal{U}$.

2. (b)

   For all $(q,q^{\prime })\in \mathcal{U}\times {\mathrm{\mathbb{P}}}^r$ if $\mathcal{S}(X,q^{\prime })=\mathcal{S}(X,q)$, then $q^{\prime }=q$.

Theorem 2.

Fix an integer $d\ge 2$ and let $X\subset {\mathrm{\mathbb{P}}}^d$ be the rational normal curve. Take any $q\in {\mathrm{\mathbb{P}}}^d$ such that $\mathcal{S}(X,q)$ is not a singleton. Then $W_q=\{q\}$ . Moreover, if $\mathcal{S}(X,q)=\mathcal{S}(X,q^{\prime })$ for some $q^{\prime }\in {\mathrm{\mathbb{P}}}^d$ , then $q^{\prime }=q$.

Take a non-degenerate $X\subset {\mathrm{\mathbb{P}}}^r$ and $q\in {\mathrm{\mathbb{P}}}^r$. For any integer $t>0$ the $t$-secant variety ${\sigma }_t(X)$ of $X$ is the closure in ${\mathrm{\mathbb{P}}}^r$ of the union of all linear spaces $〈S〉$ with $S\subset X$ and $|S|=t$. The border rank or border $X$-rank $b_X(q)$ of $q\in {\mathrm{\mathbb{P}}}^r$ is the minimal integer $b\ge 1$ such that $q\in {\sigma }_b(X)$. We say that a finite set $A\subset {\mathrm{\mathbb{P}}}^r$ irredundantly spans $q$ if $q\in 〈A〉$ and $q\notin 〈A^{\prime }〉$ for any $A^{\prime }\mathrm{⊊}A$. We use Theorem 2 to prove the following result for the order $d$ Veronese embedding of ${\mathrm{\mathbb{P}}}^n$.

Theorem 3.

Fix integers $n,d,b,k$ , such that $n\ge 2$, $d\ge 8$, $4\le 2b\le d$ and $d+2-b\le k\le 2d-2$ . Let ${\nu }_d:{\mathrm{\mathbb{P}}}^n\to {\mathrm{\mathbb{P}}}^r$, $r=(\begin{array}{c}n+d\\ n\end{array})-1$, be the order$d$ Veronese embedding. Let $L\subset {\mathrm{\mathbb{P}}}^n$ be a line. Set $Y:={\nu }_d(L)$. Fix $q^{\prime }\in 〈Y〉$ such that $b_Y(q^{\prime })=b$ and $r_Y\left(q^{\prime }\right)=d+2-b.$ Fix a general $U\subset {\mathrm{\mathbb{P}}}^n$ such that $|U|=k-d-2+b$ . Let $q\in {\mathrm{\mathbb{P}}}^r$ be any point irredundantly spanned by $\{q^{\prime }\}\cup {\nu }_d(U)$. Then:

1. (1)

   $r_X(q)=k$ and $\mathcal{S}(X,q)\supseteq {\left\{E\cup U\right\}}_{E\in \mathcal{S}(Y,q^{\prime })}$.

2. (2)

   If $k\le 2d-3$, then $\mathcal{S}(X,q)={\left\{E\cup U\right\}}_{E\in \mathcal{S}(Y,q^{\prime })}$ and $W_q=〈U\cup \{q^{\prime }\}〉$.

In Section 4 we consider the following problem. For any positive integer $t$ let $\mathcal{S}(X,q,t)$ be the set of all $S\subset X$ such that $|S|=t$ and $S$ irredundantly spans $q$. We have $\mathcal{S}(X,q,t)=ø$ for all $t<r_X(q)$ and $\mathcal{S}\left(X,q,r_X\right)=\mathcal{S}\left(X,q\right)\ne ø$. By the definition of irredundantly spanning set we have $\mathcal{S}(X,q,t)=\mathrm{ø}$ for all $t\ge r+2$. Since $X$ is integral and non-degenerate, for all $(X,q)$ we have $\mathcal{S}(X,q,r+1)\ne \mathrm{ø}$ and $\mathcal{S}(X,q,r+1)$ contains a general subset of $X$ with cardinality $r+1$. There are easy examples of triples $(X,q,t)$ such that $r>t>r_X(q)$ and $\mathcal{S}(X,q,t)=ø$ (Remark 3). It easy to check that $\mathcal{S}(X,q,t)\ne ø$ for all $t$ such that $r+1-\dim X\le t\le r$ (Lemma 2). Set $W{(X)}_{q,t}:={\cap }_{S\in \mathcal{S}(X,q,t)}〈S〉$, with the convention $W{(X)}_{q,t}:={\mathrm{\mathbb{P}}}^r$ if $\mathcal{S}(X,q,t)\ne ø$. We often write $W_{q,t}$ instead of $W{(X)}_{q,t}$.

In Section 4 we prove the following result.

Proposition 1.

Let $X\subset {\mathrm{\mathbb{P}}}^r$, $r\ge 4$, be an integral and non-degenerate curve. Then there exists a non-empty Zariski open subset $\mathcal{U}$ of ${\mathrm{\mathbb{P}}}^r$ such that $W_{q,t}=\{q\}$ for all $q\in \mathcal{U}$ and all $\lfloor (r+2)/2\rfloor \le t\le r.$

In Section 5 we briefly discuss the case of real algebraic subvarieties of ${\mathrm{\mathbb{P}}}^r(ℝ)$. In particular we show that a statement similar to Theorem 1 over $\mathrm{ℝ}$ is true if we take as $\mathcal{U}$ a non-zero open subset of ${\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$, for the euclidean topology (Theorem 4), but it fails if we ask for a non-empty open subset of ${\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$ for the Zariski topology (Remark 5).

We thanks a referee for useful comments.

2. Preliminary observations

The cactus rank or cactus $X$-rank $c_X(q)$ of $q\in {\mathrm{\mathbb{P}}}^r$ is the minimal degree of a zero-dimensional scheme $Z\subset X$ such that $q\in 〈Z〉$. Let $\mathcal{Z}\left(X,q\right)$ denote the set of all zero-dimensional schemes $Z\subset X$ such that $\mathrm{deg}(Z)=c_X(q)$ and $q\in 〈Z〉$.

Remark 2.

Let $X\subset {\mathrm{\mathbb{P}}}^d$, $d\ge 2$, be a degree $d$ rational normal curve. We use [18, §1.3] and [14] for the following observations. Fix $q\in {\mathrm{\mathbb{P}}}^r$.

(i) We have $b_X(q)=c_X(q)$ ([18, Lemma 1.38]) and $|\mathcal{Z}(X,q)|=1$ ([18, Part (i) of Theorem 1.43]).

(ii) If $c_X(q)<r_X(q)$, then $c_X(q)+r_X(q)=d+2$ and $\mathcal{S}(X,q)$ is infinite. Let $Z$ be the only element of $\mathcal{Z}(X,q)$, $d+2-c_X(q)$ is the minimal degree of a scheme $A\subset X$ such that $q\in 〈A〉$ and $A\mathrm{⊉}Z$.

(iii) If $r_X(q)>c_X(q)$, then $\{q\}=〈Z〉\cap 〈S〉,$ where $\{Z\}=\mathcal{Z}(X,q)$ and $S$ is any element of $\mathcal{S}(X,q)$ (this also follows from the fact that $h^1({\mathrm{\mathbb{P}}}^1,L)=0$ for any line bundle $L$ on ${\mathrm{\mathbb{P}}}^1$ with $\mathrm{deg}(L)\ge -1$, as in the proof of Claim 1).

(iv) then $\text{If}{r}_X(q)>b_X(q)\mathrm{,}\mathrm{then dim}\mathcal{S}(X,q)=d+3-2b$ ([14, eq. (9)]).

(v) If $d$ is odd and $r_X(q)=(d+1)/2$ (i.e. $r_X(q)=b_X(q)$ is the generic rank), then $\mathcal{S}(X,q)=\mathcal{Z}(X,q)$ and $\left|\mathcal{S}(X,q)\right|=1$ ([18, Theorem 1.43]).

(vi) Assume $d$ even and $r_X(q)=d/2+1$ and so $q$ has the generic rank and $b_X(q)=r_X(q)$, but we do not assume that $q$ is general in ${\mathrm{\mathbb{P}}}^d$. Fix $S,S^{\prime }\in \mathcal{S}(X,q)$ such that $S\ne S^{\prime }.$

Claim 1.

$〈S〉\cap 〈S^{\prime }〉=\{q\}$.

Proof of Claim 1. Since $S\ne S^{\mathit{\prime }}$ and $S\in \mathcal{S}(X,q)$, we have $q\notin 〈S\cap S^{\mathit{\prime }}〉$. The Grassmann’s formula gives $h^1({\mathrm{\mathbb{P}}}^1,{\mathcal{I}}_{S\cup S^{\prime }}(d))>0$. Since $h^1({\mathrm{\mathbb{P}}}^1,L)=0$ for any line bundle $L$ on ${\mathrm{\mathbb{P}}}^1$ with $\mathrm{deg}(L)\ge -1$ and $q\notin X$, we have $S\cap S^{\mathit{\prime }}=ø$ and $h^1({\mathrm{\mathbb{P}}}^1,{\mathcal{I}}_{S\cup S^{\prime }}(d))=1$. Thus the Grassmann’s formula implies $\dim \left(〈S〉{{\displaystyle \cap }}^{\text{}}〈S^{\prime }〉\right)=0$, proving Claim 1.

Obviously Claim 1 implies $W_q=\{q\}$ in this case, which by [18, Part (i) of Theorem 1.43] is the only case in which $r_X(q)=b_X(q)$ and $\mathcal{Z}(X,q)$ is not a singleton.

Note that (iii) implies that each $q\in {\mathrm{\mathbb{P}}}^r$ with $c_X(q)\ne r_X(q)$ is uniquely determined by the zero-dimensional scheme evincing its cactus rank and by one single set evincing its rank (any $S\in \mathcal{S}(X,q)$ would do the job). Obviously part (i) implies that most $q\in {\mathrm{\mathbb{P}}}^r$ (the ones with $r_X(q)=b_X(q)$ ) are not uniquely determined by $\mathcal{S}(X,q)$. By parts (i) and (ii) for each $q\in {\mathrm{\mathbb{P}}}^r$ such that $r_X(q)=b_X(q)$ there are exactly ${\infty }^t$, $t:=r_X(q)-1$, points $o\in {\mathrm{\mathbb{P}}}^r$ with $\mathcal{S}(X,o)=\mathcal{S}(X,q)$.

In the proof of Theorem 3 we use the following result ([5, Theorem 1], [4, Theorem 2]); we use the assumption $d\ge 6$ to have $4d-5\ge 3d+1$ and hence to apply a small part of [5, Theorem 1].

Lemma 1

([5, Theorem 1], [4, Theorem 2]). Fix an integer $d\ge 6$ . Let $S\subset {\mathrm{\mathbb{P}}}^n$, $n\ge 2$, be a finite set such that $|S|\le 4d-5$ . We have $h^1({\mathcal{I}}_S(d))>0$ if and only if there is $F\subseteq S$ in one of the following cases:

1. $|F|=d+1$ and $F$ is contained in a line;

2. $|F|=2d+2$ and $F$ is contained in a reduced conic $D$ ; if $D=L_1\cup L_2$ with each $L_i$ a line we have $L_1\cap L_2\notin F$ and $|F\cap L_1|=|F\cap L_2|=d+1$;

3. $|F|=3d$, $F$ is contained in the smooth part of a reduced plane cubic $C$ and $F$ is the complete intersection of $C$ and a degree $d$ hypersurface;

4. $|F|=3d+1$ and $F$ is contained in a plane cubic.

3. Proofs of Theorems 1–3

Proof of Theorem 1.

To prove part (b) it is sufficient to prove part (a), because $\mathcal{S}(X,q^{\prime })=\mathcal{S}(X,q)$ implies $\{q^{\prime }\}\subseteq W{q}^{\prime }=W_q$ and $W_q=\{q\}$ for $q\in \mathcal{U}$.

Since part (a) is trivial in the case $r=2$, we assume $r\ge 4$. Since no non-degenerate curve is defective ([23, Corollary 1.5 and Remark 1.6]), there is a non-empty Zariski open subset $\mathcal{V}\subset {\mathrm{\mathbb{P}}}^r$ such that $r_X(q)=r/2+1$ and $\dim \mathcal{S}(X,q)=1$ for all $q\in \mathcal{V}$.

For each set $S\subset X$ such that $|S|=r/2+1$ and $\dim 〈S〉=r/2$ let ${\ell }_S:{\mathrm{\mathbb{P}}}^r\backslash 〈S〉\to {\mathrm{\mathbb{P}}}^{r/2-1}$ denote the linear projection from $〈S〉$. For a general $S$ we have $〈S〉\cap X=S$ (scheme-theoretically) by Bertini’s theorem and the trisecant lemma ([24, Corollary 2.2]) and ${\ell }_{S|X\backslash S}$ is birational onto its image, again by the trisecant lemma and the assumption $r\ge 4$.

Fix a general $S\subset X$ such that $|S|=r/2+1$. Let $X_S\subset {\mathrm{\mathbb{P}}}^{r/2-1}$ be the closure of ${\ell }_S(X\backslash S)$ in ${\mathrm{\mathbb{P}}}^{r/2-1}$. There is a finite set $E\subset X_S$ containing $X_S\backslash {\ell }_S(X\backslash S)$ and such that for each $p\in X_S\backslash E$ there is a unique $o\in X\backslash S$ such that ${\ell }_S(o)=p$. For any set $A\subset X_S\backslash E$ let $A_S\subset X\backslash S$ denote the only set such that ${\ell }_S(A_S)=A$. Any general $A\subset X_S\backslash E$ such that $|A|=r/2+1$ is linearly dependent, but each proper subset of $A$ is linearly independent. Thus $〈S〉\cap 〈A_S〉$ is a single point, $q_{S,A}$, and $q_{S,A}\notin 〈B〉$ for any $B\mathrm{⊊}{A}_S$. For a general $A$ we get as $A_S$ a general subset of $X$ with cardinality $r/2+1$. Thus for a general $A$ we have $q_{S,A}\notin S^{\prime }$ for any $S^{\prime }\mathrm{⊊}S$. We start with $S\in \mathcal{S}(X,o)$ for a general $o\in {\mathrm{\mathbb{P}}}^r$. Thus $r_X(q)=r/2+1$ for a general $q\in 〈S〉$. Thus for a general $A$ we get $S\in \mathcal{S}(q_{S,A})$ and $A_S\in \mathcal{S}(q_{S,A})$. By construction we have $\left\{q_{S,A}\right\}=〈S〉\cap 〈A_S〉$. For a general $A$ the point $q_{S,A}$ is general in $〈S〉$. By the generality of $S$ we get that the points $q_{S,A}$ ’s (with $(S,A)$ varying, but general), cover a non-empty Zariski open subset of ${\mathrm{\mathbb{P}}}^r$.□

Proof of Theorem 2.

Set $b:=b_X(q)$. Since $\mathcal{S}(X,q)$ is not a singleton, we have $q\notin X$ and hence $b\ge 2$. Part (vi) of Remark 2 covers the case $r_X(q)=b$ and hence we may assume $r_X(q)>b$. Thus $r_X(q)=d+2-b$. By part (v) of Remark 2 we have $\mathcal{S}\left(X,q\right)\ge 2$. We will prove the stronger assumption that $\{q\}={\cap }_{A\in \Gamma }〈A〉$, where $\mathit{\Gamma }$ is any irreducible family contained in $\mathcal{S}(X,q)$ and with $\dim \mathit{\Gamma }=d+3-2b$; we do not assume that $\mathit{\Gamma }$ is closed in $\mathcal{S}(X,q)$.

(a) First assume $b=2$. We use the proof of [20, Proposition 5.1]. Fix $a\in {\mathrm{\mathbb{P}}}^d\backslash \{q\}$. Let $H\subset {\mathrm{\mathbb{P}}}^d$ be a general hyperplane containing $q$. Since $q\notin X$, Bertini’s and Bezout’s theorems give that $X\cap H$ is formed by $d$ distinct points. Since $X$ is connected, the exact sequence

$$0\to {\mathcal{I}}_X\to {\mathcal{I}}_X(1)\to {\mathcal{I}}_{X{{\displaystyle \cap }}^{\text{}}H,H}(1)\to 0$$

gives that $X\cap H$ spans $H$. Thus $q\in 〈X\cap H〉$. Since $r_X(q)=d$, we get $X\cap H\in \mathcal{S}(X,q)$. The generality of $H$ gives $a\notin H$, concluding the proof that $W_q=\{q\}$.

(b) Step (a) and part (i) (resp. part (vi)) of Remark 2 for the case $d$ odd (resp. $d$ even) and $q$ with generic rank cover all cases with $d\le 4$. Thus we may assume $d\ge 5$ and use induction on $d$. Fix a general $o\in X$. Let ${\ell }_o:{\mathrm{\mathbb{P}}}^d\backslash \{o\}\to {\mathrm{\mathbb{P}}}^{d-1}$ denote the linear projection from $o$. Let $Y\subset {\mathrm{\mathbb{P}}}^{d-1}$ denote the closure of ${\ell }_o(X\backslash \{o\})$ in ${\mathrm{\mathbb{P}}}^{d-1}$. $Y$ is a rational normal curve of ${\mathrm{\mathbb{P}}}^{d-1}$. Set $q^{\prime }:={\ell }_o(q)$ and $Z^{\prime }:={\ell }_o(Z)$ (by the generality of $o$ we have $o\notin 〈Z〉$ and hence $Z^{\prime }$ is well-defined, $\mathrm{deg}(Z^{\prime })=b$ and $\dim 〈Z^{\prime }〉=b-1$ ). The generality of $o$ also implies that $q\notin 〈Z^{″}\cup \{o\}〉$ for any $Z^{″}\mathrm{⊊}Z$ (here we use that $X$ is a smooth curve and hence $Z$ has only finitely many subschemes). Thus $q^{\prime }\in 〈Z^{\prime }〉$ and $q^{\prime }\notin 〈Z^{″}〉$ for all $Z^{″}\mathrm{⊊}{Z}^{\prime }$. Since $Y$ is a degree $d-1$ rational normal curve and $b\le d/2$, parts (i) and (ii) of Remark 2 imply $b_Y(q^{\prime })=b$ and $\mathcal{Z}(Y,q^{\prime })=\{Z^{\prime }\}$. Fix an irreducible family $\mathit{\Gamma }\subseteq \mathcal{S}(X,q)$ such that $\dim \mathit{\Gamma }=d+3-2b$ (it exists by part (v) of Remark 2). Let $\mathcal{B}$ denote the set of all $A\in \mathit{\Gamma }$ such that $o\in A$. By part (v) of Remark 2 and the generality of $o$ we have $\mathcal{B}\ne 0$ and $\dim \mathcal{B}=d+2-2b$. Set $\mathcal{A}:={\left\{{\ell }_o(B\backslash \{o\})\right\}}_{B\in \mathcal{B}}$. Since $Y$ is a rational normal curve, parts (i) and (ii) of Remark 2 imply $\mathcal{A}\subseteq \mathcal{S}(Y,q^{\prime })$. We have $\dim \mathcal{A}=(d-1)+3-2b$. The inductive assumption gives $\{q^{\prime }\}={\cap }_{A\in \mathcal{A}}〈A〉$. Thus $〈\left\{o,q\right\}〉={\cap }_{B\in \mathcal{B}}〈B〉$. Since $\dim \mathit{\Gamma }=\dim \mathcal{S}(X,q)=d+3-2b$ (part (v) of Remark 2) and $o$ is general in $X$, there is $S\in \mathit{\Gamma }$ such that $o\notin S$. Thus ${\cap }_{A\in \mathit{\Gamma }}〈A〉=\{q\}$.□

Proof of Theorem 3.

By Autarky ([19, Exercise 3.2.2.2]) we may assume $U\ne ø$. Since $U$ is general in ${\mathrm{\mathbb{P}}}^n$, we have $\dim 〈{\nu }_d(U)\cup Y〉=\min \left\{r,\dim 〈Y〉+|U|\right\}$. Since $\dim 〈Y〉=d$ and $d+|U|<r$, we have $〈{\nu }_d(U)〉{{\displaystyle \cap }}^{\text{}}〈Y〉=ø$. By Theorem 2 we have $W{(Y)}_{q^{\prime }}=\{q^{\prime }\}$. Take $E\in \mathcal{S}(Y,q^{\prime })$ and set $A:=U\cup E$. The set $\{q^{\prime }\}\cup U$ irredundantly spans $q$ and $〈Y〉{{\displaystyle \cap }}^{\text{}}〈{\nu }_d(U)〉=ø$. Hence we have $E\cap U=ø$ and $|A|=k$. Since $|A|=k$ and $q\in 〈{\nu }_d(A)〉$, we have $r_X(q)\le k$. Since $U$ is general in ${\mathrm{\mathbb{P}}}^n$, we have $h^0({\mathcal{I}}_A(t))=\max \left\{0,h^0({\mathcal{I}}_E(t))-|U|\right\}$ for all $t\in \mathrm{\mathbb{N}}$ ; to use this equality we need to fix one element, $E$, of $\mathcal{S}\left(Y,q^{\prime }\right)$, before choosing a general $U$.

Note that we have $W_q=〈{\nu }_d(U){\displaystyle \cup }\{q^{\prime }\}〉$ for any $q$ such that $\mathcal{S}(X,q)={\left\{E\cup U\right\}}_{E\in \mathcal{S}(Y,q^{\prime })}$ by Theorem 2. Assume either $r_X(q)<k$ or $k\le 2d-3$ and the existence of $B\in \mathcal{S}(X,q)\backslash {\left\{E\cup U\right\}}_{E\in \mathcal{S}(Y,q^{\prime })}$. In the former case take $B\in \mathcal{S}(X,q)$. Set $S:=A\cup B$. In both cases we have $|B|\le |A|$ and $|A|+|B|\le 4d-5$. Since $h^1({\mathcal{I}}_S(d))>0$ ([6, Lemma 1]) there is $F\subseteq S$ in one of the cases listed in Lemma 1.

(a) Assume the existence of a plane cubic $T\subset {\mathrm{\mathbb{P}}}^n$ such that $|T\cap S|\ge 3d$.

(a1) Assume $n=2$. Thus $T$ is an effective divisor of ${\mathrm{\mathbb{P}}}^2$. Consider the residual exact sequence of $T$ in ${\mathrm{\mathbb{P}}}^2$:

(1)$$0\to {\mathcal{I}}_{S\backslash S\cap T}(d-3)\to {\mathcal{I}}_S(d)\to {\mathcal{I}}_{S\cap T,T}(d)\to 0$$

Since $|S\backslash S\cap T|\le 4d-5-3d=d-5$, we have $h^1({\mathcal{I}}_{S\backslash S\cap T}(d-3))=0$. Thus either [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] give $A\backslash A\cap T=B\backslash B\cap T$. Assume for the moment $L\mathrm{⊈}T$. Bezout gives $\left|L\cap T\right|\le 3$. Since $U$ is general and $h^0\left({\mathcal{O}}_{{\mathrm{\mathbb{P}}}^2}(3)\right)=10$, we have $|U\cap T|\le 9$. Thus $|B\cap T|\ge 3d-12>12\ge |T\cap A|$ and hence $|B|>|A|$, a contradiction. Now assume $L\subset T$. Since $h^0({\mathcal{O}}_{{\mathrm{\mathbb{P}}}^2}(2))=6$ and $U\cap L=ø$, we get $\left|A\cap T\right|\le d+8-b$. Thus $\left|B\cap T\right|\ge 2d-5+b$ and again $|B|>|A|$, a contradiction.

(a2) Assume $n>2$. Let $M\subset {\mathrm{\mathbb{P}}}^n$ be a general hyperplane containing the plane $〈T〉$ (so $M=〈T〉$ if $n=3$ ). Since $S$ is a finite set and $M$ is a general hyperplane containing $〈T〉$, we have $S\cap M=S\cap 〈T〉$. Consider the residual exact sequence of $M$ in ${\mathrm{\mathbb{P}}}^n$:

(2)$$0\to {\mathcal{I}}_{S\backslash S\cap M}(d-1)\to {\mathcal{I}}_S(d)\to {\mathcal{I}}_{S\cap M,M}(d)\to 0$$

Since $|S\backslash A\cap M|\le 4d-5-3d=d-5$, we have $h^1({\mathcal{I}}_{S\backslash S\cap M}(d-1))=0$. Thus either [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] give $A\backslash A\cap M=B\backslash B\cap M$. Since no $4$ points of $U$ are coplanar, we have $\left|A\cap M\right|\le d+5-b<3d-d-2+b$. Thus $|B|>|A|$, a contradiction.

(b) Assume the existence of a plane conic $D$ such that $\left|S\cap D\right|\ge 2d+2$.

(b1) Assume $n=2$. Consider the residual exact sequence of $D$ in ${\mathrm{\mathbb{P}}}^2$:

(3)$$0\to {\mathcal{I}}_{S\backslash S\cap D}(d-2)\to {\mathcal{I}}_S(d)\to {\mathcal{I}}_{S\cap D,D}(d)\to 0$$

First assume $h^1({\mathcal{I}}_{S\backslash S\cap D}(d-2))>0$. Since $\left|S\backslash S\cap D\right|\le 4d-5-2d-2=2(d-3)-1$, there is a line $R\subset {\mathrm{\mathbb{P}}}^2$ such that $\left|R\cap (S\backslash S\cap D)\right|\ge d-1$ ([9, Lemma 34]). Thus $\left|S\cap (D\cup R)\right|\ge 3d+1$. Step (a1) gives a contradiction. Now assume $h^1({\mathcal{I}}_{S\backslash S\cap D}(d-2))=0$. Either [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] give $A\backslash A\cap D=B\backslash B\cap D$. Assume for the moment $L\mathrm{⊈}D$. Thus $\left|L\cap D\right|\le 2$. Since $U$ is general and $h^0({\mathcal{O}}_{{\mathrm{\mathbb{P}}}^2}(2))=6$, we have $\left|U\cap D\right|\le 5$. Thus $\left|B\cap D\right|>\left|A\cap D\right|$ and so $|B|>|A|$, a contradiction. Now assume $L\subset D$. Write $D=L\cup R$ with $R$ a line. Set $\{o\}:=L\cap R$. Since $U\cap L=ø$, $\left|L\cap R\right|=1$ and ${\left|U\cap R\right|}^{\prime }\le 2$ for each line $R^{\prime }$, we have $\left|A\cap D\right|\le d+4-b$. Let $Z^{\prime }\subset L$ be the only degree $b$ zero-dimensional scheme evincing the cactus rank of $q^{\prime }$ with respect to the rational normal curve ${\nu }_d(L)$ (part (i) of Remark 2). Set $Z^{″}:=Z^{\prime }\cup U$ and $Z:=Z^{″}\cup B$. Since $A\backslash A\cap D=B\backslash B\cap D$, we have $Z=Z^{\prime }\cup (U\cap R)\cup (B\cap D)\cup (U\backslash U\cap D)$. Since $q^{\prime }\in 〈{\nu }_d(Z^{\prime })〉$, we have $q\in 〈{\nu }_d\left(Z^{″}\right)〉\cap 〈{\nu }_d(B)〉$. Thus $h^1({\mathcal{I}}_Z(d))>0$. Since $h^1({\mathcal{I}}_{U\backslash S\cap D}(d-2))=h^1({\mathcal{I}}_{S\backslash S\cap D}(d-2))=0$, the residual sequence (3) of $D$ in ${\mathrm{\mathbb{P}}}^2$ with $Z$ instead of $S$ gives $h^1(D,{\mathcal{I}}_{Z\cap D,D}(d))>0$. Since $D=R\cup L$, using either [16, Corollaire 2] or the residual exact sequences of $R$ and $L$ in ${\mathrm{\mathbb{P}}}^2$ we get that we are in one of the following cases:

1. $\mathrm{deg}(Z\cap L)\ge d+2$;

2. $\mathrm{deg}(Z\cap R)\ge d+2$;

3. $\mathrm{deg}(Z\cap R)=\mathrm{deg}(Z\cap L)=d+1$ and $o\notin Z_{\text{red}}$.

Recall that $A\backslash A\cap B=B\backslash B\cap D$ (and hence) $\left|B\cap D\right|\le \left|A\cap D\right|$ and that $|A\cap D|\le d+4-b$.

(b1.1) Assume $\mathrm{deg}(Z\cap L)\ge d+2$. Since $Z\cap L=Z^{\prime }\cup (B\cap L)$, we get $\left|B\cap L\right|\ge d+b-2$. Consider the residual exact sequence of $L$ in ${\mathrm{\mathbb{P}}}^2$:

(4)$$0\to {\mathcal{I}}_{S\backslash S\cap L}(d-1)\to I_S(d)\to {\mathcal{I}}_{S\cap L,L}(d)\to 0$$

First assume $h^1({\mathcal{I}}_{S\backslash S\cap L}(d-1))>0$. Since $h^1({\mathcal{I}}_{S\backslash S\cap (R\cup L)}(d-2))=0$, the residual exact sequence of $R$ gives $h^1(R,{\mathcal{I}}_{S\cap R\backslash S\cap R\cap L}(d-1))>0$. Thus $\left|S\cap R\backslash S\cap \{o\}\right|\ge d+1$. Since $\left|U\cap R\right|\le 2$, we get $\left|B\cap R\backslash B\cap \{o\}\right|\ge d-1$ and hence $|B|>|A|$ (because $d\ge 5$), a contradiction.

Now assume $h^1({\mathcal{I}}_{S\backslash S\cap L}(d-1))=0$. By [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] we get $A\backslash A\cap L=B\backslash B\cap L$. Since $\left|B\cap L\right|\ge d+2-b=\left|A\cap L\right|$, we get $|B\cap L|=d+2-b$. In this case part (1) of Theorem 3 is proved. To prove part (2) we need to prove that $B\cap L\in \mathcal{S}(Y,q^{\prime })$. Since $\left|B\cap L\right|=d+2-b$ and $\mathrm{deg}(Z\cap L)\ge d+2$, we get $Z_1\cap B=ø$ and $\mathrm{deg}(Z\cap B)=d+2$. Thus $〈{\nu }_d(B\cap L)〉\cap 〈{\nu }_d(Z^{\prime })〉$ is a single point, $q^{″}$, and $B\cap L\in \mathcal{S}(Y,q^{″})$. Since $U$ is general, and $|U|\le \left(\begin{array}{c}d+2\\ 2\end{array}\right)-d$, we have $〈{\nu }_d(U)〉\cap 〈{\nu }_d(L)〉=ø$. Since $B\backslash B\cap L=A\backslash A\cap L=U$, we get $q^{″}=q^{\prime }$, proving part (2) in this case.

(b1.2) Assume $\mathrm{deg}(Z\cap R)=\mathrm{deg}(Z\cap L)=d+1$ and $o\notin Z_{\text{red}}$ (i.e. $o\notin Z_{\text{red}}^{\prime }$). Since $\mathrm{deg}(Z^{\prime })=b$ and $\mathrm{deg}(Z^{″}\cap D)=b+\left|U\cap R\right|\le b+2$, we get $\left|L\cap B\right|\ge d+1-b$, $\left|R\cap B\right|\ge d-1$ and $o\notin B$. Thus $\left|B\cap D\right|\ge 2d+2-b$. Since $\left|B\cap D\right|\le \left|A\cap D\right|\le d+4-b$, we obtain a contradiction.

(b1.3) Assume $\mathrm{deg}(Z\cap R)\ge d+2$. Since $\left|U\cap R\right|\le 2$ with strict inequality if $o\in Z_{\text{red}}^{\prime }$ and every point of $〈{\nu }_d(R)〉$ has rank $\le d$ by Sylvester’s theorem, we get $\left|U\cap R\right|+\mathrm{deg}(Z^{\prime }\cap R)=2$, $\left|B\cap R\right|=d$ and $U\cap B\cap R=ø$. If $h^1({\mathcal{I}}_{S\backslash S\cap R}(d-1))=0$, we have $A\backslash A\cap R=B\backslash B\cap R$ by [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] and hence $|B|>|A|$, a contradiction. Now assume $h^1({\mathcal{I}}_{S\backslash S\cap R}(d-1))>0$. Since $h^1({\mathcal{I}}_{S\backslash S\cap D}(d-2))=0$ in this part of the proof, the residual exact sequence of $L$ gives $h^1(L,{\mathcal{I}}_{L\cap S\backslash L\cap S\cap R}(d-1))>0$ and hence $\left|S\cap (L\backslash L\cap R)\right|\ge d+1$. Thus $\left|B\cap (L\backslash L\cap R)\right|\ge b-1$. We get $\left|B\cap D\right|\ge d+b-1$. Since $A\backslash A\cap D=B\backslash B\cap D=U\backslash U\cap D$, we get $d+b-1\le d+4-b$ and hence $b=2$ and $\left|B\cap L\right|\le 2$. Since $\left|U\cap R\right|+\mathrm{deg}(Z^{\prime }\cap R)\ge 2$ and $q^{\prime }$ uniquely determines $Z^{\prime }$, $R$ is uniquely determined by $Z^{\prime }$ and the set $\left|R\cap U\right|$. If $o\in Z_{\text{reg}}^{\prime }$$R$ is uniquely determined by $q^{\prime }$ and one point of $R\backslash \{o\}$. Since we took a general $U$ after fixing $q^{\prime }$, we have $\left|U\cap R\right|\le 1$ if $R\cap L\in Z_{\text{reg}}^{\prime }$. Hence (varying the points of $U\backslash U\cap R$ (if $U\mathrm{⊈}R$ ) we may (after fixing $q^{\prime }$ ) assume that $U\backslash U\cap R$ is general in ${\mathrm{\mathbb{P}}}^n\backslash D$. Since $A\backslash A\cap D=B\backslash B\cap D=U\backslash U\cap D$, $\left|U\backslash U\cap D\right|\le \left(\begin{array}{c}d+2\\ 2\end{array}\right)-2d-1$ and $U\backslash U\cap D$ is general, we have $〈{\nu }_d(U\backslash U\cap D)〉\cap 〈{\nu }_d(D)〉$. Thus there is a unique $q^{″}\in 〈{\nu }_d(D)〉$ such that $q\in 〈\left\{{\nu }_d(U\backslash U\cap R),q^{″}\right\}〉$. We have $q^{″}\in 〈{\nu }_d(A\cap D)〉\cap 〈{\nu }_d(B\cap D)〉\cap 〈\left\{q^{\prime },{\nu }_d(U\cap R)\right\}〉$. Thus is it sufficient to prove parts (1) and (2) of the theorem for $q^{″}$, $A\cap D$ and $B\cap D$ instead of $q$, $A$ and $B$, i.e. in the rest of this step we assume $U=U\cap R$. Since $|B|\le |A|$, we have $\left|B\cap L\right|\le 2$. Thus $h^1({\mathcal{I}}_{Z^{\prime }\cup (L\cap B)}(d-1))=0$. if $o\notin Z_{\text{red}}^{\prime }\cap B$, [7, Lemma 5.1] gives a contradiction, because $Z^{\prime }\mathrm{⊈}B$. Now assume $o\in Z_{\text{red}}^{\prime }\cap B$. Since $o\in Z_{\text{red}}^{\prime }$ and $Z^{\prime }$ is uniquely determined by $q^{\prime }$, we observed that $\left|U\cap R\right|\le 1$. Thus (under the assumption $U\subset D$), we have $|U|\le 1$. Since $\left|B\cap R\right|=d,$$o=L\cap R\in B$ by assumption and $Z^{\prime }\mathrm{⊈}R$, we get $\mathrm{deg}(Z\cap R)\le d+1$, a contradiction.

(b2) Assume $n>2$. Let $M\subset {\mathrm{\mathbb{P}}}^n$ be a general hyperplane containing the plane $〈D〉$. Thus $S\cap M=S\cap 〈D〉$. Since $U$ is general, no $4$ points of $U$ are coplanar. Thus $\left|U\cap M\right|=\left|U\cap 〈D〉\right|\le 3$.

(b2.1) Assume $h^1(I_{S\backslash S\cap M}(d-1))>0$. Since $\left|S\backslash S\cap M\right|\le |A|+|B|-2d-2\le 2(d-1)+1$, there is a line $R^{\prime }\subset {\mathrm{\mathbb{P}}}^n$ such that $\left|R^{\prime }\cap (S\backslash S\cap M)\right|\ge d+1$. If $R^{\prime }\subset 〈D〉$, then $R^{\prime }\cup D$ is a plane cubic and we may apply step (a1). Thus we may assume $R^{\prime }\mathrm{⊈}〈D〉$. Let $N\subset {\mathrm{\mathbb{P}}}^n$ be a general hyperplane containing $N$. Since $S$ is a finite set, the generality of $M$ and $N$ gives $S\cap (M\cup N)=S\cap (D\cup R^{\prime })$. Consider the residual exact sequence

(5)$$0\to {\mathcal{I}}_{S\backslash S\cap (M\cup N)}(d-2)\to {\mathcal{I}}_S(d)\to {\mathcal{I}}_{S\cap (M\cup N),M\cup N}(d)\to 0$$

of $M\cup N$ in ${\mathrm{\mathbb{P}}}^n$. Since $\left|S\backslash S\cap (M\cup N)\right|\le |A|+|B|-3d-3\le d-1$, we have $h^1(I_{S\backslash S\cap (M{{\displaystyle \cup }}^{\text{}}N)}(d-2))=0$. Thus either [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] give $A\backslash A\cap (M\cup N)=B\backslash B\cap (M\cup N)$. We have $A\cap (M\cup N)\subseteq E\cup (U\cap (M\cup N))$ and hence $\left|A\backslash A\cap (M\cup N)\right|\ge k-d-2+b-5$. Since $\left|A\cap (M\cup N)\right|\le d+7-b$, we get $\left|B\cap (M\cup N)\right|\ge 3d+3-d-7+b=2d-4+b$. Since $\left|B\backslash B\cap (M\cup N)\right|\ge k-d-2+b-5$, we get a contradiction.

(b2.2) Assume $h^1({\mathcal{I}}_{S\backslash S\cap M}(d-1))=0$. Either [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] give $A\backslash A\cap M=B\backslash B\cap M$. Since $|U\cap M|=|U\cap 〈D〉|\le 3$, we have $\left|U\backslash U\cap M\right|\ge k-d-5+b$. Assume for the moment $L\mathrm{⊈}〈D〉$. We get $\left|E\cap M\right|\le 1$ and hence $\left|A\backslash A\cap M\right|\ge k-4$. Since $A\backslash A\cap M=B\backslash B\cap M$, we get $\left|S\cap M\right|\le |A|+|B|-2k-8$ and hence $2d+2\le 8$, a contradiction. Now assume $L\subset 〈D〉$. If $L\mathrm{⊈}D$ we get (since $\left|L\cap D\right|\le 2$ ) $\left|S\cap M\right|\ge 3d+b$. Since $A\backslash A\cap M=B\backslash B\cap M$ and $\left|U\backslash U\cap M\right|\ge k-d-1+b$, we get $|S|\ge 2d+2b+k-1$, a contradiction.

(c) Assume the existence of a line $R\subset {\mathrm{\mathbb{P}}}^n$ such that $\left|R\cap S\right|\ge d+2$. Let $M\subset {\mathrm{\mathbb{P}}}^n$ be a general hyperplane containing $R$ (so $M=R$ if $n=2$ ). Since $S$ is a finite set and $M$ is a general hyperplane containing $R$, we have $M\cap S=R\cap S$. Since $U$ is a general subset of ${\mathrm{\mathbb{P}}}^n$ with cardinality $k-d-2+b$, no $3$ of its points are collinear (and hence $\left|U\cap R\right|\le 2$) and $U\cap L=ø$. Let $M\subset {\mathrm{\mathbb{P}}}^n$ be a general hyperplane containing $R$ (so $M=R$ if $n=2$ ). Since $S$ is a finite set and $M$ is a general hyperplane containing $R$, we have $M\cap S=R\cap S$.

(c1) Assume $h^1({\mathcal{I}}_{S\backslash S\cap M}(d-1))>0$. Since $\left|S\backslash S\cap M\right|\le |A|+|B|-d-2\le 3(d-1)-1$, either there is a line $R_1$ such that $\left|R_1\cap (S\backslash S\cap M)\right|\ge d+1$ or there is a conic $D_1$ such that $\left|D_1\cap (S\backslash S\cap M)\right|\ge 2d$. If $R$ and $R_1$ (resp. $R$ and $D_1$) are contained in a plane, and in particular if $n=2$, step (b) (resp. step (a)) gives a contradiction, because $\left|S\cap (R\cup R_1)\right|\ge 2d+3$ (resp. $\left|S\cap (R\cup D_1)\right|\ge 3d+2$). Thus we may assume that this is not the case and in particular we may assume $n>2$. Let $N$ be a general hyperplane containing $R_1$ (resp. $D_1$). We use the residual exact sequence (5). Note that $S\cap (M\cup N)=S\cap (R\cup R_1)$ (resp. $S\cap (M\cup N)=S\cap (R\cup 〈D_1〉)$.

(c1.1) Assume $h^1({\mathcal{I}}_{S\backslash S\cap (M\cup N)}(d-2))>0$. We exclude the existence of $D_1$, because $\left|S\cap (R\cup D_1)\right|\ge 3d+2$ and hence $\left|S\backslash S\cap \text{}(M\cup N)\right|\le d-1$. Thus in this case we may assume the existence of $R_1$. Since $\left|S\cap (R\cup R_1)\right|\ge 2d+3$, we have $\left|S\backslash S\cap (M\cup N)\right|\le |A|+|B|-2d-3\le 2(d-2)+1$. By [9, Lemma 34] there is a line $R_2$ such that $\left|R_2\cap S\backslash S\cap (M\cup N)\right|\ge d$. Let $M^{\prime }$ be a general hyperplane containing $R_2$. Consider the residual exact sequence of $M^{\prime }\cup M\cup N$. We have $h^1({\mathcal{I}}_{S\backslash S\cap (M\cup N\cup M^{\prime })}(d-3))=0$, because $\left|S\backslash S\cap (M\cup N\cup M^{\prime })\right|\le 2k-d-2-d-1-d\le d-4$. Either [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] give $A\backslash A\cap (M\cup N\cup M^{\prime })=B\backslash B\cap (M\cup N\cup M^{\prime })$. Since $M$, $N$ and $M^{\prime }$ are general, we have $S\cap (M\cup N\cup M^{\prime })=S\cap (R\cup R_1\cup R_2)$. Since $U$ is general, no 3 of the points of $U$ are collinear. Thus $\left|U\cap (R\cup R_1\cup R_2)\right|\le 6$. Hence $\left|A\backslash A\cap (M\cup N\cup M^{\prime })\right|\ge k-d-8+b$. Since $A\backslash A\cap (M\cup N\cup M^{\prime })=B\backslash B\cap (M\cup N\cup M^{\prime })$, we get $\left|S\cap (M\cup N\cup M^{\prime })\right|\le 2k-2k+2d+16-2b$. Hence $2d+16-2b\ge 3d+3$, a contradiction.

(c1.2) Assume $h^1({\mathcal{I}}_{S\backslash S{{\displaystyle \cap }}^{\text{}}(M{{\displaystyle \cup }}^{\text{}}N)}(d-2))=0$. Either [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] give $A\backslash A\cap (M\cup N)=B\backslash B\cap (M\cup N)$. Since $U\cap (M\cup N)=U\cap (R\cup R_1)$, we have $\left|U\backslash U\cap (M\cup N)\right|\ge k-d-6+b$. Assume for the moment $L\notin \left\{R,R_1\right\}$. We get $\left|L\cap (M\cup N)\right|\le 2$. Thus $\left|A\backslash A\cap (M\cup N)\right|\ge k+b-8$. Since $A\backslash A\cap (M\cup N)=B\backslash B\cap (M\cup N)$, we get $\left|S\cap (M\cup N)\right|\le 16-b<2d+3$ (even when instead of $|S|$ we take $2k$ ). Thus we may assume that either $L=R$ or $L=R^{\prime }$. In both cases, writing $D:=R\cup R^{\prime }$ we are in the case solved in step (b1).

(c2) Assume $h^1({\mathcal{I}}_{S\backslash S{{\displaystyle \cap }}^{\text{}}M}(d-1))=0$. Either [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] give $A\backslash A\cap M=B\backslash B\cap M$.

(c2.1) Assume $R=L$. We get $U=A\backslash A\cap L=B\backslash B\cap L$. Thus $B=U\cup (B\cap L)$. Since $〈{\nu }_d(U)〉\cap 〈Y〉=ø$, $q\in 〈v_d(U)\cup Y〉$, $q\notin 〈{\nu }_d(U)〉,$$q\notin 〈{\nu }_d(Y)〉$ (because $U\ne ø$) and $〈{\nu }_d(U)〉\cap 〈Y〉=ø$, there are uniquely determined $q_1\in 〈{\nu }_d(U)〉$ and $q_2\in 〈Y〉$ such that $q\in 〈\left\{q_1,q_2\right\}〉$. The uniqueness of $q_2$ gives $q_2=q^{\prime }$. Since $〈{\nu }_d(U)〉\cap 〈Y〉=ø$ and $q\in 〈{\nu }_d(A)〉{{\displaystyle \cap }}^{\text{}}〈{\nu }_d(B)〉,$ we get $q^{\prime }\in 〈{\nu }_d(B\cap L)〉$. Thus $\left|B\cap L\right|\ge r_Y(q^{\prime })=\left|A\cap L\right|$. Since $|B|\le |A|$ and $A\backslash A\cap L=B\backslash B\cap L$, we get $|B|=|A|$ and $B=U\cup F$ with $F\cap U=ø$ and $F\in \mathcal{S}(Y,q^{\prime })$. Thus the theorem is true in this case.

(c2.2) Assume $R\ne L$. Since $\left|L\cap R\right|\le 1$, we get $\left|E\cap R\right|\le 1$. Since $\left|U\cap R\right|\le 2$, we get $\left|A\cap R\right|\le 3$ and hence $\left|B\cap R\right|\ge d-1>\left|A\cap R\right|$. Since $A\backslash A\cap R=B\backslash B\cap R$, we get $|B|>|A|$, a contradiction.□

4. Irredundantly spanning sets

Lemma 2.

If $r+1-\dim X\le t\le r$, then $\mathcal{S}(X,q,t)\ne ø$ .

Proof. The case $t=r+1-\dim X$ is an obvious consequence of the proof of [20, Proposition 5.1]. Assume $r+2-\dim X\le t\le r$. Let $Y\subset {\mathrm{\mathbb{P}}}^r$ be the intersection of $X$ and $(t+\dim X-r-1)$ general quadric hypersurfaces. By Bertini’s theorem $Y$ is an integral and non-degenerate subvariety of ${\mathrm{\mathbb{P}}}^r$. Thus for any $q$ we have $\mathcal{S}\left(X,q,t\right)\supseteq \mathcal{S}\left(Y,q,t\right)$. Since $t=r+1-\dim Y$, we get $\mathcal{S}\left(Y,q,t\right)\ne ø$. □

Remark 3.

Let $X\subset {\mathrm{\mathbb{P}}}^d$, $d\ge 4$, be a rational normal curve. Fix $q\in {\mathrm{\mathbb{P}}}^d$ such that $r_X(q)=2$. Since any subset of $X$ with cardinality at most $d+1$ is linearly independent, the definition of irredundantly spanning set gives $\mathcal{S}(X,q,t)=ø$ for all $t$ such that $3\le t\le d-1$.

Proof of Proposition 1.

Since a finite intersection of non-empty Zariski open subsets of ${\mathrm{\mathbb{P}}}^r$ is open and non-empty and the interval $\lfloor (r+2)/2\rfloor \le t\le r$ contains only finitely many integers, it is sufficient to prove the statement for a fixed $t$. The case $t=r$ is true by Remark 3. The case $r$ even and $t=r/2+1$ is true by Theorem 2. Thus when $r$ is even we may assume $r/2+2\le t\le r$. Since we saw that the case $r=t$ is always true, we proved the proposition for $r=4$. Thus we may assume $r\ge 5$ and that the proposition is true for all curves in a lower dimensional projective space. Fix a general $p\in X$ and call $\ell :{\mathrm{\mathbb{P}}}^r\backslash \{p\}\to {\mathrm{\mathbb{P}}}^{r-1}$ the linear projection from $p$. Let $Y\subset {\mathrm{\mathbb{P}}}^{r-1}$ be the closure of $\ell \left(X\backslash \{p\}\right)$ in ${\mathrm{\mathbb{P}}}^{r-1}$. $Y$ is an integral and non-degenerate curve. Since $p$ is general in $X$, it is a smooth point of $X$ and hence ${\ell }_{|X\backslash \{p\}}$ extends to a surjective morphism $\mu :X\to Y$ with $\mu (p)$ associated to the tangent line of $X$ at $p$. Thus $Y=\mu (X)$. By the trisecant lemma ([24, Corollary 2.2]) and the generality of $p$ we have $\mathrm{deg}(L\cap X)\le 2$ for every line $L\subset {\mathrm{\mathbb{P}}}^r$ such that $p\in L$. Hence ${\ell }_{|X\backslash \{p\}}$ is birational onto its image and there is a finite set $F\subset X$ containing $p$ such that ${\mu }_{|X\backslash F}$ induces an isomorphism between $X\backslash F$ and $Y\backslash \mu (F)$. Fix the integer $t$ such that $\lfloor (r+2)/2\rfloor \le t\le r$ and write $z:=t-1$. By the inductive assumption and, if $r$ is odd and $t=\lfloor (r+2)/2\rfloor$, Theorem 2 applied to the projective space ${\mathrm{\mathbb{P}}}^{r-1}$ there is a non-empty Zariski open subset $\mathcal{V}$ of ${\mathrm{\mathbb{P}}}^{r-1}$ such that $W{(Y)}_{q,z}=\{q\}$ for all $q\in \mathcal{V}$. Fix $a\in \mathcal{V}$ and finitely many $S_i\in \mathcal{S}(Y,a,z)$, $1\le i\le e$, such that $\{a\}={\cap }_{i=1}^e〈S_i〉$. Restricting if necessary $\mathcal{V}$ we may assume that (for a choice of sufficiently general $S_1(a),\dots ,S_e(a)$) we have $S_i(a)\cap \mu (F)=ø$ for all $i$ and all $a$. Hence there is a unique $A_i(a)\subset X\backslash F$ such that $\mu (A_i(a))=S_i(a)$. Since $p\in F$, $B_i(a):=A_i(a){{\displaystyle \cup }}^{\text{}}\{p\}$ has cardinality $t$, $1\le i\le e$. Set ${\mathcal{U}}_p:={\ell }^{-1}(\mathcal{V})\subset {\mathrm{\mathbb{P}}}^r\backslash \{p\}$. For each $a\in \mathcal{V}$, set $L_a:=\{p\}{{\displaystyle \cup }}^{\text{}}{\ell }^{-1}(a)$. Each $L_a$ is a line containing $p$, ${\mathcal{U}}_p$ is the union of all $L_a\backslash \{p\}$, $a\in \mathcal{V}$, and $L_a={\cap }_{i=1}^e〈B_i(a)〉$. Fix $a\in \mathcal{V}$ and $b\in L_a\backslash \{p\}$. Note that each $B_i(a)$ irredundantly spans $b$. Fix another general $o\in X$, $o\ne p$. We get in the similar way a set ${\mathcal{U}}_o$. It is easy to check that $W_{q,t}=\{q\}$ for all $q\in {\mathcal{U}}_0{{\displaystyle \cap }}^{\text{}}{\mathcal{U}}_p$. Thus we may take $\mathrm{=}{\mathcal{U}}_p\cap {\mathcal{U}}_0$. □

5. Real varieties and real ranks

Up to now we worked over an algebraically closed field $\mathbb{K}$ with characteristic zero. In this section we take $\mathbb{K}\mathrm{=ℂ}$, but we consider varieties $X\subset {\mathrm{\mathbb{P}}}^r$ defined over $\mathrm{ℝ}$. Not only we fix the real structure of $X$ but we assume that the embedding $X\hookrightarrow {\mathrm{\mathbb{P}}}^r$ is defined over $\mathrm{ℝ}$. We call $X(\mathrm{ℂ})$ and ${\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$ the set of all complex points of $X$ and ${\mathrm{\mathbb{P}}}^r$. For any $q\in {\mathrm{\mathbb{P}}}^r(\mathrm{ℂ})$ we have defined the $X$ -rank $r_X(q)$ and the set $\mathcal{S}(X,q)$. In this section we write $r_{X(\mathrm{ℂ})}(q)$ instead of $r_X(q)$ and $\mathcal{S}(X(\mathrm{ℂ}),q)$ instead of $\mathcal{S}(X,q)$. Since $X$ is defined over $\mathrm{ℝ}$, the set $X(\mathrm{ℝ})$ of its real points is well-defined. Since the embedding $X\hookrightarrow {\mathrm{\mathbb{P}}}^r$ is defined over $\mathrm{ℝ}$, we have $X(\mathrm{ℝ})=X(\mathrm{ℂ}){{\displaystyle \cap }}^{\text{}}{\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$. Easy examples show that a nice $X$ defined over $\mathrm{ℝ}$ may have $X(\mathrm{ℝ})=ø$. For instance take the smooth plane conic $C:=\left\{x_0^2+x_1^2+x_3^2=0\right\}$ (we have $C(\mathrm{ℂ})\cong {\mathrm{\mathbb{P}}}^1(\mathrm{ℂ})$). Felix Klein proved that for every integer $g\ge 0$ there is a smooth curve $X(\mathrm{ℂ})$ of genus $g$ defined over $\mathrm{ℝ}$ and with $X(\mathrm{ℝ})=ø$ ([17, Proposition 3.1]). Thus the assumption that $X(ℝ)$ is large is necessary. We assume that $X$ has a smooth point defined over $\mathrm{ℝ}$ (in symbols, we assume $X_{\mathrm{reg}}(\mathrm{ℝ})\ne ø$ ). Set $n:=\dim X={\dim }_{\mathrm{ℂ}}X(\mathrm{ℂ})$. The sets ${\mathrm{\mathbb{P}}}^r(\mathrm{ℂ})$ and $X(\mathrm{ℂ})$ also have a euclidean topology. With the euclidean topology $X_{\mathrm{reg}}(\mathrm{ℝ})$ is a topological (and $C^{\infty }$ ) manifold with pure dimension $n$ and the assumption $X_{\mathrm{reg}}(\mathrm{ℝ})\ne ø$ says that this manifold is non-empty. The assumption $X_{\mathrm{reg}}(\mathrm{ℝ})\ne ø$ is equivalent to assuming that $X(\mathrm{ℝ})$ is Zariski dense in $X(\mathrm{ℂ})$, because $\mathrm{Sing}(X(\mathrm{ℂ}))$ is a union of complex varieties of dimension $<n$. For any set $S\subset {\mathrm{\mathbb{P}}}^r(\mathrm{ℂ})$ let ${〈S〉}_{\mathrm{ℂ}}$ be the complex linear projective subspace of ${\mathrm{\mathbb{P}}}^r(\mathrm{ℂ})$ spanned by $S$, i.e. the linear space that in the previous sections we called $〈S〉$. For any $S\subset {\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$ we write ${〈S〉}_{\mathrm{ℝ}}$ for the minimal real projective subspace of ${\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$ containing $S$. Since $S\subset {\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$ we have ${〈S〉}_{\mathrm{ℝ}}={〈S〉}_{ℂ}\cap {\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$. Since $X(\mathrm{ℝ})=X(\mathrm{ℂ}){{\displaystyle \cap }}^{\text{}}{\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$, $X(\mathrm{ℝ})$ is Zariski dense in $X(\mathrm{ℂ})$ and $X(\mathrm{ℂ})$ spans ${\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$. Thus for each $q\in {\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$ the $X(\mathrm{ℝ})$-rank (i.e. the minimal cardinality of a set $S\subset X(\mathrm{ℝ})$ such that $q\in {〈S〉}_{\mathrm{ℝ}}$ ) is a well-defined integer. For any $q\in {\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$ let $\mathcal{S}(X(\mathrm{ℝ}),q)$ denote the set of all $S\subset X(\mathrm{ℝ})$ such that $q\in {〈S〉}_{\mathrm{ℝ}}$ and $|S|=r_{X(\mathrm{ℝ})}(q)$. The interested reader may find the definition of a real semialgebraic set in [12, §2.1]. The set $\mathcal{S}(X(\mathrm{ℝ}),q)$ is semialgebraic ([12, Proposition 2.2.7]). Set

$$W_q(X(\mathrm{ℝ})):={{{\displaystyle \cap }}^{\text{}}}_{S\in \mathcal{S}(X(\mathrm{ℝ}),q)}{〈S〉}_{\mathrm{ℝ}}.$$

We always have $r_{X(\mathrm{ℝ})}(q)\ge r_{X(\mathrm{ℂ})}(q)$ and in many cases the inequality is strict. For instance, when $X\subset {\mathrm{\mathbb{P}}}^d$, $d\ge 3$, is a degree $d$ rational normal curve for each integer $t$ such that $\lfloor (d+2)/2\rfloor <t\le d$ there is $q\in {\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$ such that $r_{X(\mathrm{ℂ})}(q)=\lfloor (d+2)/2\rfloor$ and $r_{X(\mathrm{ℝ})}(q)=t$ [10,15]. See [11] for definitions and many examples when $X(ℂ)$ is a smooth curve and [1,2,21,22] for tensors and symmetric tensors. When $r_{X(\mathrm{ℝ})}(q)=r_{X(\mathrm{ℂ})}(q)$ we have $\mathcal{S}(X(\mathrm{ℝ}),q)\subseteq \mathcal{S}(X(\mathrm{ℂ}),q)$ and hence $W_q(X(\mathrm{ℝ}))\supseteq W_q(X(\mathrm{ℂ})){{\displaystyle \cap }}^{\text{}}{\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$. We give below an example with $r_{X(\mathrm{ℂ})}(q)=r_{X(\mathrm{ℝ})}(q)=2$, $W_q(X(\mathrm{ℝ}))$ a real line and $W_q(X(\mathrm{ℂ}))=\{q\}$ (see Example 1).

Theorem 4.

Fix an even integer $r\ge 2$ . Let $X\subset {\mathrm{\mathbb{P}}}^r$ be an integral and non-degenerate curve defined over $\mathrm{ℝ}$ and with $X_{\mathrm{reg}}(\mathrm{ℝ})\ne ø$ . There is a non-empty euclidean open subset $\mathcal{U}\subset {\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$ such that $r_{X(\mathrm{ℝ})}(q)=r/2+1$ for all $q\in \mathcal{U}$ and $\{q\}={\cap }_{S\in \mathcal{S}(X(\mathrm{ℝ}),q)}{〈S〉}_{\mathrm{ℝ}}$ for all $q\in U$.

Note that we also get $\{q\}={\cap }_{S\in \mathcal{S}(X(\mathrm{ℝ}),q)}{〈S〉}_{\mathrm{ℂ}}$, because${{{\displaystyle \cap }}^{\text{}}}_{S\in \mathcal{S}(X(\mathrm{ℝ}),q)}{〈S〉}_{\mathrm{ℂ}}$ is defined over $\mathrm{ℝ}$ and hence its dimension as a complex projective space is the dimension of the real projective space $\left({{{\displaystyle \cap }}^{\text{}}}_{S\in \mathcal{S}(X(\mathrm{ℝ}),q)}{〈S〉}_{\mathrm{ℂ}}\right){{\displaystyle \cap }}^{\text{}}{\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})={{{\displaystyle \cap }}^{\text{}}}_{S\in \mathcal{S}(X(\mathrm{ℝ}),q)}{〈S〉}_{\mathrm{ℝ}}$.

Remark 4.

We recall that the Zariski topology of ${\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$ (i.e. the topology in which the closed sets are the intersection with ${\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$ of a Zariski closed subset of ${\mathrm{\mathbb{P}}}^r(\mathrm{ℂ})$ ) may be defined by taking as closed subsets the zero-loci of real homogeneous polynomials. Non-empty euclidean open subsets of ${\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$ are Zariski dense. To show that in Theorem 4 we cannot take as $\mathcal{U}$ a Zariski open subset of ${\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$ it is sufficient to find a curve $X\subset {\mathrm{\mathbb{P}}}^r$ with $X_{\mathrm{reg}}(\mathrm{ℝ})\ne ø$ and with two different typical ranks. By [10] one can take the rational normal curve of ${\mathrm{\mathbb{P}}}^r$, $r\ge 4$.

Before proving Theorem 4 we describe in the next remark the topology of the real part $X(\mathrm{ℝ})$ of an integral projective curve defined over $\mathrm{ℝ}$.

Remark 5.

Let $X(\mathrm{ℂ})$ be an integral projective curve defined over $\mathrm{ℝ}$. Let $\eta :Y(\mathrm{ℂ})\to X(\mathrm{ℂ})$ denote the normalization map. Both $Y(\mathrm{ℂ})$ and $\eta$ are defined over $\mathrm{ℝ}$ and hence $Y(\mathrm{ℝ})$ is well-defined and $\eta (Y(\mathrm{ℝ}))\subseteq X(\mathrm{ℝ})$. Since $\eta$ is an isomorphism over $X_{\mathrm{reg}}(\mathrm{ℂ})$, $X_{\mathrm{reg}}(\mathrm{ℝ})$ is essentially $Y(\mathrm{ℝ})$ minus a finite set. Call $g$ the genus of $Y(\mathrm{ℂ})$. F. Klein described the possible real parts $Y(\mathrm{ℝ})$ of genus $g$ smooth curve defined over $ℝ$ ([17, Proposition 3.1]). Topologically $Y(\mathrm{ℝ})$ is the union of $k$ pairwise disjoint circles, with $k$ an integer between $0$.and $g+1$. Thus the topological space $X(\mathrm{ℝ})$ is obtained from $Y(\mathrm{ℝ})$ by an equivalence relation which only identifies finitely many finite subsets of $Y(\mathrm{ℝ})$ and then, sometimes, one adds to $\eta (Y(\mathrm{ℝ}))$ finitely many isolated real points of $\mathrm{Sing}(X(\mathrm{ℂ}))$, each of them the image of two complex conjugate points of $Y(\mathrm{ℂ})\backslash Y(\mathrm{ℝ})$. Thus $X(\mathrm{ℝ})$ is finite (and hence not Zariski dense in $X(\mathrm{ℂ})$) if and only if $Y(\mathrm{ℝ})=ø$, i.e. if and only if $X_{\mathrm{reg}}(\mathrm{ℝ})=ø$.

Proof of Theorem 4.

Since $X_{\mathrm{reg}}(\mathrm{ℝ})\ne ø$, there is a set $J\subset X(\mathrm{ℝ})$ homeomorphic to a non-empty open interval of $\mathrm{ℝ}$ for the euclidean topology (Remark 5). Since $J$ is infinite, it is Zariski dense in $X(\mathrm{ℂ})$. As in the proof of Theorem 1 let $\mathcal{V}\subset {\mathrm{\mathbb{P}}}^r(\mathrm{ℂ})$ be a non-empty Zariski open subset such that $r_{X(\mathrm{ℂ})}(q)=r/2+1$ for all $q\in \mathcal{V}$. The set $\sigma (\mathcal{V})$ is Zariski open in ${\mathrm{\mathbb{P}}}^r(\mathrm{ℂ})$. Since $X(\mathrm{ℂ})$ is defined over $\mathrm{ℝ}$, we have $r_{X(\mathrm{ℂ})}(q)=r/2+1$ for all $q\in \mathcal{V}$. Set ${\mathcal{V}}^{\prime }:=(\mathcal{V}\cup \sigma (\mathcal{V}))\cap {\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$. The set ${\mathcal{V}}^{\prime }$ is a non-empty Zariski open subset of ${\mathrm{\mathbb{P}}}^r(\mathrm{ℂ})$. Call $J_{r/2+1}$ the set of all subset $S\subset J$ such that $|S|=r/2+1$ and ${〈S〉}_{\mathrm{ℂ}}\cap \left(\mathcal{V}{{\displaystyle \cup }}^{\text{}}\sigma (\mathcal{V})\right)$. Since $r_{X(\mathrm{ℂ})}(q)=r/2+1$ for each $q\in \mathcal{V}{{\displaystyle \cup }}^{\text{}}\sigma (\mathcal{V})$, each $S\in J_{r/2+1}$ is linearly independent. Since $\mathcal{V}{{\displaystyle \cup }}^{\text{}}\sigma (\mathcal{V})$ is open, we have $S\in J_{r/2+1}$ if and only ${〈S〉}_{\mathrm{ℝ}}{{\displaystyle \cap }}^{\text{}}{\mathcal{V}}^{\prime }\ne ø$. We get a euclidean open subset ${\mathcal{U}}_1$ of ${\mathcal{V}}^{\prime }$ taking the interior of the union of all sets ${〈S〉}_{\mathrm{ℝ}}{{\displaystyle \cap }}^{\text{}}{\mathcal{V}}^{\prime }$ for some $S\in J_{r/2+1}$. To get $\{q\}={\cap }_{S\in S(X(\mathrm{ℝ}),q)}{〈S〉}_{\mathrm{ℝ}}$ for all $q\in \mathcal{U}$ we need to restrict the euclidean open set ${\mathcal{U}}_1$ in the following way. Fix $q\in {\mathcal{U}}_1$ and take $S\in \mathcal{S}(X(\mathrm{ℝ}),q)$. We run the proof of Theorem 1 with this set $S$ and get a curve $X_S$ defined over $\mathrm{ℝ}$ and, using it, a set $A_S$ defined over $\mathrm{ℝ}$. We only need to restrict ${\mathcal{U}}_{\mathrm{1}}$ so that for $q\in \mathcal{U}$ the set $A_S$ is defined and ${〈S〉}_{\mathrm{ℂ}}{\displaystyle \cap }{〈A_S〉}_{\mathrm{ℂ}}=\{q\}$.□

Example 1.

Fix an integer $r\ge 3$. Let $Y(\mathrm{ℝ})\subset {\mathrm{\mathbb{P}}}^{r+1}(\mathrm{ℝ})$ be the degree $r+1$ rational normal curve. Let $\sigma$ denote the complex conjugation of ${\mathrm{\mathbb{P}}}^{r+1}(\mathrm{ℂ})$ and ${\mathrm{\mathbb{P}}}^r(\mathrm{ℂ})$. Fix $p_1,p_2\in Y(\mathrm{ℝ})$ such that $p_1\ne p_2$ and $p_3\in Y(\mathrm{ℂ})\backslash Y(\mathrm{ℝ})$. Set $p_4:=\sigma (p_3)$. We may take homogeneous coordinates $z_0,\dots ,z_{r+1}$ of ${\mathrm{\mathbb{P}}}^{r+1}(\mathrm{ℝ})$ and ${\mathrm{\mathbb{P}}}^{r+1}(\mathrm{ℂ})$ such that $p_3=\left[1:a_1:\cdots :a_{n+1}\right]$ with $a_i\in \mathrm{ℂ}$ for all $i$ and $a_i\notin \mathrm{ℝ}$ for at least one $i$. Set $o_1:=\left[1:\text{Re}(a_1):\cdots :\text{Re}(a_{r+1})\right]$ and $o_2:=\left[1:\text{Im}(a_1):\cdots :\text{Im}(a_{r+1})\right]$. We have $o_i\in {\mathrm{\mathbb{P}}}^{r+1}(\mathrm{ℝ})$ and $o_1\ne o_2$, because $p_3\notin {\mathrm{\mathbb{P}}}^{r+1}(\mathrm{ℝ})$. Since $r\ge 2$, $o_i\notin X(\mathrm{ℂ})$. We have $|\left\{p_1,p_2,p_3,p_4\right\}|=4$ and hence ${〈\left\{p_1,p_2,p_3,p_4\right\}〉}_{\mathrm{ℂ}}$ is a $3$. -dimensional complex linear subspace. Since $\sigma \left(\left\{p_1,p_2,p_3,p_4\right\}\right)=\left\{p_1,p_2,p_3,p_4\right\}$, the linear space ${〈\left\{p_1,p_2,p_3,p_4\right\}〉}_{\mathrm{ℂ}}$ is defined over $\mathrm{ℝ}$, i.e.${〈\left\{p_1,p_2,p_3,p_4\right\}〉}_{\mathrm{ℂ}}{{\displaystyle \cap }}^{\text{}}{\mathrm{\mathbb{P}}}^{r+1}(\mathrm{ℝ})$ is a $3$-dimensional real linear space (it is the real linear space ${〈\left\{p_1,p_2,o_1,o_2\right\}〉}_{\mathrm{ℝ}}$ ). Fix $o\in {〈\left\{p_1,p_2,o_1,o_2\right\}〉}_{\mathrm{ℝ}}$ such that $o$ is not in the linear span of any proper subset of $\left\{p_1,p_2,o_1,o_2\right\}$. Let ${\ell }_o:{\mathrm{\mathbb{P}}}^{r+1}(\mathrm{ℂ})\backslash \{o\}\to {\mathrm{\mathbb{P}}}^r(\mathrm{ℂ})$ denote the linear projection from $o$. Since $o\in {\mathrm{\mathbb{P}}}^{r+1}(\mathrm{ℝ})$, ${\ell }_o$ is defined over $\mathrm{ℝ}$ and ${\ell }_o^{-1}\left({\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})\right)={\mathrm{\mathbb{P}}}^{r+1}(\mathrm{ℝ})\backslash \{o\}$. By Sylvester’s theorem we have $o\notin {\sigma }_2\left(Y(\mathrm{ℂ})\right)$. Thus $X(\mathrm{ℂ}):={\ell }_o\left(Y(\mathrm{ℂ})\right)$ is a smooth and non-degenerate rational curve defined over $\mathrm{ℝ}$. Since $Y(\mathrm{ℝ})\ne ø$, we have $X(\mathrm{ℝ})\ne ø$. The complex linear space $V_{\mathrm{ℂ}}:={\ell }_o\left({〈\left\{p_1,p_2,o_1,o_2\right\}〉}_{\mathrm{ℂ}}\right)$ is a plane containing exactly $4$ points of $X(\mathrm{ℂ})$ (the points ${\ell }_o(p_1)$, ${\ell }_o(p_2)$, ${\ell }_o(p_3)$ and ${\ell }_o(p_4)$), because any $r+2$ points of $Y(\mathrm{ℂ})$ are linearly independent. Set $L:={〈\left\{{\ell }_o(p_1),{\ell }_o(p_2)\right\}〉}_{\mathrm{ℂ}}$ and $R:={〈\left\{{\ell }_o(p_1),{\ell }_o(p_2)\right\}〉}_{\mathrm{ℂ}}$. Since $L\ne R$ and ${\dim }_{\mathrm{ℂ}}{V}_{\mathrm{ℂ}}$, the set $L\cap R$ is a unique point, $q$. Since $\sigma (L)=L$ and $\sigma (R)=R$, we have $\sigma (q)=q$, i.e. $q\in {\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$. Since $q\notin X(\mathrm{ℂ})$ and ${\ell }_o(p_1),{\ell }_o(p_2)\in X(\mathrm{ℝ})$, we have $r_{X(\mathrm{ℝ})}(q)=2$ and hence $r_{X(\mathrm{ℂ})}(q)=2$. Since $\left\{{\ell }_o(p_1),{\ell }_o(p_2)\right\}$, $\left\{{\ell }_o(p_3),{\ell }_o(p_4)\right\}\in \mathcal{S}(X(\mathrm{ℂ}),q)$, we have $W_q(X(\mathrm{ℂ}))=\{q\}$. Using that any $r+2$ elements of $Y(\mathrm{ℂ})$ are linearly independents, we get that $\left\{{\ell }_o(p_1),{\ell }_o(p_2)\right\}$ and $\left\{{\ell }_o(p_3),{\ell }_o(p_4)\right\}$ are the only elements of $\mathcal{S}(X(\mathrm{ℂ}),q)$. Thus $W_q(X(\mathrm{ℝ}))={〈\left\{{\ell }_o(p_1),{\ell }_o(p_2)\right\}〉}_{\mathrm{ℝ}}$ is a line. Since $\mathcal{S}(X(\mathrm{ℝ}),q)=\left\{{\ell }_o(p_1),{\ell }_o(p_2)\right\}$, $q$ is $X(\mathrm{ℝ})$-identifiable. This is not the first example of some $q\in {\mathrm{\mathbb{P}}}^r(\mathrm{ℝ})$ which is identifiable over $\mathrm{ℝ}$, but not over $\mathrm{ℂ}$ [1,2].

Note

1 The author was partially supported by MIUR and GNSAGA of INdAM (Italy).