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

TL;DR

This paper investigates the problem of reconstructing homogeneous polynomials from their additive decompositions, especially in cases where the decomposition is not unique, focusing on curves and Veronese embeddings.

Contribution

It introduces the concept of the non-uniqueness set for points with multiple decompositions and analyzes its properties for curves and Veronese varieties.

Findings

Characterizes the non-uniqueness set for non-identifiable points.

Provides results specific to curves and Veronese embeddings.

Enhances understanding of polynomial reconstruction when identifiability fails.

Abstract

Let $X\subset \mathbb {P}^r$ be an integral and non-degenerate variety. For any $q\in \mathbb {P}^r$ 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 $q\in \langle S\rangle$, where $\langle \ \ \rangle$ denotes the linear span. We consider the case $|\mathcal {S} (X,q)|>1$ (i.e. when $q$ is not $X$-identifiable) and study the set $W(X)_q:= \cap _{S\in\mathcal {S}}\langle S\rangle$, which we call the non-uniqueness set of $q$. We study the case $\dim X=1$ and the case $X$ a Veronese embedding of $\mathbb {P}^n$.

Equations

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

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

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

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

$$0\to {\mathcal{I}}_{{\mathrm{Res}}_{\mathcal{R}}(\mathcal{Z})}(d-1)\to {\mathcal{I}}_{\mathcal{Z}}(d)\to {\mathcal{I}}_{\mathcal{Z}\cap \mathcal{R},\mathcal{R}}(d)\to 0$$

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

Full text

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

1. Introduction

Let $X\subset\mathbb{P}^{r}$ be an integral and non-degenerate variety. For any set $A\subset\mathbb{P}^{r}$ let $\langle A\rangle$ denote its linear span. Fix any $q\in\mathbb{P}^{r}$. The $X$-rank $r_{X}(q)$ of $X$ is the minimal cardinality of a finite set $S\subset X$ such that $q\in\langle S\rangle$. The notion of $X$-rank includes the notion of tensor rank of a tensor (take $X$ a multiprojective space and $X\subset\mathbb{P}^{r}$ 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 $X\subset\mathbb{P}^{r}$ one of its Veronese embeddings). See [2, 9, 12, 13] for a long list of applications of these notions. Let $\cal{S}(X,q)$ denote the set of all $S\subset X$ such that $|S|=r_{X}(q)$ and $q\in\langle S\rangle$. Set $W(X)_{q}:=\cap_{S\in\cal{S}(X,q)}\langle S\rangle$. The set $W(X)_{q}$ is the main actor of this paper. We often write $W_{q}$ if $X$ is clear from the context. Note that $W_{q}$ is a linear subspace of $\mathbb{P}^{r}$ containing $q$ and that if $W_{q}=\{q\}$, and $\cal{S}(X,q)=\cal{S}(X,q^{\prime})$ for some $q^{\prime}\in\mathbb{P}^{r}$, then $q^{\prime}=q$. If $W_{q}\neq\{q\}$, then there are infinitely many $q^{\prime}\in\mathbb{P}^{r}$ such that $\cal{S}(X,q^{\prime})=\cal{S}(X,q)$ (Remark 1). We will call $W_{q}$ the non-uniqueness set of $q$. We have $\dim W_{q}=r_{X}(q)-1$ if and only if $\langle S\rangle=\langle S^{\prime}\rangle$ for all $S,S^{\prime}\in\cal{S}(X,q)$. In particular $W_{q}=\{q\}$ and $q\notin X$ imply $|\cal{S}(X,q)|>1$.

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

Theorem 1**.**

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

(a) We have $\{q\}=\cap_{S\in\cal{S}(X,q)}\langle S\rangle$ for all $q\in\cal{U}$.

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

Theorem 2**.**

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

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

Theorem 3**.**

Fix integers $n,d,b,k$, such that $n\geq 2$, $d\geq 8$, $4\leq 2b\leq d$ and $d+2-b\leq k\leq 2d-2$. Let $\nu_{d}:\mathbb{P}^{n}\xrightarrow{\ \ }\mathbb{P}^{r}$, $r=\binom{n+d}{n}-1$, be the order $d$ Veronese embedding. Let $L\subset\mathbb{P}^{n}$ be a line. Set $Y:=\nu_{d}(L)$. Fix $q^{\prime}\in\langle Y\rangle$ such that $b_{Y}(q^{\prime})=b$ and $r_{Y}(q^{\prime})=d+2-b$. Fix a general $U\subset\mathbb{P}^{n}$ such that $|U|=k-d-2+b$. Let $q\in\mathbb{P}^{r}$ be any point irredundantly spanned by $\{q^{\prime}\}\cup\nu_{d}(U)$. Then:

1. (1)

$r_{X}(q)=k$, $\cal{S}(X,q)\supseteq\{E\cup U\}_{E\in\cal{S}(Y,q^{\prime})}$ and $\cal{S}(X,q)$ uniquely determines $q^{\prime}$ and $U$. 2. (2)

If $k\leq 2d-3$, then $\cal{S}(X,q)=\{E\cup U\}_{E\in\cal{S}(Y,q^{\prime})}$ and $W_{q}=\langle U\cup\{q^{\prime}\}\rangle$.

In the last section we consider the following problem. For any positive integer $t$ let $\cal{S}(X,q,t)$ be the set of all $S\subset X$ such that $|S|=t$ and $S$ irredundantly spans $q$. We have $\cal{S}(X,q,t)=\emptyset$ for all $t<r_{X}(q)$ and $\cal{S}(X,q,r_{X})=\cal{S}(X,q)\neq\emptyset$. By the definition of irredundantly spanning set we have $\cal{S}(X,q,t)=\emptyset$ for all $t\geq r+2$. Since $X$ is integral and non-degenerate, for all $(X,q)$ we have $\cal{S}(X,q,r+1)\neq\emptyset$ and $\cal{S}(X,q,r+1)$ 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 $\cal{S}(X,q,t)=\emptyset$ (Example 3). It easy to check that $\cal{S}(X,q,t)\neq\emptyset$ for all $t$ such that $r+1-\dim X\leq t\leq r$ (Lemma 2). Set $W(X)_{q,t}:=\cap_{S\in\cal{S}(X,q,t)}\langle S\rangle$, with the convention $W(X)_{q,t}:=\mathbb{P}^{r}$ if $\cal{S}(X,q,t)=\emptyset$. We often write $W_{q,t}$ instead of $W(X)_{q,t}$.

In section 4 we prove the following result.

Proposition 1**.**

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

2. Preliminary observations

The cactus rank or cactus $X$-rank $c_{X}(q)$ of $q\in\mathbb{P}^{r}$ is the minimal degree of a zero-dimensional scheme $Z\subset X$ such that $q\in\langle Z\rangle$. Let $\cal{Z}(X,q)$ denote the set of all zero-dimensional schemes $Z\subset X$ such that $\deg(Z)=c_{X}(q)$ and $q\in\langle Z\rangle$.

Remark 1**.**

Take any $q\in\mathbb{P}^{r}$ and any $S\in\cal{S}(X,q)$. For any $o\in\langle S\rangle$ the integer $r_{X}(o)$ is the minimal cardinality of a subset $S_{1}\subsetneq S$ such that $o\in\langle S_{1}\rangle$. We have $S_{1}\in\cal{S}(X,o)$. Now assume $W_{q}\supsetneq\{q\}$. Fix $S\in\cal{S}(X,q)$. Let $W^{\prime}_{q}$ be the set of all $o\in W_{q}$ not contained in some $S^{\prime}\subsetneq S$. Note that $q\in W^{\prime}_{q}$. We saw that $W^{\prime}_{q}$ is the complement in $W_{q}$ of at most $r_{X}(q)$ hyperplanes of $W_{q}$. Thus $W^{\prime}_{q}$ is an irreducible algebraic set of dimension $\dim W_{q}$. We saw that each $o\in W_{q}$ has rank equal to the minimal cardinality of a subset of $S$ spanning $q$. Thus $r_{X}(o)=r_{X}(q)$ for each $o\in W^{\prime}_{q}$. Since $W^{\prime}_{q}\subseteq\cap_{S\in\cal{S}(X,q)}\langle S\rangle$, we get $\cal{S}(X,q)\subseteq\cal{S}(X,o)$ for each $o\in W^{\prime}_{q}$. Since $q\in\langle S\rangle$, $q\notin\langle S^{\prime}\rangle$ for any $S^{\prime}\subsetneq S$ and $S\in\cal{S}(X,o)$, we get $\cal{S}(X,o)\subseteq\cal{S}(X,q)$. Thus $\cal{S}(X,q)=\cal{S}(X,o)$.

Remark 2**.**

Let $X\subset\mathbb{P}^{d}$, $d\geq 2$, be a degree $d$ rational normal curve. We use [12, §1.3] and [10] for the following observations. Fix $q\in\mathbb{P}^{r}$.

(i) We have $b_{X}(q)=c_{X}(q)$ ([12, Lemma 1.38]) and $|\cal{Z}(X,q)|=1$ ([12, Part (i) of Theorem 1.43]).

(ii) If $c_{X}(q)<r_{X}(q)$, then $c_{X}(q)+r_{X}(q)=d+2$ and $\cal{S}(X,q)$ is infinite. Let $Z$ be the only element of $\cal{Z}(X,q)$, $d+2-c_{X}(q)$ is the minimal degree of a scheme $A\subset X$, such that $q\in\langle A\rangle$ and $A\nsupseteq Z$.

(iii) If $r_{X}(q)>c_{X}(q)$, then $\{q\}=\langle Z\rangle\cap\langle S\rangle$, where $\{Z\}=\cal{Z}(X,q)$ and $S$ is any element of $\cal{S}(X,q)$ (this also follows from the fact that $h^{1}(\mathbb{P}^{1},L)=0$ for any line bundle $L$ on $\mathbb{P}^{1}$ with $\deg(L)\geq-1$, as in the proof of Claim 1 below).

(iv) If $r_{X}(q)>b_{X}(q)$, then $\dim\cal{S}(X,q)=d+3-2b$ ([10, eq. (9)]).

(v) If $r$ is odd and $r_{X}(q)=(d+2)/2$ (i.e. $r_{X}(q)=b_{X}(q)$ is the generic rank), then $\cal{S}(X,q)=\cal{Z}(X,q)$ and $|\cal{S}(X,q)|=1$ ([12, 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 $\mathbb{P}^{d}$. Fix $S,S^{\prime}\in\cal{S}(X,q)$ such that $S\neq S^{\prime}$.

Claim 1: $\langle S\rangle\cap\langle S^{\prime}\rangle=\{q\}$.

Proof of Claim 1: Since $S\neq S^{\prime}$ and $S\in\cal{S}(X,q)$, we have $q\notin\langle S\cap S^{\prime}\rangle$. The Grassmann’s formula gives $h^{1}(\mathbb{P}^{1},\cal{I}_{S\cup S^{\prime}}(d))>0$. Since $h^{1}(\mathbb{P}^{1},L)=0$ for any line bundle $L$ on $\mathbb{P}^{1}$ with $\deg(L)\geq-1$ and $q\notin X$, we have $S\cap S^{\prime}=\emptyset$ and $h^{1}(\mathbb{P}^{1},\cal{I}_{S\cup S^{\prime}}(d))=1$. The Grassmann’s formula then implies $\dim(\langle S\rangle\cap\langle S^{\prime}\rangle)=0$, proving Claim 1.

Obviously Claim 1 implies $W_{q}=\{q\}$ in this case, which by [12, Part (i) of Theorem 1.43] is the only case in which $r_{X}(q)=b_{X}(q)$ and $\cal{Z}(X,q)$ is not a singleton.

Note that (iii) implies that each $q\in\mathbb{P}^{r}$ with $c_{X}(q)\neq r_{X}(q)$ is uniquely determined by the zero-dimensional scheme evincing its cactus rang and by one single set evincing its rank (any $S\in\cal{S}(X,q)$ would do the job). Obviously part (i) implies that most $q\in\mathbb{P}^{r}$ (the ones with $r_{X}(q)=b_{X}(q)$) are not uniquely determined by $\cal{S}(X,q)$. By Remark 1 and part (i) for each $q\in\mathbb{P}^{r}$ such that $r_{X}(q)=b_{X}(q)$ there are exactly $\infty^{t}$, $t:=r_{X}(q)-1$, points $o\in\mathbb{P}^{r}$ with $\cal{S}(X,o)=\cal{S}(X,q)$. A similar remark holds for an arbitrary variety $X$ for the points $q$ such that $|\cal{S}(X,q)|=1$.

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

Lemma 1**.**

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

1. (1)

$|F|=d+1$* and $F$ is contained in a line;* 2. (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. (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. (4)

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

3. Proofs of Theorems 1, 2 and 3

Proof of Theorem 1:.

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

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

For each set $S\subset X$ such that $|S|=r/2+1$ and $\dim\langle S\rangle=r/2$ let $\ell_{S}:\mathbb{P}^{r}\setminus\langle S\rangle\xrightarrow{\ \ }\mathbb{P}^{r/2-1}$ denote the linear projection from $\langle S\rangle$. For a general $S$ we have $\langle S\rangle\cap X=S$ (scheme-theoreticaly) by Bertini’s theorem and the trisecant lemma ([15, Corollary 2.2]) and $\ell_{S|X\setminus S}$ is birational onto its image, again by the trisecant lemma and the assumption $r\geq 4$. Let $X_{S}\subset\mathbb{P}^{r/2-1}$ be the closure of $\ell_{S}(X\setminus S)$ in $\mathbb{P}^{r/2-1}$. There is a finite set $E\subset X_{S}$ containing $X_{S}\setminus\ell_{S}(X\setminus S)$ and such that for each $p\in X_{S}\setminus E$ there is a unique $o\in X\setminus S$ such that $\ell_{S}(o)=p$. For any set $A\subset X_{S}\setminus E$ let $A_{S}\subset X\setminus S$ denote the only set such that $\ell_{S}(A_{S})=A$. Any general $A\subset X_{S}\setminus E$ such that $|A|=r/2+1$ is linearly dependent, but each proper subset of $A$ is linearly independent. Thus $\langle S\rangle\cap\langle A_{S}\rangle$ is a single point, $q_{S,A}$, and $q_{S,A}\notin\langle B\rangle$ for any $B\subsetneq A_{S}$. 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 $S^{\prime}\subsetneq S$. We start with $S\in\cal{S}(X,o)$ for a general $o\in\mathbb{P}^{r}$. Thus $r_{X}(q)=r/2+1$ for a general $q\in\langle S\rangle$. Thus for a general $A$ we get $S\in\cal{S}(q_{S,A})$ and $A_{S}\in\cal{S}(q_{S,A})$. By construction we have $\{q_{S,A}\}=\langle S\rangle\cap\langle A_{S}\rangle$. For a general $A$ the point $q_{S,A}$ is general in $\langle S\rangle$. By the generality of $S$ we get that the points $q_{S,A}$’s (with $(S,A)$ varying, but general), cover a non-empty open subset of $\mathbb{P}^{r}$. ∎

Proof of Theorem 2:.

Set $b:=b_{X}(q)$. Since $\cal{S}(X,q)$ is not a singleton, we have $q\notin X$ and hence $b\geq 2$. 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 $\dim\cal{S}(X,q)\geq 2$. We will prove the stronger assumption that $\{q\}=\cap_{A\in\Gamma}\langle A\rangle$, where $\Gamma$ is any irreducible family contained in $\cal{S}(X,q)$ and with $\dim\Gamma=d+3-2b$; we do not assume that $\Gamma$ is closed in $\cal{S}(X,q)$.

(a) First assume $b=2$. We use the proof of [14, Proposition 5.1]. Fix $a\in\mathbb{P}^{d}\setminus\{q\}$. Let $H\subset\mathbb{P}^{d}$ 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

[TABLE]

gives that $X\cap H$ spans $H$. Thus $q\in\langle X\cap H\rangle$. Since $r_{X}(q)=d$, we get $X\cap H\in\cal{S}(X,q)$. 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 $d\leq 4$. Thus we may assume $d\geq 5$ and use induction on $d$. Fix a general $o\in X$. Let $\ell_{o}:\mathbb{P}^{d}\setminus\{o\}\xrightarrow{\ \ }\mathbb{P}^{d-1}$ denote the linear projection from $o$. Let $Y\subset\mathbb{P}^{d-1}$ denote the closure of $\ell_{o}(X\setminus\{o\})$ in $\mathbb{P}^{d-1}$. $Y$ is a rational normal curve of $\mathbb{P}^{d-1}$. Set $q^{\prime}:=\ell_{o}(q)$ and $Z^{\prime}:=\ell_{o}(Z)$ (by the generality of $o$ we have $o\notin\langle Z\rangle$ and hence $Z^{\prime}$ is well-defined, $\deg(Z^{\prime})=b$ and $\dim\langle Z^{\prime}\rangle=b-1$). The generality of $o$ also implies that $q\notin\langle Z^{\prime\prime}\cup\{o\}\rangle$ for any $Z^{\prime\prime}\subsetneq Z$ (here we use that $X$ is a smooth curve and hence $Z$ has only finitely many subschemes). Thus $q^{\prime}\in\langle Z^{\prime}\rangle$ and $q^{\prime}\notin\langle Z^{\prime\prime}\rangle$ for all $Z^{\prime\prime}\subsetneq Z^{\prime}$. Since $Y$ is a degree $d-1$ rational normal curve and $b\leq d/2$, parts (i) and (ii) of Remark 2 imply $b_{Y}(q^{\prime})=b$ and $\cal{Z}(Y,q^{\prime})=\{Z^{\prime}\}$. Fix an irreducible family $\Gamma\subseteq\cal{S}(X,q)$ such that $\dim\Gamma=d+3-2b$ (it exists by part (v) of Remark 2) . Let $\cal{B}$ denote the set of all $A\in\Gamma$ such that $o\in A$. By part (v) of Remark 2 and the generality of $o$ we have $\cal{B}\neq\emptyset$ and $\dim\cal{B}=d+2-2b$. Set $\cal{A}:=\{\ell_{o}(B\setminus\{o\})\}_{B\in\cal{B}}$. Since $Y$ is a rational normal curve, parts (i) and (ii) of Remark 2 imply $\cal{A}\subseteq\cal{S}(Y,q^{\prime})$. We have $\dim\cal{A}=(d-1)+3-2b$. The inductive assumption gives $\{q^{\prime}\}=\cap_{A\in\cal{A}}\langle A\rangle$. Thus $\langle\{o,q\}\rangle=\cap_{B\in\cal{B}}\langle B\rangle$. Since $\dim\Gamma=\dim\cal{S}(X,q)=d+3-2b$ (part (v) of Remark 2) and $o$ is general in $X$, there is $S\in\Gamma$ such that $o\notin S$. Thus $\cap_{A\in\Gamma}\langle A\rangle=\{q\}$. ∎

Proof of Theorem 3:.

By Autarky ([13, Exercise 3.2.2.2]) we may assume $U\neq\emptyset$. Since $U$ is general in $\mathbb{P}^{n}$, we have $\dim\langle U\cup Y\rangle=\min\{r,\dim\langle Y\rangle+|U|\}$. Since $\dim\langle Y\rangle=d$ and $d+|U|<r$, we have $\langle U\rangle\cap\langle Y\rangle=\emptyset$. By Theorem 2 we have $W(Y)_{q^{\prime}}=\{q^{\prime}\}$. Take $E\in\cal{S}(Y,q^{\prime})$ and set $A:=U\cup E$. The set $\{q^{\prime}\}\cup U$ irredundantly spans $q$ and $\langle Y\rangle\cap\langle U\rangle=\emptyset$, we have $E\cap U=\emptyset$ and hence $|A|=k$. Since $|A|=k$ and $q\in\langle\nu_{d}(A)\rangle$, we have $r_{X}(q)\leq k$. Since $U$ is general in $\mathbb{P}^{n}$, we have $h^{0}(\cal{I}_{A}(t))=\max\{0,h^{0}(\cal{I}_{E}(t))-|U|\}$ for all $t\in\mathbb{N}$; to use this equality we need to fix one element, $E$, of $\cal{S}(Y,q^{\prime})$, before choosing a general $U$.

(a) In this step we prove that $r_{X}(q)=k$ and that $\cal{S}(X,q)=\{E\cup U\}_{E\in\cal{S}(Y,q^{\prime})}$ if $k\leq 2d-3$. Note that we have $W_{q}=\langle\nu_{d}(U)\cup\{q^{\prime}\}\rangle$ for any $q$ such that $\cal{S}(X,q)=\{E\cup U\}_{E\in\cal{S}(Y,q^{\prime})}$ by Theorem 2. Assume either $r_{X}(q)<k$ or $k\leq 2d-3$ and the existence of $B\in\cal{S}(X,q)\setminus\{E\cup U\}_{E\in\cal{S}(Y,q^{\prime})}$. In the former case take $B\in\cal{S}(X,q)$. Set $S:=A\cup B$. In both cases we have $|B|\leq|A|$ and $|A|+|B|\leq 4k-5$. Since $h^{1}(\cal{I}_{S}(d))>0$ ([5, Lemma 1]) there is $F\subseteq S$ in one of the cases listed in Lemma 1.

(a1) Assume the existence of a plane cubic $T\subset\mathbb{P}^{n}$ such that $|T\cap S|\geq 3d$.

(a1.1) Assume $n=2$. Thus $T$ is an effective divisor of $\mathbb{P}^{n}$. Consider the residual exact sequence of $T$ in $\mathbb{P}^{2}$:

[TABLE]

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

(a1.2) Assume $n>2$. Let $M\subset\mathbb{P}^{n}$ be a general hyperplane containing the plane $\langle T\rangle$ (so $M=\langle T\rangle$ if $n=3$). Since $S$ is a finite set and $M$ is a general hyperplane containing $\langle T\rangle$, we have $S\cap M=S\cap\langle T\rangle$. Consider the residual exact sequence of $M$ in $\mathbb{P}^{n}$:

[TABLE]

Since $|S\setminus A\cap M|\leq 4d-5-3d=d-5$, we have $h^{1}(\cal{I}_{S\setminus S\cap M}(d-1))=0$. Thus either [6, Lemma 5.1] or [7, Lemmas 2.4 and 2.5] give $A\setminus A\cap M=B\setminus B\cap M$. Since no $4$ points of $U$ are coplanar, we have $|A\cap M|\leq d+5-b<3d-d-2+b$. Thus $|B|>|A|$, a contradiction.

(a2) Assume the existence of a plane conic $D$ such that $|S\cap D|\geq 2d+2$.

(a2.1) Assume $n=2$. Consider the residual exact sequence of $D$ in $\mathbb{P}^{2}$:

[TABLE]

First assume $h^{1}(\cal{I}_{S\setminus S\cap D}(d-2))>0$. Since $|S\setminus S\cap D|\leq 4d-5-2d-2=2(d-3)-1$, there is a line $R\subset\mathbb{P}^{2}$ such that $|R\cap(S\setminus S\cap D)|\geq d-1$ ([8, Lemma 34]). Thus $|S\cap(D\cup R)|\geq 3d+1$. Step (a1) gives a contradiction. Now assume $h^{1}(\cal{I}_{S\setminus S\cap D}(d-2))=0$. Either [6, Lemma 5.1] or [7, Lemmas 2.4 and 2.5] give $A\setminus A\cap D=B\setminus B\cap D$. Assume for the moment $L\nsubseteq D$. Thus $|L\cap D|\leq 2$. Since $U$ is general and $h^{0}(\cal{O}_{\mathbb{P}^{2}}(2))=6$, we have $|U\cap D|\leq 5$. Thus $|B\cap D|>|A\cap D|$ and so $|B|>|A|$, a contradiction. Now assume $L\subset D$. Write $D=L\cup R$ with $R$ a line. Since $U\cap L=\emptyset$, $|L\cap D|=1$ and $|U\cap R^{\prime}|\leq 2$ for each line $R^{\prime}$, we have $|A\cap D|\leq d+4-b$ and $|A\cap L|\geq b-2$. If $b\geq 4$ (resp. $b\geq 3$) we get $|A|<|B|$ (resp. $|A|\leq|B|$); we also conclude if $k\leq 2d-4$ or $|A\cap B|\geq 4$. Thus we may exclude these cases; we only need to assume that $S\nsubseteq D$ (i.e. $A\nsubseteq D$, i.e. $B\nsubseteq D$). From now on in this step (a2.1) we assume $2\leq b\leq 4$ and $A\nsubseteq D$. By Remark 2 there is a unique zero-dimensional scheme $Z_{1}\subset L$ such that $\deg(Z_{1})=b$ and $q^{\prime}\in\langle\nu_{d}(Z_{1})$. Set $Z_{2}:=Z_{1}\cup E$ and $Z:=Z_{2}\cup Z$. Since $Z_{1}$ is unique and we take $U$ general after fixing $q^{\prime}$, no line $F\subset\mathbb{P}^{n}$ spanned by $2$ points of $U$ meets $Z_{1}$. Hence $\deg(Z_{1}\cap F)\leq 2$ for each line $F\neq L$. Note that $\deg(Z)\leq 2k-d-2+2b\leq 3d-6+2b$. By [5, Lemma 1] we have $h^{1}(\cal{I}_{Z}(d))>0$. Consider the residual exact sequence of $R$ in $\mathbb{P}^{2}$:

[TABLE]

Since $B$ is a finite set, obviously $\mathrm{Res}_{R}(Z)=\mathrm{Res}_{R}(Z_{2})\cup\mathrm{Res}_{R}(B)$ and $\mathrm{Res}_{R}(B)$. We have $\mathrm{Res}_{R}(Z_{2})\subseteq Z_{1}\cup(U\setminus U\cap L)$. Since $U\cap L=\emptyset$, we have $U\setminus U\cap R=U\setminus U\cap D=A\setminus A\cap D=B\setminus B\cap D$. Since $|S\cap D|\geq 3$, $|E\cap R|\leq 1$ and $|U\cap R|\leq 2$, we have $|B\cap R|\geq d-1>1+|A\cap R|$. Since $\deg(Z)\leq 3d-6+2b$, $\deg(S\cap R)\geq d+2$ and $\deg(E\cap R)\leq 1$, we have $\deg(\mathrm{Res}_{R}(Z))\leq 2d-7+2b$. Thus the inequality $|B|\leq|A|$ gives $|B\cap L|\leq d-b$. Hence $\deg(Z\cap L)\leq d$. The generality of $U$ implies that no line intersects $\mathrm{Res}_{R}(Z)$ in a scheme of degree $\geq d+1$ and no conic intersects $\mathrm{Res}_{R}(Z)$ in a scheme of degree $\geq 2d-2$. By [11, Corollaire 2] we have $h^{1}(\cal{I}_{\mathrm{Res}_{R}(Z)}(d-1))=0$. By [6, Lemma 5.1] or [7, Lemma 2.5] we have $\mathrm{Res}_{R}(Z_{2})=B\setminus B\cap R$. Since $Z_{1}$ is not reduced, we get that $Z_{1}$ has a unique non-reduced connected component, that this connected component has degree $2$ and that $R$ meets it. The generality of $U$ gives $|U\cap R|\leq 1$. Since $\mathrm{Res}_{D}(Z_{2})=\mathrm{Res}_{D}(A)=U\setminus U\cap D$, we get $|B|>|A|$, a contradiction.

(a2.2) Assume $n>2$. Let $M\subset\mathbb{P}^{n}$ be a general hyperplane containing the plane $\langle D\rangle$. Thus $S\cap M=S\cap\langle D\rangle$. Since $U$ is general, no $4$ points of $U$ are coplanar. Thus $|U\cap M|=|U\cap\langle D\rangle|\leq 3$.

(a2.2.1) Assume $h^{1}(\cal{I}_{S\setminus S\cap M}(d-1))>0$. Since $|S\setminus S\cap M|\leq|A|+|B|-2d-2\leq 2(d-1)+1$, there is a line $R^{\prime}\subset\mathbb{P}^{n}$ such that $|R^{\prime}\cap(S\setminus S\cap M)|\geq d+1$. If $R^{\prime}\subset\langle D\rangle$, then $R^{\prime}\cup D$ is a plane cubic and we may apply step (a1). Thus we may assume $R^{\prime}\nsubseteq\langle D\rangle$. Let $N\subset\mathbb{P}^{n}$ be a general hyperplane containing $N$. Since $S$ is a finite set, the generality of $M$ and $N$ gives $S\cap(M\cup N)=S\cap(\langle D\rangle\cup R^{\prime})$. Consider the residual exact sequence

[TABLE]

of $M\cup N$ in $\mathbb{P}^{n}$. Since $|S\setminus S\cap(M\cup N)|\leq|A|+|B|-3d-3\leq d-1$, we have $h^{1}(\cal{I}_{S\setminus S\cap(M\cup N)}(d-2))=0$. Thus either [6, Lemma 5.1] or [7, Lemmas 2.4 and 2.5] give $A\setminus A\cap(M\cup N)=B\setminus B\cap(M\cup N)$. We have $A\cap(M\cup N)\subseteq E\cup(U\cap(M\cup N))$ and hence $|A\setminus A\cap(M\cup N)|\geq k-d-2+b-5$. Since $|A\cap(M\cup N)|\leq d+7-b$, we get $|B\cap(M\cup N)|\geq 3d+3-d-7+b=2d-4+b$. Since $|B\setminus B\cap(M\cup N)|\geq k-d-2+b-5$, we get a contradiction.

(a2.2.2) Assume $h^{1}(\cal{I}_{S\setminus S\cap M}(d-1))=0$. Either [6, Lemma 5.1] or [7, Lemmas 2.4 and 2.5] give $A\setminus A\cap M=B\setminus B\cap M$. Since $|U\cap M|=|U\cap\langle D\rangle|\leq 3$, we have $|U\setminus U\cap M|\geq k-d-5+b$. Assume for the moment $L\nsubseteq\langle D\rangle$. We get $|E\cap M|\leq 1$ and hence $|A\setminus A\cap M|\geq k-4$. Since $A\setminus A\cap M=B\setminus B\cap M$, we get $|S\cap M|\leq|A|+|B|-2k-8$ and hence $2d+2\leq 8$, a contradiction. Now assume $L\subset\langle D\rangle$. If $L\nsubseteq D$ we get (since $|L\cap D|\leq 2$) $|S\cap M|\geq 3d+b$. Since $A\setminus A\cap M=B\setminus B\cap M$ and $|U\setminus U\cap M|\geq k-d-1+b$, we get $|S|\geq 2d+2b+k-1$, a contradiction.

(a3) Assume the existence of a line $R\subset\mathbb{P}^{n}$ such that $|R\cap S|\geq d+2$. Since $U$ is a general subset of $\mathbb{P}^{n}$ with cardinality $k-d-2+b$, no $3$ of its points are collinear and $U\cap L=\emptyset$. Hence $|U\cap R|\leq 2$ and $U\cap L=\emptyset$. Let $M\subset\mathbb{P}^{n}$ 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 $S$, we have $M\cap S=R\cap S$. Consider the residual exact sequence (2) of $M$ in $\mathbb{P}^{n}$.

(a3.1) Assume $h^{1}(\cal{I}_{S\setminus S\cap M}(d-1))>0$. Since $|S\setminus S\cap M|\leq|A|+|B|-d-2\leq 3(d-1)-1$, either there is a line $R_{1}$ such that $|R_{1}\cap(S\setminus S\cap M)|\geq d+1$ or there is a conic $D_{1}$ such that $|D_{1}\cap(S\setminus S\cap M)|\geq 2d$. If $R$ and $R_{1}$ (resp. $R$ and $D_{1}$) are contained in a plane, and in particular if $n=2$, step (a2) (resp. step (a1)) gives a contradiction, because $|S\cap(R\cup R_{1})|\geq 2d+3$ (resp. $|S\cap(R\cup D_{1})|\geq 3d+2$). 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. $S\cap(M\cup N)=S\cap(R\cup\langle D_{1}\rangle)$.

(a3.1.1) Assume $h^{1}(\cal{I}_{S\setminus S\cap(M\cup N)}(d-2))>0$. We exclude the existence of $D_{1}$, because $|S\cap(R\cup D_{1})|\geq 3d+2$ and hence $|S\setminus S\cap(M\cup N)|\leq d-1$. Thus in this case we may assume the existence of $R_{1}$. Since $|S\cap(R\cup R_{1})|\geq 2d+3$, we have $|S\setminus S\cap(M\cup N)|\leq|A|+|B|-2d-3\leq 2(d-2)+1$. By [8, Lemma 34] there is a line $R_{2}$ such that $|R_{2}\cap S\setminus S\cap(M\cup N)|\geq d$. Let $M^{\prime}$ be a general hyperplane containing $R_{2}$. Consider the residual exact sequence of $M^{\prime}\cup M\cup N$. We have $h^{1}(\cal{I}_{S\setminus S\cap(M\cup N\cup M^{\prime})}(d-3))=0$, because $|S\setminus S\cap(M\cup N\cup M^{\prime})|\leq 2k-d-2-d-1-d\leq d-4$. Either [6, Lemma 5.1] or [7, Lemmas 2.4 and 2.5] give $A\setminus A\cap(M\cup N\cup M^{\prime})=B\setminus B\cap(M\cup N\cup M^{\prime})$. 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 $|U\cap(R\cup R_{1}\cup R_{2})|\leq 6$. Hence $|A\setminus A\cap(M\cup N\cup M^{\prime})|\geq k-d-8+b$. Since $A\setminus A\cap(M\cup N\cup M^{\prime})=B\setminus B\cap(M\cup N\cup M^{\prime})$, we get $|S\cap(M\cup N\cup M^{\prime})|\leq 2k-2k+2d+16-2b$. Hence $2d+16-2b\geq 3d+3$, a contradiction.

(a3.1.2) Assume $h^{1}(\cal{I}_{S\setminus S\cap(M\cup N)}(d-2))=0$. Either [6, Lemma 5.1] or [7, Lemmas 2.4 and 2.5] give $A\setminus A\cap(M\cup N)=B\setminus B\cap(M\cup N)$. Since $U\cap(M\cup N)=U\cap(R\cup R_{1})$, we have $|U\setminus U\cap(M\cup N)|\geq k-d-6+b$. Assume for the moment $L\notin\{R,R_{1}\}$. We get $|L\cap(M\cup N)|\leq 2$. Thus $|A\setminus A\cap(M\cup N)|\geq k+b-8$. Since $A\setminus A\cap(M\cup N)=B\setminus B\cap(M\cup N)$, we get $|S\cap(M\cup N)|\leq 16-b<2d+3$ (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 (a2.1).

(a3.2) Assume $h^{1}(\cal{I}_{S\setminus S\cap M}(d-1))=0$. Either [6, Lemma 5.1] or [7, Lemmas 2.4 and 2.5] give $A\setminus A\cap M=B\setminus B\cap M$.

(a3.2.1) Assume $R=L$. We get $U=A\setminus A\cap L=B\setminus B\cap L$. Thus $B=U\cup(B\cap L)$. Since $\langle\nu_{d}(U)\rangle\cap\langle Y\rangle=\emptyset$, $q\in\langle\nu_{d}(U)\cup Y\rangle$, $q\notin\langle\nu_{d}(U)\rangle$, $q\notin\langle\nu_{d}(Y)\rangle$ (because $U\neq\emptyset$) and $\langle\nu_{d}(U)\rangle\cap\langle Y\rangle=\emptyset$, there are uniquely determined $q_{1}\in\langle\nu_{d}(U)\rangle$ and $q_{2}\in\langle Y\rangle$ such that $q\in\langle\{q_{1},q_{2}\}\rangle$. The uniqueness of $q_{2}$ gives $q_{2}=q^{\prime}$. Since $\langle\nu_{d}(U)\rangle\cap\langle Y\rangle=\emptyset$ and $q\in\langle\nu_{d}(A)\rangle\cap\langle\nu_{d}(B)\rangle$, we get $q^{\prime}\in\langle\nu_{d}(B\cap L)\rangle$. Thus $|B\cap L|\geq r_{Y}(q^{\prime})=|A\cap L|$. Since $|B|\leq|A|$ and $A\setminus A\cap L=B\setminus B\cap L$, we get $|B|=|A|$ and $B=U\cup F$ with $F\cap U=\emptyset$ and $F\in\cal{S}(Y,q^{\prime})$. Thus the theorem is true in this case.

(a3.2.2) Assume $R\neq L$. Since $|L\cap R|\leq 1$, we get $|E\cap R|\leq 1$. Since $|U\cap R|\leq 2$, we get $|A\cap R|\leq 3$ and hence $|B\cap R|\geq d-1>|A\cap R|$. Since $A\setminus A\cap R=B\setminus B\cap R$, we get $|B|>|A|$, a contradiction.

(b) By step (a) we always have $r_{X}(q)=r_{Y}(q)$ and hence $\cal{S}(X,q)\supseteq\{U\cup E\}_{E\in\cal{S}(L,q^{\prime})}$, in which we write $\cal{S}(L,q^{\prime})$ for the set of all $E\subset L$ such that $\nu_{d}(E)\in\cal{S}(Y,q^{\prime})$. Note that $U=\cap_{E\in\cal{S}(L,q^{\prime})}\{U\cup E\}_{E\in\cal{S}(L,q^{\prime})}$. Thus $U$ is uniquely determined by $\cal{S}(Y,q^{\prime})$. By step (a) the theorem is true if $k\leq 2d-3$. Thus we may assume $k=2d-2$. We fix $E\in\cal{S}(L,q^{\prime})$ and $U\in\mathbb{P}^{n}$ such that $U\cap L=\emptyset$, $U$ has general Hilbert function and set $A:=U\cup E\in\cal{S}(X,q)$. We only need to prove that $L$ is the only line $J$ such that there are $q^{\prime\prime}\in\langle\nu_{d}(J)\rangle$ and infinitely many $S_{\alpha}\in\cal{S}(X,q)$, $\alpha\in\Gamma$, $\Gamma$ an integral quasi-projective curve, $S_{\alpha}=U^{\prime}\cup E_{\alpha}$ for all $\alpha\in\Gamma$, $E_{\alpha}\subset J$, $|E_{\alpha}|=d+2-b$, $\langle\nu_{d}(U^{\prime})\rangle\cap\langle\nu_{d}(J)\rangle=\emptyset$, $E_{\alpha}\in\cal{S}(\nu_{d}(J),q^{\prime\prime})=\cal{S}(X,q^{\prime\prime})$ (by Autarky), $\cap_{\alpha}\langle\nu_{d}(E_{\alpha})\rangle=\{q^{\prime\prime}\}$ and $J\neq L$. Since $\cap_{\alpha}\langle\nu_{d}(E_{\alpha})\rangle=\{q^{\prime}\}\notin X$, we have $\cap_{\alpha}E_{\alpha}=\emptyset$. Thus there is $E_{\alpha}$ such that $E_{\alpha}\cap A=\emptyset$ and we call $F$ this set $E_{\alpha}$. Set $B:=U^{\prime}\cup F$ and $S:=A\cup B$. Thus $h^{1}(\cal{I}_{S}(d))>0$. Since $J\neq L$, we have $|J\cap L|\leq 1$ and equality holds if $n=2$. Set $D:=J\cup L$. We have $|U|=|U^{\prime}|=d-4+b$. If $b=2$ we assume until step (b5) that $|U^{\prime}\cap L|\leq 1$ and $|U\cap J|\leq 1$.

(b1) Assume $n=2$. Consider the residual exact sequence of (3). First assume $h^{1}(\cal{I}_{S\setminus S\cap D}(d-2))=0$. By either [6, Lemma 5.1] or [7, Lemmas 2.4 and 2.5] we have $A\setminus A\cap D=B\setminus B\cap D$. At most two points of $U^{\prime}$ are contained in $L$ and the same number of points are contained in $J$. Since $h^{1}(\cal{I}_{U\cup E}(t))=0$ for all $t\geq d+1-b$, we get $h^{1}(\cal{I}_{S\setminus S\cap J}(d-1))=0$ and so $A\setminus A\cap J=U^{\prime}$, contradicting the inequality $|E\cap J|\leq|L\cap J|=1$. Thus $h^{1}(\cal{I}_{S\setminus S\cap D}(d-2))>0$. Since $S\setminus S\cap D\subseteq U\cup U^{\prime}$, $|U\cup U^{\prime}|\leq 2d-8+2b$, at most $2$ (resp. $5$, resp. $9$) points of $U$ or $U^{\prime}$ are contained in a line (resp. a conic, resp. a cubic), we get a contradiction.

(b2) Assume $n>2$ and $J\cap L\neq\emptyset$. Let $N:=\langle L\cup J\rangle$ be the plane spanned by the conic $L\cup J$. Let $M\subset\mathbb{P}^{n}$ be a general hyperplane containing $N$. Since $S$ is a finite set and $M$ is general, we have $S\cap M=S\cap N$. Note that $|U\cap N|\leq 3$ and $|U^{\prime}\cap N|\leq 3$. We use the residual exact sequence (3) of $M$ in $\mathbb{P}^{n}$. First assume $h^{1}(\cal{I}_{S\setminus S\cap M}(d-1))=0$. We get $h^{1}(M,\cal{I}_{S\cap M,M}(d))>0$. Since $S\cap M=S\cap N$, we have $h^{1}(N,\cal{I}_{S\cap N}(d))>0$. Since $|U\cap N|\leq 3$ and $|U^{\prime}\cap N|\leq 3$, as in step (b1) the residual exact sequence of $J$ in $N$ gives a contradiction. Now assume $h^{1}(\cal{I}_{S\setminus S\cap M}(d-1))>0$. Since $S\setminus S\cap M\subseteq U\cup U^{\prime}$ and $|U\cup U^{\prime}|\leq 2d-8+2b$, we get a contradiction by Lemma 1.

(b3) Assume $J\cap L=\emptyset$ and $n=3$. Let $Q\subset\mathbb{P}^{3}$ be a general quadric surface containing $I\cup J$. Since $\cal{I}_{J\cup L}(2)$ is globally generated, we have $Q\cap S=S\cap(I\cup J)$. Our assumption on $|U\cap J|$ and on $|U^{\prime}\cap L|$ gives $h^{1}(Q,\cal{I}_{S\cap Q}(d))=0$. The residual exact sequence of $Q$ in $\mathbb{P}^{3}$ gives $h^{1}(\cal{I}_{S\setminus S\cap Q}(d-2))>0$. Since $S\setminus S\cap Q\subseteq U\cup U^{\prime}$ and both $U$ and $U^{\prime}$ are general (each of them have no $3$ points collinear, no $6$ points in a conic), Lemma 1 gives a contradiction.

(b4) Assume $J\cap L=\emptyset$ and $n>3$. We use induction on $n$, assuming the result in $\mathbb{P}^{n-1}$. We have $\dim\langle J\cup L\rangle=3$. Take a general hyperplane $M\subset\mathbb{P}^{n}$ containing $\langle J\cup J\rangle$ and mimic the proof of step (b2).

(b5) Now assume $b=2$ and that either $|U\cap J|\geq 2$ or $|U^{\prime}\cap L|\geq 2$. Since $U$ and $U^{\prime}$ have general Hilbert function, each of the occurring inequalities is an equality. Instead of $E$ we take the unique degree $2$ scheme $Z_{1}\subset L$ evincing the cactus rank of $q^{\prime}$ and $Z_{2}$ for the corresponding one for the point $q^{\prime\prime}$ such that $\{q^{\prime\prime}\}:=\langle\nu_{d}(J)\rangle\cap\langle\nu_{d}(U)\cup\{q\}\rangle$. Set $Z:=Z_{1}\cup Z_{2}\cup U\cup U^{\prime}$. We have $\deg(Z)\leq|U|+|U^{\prime}|+4\leq 2d+4$. By [5, Lemma 1] we have $h^{1}(\cal{I}_{Z}(d))>0$. Since $\deg(Z)<3d$, either there is a line $R$ such that $\deg(R\cap Z)\geq d+2$ or there is a conic $D^{\prime}$ such that $\deg(D^{\prime}\cap Z)\geq 2d+2$ (if $n=2$ this is true by [11, Corollaire 2]; if $n>2$ we take a general hyperplane containing $L$ and get a contradiction). Both possibilities are excluded by the assumptions on $U$ and $U^{\prime}$ (at most $2$ points of $U^{\prime}$ in a line and none in $J$, at most $5$ points in a conic). ∎

4. Irredundantly spanning sets

Lemma 2**.**

If $r+1-\dim X\leq t\leq r$, then $\cal{S}(X,q,t)\neq\emptyset$.

Proof.

The case $t=r+1-\dim X$ is an obvious consequence of the proof of [14, Proposition 5.1]. Assume $r+2-\dim X\leq t\leq r$. Let $Y\subset\mathbb{P}^{r}$ 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 $\mathbb{P}^{r}$. Thus for any $q$ we have $\cal{S}(X,q,t)\supseteq\cal{S}(Y,q,t)$. Since $t=r+1-\dim Y$, we get $\cal{S}(Y,q,t)\neq\emptyset$. ∎

Remark 3**.**

Let $X\subset\mathbb{P}^{d}$, $d\geq 4$, be a rational normal curve. Fix $q\in\mathbb{P}^{d}$ 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 $\cal{S}(X,q,t)=\emptyset$ for all $t$ such that $3\leq t\leq d-1$.

Proof of Proposition 1:.

Since a finite intersection of non-empty Zariski open subsets of $\mathbb{P}^{r}$ is open and non-empty and the interval $\lfloor(r+2)/2\rfloor\leq t\leq r$ contains only finitely many integers, it is sufficient to prove the statement for a fixed $t$. The case $t=r$ is true by Lemma 3. The case $r$ even at $t=r/2+1$ is true by Theorem 2. Hence all cases for $r=4$ are true. Thus we may assume $r\geq 5$ and that the proposition is true for all curves in a lower dimensional projective space. If $r$ is even we may also assume $t\neq r/2+1$. Fix a general $p\in X$ and call $\ell:\mathbb{P}^{r}\setminus\{p\}\xrightarrow{\ \ }\mathbb{P}^{r-1}$ the linear projection from $p$. Let $Y\subset\mathbb{P}^{r-1}$ be the closure of $\ell(X\setminus\{p\})$ in $\mathbb{P}^{r-1}$. $Y$ is an integral and non-degenerate curve. Since $p$ is general in $X$, it is a smooth point of $X$ and hence $\ell_{|X\setminus\{p\}}$ exstend to a surjective morphism $\mu:X\xrightarrow{\ \ }Y$ with $\mu(p)$ associated to the tangent line of $X$ at $p$. Thus $Y=\mu(Y)$. By the trisecant lemma ([15, Corollary 2.2]) and the generality of $p$ we have $\deg(L\cap X)\leq 2$ for every line $L\subset\mathbb{P}^{r}$ such that $p\in L$. Hence $\ell_{|X\setminus\{p\}}$ is birational onto its image and there a finite set $F\subset X$ containing $p$ such that $\mu_{|X\setminus F}$ induces an isomorphism between $X\setminus F$ and $Y\setminus\mu(F)$. Fix the integer $t$ such that $\lfloor(r+2)/2\rfloor\leq t\leq r$ 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 $\mathbb{P}^{r-1}$ there is a non-empty open subset $\cal{V}$ of $\mathbb{P}^{r-1}$ such that $W(Y)_{q,z}=\{q\}$ for all $q\in\cal{V}$. Fix $a\in\cal{V}$ and finitely many $S_{i}\in\cal{S}(Y,a,z)$, $1\leq i\leq e$, such that $\{a\}=\cap_{i=1}^{e}\langle S_{i}\rangle$. Restricting if necessary $\cal{V}$ we may assume that (for a choice of sufficiently general $S_{1}(a),\dots,S_{e}(a)$) we have $S_{i}(a)\cap\mu(F)=\emptyset$ for all $i$ and all $a$. Hence there is a unique $A_{i}(a)\subset X\setminus F$ such that $\mu(A_{i}(a))=S_{i}(a)$. Since $p\in F$, $B_{i}(a):=A_{i}(a)\cup\{p\}$ has cardinality $t$, $1\leq i\leq e$. Set $\cal{U}_{p}:=\ell^{-1}(\cal{V})\subset\mathbb{P}^{r}\setminus\{p\}$. For each $a\in\cal{V}$, set $L_{a}:=\{p\}\cup\ell^{-1}(a)$. Each $L_{a}$ is a line containing $p$, $\cal{U}_{p}$ is the union of all $L_{a}\setminus\{p\}$, $a\in\cal{V}$, and $L_{a}=\cap_{i=1}^{e}\langle B_{i}(a)\rangle$. Fix $a\in\cal{V}$ and $b\in L_{a}\setminus\{p\}$. Note that each $B_{i}(a)$ irredundantly spans $b$. Fix another general $o\in X$, $o\neq p$. We get in the similar way a set $\cal{U}_{o}$. It is easy to check that $W_{q,t}=\{q\}$ for all $q\in\cal{U}_{o}\cap\cal{U}_{p}$. Thus we may take $\cal{U}=\cal{U}_{p}\cap\cal{U}_{o}$.∎