Dependent subsets of embedded projective varieties

TL;DR

This paper investigates the properties of embedded projective varieties, establishing bounds on the maximal size of zero-dimensional schemes that are linearly independent within the variety, with implications for linear normality.

Contribution

It provides new bounds on the maximal size of zero-dimensional schemes that are smoothable and linearly independent in embedded projective varieties, linking these bounds to the variety's dimension and embedding dimension.

Findings

If $ ho (X)'' extgreater= ceil (r+2)/2 ceil$, then $X$ is linearly normal.

The value of $ ho (X)''$ is strictly less than $2 ext{ceil}((r+1)/(n+1))$, except when $n=r$ or $X$ is a rational normal curve.

Abstract

Let $X\subset \mathbb {P}^r$ be an integral and non-degenerate variety. Set $n:= \dim (X)$. Let $\rho (X)''$ be the maximal integer such that every zero-dimensional scheme $Z\subset X$ smoothable in $X$ is linearly independent. We prove that $X$ is linearly normal if $\rho (X)''\ge \lceil (r+2)/2\rceil$ and that $\rho (X)'' < 2\lceil (r+1)/(n+1)\rceil$, unless either $n=r$ or $X$ is a rational normal curve.

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 defined over an algebraically closed field with characteristic zero. Set $n:=\dim X$. We recall that a zero-dimensional scheme $Z\subset X$ is said to be smoothable in $X$ if it is a flat limit of a family of finite subsets of $X$ with cardinality $\deg(Z)$ (see [5] for a discussion of it). If $X$ is smooth (or if $Z$ is contained in the smooth locus of $X$) $Z$ is smoothable in $X$ if and only if it is smoothable in $\mathbb{P}^{r}$ and the notion of smoothability in $\mathbb{P}^{r}$ does not depend on the choice of the embedding of $Z$ in a projective space ([5, Proposition 2.1]). Let $\rho(X)$ (resp. $\rho(X)^{\prime}$, resp. $\rho(X)^{\prime\prime}$) denote the maximal integer $t>0$ such that each zero-dimensional scheme (resp. each finite set, resp. each zero-dimensional scheme smoothable in $X$) $Z\subset X$ with $\deg(Z)=t$ is linearly independent. Obviously $\rho(X)\leq\rho(X)^{\prime\prime}\leq\rho(X)^{\prime}$. Since $X$ is embedded in $\mathbb{P}^{r}$, we have $\rho(X)\geq 2$. The integers $\rho(X)$, $\rho(X)^{\prime}$ and $\rho(X)^{\prime\prime}$ have been used in several papers connected to $X$-rank, the symmetric tensor rank, i.e. the additive decomposition of polynomials, and the tensor rank ([5], [6]).

For any $q\in\mathbb{P}^{r}$ the $X$-rank $r_{X}(q)$ of $q$ is the minimal positive integer $t$ such that $q\in\langle S\rangle$ for some finite subset $S\subset X$ with $\sharp(S)=t$, where $\langle\ \ \rangle$ denote the linear span. For any positive integer $t$ the $t$-secant variety $\sigma_{t}(X)$ of $X$ is the closure in $\mathbb{P}^{r}$ of the union of all $\langle S\rangle$ with $S$ a finite subset of $X$ with cardinality $t$.

The border $X$-rank $b_{X}(q)$ of $q\in\mathbb{P}^{r}$ is the minimal integer $k$ such that $q\in\sigma_{k}(X)$. The generic rank $r_{X,\mathrm{gen}}$ is the minimal integer $k>0$ such that $\sigma_{k}(X)=\mathbb{P}^{r}$. There is a non-empty open subset $U\subset\mathbb{P}^{r}$ such that $r_{X}(q)=r_{X,\mathrm{gen}}$ for all $q\in U$.

In this paper we prove that if $\rho(X)^{\prime\prime}$ is large, then $X$ is linearly normal and that $\rho(X)^{\prime\prime}$ cannot be very large for $n>1$ (Theorem 1.1).

We prove the following results.

Proposition 1.1**.**

Assume that $X$ is a curve and $\rho(X)^{\prime\prime}\geq\lceil(r+2)/2\rceil$. Then $X$ is linearly normal.

Proposition 1.2**.**

Assume $n:=\dim X\geq 2$, $r_{X,\mathrm{gen}}=\lceil(r+1)/(n+1)\rceil$ and $\rho(X)^{\prime\prime}>\lceil(r+1)/(n+1)\rceil$. Then $X$ is linearly normal.

Theorem 1.3**.**

Let $X\subset\mathbb{P}^{r}$ be an integral and non-degenerate variety. Set $n:=\dim X$. We have $\rho(X)^{\prime\prime}\geq 2\lceil(r+1)/(n+1)\rceil$ if and only if either $r=n$ (i.e. $X=\mathbb{P}^{r}$) or $n=1$, $r$ is odd and $X$ is a rational normal curve.

If $n=r$ we have $\rho(X)^{\prime}=\rho(X)=2$. If $X$ is a rational normal curve we have $\rho(X)=\rho(X)^{\prime}=r+1$. This is the only case with $\rho(X)^{\prime}=r+1$ (Lemma 2.3). Theorem 1.2 implies that $\rho(X)^{\prime\prime}<2\lceil(r+1)/(n+1)\rceil$ if $(r+1)/(n+1)\notin\mathbb{Z}$.

The example of a general linear projection in $\mathbb{P}^{4}$ of the Veronese surface shows that in Proposition 1.2 it is not sufficient to assume that $\rho(X)^{\prime\prime}\geq\lceil(r+2)/(n+1)\rceil$.

We point out that to get our results we only use a small family of zero-dimensional schemes, each of them with connected components of degree $1$ or $2$, but that this family contains a complete family covering $X$: each $p\in X$ is contained in some scheme $Z$ of the family.

2. Preliminaries

We recall that if $q\in\sigma_{k}(X)$ and $k\leq\rho(X)^{\prime\prime}$ there is a zero-dimensional scheme $Z\subset X$ smoothable in $X$ and such that $\deg(Z)\leq b$ and $q\in\langle Z\rangle$ ([6, Lemma 2.6, Theorem 1.18] and [5, Proposition 2.5]).

For any $q\in\mathbb{P}^{r}$ let $\cal{S}(X,q)$ be the set of all $S\subset X$ such that $|S|=r_{X}(q)$ and $q\in\langle S\rangle$.

Remark 2.1**.**

Let $X\subset\mathbb{P}^{r}$ be a smooth variety with $\dim X\leq 2$. Every zero-dimensional scheme of $X$ is smoothable ([8]) and hence $\rho(X)^{\prime\prime}=\rho(X)$. Easy examples show that we may have $\rho(X)<\rho(X)^{\prime}$ for a smooth curve (Examples 4.2 and 4.3).

Remark 2.2**.**

([6, Theorem 1.17]) Fix $q\in\mathbb{P}^{r}$ and $A,B\in\cal{S}(X,q)$. Set $x:=r_{X}(q)$ and assume $\rho(X)^{\prime}\geq 2a$. Since $|A\cup B|\leq 2a$, $A\cup B$ is linearly independent. Thus $A=B$ ([4, Lemma 1]). Hence $|\cal{S}(X,q)|=1$.

The following extremal case is the only result in which we are able to use only $\rho(X)^{\prime}$ instead of $\rho(X)^{\prime\prime}$.

Lemma 2.3**.**

The following conditions are equivalent;

1. (1)

$X$* is a rational normal curve;* 2. (2)

$\rho(X)^{\prime}=r+1$; 3. (3)

$\rho(X)^{\prime}>r$;

Proof.

It is sufficient to prove that (3) implies (1).

First assume $n=1$. Let $H\subset\mathbb{P}^{r}$ be a general linear hyperplane. Since $X\cap H$ is formed by $\deg(X)$ points, if $\rho(X^{\prime})>r$ we have $\deg(X)=r$ and hence $X$ is a rational normal curve.

Now assume $n>1$. Take a general linear subspace $V\subset\mathbb{P}^{r}$ with codimension $n-1$. The scheme $X\cap V$ is an integral curve spanning $V$. We have $\rho(X)^{\prime}\leq\rho(X\cap V)^{\prime}\leq r-n+1$ by the case $n=1$ just proved. ∎

3. The proofs

Proof of Proposition 1.1:.

Assume that $X$ is not linearly normal. Thus there is a non-degenerate variety $Y\subset\mathbb{P}^{r+1}$ such that $X$ is an isomorphic linear projection of $Y$ from some $o\in\mathbb{P}^{r+1}\setminus Y$. Set $b:=b_{Y}(o)$. Each secant variety of a curve has the expected dimension ([1, Remark 1.6]). Thus $r_{Y,\mathrm{gen}}=\lceil(r+2)/2\rceil$. Hence $b\leq\lceil(r+2)/2\rceil$. Let $\ell:\mathbb{P}^{r+1}\setminus\{o\}\xrightarrow{\ \ }\mathbb{P}^{r}$ denote the linear projection from $o$. By assumption $o\notin Y$ and $\ell_{|Y}$ is an embedding with $\ell(Y)=X$. Let $W\subset Y$ be any zero-dimensional scheme. Since $\ell_{|Y}:Y\xrightarrow{\ \ }X$ is an isomorphism, $W$ is smoothable in $Y$ if and only if $\ell(W)$ is smoothable in $X$ and any degree $b$ smoothable zero-dimensional scheme is the image of a unique degree $b$ zero-dimensional scheme. Thus $\rho(Y)^{\prime\prime}\geq\rho(X)^{\prime\prime}$. The image in $\mathbb{P}^{r}$ of a linear subspace $V\subset\mathbb{P}^{r+1}$ has either dimension $\dim V$ (case $o\notin V$) or dimension $\dim V-1$ (case $o\in V$). Since $\rho(Y)^{\prime\prime}\geq\rho(X)^{\prime\prime}\geq\lceil(r+2)/2\rceil=r_{Y,\mathrm{gen}}$ and $b\leq r_{Y,\mathrm{gen}}$, there is a smoothable zero-dimensional scheme $W\subset Y$ such that $o\in\langle W\rangle$ and $\deg(W)=b$. Since $\ell(W)$ is not linearly independent, we have $\rho(X)^{\prime\prime}\leq b-1$, a contradiction. ∎

Proof of Proposition 1.2:.

Assume that $X$ is not linearly normal. Thus there is a non-degenerate variety $Y\subset\mathbb{P}^{r+1}$ such that $X$ is an isomorphic linear projection of $Y$ from some $o\in\mathbb{P}^{r+1}\setminus Y$. Set $b:=b_{Y}(o)$ and $a:=r_{X,\mathrm{gen}}=\lceil(r+1)/(n+1)\rceil$. Let $\ell:\mathbb{P}^{r+1}\setminus\{o\}\xrightarrow{\ \ }\mathbb{P}^{r}$ denote the linear projection from $o$. By assumption $o\notin Y$ and $\ell_{|Y}$ is an embedding with $\ell(Y)=X$. As in the proof of Proposition 1.1 we have $\rho(Y)^{\prime\prime}\geq\rho(X)^{\prime\prime}$ and to get a contradiction it is sufficient to prove that $b\leq\rho(X)^{\prime\prime}$. Assume $b>\rho(X)^{\prime\prime}$, i.e. assume $b\geq a+2$. Since $b>a$, we have $o\notin\sigma_{a}(Y)$. Hence $\ell_{|\sigma_{a}(Y)}:\sigma_{a}(Y)\xrightarrow{\ \ }\mathbb{P}^{r}$ is a finite map. Since $\ell(\sigma_{a}(Y))=\sigma_{a}(X)$, we get $\dim\sigma_{a}(Y)=r$. Since $\dim\sigma_{a+1}(Y)>\dim\sigma_{a}(Y)$ ([1, Proposition 1.3]), we get $\sigma_{a+1}(Y)=\mathbb{P}^{r+1}$. Thus $b\leq a+1$, a contradiction.∎

Lemma 3.1**.**

Assume $\rho(X)^{\prime\prime}\geq 2\lceil(r+1)/(n+1)\rceil$. Then $X$ is not defective, $r+1\equiv 0\pmod{n+1}$ and for a general $q\in\mathbb{P}^{r}$ we have $|\cal{S}(X,q)|=1$.

Proof.

Set $a:=\lceil(r+1)/(n+1)\rceil$. By Terracini’s lemma ([1, Corollary 1.11]) to prove that $X$ is not defective and that $r+1\equiv 0\pmod{n+1}$ it is sufficient to prove that $\dim\langle T_{p_{1}}X\cup\cdots\cup T_{p_{a}}X\rangle=a(n+1)-1$ for a general $(p_{1},\dots,p_{a})\in X^{a}$. Since each $p_{i}$ is a smooth point of $X$, each degree $2$ connected zero-dimensional scheme $v_{i}\subset X$ such that $(v_{i})_{\mathrm{red}}=\{p_{i}\}$ is smoothable. Apply [3, Lemma 1]. Thus $X$ is not defective. Since $r+1=a(n+1)$ and $X$ is not defective, we have $\sigma_{a}(X)=\mathbb{P}^{r}$ and $\cal{S}(X,q)$ is finite for a general $q\in\cal{S}(X,q)$. Fix any $q\in\mathbb{P}^{r}$ such that $x:=r_{X}(q)$. We have $|\cal{S}(X,q)|=1$ if $\rho(X)^{\prime}\geq 2x$ by Remark 2.2. We proved that $a\geq x$. ∎

The (smooth) $n$-dimensional varieties $X\subset\mathbb{P}^{2n+1}$ such that $\sigma_{2}(X)=\mathbb{P}^{2n+1}$ and $|\cal{S}(X,q)|=1$ are classically called OADP (or varieties with only one apparent double point), because projecting them from a general point of $X$ one gets a variety with a unique singular point ([7]). They are always linearly normal ([7, Remark 1.2]. In [7] there are also older references and the classification of the smooth ones with dimension up to $3$ ([10], [7, Theorem 7.1]). Thus the thesis of Lemma 3.1 is a generalization of this concept to the case in which $(r+1)/(n+1)$ is an integer $>2$. But the assumption “$\rho(X)^{\prime\prime}\geq 2\lceil(r+1)/(n+1)\rceil$ ” of the lemma is too strong to be interesting for the classification of extremal varieties. Just assuming $\rho^{\prime}(X)>2$ excludes all $X$ containing lines and hence all smooth OADP’s of dimension $2$ and $3$.

Corollary 3.2**.**

Assume $n:=\dim X\geq 2$ and $\rho(X)^{\prime\prime}\geq 2\lceil(r+1)/(n+1)\rceil$. Then $X$ is linearly normal, non-defective, $r\equiv 0\pmod{n+1}$, $\rho(X)^{\prime\prime}=2(r+1)/(n+1)$ and $|\cal{S}(X,q)|=1$ for a general $q\in\mathbb{P}^{r}$.

Proof of Corollary 3.2:.

By Proposition 1.2 it is sufficient to prove that $X$ is non-defective. Apply Lemma 3.1. ∎

Proof of Theorem 1.3:.

Assume the existence of $X$ with $\rho(X)^{\prime\prime}\geq 2\lceil(r+1)/(n+1)\rceil$. We may assume $n<r$, i.e. $X\neq\mathbb{P}^{n}$.

First assume $n=1$. Lemma 2.3 gives that $X$ is a rational normal curve, that $r$ is odd and that $\rho(X)=\rho(X)^{\prime}=r+1$.

Now assume $n\geq 2$. By Lemma 3.1 $a:=(r+1)/(n+1)$ is an integer. We may assume $a\geq 2$, i.e. $r\neq n$. Fix a general $S\subset X$ such that $|S|=a-1$. Since $S$ is general, each $p\in S$ is a smooth point of $X$. We saw in the proof of Proposition 1.2 that $V:=\langle\cup_{p\in S}T_{p}X\rangle$ has dimension $(a-1)(n+1)-1$. Fix $o\in X\setminus S$.

Claim 1: $o\notin V$.

Proof of Claim 1: Assume $o\in V$. We saw in the proof of Proposition 1.2 that there are connected degree $2$ zero-dimensional schemes $v_{p}\subset X$ such that $(v_{p})_{\mathrm{red}}\{p\}$ and $o\in\langle Z\rangle$, where $Z:=\cup_{p\in S}v_{p}$. Since $o\notin S$, we have $o\nsubseteq Z$. Thus the scheme $Z\cup\{o\}$ is linearly dependent. Since $\deg(Z\cup\{o\})=2a-1<\rho(X)^{\prime\prime}$ and $Z\cup\{o\}$ is smoothable, we got a contradiction. Let $\ell:\mathbb{P}^{r}\setminus V\xrightarrow{\ \ }\mathbb{P}^{n}$ denote the linear projection from $V$. By Claim 1 $\ell_{1}:X\setminus S\xrightarrow{\ \ }\mathbb{P}^{n}$ is a morphism. Fix $o\in X\setminus S$ and assume the existence of $o^{\prime}\in X\setminus S$ such that $o\neq o^{\prime}$ and $\ell_{1}(o)=\ell_{1}(o^{\prime})$. Thus $o^{\prime}\in\langle\{o\}\cup Z\rangle$. Hence $\{o,o^{\prime}\}\cup Z$ is linearly dependent. The zero-dimensional scheme $\{o,o^{\prime}\}\cup Z$ is smoothable and it has degree $2a\leq\rho(X)^{\prime\prime}$, a contradiction.

Thus $\ell_{1}:X\setminus S\xrightarrow{\ \ }\mathbb{P}^{n}$ is an injective morphism between two quasi-projective varieties. Since $\mathbb{P}^{n}$ is smooth (it would be sufficient to assume that the target, $X$, is normal or even less (weakly normal)) and we are in characteristic zero, $\ell_{1}$ is an open map which is an isomorphism onto its image ([9]). Since $X$ is smooth at each point of $S$, $X$ is smooth. Hence the map $\ell_{1}:X\setminus S\xrightarrow{\ \ }\mathbb{P}?n$ extends over $S$. Since $X$ is smooth at each point of $S$, $\ell_{1}$ lifts to a morphism $u:\tilde{X}\xrightarrow{\ \ }X$, where $\tilde{X}$ is the blowing-up of $X$ at all points of $S$. We first get that $X\cong\mathbb{P}^{n}$ as an abstract variety and then (since any morphism $\mathbb{P}^{n}\xrightarrow{\ \ }\mathbb{P}^{n}$ injective outside a finite set is an isomorphism) that $u$ does not exist when $n>1$.∎

4. Elementary examples

By Proposition 1.1 to complete the picture for curves we need to describe the linearly normal curves with very high $\rho(X)^{\prime}$, $\rho(X)$ and $\rho(X)^{\prime\prime}$.

Remark 4.1**.**

Let $X$ be an integral projective curve. To compute $\rho(X)^{\prime\prime}$ we recall that every Cartier divisor of $X$ is smoothable. Let $\cal{F}$ be any torsion free sheaf of $X$. Duality gives $h^{1}(\cal{F})=\dim\mathrm{Hom}(\cal{F},\omega_{X}))$, ([2, 1.1 at p. 5]). Thus for any zero-dimensional scheme $Z\subset X$ we have $h^{1}(\cal{I}_{Z}(1))=\dim\mathrm{Hom}(\cal{I}_{Z},\omega_{X}(1))$. If $d\geq 3g-2$ we have $\rho(X)=d-2g+2$. We have $\rho(X)^{\prime}=d-2g+2$ if and only if $X$ is Gorenstein, i.e. $\omega_{X}$ is locally free. For lower $d$ the integers $\rho(X)$, $\rho(X)^{\prime}$ and $\rho(X)^{\prime\prime}$ depends both from the Brill-Noether theory of the special line bundles on $X$ and the choice of the very ample line bundle $\cal{O}_{X}(1)$, not just the integers $d$ and $g$.

Example 4.2**.**

Fix integers $r,a$ such that $2\leq a\leq r+1$. Here we prove the existence of a smooth and non-degenerate curve $X\subset\mathbb{P}^{r}$ such that $\rho(X)^{\prime}=\rho(X)=a$. If $a=r+1$ we know that $X$ is a rational normal curve. See the case $g=1$ and $d=r+1$ of Remark 4.1 for the case $a=r$. Now assume $2\leq a\leq r-1$.

(a) We first cover the case $2a\leq r+1$. In this range we construct a smooth rational curve $X$ with $\rho(X)=\rho(X)^{\prime}=a$, but of course $X$ is not linearly normal. Let $Y\subset\mathbb{P}^{r+1}$ be a rational normal curve. Fix a set $S\subset X$ such that $|S|=a+1$ and take any $o\in\langle S\rangle$ such that $o\notin\langle S^{\prime}\rangle$ for any $S^{\prime}\subsetneq S$. Let $\ell:\mathbb{P}^{r+1}\setminus\{o\}\xrightarrow{\ \ }\mathbb{P}^{r}$ denote the linear projection from $o$. Since $a\geq 2$ and $\rho(Y)=r+2$, we have $o\notin Y$. Hence $\ell_{|Y}$ is a morphism. Set $X:=\ell(Y)$.

Claim 1: $\ell_{|Y}$ is an embedding.

Proof of Claim 1: It is sufficient to prove that for any zero-dimensional scheme $A\subset Y$ with $\deg(A)\leq 2$ we have $o\notin\langle A\rangle$. Assume the existence of a zero-dimensional scheme $A\subset Y$ with $\deg(A)\leq 2$ and $o\in\langle A\rangle$. Since $o\notin Y$, we have $\deg(A)=2$. Since $o\notin\langle S^{\prime}\rangle$ for any $S^{\prime}\subsetneq S$ and $|S|=a+1>2$, we have $A\nsubseteq S$. Since $o\in\langle A\rangle\cap\langle S\rangle$, $A\cup S$ is linearly dependent. Since $\deg(A\cup S)\leq a+3$ and $\rho(Y)=r+1$, we get a contradiction.

By Claim 1 $X$ is a smooth rational curve and $\deg(X)=r+1$.

Claim 2: We have $\rho(X)=\rho(X)^{\prime}=a$.

Proof of Claim 2: Since $\ell_{|X}$ is an embedding, we have $|\ell(S)|=a+1$. Since $o\in\langle S\rangle$, $\ell(S)$ is linear dependent and hence $\rho(X)^{\prime}\leq a$. Assume $\rho(X)^{\prime}<a$ and take a zero-dimensional scheme $Z\subset X$ such that $\deg(Z)\leq a$ and $Z$ is linearly dependent. Le $W\subset Y$ be the only scheme such that $\ell(W)=Z$. Since $Z$ is linearly dependent, we have $o\in\langle W\rangle$. Since $\deg(W)\leq a$ and $o\notin\langle S^{\prime}\rangle$ for any $S^{\prime}\subsetneq S$, we have $W\nsubseteq S$. Thus $S\cup W$ is linearly dependent. Hence $2a+1\geq r+2$, a contradiction.

(b) Now assume $2a\geq r$, $a\leq r-1$ and $a\geq 2$. Fix a smooth curve $C$ of genus $g$ and a zero-dimensional scheme $A\subset X$ such that $\deg(A)=a+1$. Since $\deg(\omega_{C}(A))=2g+a+1-2\geq 2g+1$, $\omega_{C}(A)$ is very ample. By Riemann-Roch we have $h^{0}(\omega_{C}(A))=g+a=r+1$ and $A$ is the only zero-dimensional scheme $Z\subset C$ such that $h^{1}(\omega_{C}(A-Z))>0$ and $\deg(Z)\leq a+1$. Let $f:C\xrightarrow{\ \ }\mathbb{P}^{r}$, $r=g+a$, denote the embedding induced by the complete linear system $|\omega_{C}(A)$. Set $X:=f({C})$. We have $\rho(X)=a$. We have $\rho(A)^{\prime}=a$ if and only if $A$ is reduced.

(c) Take in part (a) instead of $S$ a zero-dimensional scheme $S_{1}$ such that $\deg(S_{1})=a+1$ and $S_{1}$ is not reduced. Taking the linear projection from $o$ we get an example with $\rho(X)<\rho(X)^{\prime}$.

Example 4.3**.**

Fix integers $r,a$ such that $2\leq a\leq r+1$. Here we prove the existence of a smooth and non-degenerate curve $X\subset\mathbb{P}^{r}$ such that $\rho(X)^{\prime}=\rho(X)=a$. If $a=r+1$ we know that $X$ is a rational normal curve. The case $g=1$, $d=r+1$ covers the the case $a=r$. Now assume $2\leq a\leq r-1$.

(a) We first cover the case $2a\leq r+1$. In this range $X$ we construct a smooth rational curve $X$ with $\rho(X)=\rho(X)^{\prime}=a$, but of course $X$ is not linearly normal. Let $Y\subset\mathbb{P}^{r+1}$ be a rational normal curve. Fix a set $S\subset X$ such that $|S|=a+1$ and take any $o\in\langle S\rangle$ such that $o\notin\langle S^{\prime}\rangle$ for any $S^{\prime}\subsetneq S$. Let $\ell:\mathbb{P}^{r+1}\setminus\{o\}\xrightarrow{\ \ }\mathbb{P}^{r}$ denote the linear projection from $o$. Since $a\geq 2$ and $\rho(Y)=r+2$, we have $o\notin Y$. Hence $\ell_{|Y}$ is a morphism. Set $X:=\ell(X)$.

Claim 1: $\ell_{|Y}$ is an embedding.

Proof of Claim 1: It is sufficient to prove that for any zero-dimensional scheme $A\subset Y$ with $\deg(A)\leq 2$ we have $o\notin\langle A\rangle$. Assume the existence of a zero-dimensional scheme $A\subset Y$ with $\deg(A)\leq 2$ and $o\in\langle A\rangle$. Since $o\notin Y$, we have $\deg(A)=2$. Since $o\notin\langle S^{\prime}\rangle$ for any $S^{\prime}\subsetneq S$ and $|S|=a+1>2$, we have $A\nsubseteq S$. Since $o\in\langle A\rangle\cap\langle S\rangle$, $A\cup S$ is linearly dependent. Since $\deg(A\cup S)\leq a+3$ and $\rho(Y)=r+1$, we get a contradiction.

By Claim 1 $X$ is a smooth rational curve and $\deg(X)=r+1$.

Claim 2: We have $\rho(X)=\rho(X)^{\prime}=a$.

Proof of Claim 2: Since $\ell_{|X}$ is an embedding, we have $|\ell(S)|=a+1$. Since $o\in\langle S\rangle$, $\ell(S)$ is linear dependent and hence $\rho(X)^{\prime}\leq a$. Assume $\rho(X)^{\prime}<a$ and take a zero-dimensional scheme $Z\subset X$ such that $\deg(Z)\leq a$ and $Z$ is linearly dependent. Le $W\subset Y$ be the only scheme such that $\ell(W)=Z$. Since $Z$ is linearly dependent, we have $o\in\langle W\rangle$. Since $\deg(W)\leq a$ and $o\notin\langle S^{\prime}\rangle$ for any $S^{\prime}\subsetneq S$, we have $W\nsubseteq S$. Thus $S\cup W$ is linearly dependent. Hence $2a+1\geq r+2$, a contradiction.

(b) Now assume $2a\geq r$, $a\leq r-1$ and $a\geq 2$. Set $g:=r+1-a$. Fix a smooth curve $C$ of genus $g$ and a zero-dimensional scheme $A\subset X$ such that $\deg(A)=a+1$. Since $\deg(\omega_{C}(A))=2g+a+1-2\geq 2g+1$, $\omega_{C}(A)$ is very ample. By Riemann-Roch we have $h^{0}(\omega_{C}(A))=g+a=r+1$. Let $f:C\xrightarrow{\ \ }\mathbb{P}^{r}$, be the embedding induced by $|\omega_{X}(A)|$ Set $X:=f({C})$ and $Z:=f(A)$. By Rieman-Roch $Z$ is linearly dependent and $Z$ is the only zero-dimensional scheme $W\subset X$ such that $\deg(W)\leq a+1$ and $W$ linearly dependent. Thus $\rho(X)=a$. We have $\rho(X)^{\prime}=a$ if and only if $A$ is a reduced set.

(c) Take in part (a) instead of $S$ a zero-dimensional scheme $S_{1}$ such that $\deg(S_{1})=a+1$ and $S_{1}$ is not reduced. Taking the linear projection from $o$ we get smooth rational curves with $\rho(X)<\rho(X)^{\prime}$.

(d) Take $C$ as in step (b) and a zero-dimensional scheme $A_{1}\subset\mathbb{P}^{r}$ such that $\deg(A_{1})=a+1$ and $A_{1}$ is not reduced. As in step (b) we get an embedding $X\subset\mathbb{P}^{r}$ of $C$ such that $\rho(X)=a$ and the image of $A_{1}$ is the only linearly dependent degree $a+1$ subscheme of $X$. Thus $\rho(X)^{\prime}>a$.