The stratification by rank for homogeneous polynomials with border rank 5 which essentially depend on 5 variables

TL;DR

This paper classifies the symmetric tensor ranks of degree d homogeneous polynomials with border rank 5, depending on the structure of an associated zero-dimensional scheme, for degrees d ≥ 9.

Contribution

It extends previous classifications to higher degrees, providing a detailed stratification of ranks based on the geometry of associated schemes.

Findings

Five possible ranks for polynomials depending on the scheme structure.

Ranks are uniquely determined by the number of connected components of the associated scheme.

Each family of polynomials is characterized by the degrees of the scheme's components.

Abstract

We give the stratification by the symmetric tensor rank of all degree $d \ge 9$ homogeneous polynomials with border rank $5$ and which depend essentially on at least 5 variables, extending previous works (A. Bernardi, A. Gimigliano, M. Id\`{a}, E. Ballico) on lower border ranks. For the polynomials which depend on at least 5 variables only 5 ranks are possible: $5$, $d+3$, $2d+1$, $3d-1$, $4d-3$, but each of the ranks $3d-1$ and $2d+1$ is achieved in two geometrically different situations. These ranks are uniquely determined by a certain degree 5 zero-dimensional scheme $A$ associated to the polynomial. The polynomial $f$ depends essentially on at least 5 variables if and only if $A$ is linearly independent (in all cases $f$ essentially depends on exactly 5 variables). The polynomial has rank $4d-3$ (resp $3d-1$, resp. $2d+1$, resp. $d+3$, resp. $5$) if $A$ has $1$ (resp. $2$, resp. $3$, resp. $4$, resp. $5$) connected components. The assumption $d\ge 9$ guarantees that each polynomial has a uniquely determined associated scheme $A$. In each case we describe the dimension of the families of the polynomials with prescribed rank, each irreducible family being determined by the degrees of the connected components of the associated scheme $A$.

Equations

$$0\to {\mathcal{I}}_{\text{mbox}{Res}_D(Z)}\otimes \mathcal{L}(-D)\to {\mathcal{I}}_Z\otimes \mathcal{L}\to {\mathcal{I}}_{Z\cap D}\otimes {\mathcal{L}}_{|D}\to 0.$$

$$0\to {\mathcal{I}}_{W_i}(d-i)\to {\mathcal{I}}_{W_{i-1}}(d+1-i)\to {\mathcal{I}}_{W_{i-1}\cap H_i,H_i}(d+1-i)\to 0$$

Full text

1 Introduction

Let $\mathbb{C}[x_{0},\dots,x_{m}]_{d}$ be the set of all homogeneous degree $d$ polynomials in the variables $x_{0},\dots,x_{m}$ with complex coefficients. For any $f\in\mathbb{C}[x_{0},\dots,x_{m}]_{d}\setminus\{0\}$ the rank of $f$ (or the symmetric tensor rank of $f$) is the minimal integer $r>0$ such that $f=\sum_{i=1}^{r}\ell_{i}^{d}$ with $\ell_{i}\in\mathbb{C}[x_{0},\dots,x_{m}]_{1}$. These additive decompositions of $f$ as sums of $d$-powers of linear forms are useful even to decompose symmetric tensors and hence they appear in some applications (ACCF , cm , DM , Ch , dLC , McC , Co1 , JS , l ). A different notion is the notion of border rank of $f$ or approximate rank of $f$ (the minimal integer $r$ such that $f$ is the limit of a family of homogeneous polynomials of rank $\leq r$). Both notions may be translated in the setting of projective geometry and algebraic geometry and it is in this setting that it was done the classification of all $f$ with border rank $\leq 3$ ((bgi, , Theorems 32 and 37)) and of border rank $4$ (bb2 ). In this paper we give the classification of all $f$ with border rank $5$ AND which depend essentially on at least 5 variables (in each case $P$ depends on exactly 5 variables, because it is contained in the degree $d$ Veronese embedding of a $4$-dimensional linear subspace of $\mathbb{P}^{m}$). This is a strong condition and in Remark 9 and Proposition 6 we explain our knowledge of the rank for polynomials depending on fewer variables.

Now we explain the translation of the additive decomposition of homogeneous polynomials in terms of projective geometry. For all positive integers $m,d$ let $\nu_{d}:\mathbb{P}^{m}\to\mathbb{P}^{r}$, $r:={m+d}\choose{m}-1$, denote the Veronese embedding induced by the vector space $\mathbb{C}[x_{0},\dots,x_{m}]_{d}$. Set $X_{m,d}:=\nu_{d}(\mathbb{P}^{m})$. For any $P\in\mathbb{P}^{r}$ the rank (or symmetric tensor rank or symmetric rank) $r_{m,d}(P)$ of $P$ is the minimal cardinality of a set $S\subset X_{m,d}$ such that $P\in\langle\nu_{d}(S)\rangle$, where $\langle\ \ \rangle$ denote the linear span. For any integer $t\geq 1$ the $t$-secant variety $\sigma_{t}(X_{m,d})$ of $X_{m,d}$ is the closure in $\mathbb{P}^{r}$ of the union of all linear spaces $\langle\nu_{d}(S)\rangle$ with $S\subset\mathbb{P}^{m}$ and $\sharp(S)\leq t$ (to get the closure it is the same if we use the euclidean topology or the Zariski topology). This approximation makes sense for homogeneous polynomials, too: the homogeneous polynomial $f$ is a limit of a family of homogeneous polynomials with rank $\leq t$ if and only if its associated point in $\mathbb{P}^{r}$ is contained in $\sigma_{t}(X_{m,d})$. Notice that “ closure ” in the definition of $\sigma_{t}(X_{m,d})$ is a good way to formalize the approximation by points with rank $\leq t$. Set $\sigma_{0}(X_{m,d})=\emptyset$. The border rank $b_{r,m}(P)$ of $P$ is the minimal integer $t>0$ such that $P\in\sigma_{t}(X_{m,d})$, i.e. the only integer $b>0$ such that $P\in\sigma_{b}(X_{m,d})\setminus\sigma_{b-1}(X_{m,d})$. The cactus rank (br , rs ) (introduced in ik with the name scheme-rank) $z_{m,d}(P)$ of $P$ is the minimal integer $z$ such that there is a zero-dimensional scheme $Z\subset\mathbb{P}^{m}$ with $\deg(Z)=z$ and $P\in\langle\nu_{d}(Z)\rangle$. We always have $z_{m,d}(P)\leq r_{m,d}(P)$. If $d\geq b_{r,m}(P)-1$, then $z_{r,m}(P)\leq b_{r,m}(P)$ and often equality holds. We are interested in points with border rank $5$ and we assume that $d$ is not too low, e.g., we assume $d\geq 9$. Since $d\geq 4$ and $b_{m,d}(P)=5$, we have $z_{m,d}(P)\leq 5$ and hence we get at least one scheme $A$ with $\deg(A)\leq 5$ and $P\in\langle\nu_{d}(A)\rangle$. Now assume $m\geq 4$, $d\geq 4$ and that $P\notin\langle\nu_{d}(M)\rangle$ for any linear subspace $M$ of $\mathbb{P}^{m}$ with $\dim(M)\leq 3$, i.e. assume that the polynomial $f$ associated to $P$ does not depend on at most 4 variables, up to a linear change of coordinates. Since $P\in\langle\nu_{d}(A)\rangle$, the condition “  $P\notin\langle\nu_{d}(M)\rangle$ for any linear subspace $M$ of $\mathbb{P}^{m}$ with $\dim(M)\leq 3$ ” implies $\dim(\langle A\rangle)=4$, i.e. $A$ is linearly independent. Assume that $A$ has $s\geq 1$ connected components, say $A=A_{1}\sqcup\cdots\sqcup A_{s}$. We say that $A$ has type $(s;b_{1},\dots,b_{s})$. We have $5=\deg(A)=b_{1}+\cdots+b_{s}$. In our case we have $z_{m,d}(P)=5$ if and only if $b_{m,d}(P)=5$ and the scheme $A$ evincing the cactus rank is unique ((bgl, , Corollary 2.7) or (bb2, , Lemma 4.2 and text after Remark 2.7)). Therefore if $d\geq 9$ we may define the type of $P$ as the type of $A$. We prove the following result.

Theorem 1.1

Fix integers $m\geq 4$ and $d\geq 9$. Let $P\in\mathbb{P}^{r}$, $r={m+d}\choose{m}-1$, be a point with border rank $5$. Assume that there is no 3-dimensional linear subspace $H\subset\mathbb{P}^{m}$ such that $P\in\langle\nu_{d}(H)\rangle$.

• (i)

If $P$ has type $(1;5)$, then $r_{m,d}(P)=4d-3$.

• (ii)

If $P$ has type $(2;3,2)$, then $r_{m,d}(P)=3d-1$.

• (iii)

If $P$ has type $(2;4,1)$, then $r_{m,d}(P)=3d-1$.

• (iv)

If $P$ has type $(3;3,1,1)$, then $r_{m,d}(P)=2d+1$.

• (v)

If $P$ has type $(3;2,2,1)$, then $r_{m,d}(P)=2d+1$.

• (vi)

If $P$ has type $(4;2,1,1,1)$, then $r_{m,d}(P)=d+3$.

• (vii)

If $P$ has type $(5;1,1,1,1,1)$, then $r_{m,d}(P)=5$.

In each case (i),…,(vii) the description of each $A$ easily give the dimension, $5m+5-s$, of all points $P$ in that case and that they are parametrized by an irreducible variety (see Remark 8). Each case occurs (Remark 7).

Case (vii) is obvious. Case (i) is the hard one and we use its proof (called step (b) in the proof of Theorem 1.1) to prove the other cases and (for the easiest cases) more informations that just the integer $r_{m,d}(P)$. For each $P\in\mathbb{P}^{r}$ let $\mathcal{S}(P)$ be the set of all sets $S\subset\mathbb{P}^{m}$ evincing the rank of $P$, i.e. the set of all subset $S\subset\mathbb{P}^{m}$ such that $P\in\langle\nu_{d}(S)\rangle$ and $\sharp(S)=r_{m,d}(P)$ (see Proposition 1 (case arbitrary cactus rank $b\geq 2$ and type $(b-1;2,1,\dots)$, Proposition 2 (case $s=3$, $b_{1}=b_{2}=2$, $b_{3}=1$), Proposition 3 (case arbitrary cactus rank $b\geq 3$ and type $(b-2;3,1,\dots)$ and Proposition 4 (case $s=2$, $b_{1}=3$, $b_{2}=2$)). In each case $P$ depends on exactly $5$ variables, because $P\in\langle\nu_{d}(A)\rangle$.

Remark 1

If $d\geq 9$, then for each $P\in\mathbb{P}^{r}$ with border rank $5$ there is a unique degree 5 scheme $A$ evincing the cactus rank of $P$ ((bgl, , Corollary 2.7) or (bb2, , Lemma 4.2 and text after Remark 2.7)). Fix any degree $5$ zero-dimensional scheme $A$ achieving the cactus rank of some $P\in\mathbb{P}^{r}$ with border rank $5$. Fix any $Q\in\langle\nu_{d}(A)\rangle$ and let $E\subseteq A$ be a minimal subscheme of $A$ such that $P\in\langle\nu_{d}(E)\rangle$. If $d\geq 9$ we get that $b_{m,d}(Q)=z_{m,d}(Q)=\deg(E)$ and that $E$ is the only scheme evincing the cactus rank of $Q$. $Q$ depends essentially on 5 variables if and only if $E=A$ and $\dim(\langle A\rangle)=\mathbb{P}^{4}$. In each step of the proof of Theorem 1.1 we give a classification of all $A$ appearing as cactus rank and an easy way to write $A$. Therefore, using $A$, one may easily give $P\in\mathbb{P}^{r}$ with a prescribed rank, among the possible ones $4d-3$, $3d-1$, $2d+1$, $d+3$, $5$. Without knowing $A$ it may be possible to detect the value of $r_{m,d}(P)$ for a specific $P$, just because Theorem 1.1 list all possible ranks and the difference between two different ranks is huge for large $d$.

Remark 2

Take the set-up of Theorem 1.1, but only assume $d\geq 4$. Fix $P\in\mathbb{P}^{r}$ with border rank $5$ and assume that there is no 3-dimensional linear subspace $H\subset\mathbb{P}^{m}$ such that $P\in\langle\nu_{d}(H)\rangle$. Since $d\geq 4$, there is at least one scheme $A$ evincing the cactus rank of $P$. Call $(s;b_{1},\dots,b_{s})$, $b_{1}+\cdots+b_{s}=5$, the type of $A$. The classification of the possible connected components $A_{1},\dots,A_{s}$ of $A$ is done in the corresponding step of the proof of Theorem 1.1, because it does not depend on $d$. At the beginning of each step we prove an upper bound for the integer $r_{m,d}(P)$ (the hard part was then to check the opposite inequality). This part of the inequality does not depend on $d$. Hence in each case $r_{m,d}(P)$ is at most the value listed in Theorem 1.1. In particular we get $r_{m,d}(P)\leq 4d-3$.

We work over an algebraically closed field $\mathbb{K}$ with $\mbox{char}(\mathbb{K})=0$. Any easy weak form of Lefschetz’s principle would allow to deduce this case from the case $\mathbb{K}=\mathbb{C}$.

2 Preliminaries

Let $X$ be a projective variety, $D$ an effective Cartier divisor of $X$ and $Z\subset X$ a zero-dimensional scheme. The residual scheme $\mbox{Res}_{D}(Z)$ of $Z$ with respect to $D$ (or with respect to the inclusion $D\subset X$) is the closed subscheme of $X$ with $\mathcal{I}_{Z}:\mathcal{I}_{D}$ as its ideal sheaf. For every line bundle $\mathcal{L}$ on $X$ we have a residual exact sequence of coherent sheaves

[TABLE]

The case $m=2$ of the next lemma is a very particular case of (ep, , Corollaire 2); the general case easily follows by induction on $m$, taking a hyperplane $H\subset\mathbb{P}^{m}$ with maximal $\deg(Z\cap H)$ as in step (b) of the proof of Theorem 1.1 (case $g\leq 2$, $g^{\prime}=1$); if $\deg(Z)\leq 2d+1$, then the lemma is (bgi, , Lemma 34).

Lemma 1

Fix an integer $d\geq 6$. Let $Z\subset\mathbb{P}^{m}$, $m\geq 2$, be a zero-dimensional scheme with $\deg(Z)\leq 3d+1$ and $h^{1}(\mathcal{I}_{Z}(d))>0$. Then either there is a line $L\subset\mathbb{P}^{m}$ with $\deg(L\cap Z)\geq d+1$ or there is a conic $T\subset\mathbb{P}^{m}$ with $\deg(T\cap Z)\geq 2d+2$ or there is a plane cubic $F$ with $\deg(F\cap Z)\geq 3d$.

Remark 3

Take the set-up of Lemma 1 and assume the existence of a plane conic $T$ such that $\deg(T\cap Z)\geq 2d+2$, but that there is no line $L\subset\mathbb{P}^{m}$ with $\deg(L\cap Z)\geq d+2$. When $Z$ has many reduced connected components, it is obvious that $T$ must be reduced. Assume that $T$ is reduced, but reducible, say $T=D\cup R$, with $D$ and $R$ lines. Set $\{o\}:=D\cap R$. Since $\deg(D\cap Z)\leq d+1$ and $\deg(R\cap Z)\leq d+1$, we get $\deg(D\cap Z)=\deg(R\cap Z)=d+1$ and that $Z$ is a Cartier divisor of the nodal curve $T$ (it is a general property of nodal curves with smooth irreducible component: for any zero-dimensional scheme $E\subset D\cup R$ we have $\deg(E\cap D)+\deg(E\cap R)-1\leq\deg(E)\leq\deg(E\cap D)+\deg(E\cap R)$ and $\deg(E)=\deg(E\cap D)+\deg(E\cap R)$ if and only if $E$ is a Cartier divisor of $D\cup R$). Now assume that existence of a plane cubic $F$ with $\deg(F\cap Z)\geq 3d$. $F$ is not reduced if and only if there is a line $L\subset F$ appearing in $F$ with multiplicity at least two. To get that $F$ is reduced it is sufficient to assume that $\deg(R\cap Z)\leq d+1$ for each line $R$ and that $Z$ has at least $2d+2$ reduced connected components.

Remark 4

Fix integers $m\geq 2$ and $d\geq 3$. Let $Z\subset\mathbb{P}^{m}$ be a subscheme such that $Z$ spans $\mathbb{P}^{m}$, $\deg(Z)\leq d+1+m$ and $h^{1}(\mathcal{I}_{Z}(d)>0$. Then $\deg(Z)=d+1+m$ and there is a line $L\subset\mathbb{P}^{m}$ such that $\deg(Z\cap L)=d+2$ (a quick proof: use induction on $m$ and the proof of Theorem 1.1 below using the hyperplanes $H_{i}$; we get $g\leq 2$).

Remark 5

Let $T\subset\mathbb{P}^{2}$ be a reduced curve of degree $t<d$. It is connected and the projective space $\langle\nu_{d}(T))\rangle$ has dimension $x:={{d+2}\choose{2}}-{{d-t+2}\choose{2}}-1$. Every point of $\langle\nu_{d}(T)\rangle$ has rank at most $x$ with respect to the curve $\nu_{d}(T)$ (the proof of (lt, , Proposition 5.1) works verbatim for reduced and connected curves). Hence $\sharp(B\cap T)\leq x$. If $t=1$ (resp. $t=2$, resp $t=3$) then $x=d$ (resp. $2d$, resp. $3d-1$).

3 Proof of Theorem 1.1 and related result

Step (b) of the proof of Theorem 1.1 is by far the most difficult part of this paper and the one which may be used elsewhere. We outline here the proof of Theorem 1.1 and of the other results of this section.

3.1 Outline of the proof of Theorem 1.1 and of the other results of this section

We first give the classification of all possible degree $5$ schemes $A$. The key step (called step (b)) is the one in which we handle the case $s=1$ (it is an enhanced version of the proof of (bb2, , Proposition 5.19)). This step is subdivided into several substeps and sub-substeps (up to, e.g., step (b2.2.2.1), which is the first step of step (b2.2.2), which is the second step of step (b2.2), which is the second step of step (b2), which is the second step of step (b)). There are several good reasons for the splitting of the proof in this way. In the proof of the other cases and in the proof of Propositions 1, 2, 3, 4 we need to quote a specific substep or to modify it a little bit; in this way all proofs are detailed and complete, but with no duplication, and each part of the proof has its own label, making easier to quote it in future works. In step (c) of the proof of Theorem 1.1 we reduce some cases to Propositions 1, 2, 3 and 4. Proposition 1 covers the case $b_{1}=2$ and $b_{2}=b_{3}=b_{4}=1$. Proposition 2 covers the case $b_{1}=b_{2}=2$ and $b_{3}=1$. Proposition 4 covers the case $b_{1}=3$ and $b_{2}=b_{3}=1$. Proposition 3 covers the case $b_{1}=3$ and $b_{2}=2$. In step (d) of the proof of Theorem 1.1 we prove the case $s=2$, $b_{1}=4$, $b_{2}=1$(this is far easier than in step (b)). Proposition 1 covers all points $P$ with $b_{m,d}(P)=z_{m,d}(P)=b$ evinced by a scheme of type $(b-1;2,\dots,1)$ (it also describes all $B\in\mathcal{S}(P)$, i.e. all sets $S\subset\mathbb{P}^{m}$ such that $P\in\langle\nu_{d}(S)\rangle$ and $\sharp(S)=r_{m,d}(P)$). Proposition 2 gives the rank and a description of all $B\in\mathcal{S}(P)$ when $s=3$, $b_{1}=b_{2}=2$ and $b_{3}=1$. Proposition 4 describes the case $s=2$, $b_{1}=3$, $b_{2}=2$. The proof of Theorem 1.1 may be seen as an iteration (several times) of the proofs in (bb2, , §5.2).

Proof (Proof of Theorem 1.1)

It is a general result that every zero-dimensional scheme evincing the cactus rank of some $P\in\mathbb{P}^{r}$ is Gorenstein ((bb+, , Lemma 2.3)). There is a scheme, $A$, evincing the border rank $b_{m,d}(P)$ of $P$, because $d\geq b_{m,d}(P)-1$ ((bgl, , Lemma 2.6)). This scheme is unique, because $d\geq 9=2b_{m,d}(P)-1$ ((bgl, , Corollary 2.7 and its proof) or (bb2, , Lemma 4.2 and the text after Remark 2.7)). It also evinces the cactus rank of $P$, i.e. $z_{m,d}(P)=5$, because bgi and bb2 give the classifications for all points with cactus rank $\leq 4$ and they have border rank $\leq 4$. Since $P\notin\langle\nu_{d}(H)\rangle$ for any 3-dimensional linear subspace $H\subset\mathbb{P}^{m}$, $\deg(A)=5$ and $P\in\langle\nu_{d}(A)\rangle$, we have $\dim(\langle A\rangle)=4$, i.e. $A$ is linearly independent. In particular it is in linearly general position in $\langle A\rangle$. The scheme $A$ is not the double $2O$ of one point of $\langle A\rangle$ (i.e. the closed subscheme of the projective space $\langle A\rangle$ with $(\mathcal{I}_{O,\langle A\rangle})^{2}$ as its ideal sheaf), because the proof of (bgi, , Theorem 32) shows that for each $P\in\langle\nu_{d}(2O)\rangle$ there is a line $L\subset\mathbb{P}^{m}$ such that $P\in\langle\nu_{d}(L)\rangle$; alternatively, use that $A$ is Gorenstein by (bb+, , Lemma 2.3). By (eh, , Theorem 1.3) $A$ is curvilinear.

By assumption $A$ is linearly independent and it spans a 4-dimensional linear subspace of $\mathbb{P}^{m}$. By concision ((l, , Exercise 3.2.2.2)), it is sufficient to do the case $m=4$. Write $A=A_{1}\sqcup\cdots\sqcup A_{s}$ with $\deg(A_{i})=b_{i}$, $b_{i}\geq b_{j}$ if $i\leq j$, and set $\{O_{i}\}:=(A_{1})_{red}$. Since $A$ evinces the cactus rank of $P$, then $P\notin\langle\nu_{d}(A^{\prime})\rangle$ for any $A^{\prime}\subsetneq A$. Let $B\subset\mathbb{P}^{m}$ any set evincing the rank of $P$. Set $W_{0}:=A\cup B$. If $A\neq B$, then $h^{1}(\mathcal{I}_{W_{0}}(d))>0$ ((bb1, , Lemma 1)). If $s\neq 5$, then $A\neq B$.

(a) If $A$ is reduced, i.e. if $s=5$, then $r_{4,d}(P)\leq 5$. Since we assumed border rank 5, we get $r_{4,d}(P)=5$.

(b) Now assume $s=1$ and hence $b_{1}=5$. The scheme $A_{1}$ is curvilinear and unramified, because $\mathbb{P}^{4}=\mathbb{P}(W)$ with $\dim(W)=5$, $\langle A_{1}\rangle=\mathbb{P}^{4}$ and we may apply (eh, , Theorem 1.3).

Claim: $A_{1}$ is contained in a rational normal curve $C\subset\mathbb{P}^{4}$.

Proof of the Claim: Since $A_{1}$ is connected and curvilinear for each $i=1,\dots,5$, $A_{1}$ has a unique subscheme $A_{1}[i]$ of $A_{1}$ such that $\deg(A_{1}[i])=i$. Since $A_{1}$ is linearly independent, then $\dim(\langle A_{1}[i]\rangle)=i-1$. For a parametrization $t\mapsto(t,t^{2},t^{3},t^{4})$ of $C$ take the flag of linear subspaces $\langle A_{1}[1]\rangle\subset\langle A_{1}[2]\rangle\subset\cdots\subset\langle A_{1}[5]\rangle$.

The existence of the rational normal curve $C$ also gives that $A_{1}$ is unique, up to a projective transformation. The curve $\nu_{d}(C)$ is a degree $4d$ rational normal curve in its linear span. By Sylvester’s theorem (cs , (lt, , Theorem 4.1), (bgi, , Theorem 23)) the rank of $P$ with respect to $\nu_{d}(C)$ is $4d-3$. Hence $r_{4,d}(P)\leq 4d-3$. Assume $\sharp(B)\leq 4d-4$ and hence $\deg(W_{0})\leq 4d+1$. Let $H_{1}$ be a hyperplane such that $e_{1}:=\deg(W_{0}\cap H_{1})$ is maximal. Set $W_{1}:=\mbox{Res}_{H_{1}}(W_{0})$. Fix an integer $i\geq 2$ and assume to have defined the integers $e_{j}$, the hyperplanes $H_{j}$ and the scheme $W_{j}$, $1\leq j<i$. Let $H_{i}$ be any hyperplane such that $e_{i}:=\deg(H_{i}\cap W_{i-1})$ is maximal. Set $W_{i}:=\mbox{Res}_{H_{i}}(W_{i-1})$. We have $e_{i}\geq e_{i+1}$ for all $i$. For each integer $i>0$ we have the residual exact sequence

[TABLE]

Since $h^{1}(\mathcal{I}_{W_{0}}(d))>0$, there is an integer $i>0$ such that $h^{1}(H_{i},\mathcal{I}_{W_{i-1}\cap H_{i},H_{i}}(d+1-i))>0$. We call $g$ the minimum such an integer. Since any zero-dimensional scheme with degree $4$ of $\mathbb{P}^{4}$ is contained in a hyperplane, if $e_{i}\leq 3$, then $W_{i}=\emptyset$ and $e_{j}=0$ if $j>i$. Hence $e_{d+2}=0$ and $e_{d+1}\leq 1$. Since $h^{1}(\mathcal{O}_{\mathbb{P}^{4}}(d))=h^{1}(\mathcal{I}_{Q})=0$ for any $Q\in\mathbb{P}^{4}$, we get $g\leq d$. By (bgi, , Lemma 34) either $e_{g}\geq 2(d+1-g)+2$ or there is a line $L\subset H_{g}$ such that $\deg(L\cap W_{i-1})\geq d+3-g$. Assume for the moment $g\geq 2$ and $e_{g}\leq 2(d+1-g)+1$. Since $e_{g}>0$, $W_{g-2}$ spans $\mathbb{P}^{4}$. Hence $e_{g-1}\geq d+5-g$. This inequality is obvious if $e_{g}\geq 2(d+1-g)+2$ and $g\leq d-1$, because $2(d+1-g)+2\geq d+5-g$ if $g\leq d-1$. Now assume $g=d$; if $e_{d}\leq 4$ and there is no line $L\subset H_{d}$ with $\deg(L\cap W_{d-1})\geq 3$, then $e_{d}=4$ and $W_{d-1}\cap H_{d}$ is contained in a plane; hence even in this case we get $e_{g-1}\geq 5=d+5-g$. Hence if $g\geq 2$, then $e_{g-1}\geq d+5-g$. Since $e_{i}\geq e_{i+1}$ for all $i$, we get that $4d+1\geq\deg(W_{0})\geq g(d+5-g)-2$. Set $\psi(t)=t(d+5-t)-2$. The function $\psi:\mathbb{R}\to\mathbb{R}$ is increasing if $t\leq(d+5)/2$, decreasing if $t\geq(d+5)/2$ and strictly monotone in any interval not containing $(d+5)/2$. Since $\psi(4)=4d+2>4d+1$ and $\psi(d)=5d-2>4d+1$, we get $1\leq g\leq 3$.

(b1) Assume $g=3$. We have $e_{3}\geq d$ and $e_{1}\geq e_{2}\geq d+2$. Hence $\deg(W_{3})\leq d-3$ and $e_{1}\leq 2d-1$. Since $h^{1}(\mathcal{I}_{W_{4}}(d-3))=0$, (bb2, , Lemma 5.1) gives $W_{0}\subset H_{1}\cup H_{2}\cup H_{3}$. Since $e_{1}\geq e_{2}\geq e_{3}$ and $e_{1}+e_{2}+e_{3}\leq 4d+1$, we have $e_{3}\leq(4d+1)/3\leq 2(d-2)+1$. By (bgi, , Lemma 34) there is a line $L\subset H_{3}$ such that $\deg(L\cap W_{3})\geq d$. Taking instead of $H_{3}$ any hyperplane $M\subset L$ we get $W_{0}\subset H_{1}\cup H_{2}\cup M$. In particular we have $B\subset H_{1}\cup H_{2}\cup L$. Since $A$ is curvilinear, we even get $W_{0}\subset H_{1}\cup H_{2}\cup L$. Set $Z_{0}:=W_{0}$. Let $N_{1}\subset\mathbb{P}^{4}$ be a hyperplane containing $L$ and such that $f_{1}:=\deg(W_{0}\cap N_{1})$ is maximal among the hyperplanes containing $L$. We have $f_{1}\geq e_{3}+2\geq d+2$. Set $Z_{1}:=\mbox{Res}_{N_{1}}(Z_{0})$. Let $N_{2}\subset\mathbb{P}^{4}$ be a hyperplane such that $f_{2}:=\deg(N_{2}\cap Z_{1})$ is maximal. Set $Z_{2}:=\mbox{Res}_{N_{2}}(Z_{1})$. For any integer $i\geq 3$ let $N_{i}\subset\mathbb{P}^{4}$ be a hyperplane such that $f_{i}:=\deg(Z_{i-1}\cap N_{i})$ is maximal. Set $Z_{i}:=\mbox{Res}_{N_{i}}(Z_{i-1})$. Let $g^{\prime}$ be the minimal positive integer $i$ such that $h^{1}(N_{i},\mathcal{I}_{N_{i}\cap Z_{i-1},N_{i}}(d+1-i))>0$ (it exists by the residual exact sequence (2) obtained using the hyperplanes $N_{i}$). Now we only have that $f_{i}\geq f_{i+1}$ if $i\geq 2$, but we also have $\deg(Z_{0})-f_{1}\leq 3d-1$. Hence $g^{\prime}\leq 3$.

(b1.1) Assume $g^{\prime}=3$. We have $f_{3}\leq(3d-1)/2\leq 2(d-2)+1$. Hence there is a line $R\subset N_{3}$ such that $\deg(R\cap Z_{2})\geq d$. Since $B\cap L\cap W_{1}=\emptyset$, we have $R\neq L$. Set $Z^{\prime}_{0}:=Z_{0}$. Take any hyperplane $N^{\prime}_{1}$ containing $L\cup R$ and with $f^{\prime}_{i}:=\deg(N^{\prime}_{1}\cap Z_{0})$ maximal among the hyperplanes containing $L\cup R$. Set $Z^{\prime}_{1}:=\mbox{Res}_{N^{\prime}_{1}}(Z^{\prime}_{0})$. Define as above the integers $f^{\prime}_{i}$, the hyperplanes $N^{\prime}_{i}$ and the schemes $W^{\prime}_{i}$ with $f^{\prime}_{i}\geq f^{\prime}_{i+1}$ for all $i\geq 2$ and $f^{\prime}_{1}\geq 2d-1$. Let $g^{\prime\prime}$ be the minimal positive integer $i$ such that $h^{1}(N^{\prime}_{i},\mathcal{I}_{N^{\prime}_{i}\cap Z^{\prime}_{i-1},N^{\prime}_{i}}(d+1-i))>0$ (it exists by the residual exact sequence (2) obtained using the hyperplanes $N^{\prime}_{i}$). Since $f^{\prime}_{1}\geq 2d-1$, we have $g^{\prime\prime}\leq 2$.

(b1.1.1) Assume for the moment $g^{\prime\prime}=2$. We get $f^{\prime}_{2}\geq d+1$ and the existence of a line $D\subset N^{\prime}_{2}$ such that $\deg(D\cap Z^{\prime}_{2})\geq d+1$ ((bgi, , Lemma 34)).

First assume that $L\cup D\cup R$ is contained in a hyperplane $M$. In this case we get $e_{1}\geq\deg(M\cap W_{0})\geq 3d-1$. Hence $e_{3}+e_{2}\leq d+2$, a contradiction. Now assume that $L\cup D\cup R$ is not contained in a hyperplane. Since $h^{0}(\mathcal{O}_{\mathbb{P}^{4}}(2))=15$ and $h^{0}(\mathcal{O}_{L\cup R\cup D}(2))\leq 9$, we have $h^{0}(\mathcal{I}_{L\cup R\cup D}(2))\geq 6$. Since $L\cup D\cup R$ is not contained in a hyperplane, either these lines are pairwise disjoint and they are not contained in a plane or two of them meets and the other one is disjoint from the plane spanned by the first two lines. Therefore $L\cup R\cup D$ is the scheme-theoretic base locus of $|\mathcal{I}_{L\cup R\cup D}(2)|$. Let $Q$ be a general hyperquadric containing $L\cup D\cup R$. A dimensional count gives that $Q$ is irreducible. First assume $h^{1}(\mathcal{I}_{\mbox{Res}_{Q}(W_{0})}(d-2))=0$. By (bb2, , Lemma 5.1) we get $A\cup B\subset Q$. Since $L\cup R\cup D$ is the scheme-theoretic base locus of $|\mathcal{I}_{L\cup R\cup D}(2)|$ and $A$ is curvilinear, we get $A\subset L\cup R\cup D$, contradicting the fact that at most two of these lines contain $O$.

Now assume $h^{1}(\mathcal{I}_{\mbox{Res}_{Q}(W_{0})}(d-2))>0$. Since $\deg(\mbox{Res}_{Q}(W_{0}))\leq d+2$, we get the existence of a line $R^{\prime}$ such $\deg(R^{\prime}\cap\mbox{Res}_{Q}(W_{0}))\geq d$. We have $h^{0}(\mathcal{I}_{L\cup D\cup R\cup R^{\prime}}(2))\geq 3$ and we take a general $Q_{1}\in|\mathcal{I}_{L\cup R^{\prime}\cup R\cup D}(2)|$. As above we find $A\cup B\subset Q_{1}$. Since $e_{1}\leq 2d-1$, no 3 of the lines $L$, $R$, $R^{\prime}$ and $D$ are contained in a plane. Hence at most two of these lines contain $O$ and at most one of the lines is tangent to $C$ at $O$. We get $\deg(B)\geq-3+\deg(W_{0}\cap L)+\deg(W_{0}\cap R)+\deg(W_{0}\cap R^{\prime})+\deg(W_{0}\cap D)\geq-3+d+d+d+1+d=4d-2$, a contradiction.

(b1.1.2) Now assume $g^{\prime\prime}=1$. Since $A$ is connected and not contained in a hyperplane, (bb2, , Lemma 5.1) gives $h^{1}(\mathcal{I}_{Z^{\prime}_{1}}(d-1))>0$. We have $\deg(Z^{\prime}_{1})\leq 3d-1$. By Lemma 1 either there is a line $R\subset\mathbb{P}^{4}$ such that $\deg(R\cap Z^{\prime}_{1})\geq d+1$ or there is a reduced conic $T\subset\mathbb{P}^{4}$ such that $\deg(T\cap Z^{\prime}_{1})\geq 2d$. In the latter case we would have $e_{1}\geq 2d+1$, a contradiction. Hence there is a line $R\subset\mathbb{P}^{4}$ such that $\deg(R\cap Z^{\prime}_{1})\geq d+1$. Since $R\cap(B\setminus R\cap L)\supseteq R\cap(B\setminus B\cap N^{\prime}_{1})\neq\emptyset$, then $R\neq L$ and hence at most one of the lines $R,L$ is tangent to $C$ at $O$. If $O\notin(L\cap R)$, then $\deg(W_{0}\cap(L\cup R))\geq 2d$ and hence $e_{1}\geq 2d$, a contradiction. Now assume $O\in(L\cap R)$. At most one of the lines $L,R$ is tangent to $C$ at $O$. We get $\deg(W_{0}\cap(L\cup R))\geq 2d-1$. By concision ((l, , Exercise 3.2.2.2)) $B$ spans $\mathbb{P}^{4}$. Hence there is a hyperplane $M\supset L\cup R$ containing a point of $B\setminus B\cap(L\cup R)$. Hence $e_{1}\geq 2d$, a contradiction.

(b1.2) Assume $g^{\prime}=2$. Since $f_{2}\leq e_{1}\leq 2d-1$, there is a line $R\subset N_{2}$ such that $\deg(R\cap Z_{1})\geq d+1$. Since $R\cap(B\setminus B\cap L)\subseteq R\cap B\cap Z_{1}$ and $\deg(R\cap B\cap Z_{1})\leq\deg(R\cap(B\setminus B\cap L))+\deg(R\cap A)\leq\deg(R\cap(B\setminus B\cap L))+2$, we get $R\neq L$ and hence $e_{1}\geq-1+\deg(L\cap W_{0})+\deg(R\cap W_{0})\geq 2d$, a contradiction.

(b2) Assume $g=2$. Since $e_{2}\leq e_{1}$ and $e_{1}+e_{2}\leq 4d+1$, we have $e_{2}\leq 2d$. By Lemma 1 either there is a line $L\subset H_{2}$ such that $\deg(L\cap W_{1})\geq d+1$ or $e_{2}=2d$ and there is a conic $T\subset H_{2}$ such that $W_{1}\cap H_{2}\subset T$, but there is no line $L\subset T$ with $\deg(L\cap W_{0})\geq d+1$; in the latter case $T$ is reduced and if $T$ is reducible, then $\deg(J\cap W_{1})=d$ for every component of $T$ (Remark 3).

(b2.1) Assume $e_{2}=2d$ and the existence of a conic $T\subset H_{2}$ such that $W_{1}\cap H_{2}\subset T$. Since $A$ spans $\mathbb{P}^{4}$, then $A\nsubseteq T$ and hence $O\in T$. Since $B$ spans $\mathbb{P}^{4}$ ((l, , Exercise 3.2.2.2)), there is $o\in B\setminus B\cap\langle T\rangle$. Let $M$ be the plane spanned by $o$ and $T$. Since $e_{1}\leq 2d+1$, we get $e_{1}=2d+1$, $\sharp(B)=4d-4$, $O\notin B$ and that $W_{0}\cap M=\{o\}\sqcup(W_{1}\cap T)$ with $\deg(W_{1}\cap T)=2d$. Therefore $B=(B\cap H_{1})\sqcup(B\cap T)$ and no point of $B\cap H_{1}$ is contained in the plane $\langle T\rangle$. Since $A\nsubseteq M$, (bb2, , Lemma 5.1) gives $h^{1}(\mathcal{I}_{\mbox{Res}_{M}(W_{0})}(d-1))>0$. Lemma 1 and Remark 3 give that either there is a line $R$ with $\deg(R\cap\mbox{Res}_{M}(W_{0}))\geq d+1$ or $\deg(\mbox{Res}_{M}(W_{0}))=2d$ and $\mbox{Res}_{M}(W_{0})$ is contained in a reduced conic $T^{\prime}$. The plane $M$ and hence $R$ or $T$ depends on the choice of $o$ (call them $M_{o},R_{o},T_{o}$).

(b2.1.1) First assume that $\deg(\mbox{Res}_{M}(W_{0}))=2d$ and that $\mbox{Res}_{M}(W_{0})$ is contained in a reduced conic $T^{\prime}$, but there is no line $R$ with $\deg(R\cap\mbox{Res}_{M}(W_{0}))\geq d+1$; hence if $T^{\prime}$ is reducible, then $\deg(J\cap\mbox{Res}_{M}(W_{0}))=d$ for each component of $T^{\prime}$ and the singular point of $T^{\prime}$ is not contained in $\mbox{Res}_{M}(W_{0})$. Fix $q\in T^{\prime}\cap B$ and take the hyperplane $M_{q}:=\langle\{q\}\cup T\rangle$. We have $\mbox{Res}_{M_{q}}(B)=((T^{\prime}\cap B)\setminus\{q\})\sqcup\{o\}$. This set is not contained in a line (even if $T^{\prime}$ is reducible), because $d\geq 9$. This set is contained in a conic only if $o\in T^{\prime}$. Hence (applying (bb2, , Lemma 5.1) to $M_{q}$) we get $B\subset T\cup T^{\prime}$. As in step (b2.1) we have $O\in T^{\prime}$. Hence $O\in T\cap T^{\prime}$. If one of the two conics is reducible, then taking a connected union $J\subset T\cup T^{\prime}$ of 3 lines we get $e_{1}\geq 2d+2$, a contradiction. Hence both conics $T,T^{\prime}$ are smooth. Since $\langle B\rangle=\mathbb{P}^{4}$, we have $\langle T\rangle\cap\langle T^{\prime}\rangle=\{O\}$. Hence at most one of the two conics is tangent to $C$ at $O$. If $T^{\prime}$ is not tangent to $C$ at $O$ we get $\sharp(B\cap T)\geq 2d-2$ and $\sharp(B\cap T^{\prime})\geq 2d-1$, a contradiction (and similarly if $T$ is not tangent to $C$ at $O$).

(b2.1.2) Now assume the existence of the line $R$. Since $e_{1}=2d+1$, then $T\cup R$ spans $\mathbb{P}^{4}$, i.e. $R\cap\langle T\rangle=\emptyset$. In particular $O\notin R\cap T$. First assume $O\notin R$. We have $\sharp(B\cap R)=d+1$, contradicting the case $t=1$ of Remark 5. Now assume $O\notin T$ and hence $\sharp(B\cap T)=2d$. Since $O\in R$, we get $O\notin\langle R\rangle$ and either $\sharp(B\cap(R\setminus\{O\}))=d$ or $R$ is the tangent line to $C$ at $O$. Since $h^{0}(\mathcal{I}_{T\cup R}(2))=7$, there is $Q\in|\mathcal{I}_{T\cup R}(2)|$ with $\deg(Q\cap W_{0})\geq 3d+2$ (e.g., if $R$ is not tangent to $C$ at $O$ it is sufficient to take $Q$ containing $R\cup T$ and the degree two zero-dimensional subscheme $\eta$ of $A$). Since $\deg(Q\cap\mbox{Res}_{Q}(W_{0}))\leq d-1$, we have $h^{1}(\mathcal{I}_{\mbox{Res}_{Q}(W_{0})}(d-2))=0$ and hence $Q\supset W_{0}$ ((bb2, , Lemma 5.1)), while we may even find $Q\in|\mathcal{I}_{T\cup R\cup\eta}(2)|$ not containing the degree $4$ subscheme of $A$, because $A$ is curvilinear.

(b2.2) Assume the existence of a line $L\subset H_{2}$ such that $\deg(L\cap W_{1})\geq d+1$. The case $t=1$ of Remark 5 gives $O\in L$, $d-1\leq\sharp(B\cap L)\leq d$ and that $\sharp(B\cap L)=d$ and $O\notin B$ if $L$ is not the tangent line to $C$ at $O$. Let $N_{1}\subset\mathbb{P}^{4}$ be a hyperplane containing $L$ and such that $f_{1}:=\deg(W_{0}\cap L)$ is maximal among the hyperplanes containing $L$. Since $A$ spans $\mathbb{P}^{4}$, we have $f_{1}\geq d+3$ and hence $e_{1}\geq d+3$. Set $Z_{0}:=W_{0}$ and $Z_{1}:=\mbox{Res}_{N_{1}}(Z_{0})$. Let $N_{2}$ be a hyperplane such that $f_{2}:=\deg(N_{2}\cap Z_{1})$. Set $Z_{1}:=\mbox{Res}_{N_{2}}(Z_{1})$. Define inductively the hyperplane $N_{i}$, $i\geq 3$, the schemes $Z_{i}:=\mbox{Res}_{N_{i}}(Z_{i-1})$ and the integer $f_{i}:=\deg(N_{i}\cap Z_{i-1})$ with $f_{i}$ maximal among all hyperplanes. We have $f_{i}\geq f_{i+1}$ for all $i\geq 3$ and if $f_{i}\leq 3$, then $f_{i+1}=0$ and $Z_{i}=\emptyset$. We have the residual exact sequence similar to (2) with $N_{i}$ instead of $H_{i}$, $Z_{i-1}$ instead of $W_{i-1}$ and $Z_{i}$ instead of $W_{i}$. Hence there is an integer $i>0$ such that $h^{1}(N_{i},\mathcal{I}_{N_{i}\cap Z_{i-1},H_{i}}(d+1-i))>0$ and we call $g^{\prime}$ the first such an integer. Since $\sum_{i\geq 2}f_{i}\leq 3d-2$ and $f_{i+1}=0$ if $f_{i}\leq 3$, we have $g^{\prime}\leq d+1$. Assume for the moment $g^{\prime}\leq d+1$. Hence either $f_{g^{\prime}}\geq 2(d+1-g^{\prime})+2$ or there is a line $R\subset N_{g^{\prime}}$ such that $\deg(R\cap Z_{i-1})\geq d+3-g^{\prime}$. Assume for the moment $3\leq g^{\prime}\leq d$. We get $f_{g^{\prime}-1}\geq d+5-g^{\prime}$ ((bgi, , Lemma 34)). Hence $3d-2\geq\sum_{i\geq 2}f_{i}\geq(g^{\prime}-1)(d+5-g^{\prime})-1$. As in step (b) we also exclude the case $g^{\prime}=d$. We get $g^{\prime}\leq 3$ if $g\leq d$.

Now assume $g^{\prime}=d+1$, i.e. assume $\deg(W_{2})\geq 2$. Since $f_{1}\geq d+3$ and $f_{i}\geq 4$ for $2\leq i\leq d$, we get $4d+1\geq d+3+4(d-1)+2$, a contradiction.

(b2.2.1) Assume $g^{\prime}=3$. Since $f_{1}\geq d+3$, $f_{2}\geq f_{3}$ and $f_{1}+f_{2}+f_{3}\leq 4d+1$, we get $f_{3}\leq(3d-2)/2\geq 2(d-2)+1$ (since $d\geq 4$). By (bgi, , Lemma 34) there is a line $R\subset N_{3}$ such that $\deg(R\cap Z_{2})\geq d$. Since $f_{3}>0$, we also see that $Z_{1}$ spans $\mathbb{P}^{4}$. Hence $f_{2}\geq d+2$. Therefore $f_{1}\leq 2d-1$. Let $M_{1}$ be a hyperplane containing $L\cup R$ and with $h_{1}:=\deg(M_{1}\cap 0)$ maximal. We have $f_{1}\geq h_{1}$. If $L\cap R=\emptyset$, then $h_{1}\geq\deg(L\cap W_{0})+\deg(R\cap W_{0})\geq 2d+1$, a contradiction. Now assume $L\cap R\neq\emptyset$. We have $R\neq L$, because $R\cap(B\setminus B\cap L)\neq\emptyset$. Since $L\cap R$ is scheme-theoretically a point, we have $\deg(W_{0}\cap(L\cup R))\geq\deg(W_{0}\cap L)+\deg(W_{0}\cap R)-1$. Hence $h_{1}\geq 1+\deg(W_{0}\cap L)+\deg(W_{0}\cap R)-1\geq 2d$, a contradiction.

(b2.2.2) Assume $g^{\prime}=2$. Since $e_{2}\geq d+1$, we have $e_{1}\leq 3d$. Since $f_{1}\geq d+3$, we have $f_{2}\leq 3d-2=3(d-1)+1$. By Lemma 1 and Remark 3 either there is a line $R\subset N_{2}$ with $\deg(R\cap Z_{1})\geq d+1$ or there is a reduced conic $T\subset N_{2}$ with $\deg(Z_{2}\cap T)\geq 2d$ or there is reduced plane cubic $F\subset N_{2}$ with $\deg(F\cap N_{2})\geq 3d-3$.

(b2.2.2.1) Assume the existence of a reduced plane cubic $F\subset N_{2}$ with $\deg(F\cap N_{2})\geq 3d-3$ and take any hyperplane $M$ containing $F$ and another point of $B$ (it exists, because $\langle B\rangle=\mathbb{P}^{4}$ by (l, , Exercise 3.2.2.2)). By (bb2, , Lemma 5.1) we have $h^{1}(\mathcal{I}_{\mbox{Res}_{M}(W_{0})}(d-1))>0$. Since $\deg(\mbox{Res}_{M}(W_{0}))\leq d+4\leq 2(d-1)+1$ ($d\geq 5$), there is a line $R\subset\mathbb{P}^{4}$ such that $\deg(R\cap\mbox{Res}_{M}(W_{0}))\geq d+1$. The case $t=1$ of Remark 5 implies $O\in R$. Since $\deg(R\cap A)\leq 2$, we have $\sharp(B\setminus B\cap M)\geq d-1$ with equality only if $R$ is the tangent line of $C$ at $O$. Assume for the moment $O\notin F$. We get $\sharp(B\cap M)\geq 1+\sharp(B\cap F)\geq 3d-2$ and hence $\sharp(N)\geq 3d-3$, a contradiction. Now assume $O\in F$. Since $\langle F\cup R\rangle\neq\mathbb{P}^{4}$, we get $e_{1}\geq(3d-3)+d-1$, a contradiction.

(b2.2.2.2) Assume the existence of a reduced conic $T\subset N_{2}$ with $\deg(Z_{2}\cap T)\geq 2d$; if $T$ is reducible also assume that $\deg(J\cap Z_{2})=d$ for every component $J$ of $T$ (Remark 1). First assume that $L$ is an irreducible component of $T$; since $L\subset N_{1}$, we have $\deg(L\cap Z_{1})\leq\deg(L\cap A)\leq 2$, a contradiction. Now assume that $L$ is not an irreducible component of $T$, but that $L\subset\langle T\rangle$; this case is done as in step (b2.2.2.1). Now assume $L\cap\langle T\rangle=\emptyset$ and hence $O\notin\langle T\rangle$. Let $Q$ be a quadric hypersurface containing $T\cup L$ and at least one other point of $B$ (it exists, because $h^{0}(\mathcal{I}_{L\cup T}(2))=7$. Since $\deg(W_{0})-\deg(W_{0}\cap Q)\leq d-2$, then $h^{1}(\mathcal{I}_{\mbox{Res}_{Q}(W_{0})}(d-2))=0$ and hence (bb2, , Lemma 5.1) gives $W_{0}\subset Q$. Hence $h^{0}(\mathcal{I}_{A\cup L\cup T}(2))\geq 6$. Since $O\notin\langle T\rangle$, we have $h^{0}(\mathcal{I}_{A\cup T}(2))=5$, a contradiction.

Now assume $L\cap\langle T\rangle\neq\emptyset$ and call $N$ any hyperplane containing $T\cup L$. Either by a residual exact sequence (case $h^{1}(N,\mathcal{I}_{W_{0}\cap N}(d))=0$) or quoting (bb2, , Lemma 5.1) we get $h^{1}(\mathcal{I}_{\mbox{Res}_{N}(W_{0})}(d-1))>0$. Since $\deg(\mbox{Res}_{N}(W_{0}))\leq d+1$, there is a line $R\subset\mathbb{P}^{4}$ such that $\deg(R\cap\mbox{Res}_{N}(W_{0}))\geq d+1$. We have $L\neq R$. Remark 5 gives $O\in R$ and $\sharp(R\cap(B\setminus N))\geq d-1$ with strict inequality, unless $\sharp(B\cap L)=d$. Let $Q$ be any quadric surface containing the two conics $L\cup R$ and $T$ (it exists, because $h^{0}(\mathcal{I}_{L\cup R\cup T}(2))\geq 5$). Since $\deg(\mbox{Res}_{Q}(W_{0}))\leq 5$, we have $h^{1}(\mathcal{I}_{\mbox{Res}_{Q}(W_{0})}(d-2))=0$ and hence (bb2, , Lemma 5.1) gives $W_{0}\subset Q$. Since $A$ is curvilinear, we get $A\cup B\subset L\cup R\cup T$. Since $A\subset B\cup L\cup R\cup T$, the reduced conic must be reducible and the four lines of $L\cup R\cup T$ pass through the point $O_{1}$. Moreover, the union of any 3 of these lines span a hyperplane. Write $L\cup R\cup T=L_{1}\cup L_{2}\cup L_{3}\cup L_{4}$ with each $L_{i}$ a line. We have $\sharp(B\cap(L_{i}\setminus\{O_{1}\}))+1\leq\deg(L_{i}\deg(L_{i}\cap W_{0})\leq\sharp(B\cap(L_{i}\setminus\{O_{1}\}))+2$ and the second inequality is not strict if and only if $L_{i}$ is the tangent line of $C$. Hence the second inequality is not strict for at most one index $i$. If $T=L_{i}\cup L_{j}$ we also have $\deg(T\cap W_{0})\leq\deg(T\cap L_{i})+\deg(L_{j}\cap W_{0})$. We get $\sharp(B\setminus\{O_{1}\})\geq-5+(d+1)+(d+1)+2d=4d-3$, a contradiction.

(b2.2.3) Assume $g^{\prime}=1$.

(b2.2.3.1) Assume $\deg(L\cap W_{0})\geq d+2$. By the case $t=1$ of Remark 5 we have $\sharp(B\cap L)=d$, $O\notin B$ and $L$ is the tangent line to $C$ at $O$. The maximality property of $N_{1}$ implies $f_{1}\geq d+4$ and hence $\deg(Z_{1})\leq 3d-3$. Since $W_{0}\nsubseteq N_{1}$, then $h^{1}(\mathcal{I}_{Z_{1}}(d-1))>0$. By Lemma 1 and Remark 3 either there is a line $R\subset\mathbb{P}^{4}$ with $\deg(R\cap Z_{1})\geq d+1$ or there is a reduced conic $T\subset\mathbb{P}^{4}$ with $\deg(Z_{1}\cap T)\geq 2d$ (if $T$ is reducible we may also assume $\deg(J\cap W_{0})=d$ for each irreducible component $J$ of $T$) or there is reduced plane cubic $F\subset\mathbb{P}^{4}$ with $\deg(F\cap Z_{1})\geq 3d-3$.

(b2.2.3.1.1) Assume the existence of $F$. In particular we have $e_{1}\geq 3d-2$. Fix any plane $M\supset L$. Since $\deg((H_{1}\cap M)\cap W_{0})\geq 4d-1$, we have $h^{1}(\mathcal{I}_{\mbox{Res}_{H_{1}\cup M}(W_{0})}(d-2))=0$ and hence $W_{0}\subset H_{1}\cup N$. Recall that in this step $L$ is the tangent line to $C$ at $O$. Set $c:=\deg(A\cap H_{1})$. Since $L\notin H_{1}$, we have $c\leq 1$. Since $A\subset H_{1}\cup N$ for each hyperplane containing $L$, we get $c+2\geq 5$, a contradiction.

(b2.2.3.1.2) Assume the existence of $T$. Since $L\subset N_{1}$ and $\deg(J\cap Z_{1})>\deg(J\cap A)$ for any line $J\subset T$, the set $T\cap L$ is finite. Assume for the moment $L\subset\langle T\rangle$. Since $\deg(W_{0}\cap\langle T\rangle)\geq d+\deg(T\cap Z_{1})\geq 3d$, we get $e_{1}\geq 3d+1$ and hence $e_{2}\leq d$, a contradiction. Now assume $L\nsubseteq\langle T\rangle$ and $L\cap\langle T\rangle\neq\emptyset$. Let $M$ be the hyperplane spanned by $L\cup T$. Since $\langle T\rangle$ does not contain the tangent line, $L$, of $C$ at $O$, we have $\deg(A\cap T)\leq 1$. Hence $\sharp(B\cap(M\setminus\{O\}))\geq 3d-1$. Hence $e_{1}\geq 3d$, a contradiction.

Now assume $L\cap\langle T\rangle=\emptyset$. We have $\deg(W_{0}\cap(L\cup T))\geq 3d+2$. Since $O\in L$, we have $T\cap A=\emptyset$ and hence $A\nsubseteq T\cup L$. We have $h^{0}(\mathcal{I}_{L\cup T}(2))=7$ and $T\cup L$ is the scheme-theoretic base locus of $|\mathcal{I}_{L\cup T}(2)|$. Fix any $Q\in|\mathcal{I}_{L\cup T}(2)|$. Since $\deg(\mbox{Res}_{Q}(W_{0}))\leq d-1$, we have $h^{1}(\mathcal{I}_{\mbox{Res}_{Q}(W_{0})}(d-2))=0$ and hence $W_{0}\subset Q$. Moving $Q$ in $|\mathcal{I}_{L\cup T}(2)|$ we get $W_{0}\subset L\cup T$, a contradiction.

(b2.2.3.1.3) Assume the existence of $R$. Since $R\neq L$ and $L$ is the tangent line to $C$ at $O$, Remark 5 gives $O\in R$ and $\sharp(B\cap(R\setminus\{O\})=d$. Let $M\subset\mathbb{P}^{4}$ be a hyperplane containing $R\cup L$ and with maximal $\deg(M\cap W_{0})$. Since $A\nsubseteq M$, we have $h^{1}(\mathcal{I}_{\mbox{Res}_{M}(W_{0})}(d-1))>0$ ((bb2, , Lemma 5.1)). Since $h\geq 1+\deg(W_{0}\cap\langle R\cup L\rangle)\geq 2d+3$, we have $\deg(\mbox{Res}_{M}(W_{0}))\leq 2d-2$. Hence there is a line $D\subset\mathbb{P}^{4}$ such that $\deg(D\cap\mbox{Res}_{M}(W_{0}))\geq d+1$. Since $L\cup R\subset M$, then $D\cap(B\setminus B\cap M)\neq\emptyset$ and in particular $D\neq L$ and $D\neq L$. Since $L$ is the tangent line to $C$ at $O$, Remark 5 gives $O\in D$ and $\sharp(B\cap(D\setminus\{O\}))=d$. Let $N$ be a hyperplane containing $D\cup R\cup L$. We have $\deg(N\cap W_{0})\geq 3d+2$ and hence $h^{1}(\mathcal{I}_{\mbox{Res}_{N}(W_{0})}(d-1))=0$. Hence $A\subset N$, a contradiction.

(b2.2.3.2) Assume $\deg(L\cap W_{0})=d+1$. By (bgi, , Lemma 34) we have $f_{1}\geq 2d+2$. Since $W_{0}\nsubseteq N_{1}$, we have $h^{1}(\mathcal{I}_{Z_{1}}(d-1))>0$ ((bb2, , Lemma 5.1)). Since $\deg(Z_{1})\leq 4d+1-f_{1}\leq 2(d-1)+1$, there is a line $R\subset\mathbb{P}^{4}$ such that $\deg(R\cap Z_{1})\geq d+1$. The case $t=1$ of Remark 5 gives $O\in R$. Since $L\subset N_{1}$ and $d+1\geq 3$, we have $R\cap(B\setminus B\cap L)\neq\emptyset$ and hence $L\neq R$. Let $M$ be the hyperplane containing $L\cup R$ and with maximal $g_{1}:=\deg(M\cap W_{0})$. Since $\langle R\cup L\rangle$ is a plane, we have $g_{1}\geq 1+\deg(W_{0}\cap\langle L\cup R\rangle)\geq 2d$. Since $W_{0}\nsubseteq M$, we get $h^{1}(\mathcal{I}_{\mbox{Res}_{M}(W_{0})}(d-1))>0$. Since $\deg(\mbox{Res}_{M}(W_{0}))\leq 2d+1$, either there is a line $D\subset\mathbb{P}^{4}$ with $\deg(D\cap\mbox{Res}_{M}(W_{0}))\geq d+1$ or there is a conic $T\subset\mathbb{P}^{4}$ with $\deg(T\cap\mbox{Res}_{M}(W_{0}))\geq 2d$.

(b2.2.3.2.1) Assume the existence of $T$. First, assume $L\subset\langle T\rangle$. Since $\deg(W_{0}\cap\langle T\rangle)\geq d+\deg(T\cap Z_{1})\geq 3d$, we get $e_{1}\geq 3d+1$ and hence $e_{2}\leq d$, a contradiction. Now, assume $L\nsubseteq\langle T\rangle$ and $L\cap\langle T\rangle\neq\emptyset$. Let $M$ be the hyperplane spanned by $L\cup T$. Since $\langle T\rangle$ does not contain the tangent line, $L$, of $C$ at $O$, we have $\deg(A\cap T)\leq 1$. Hence $\sharp(B\cap(M\setminus\{O\}))\geq 3d-1$. Hence $e_{1}\geq 3d$, a contradiction. Now, assume $L\cap\langle T\rangle=\emptyset$. We have $\deg(W_{0}\cap(L\cup T))\geq 3d+1$. Since $O\in L$, we have $T\cap A=\emptyset$ and hence $A\nsubseteq T\cup L$. We have $h^{0}(\mathcal{I}_{L\cup T}(2))=7$ and $T\cup L$ is the scheme-theoretic base locus $|\mathcal{I}_{L\cup T}(2)|$. Fix any $Q\in|\mathcal{I}_{L\cup T}(2)|$. Since $\deg(\mbox{Res}_{Q}(W_{0}))\leq d$, either $h^{1}(\mathcal{I}_{\mbox{Res}_{Q}(W_{0})}(d-2))=0$ or $\deg(\mbox{Res}_{Q}(W_{0}))=d$ and there is a line $J\supset\mbox{Res}_{Q}(W_{0})$. In the former case we have $W_{0}\subset Q$ and hence $W_{0}\subset L\cup T$, a contradiction. Now assume $\deg(\mbox{Res}_{Q}(W_{0}))=d$ and $J\supset\mbox{Res}_{Q}(W_{0})$ for some line $J$. Since $\deg(A\cap J)\leq 2$, we have $J\cap(B\setminus B\cap Q)\neq\emptyset$. Hence $J\neq L$ and $J$ is not a component of $T$. Since $L\cap\langle T\rangle=\emptyset$, we have $\sharp(B\cap(L\cup T\cup J))\geq d-1+2d+d-2$, a contradiction.

(b2.2.3.2.2) Assume the existence of $D$. Remark 5 gives $O\in D\cap L$. Since $f_{1}\leq e_{1}\leq 3d$ and $g^{\prime}=1$, either there is a line $R^{\prime}\subset N_{1}$ with $\deg(R^{\prime}\cap Z_{0})\geq d+2$ or there is a conic $T^{\prime}\subset N_{1}$ with $\deg(Z_{0}\cap T^{\prime})\geq 2d+2$ or there is a plane cubic $F^{\prime}\subset N_{1}$ with $\deg(Z_{0}\cap F^{\prime})\geq 3d$ (Lemma 1). The last case cannot occur, because it would give $f_{1}\geq 3d+1$ and hence $e_{1}\geq 3d+1$.

(b2.2.3.2.2.1) Assume the existence of $R^{\prime}$. Remark 5 gives that $R^{\prime}$ is the tangent line of $C$ at $O$. Since $\deg(L\cap W_{0})=d+1$, we have $R^{\prime}\neq L$. Since $R^{\prime}\subset N_{1}$ and $D\cap(B\setminus B\cap N_{1})\neq\emptyset$, we get $D\neq R^{\prime}$. Remark 5 gives $O\in R^{\prime}$ and that $R^{\prime}$ is the tangent line of $C$ at $O$. Let $M\subset\mathbb{P}^{4}$ be any hyperplane containing $D\cup L\cup R^{\prime}$. Since at most one of the lines $L,D,R^{\prime}$ is tangent to $C$ at $O$, we have $\sharp((B\setminus\{O\})\cap M)\geq 3d$ and $e_{1}\geq\deg(W_{0}\cap M)\geq 3d+2$, a contradiction.

(b2.2.3.2.2.2) Assume the existence of $T^{\prime}$. If $O\notin T^{\prime}$, then we get $\sharp(B\cap(T^{\prime}\cup L\cup R))\geq 4d-2$, a contradiction. Hence $O\in T^{\prime}$. Since $\sharp(B)\leq 4d-4$, we also get that $\langle T^{\prime}\rangle$ contains the tangent line to $C$ at $O$. We have $\sharp(B\cap(T\setminus\{O\}))\geq 2d-3$. If either $L\subset\langle T^{\prime}\rangle$ or $R\subset\langle T\rangle$, then we get $e_{1}>\deg(W_{0}\cap\langle T\rangle)\geq 3d$, a contradiction. If $L\nsubseteq\langle T^{\prime}\rangle$ and $R\nsubseteq\langle T\rangle$, then neither $L$ nor $R$ is tangent to $C$ at $O$ and hence $\sharp((B\setminus\{O\})\cap R)\geq d$ and $\sharp((B\setminus\{O\})\cap L)\geq d$. Therefore $\sharp(B\setminus\{O\})\geq 4d-3$, a contradiction.

(b3) Assume $g=1$. Since $A\nsubseteq H_{1}$, (bb2, , Lemma 5.1) gives $h^{1}(\mathcal{I}_{W_{1}}(d-1))>0$ and in particular $\deg(W_{1})\geq d+1$. Hence $e_{1}\leq 3d$. By Lemma 1 and Remark 3 either there is a line $L\subset H_{1}$ such that $\deg(L\cap W_{0})\geq d+2$ or there is a reduced conic $T\subset H_{1}$ such that $\deg(T\cap W_{0})\geq 2d+2$ or there is a plane cubic $F\subset H_{1}$ with $\deg(W_{0}\cap F)\geq 3d$.

(b3.1) Assume the existence of $F$. Since $A$ spans $\mathbb{P}^{4}$, we get $e_{1}>\deg(F\cap W_{0})\geq 3d$, a contradiction.

(b3.2) Assume the existence of $T$. Since $T$ is a plane curve, we have $\deg(T\cap A)\leq 3$ and hence $\sharp(T\cap(B\setminus\{O\})\geq 2d-1$. Remark 5 for $t=2$ gives $O\in T$ and that $T$ and $C$ have the same tangent line at $O$. Since $A\nsubseteq H_{1}$, (bb2, , Lemma 5.1) gives $h^{1}(\mathcal{I}_{W_{1}}(d-1))>0$ and hence $e_{1}\leq 4d+1-\deg(W_{1})\leq 3d$. Since $A$ spans $\mathbb{P}^{4}$, we get $e_{1}>\deg(W_{0}\cap T)$. Hence $\deg(W_{1})\leq 2d-2$. Therefore there is a line $R\subset\mathbb{P}^{4}$ such that $\deg(R\cap W_{1})\geq d+1$. Since $R\cap(B\setminus B\cap H_{1})\neq\emptyset$, we have $R\nsubseteq H_{1}$ and in particular $R\nsubseteq\langle T\rangle$. Hence $R$ is not tangent to $C$ at $O$. Remark 5 gives $O\in R$ and $\sharp(R\cap(B\setminus\{O\}))=d$. Hence $M:=\langle T\cup R\rangle$ is a hyperplane, $e_{1}\geq\deg(M\cap W_{0})\geq\deg(T\cap W_{0})+\sharp(R\cap(B\setminus\{O\}))\geq 3d+2$, a contradiction.

(b3.3) Assume the existence of $L$. Since $A$ spans $\mathbb{P}^{4}$ we have $e_{1}\geq d+4$. Remark 5 gives $O\in L$, that $L$ is the tangent line of $C$ at $O$ and that $\sharp(B\cap(L\setminus\{O\}))=d$ (it also gives $O\notin B$). Since $A\nsubseteq H_{1}$, (bb2, , Lemma 5.1) gives $h^{1}(\mathcal{I}_{W_{1}}(d-1))>0$. We have $e_{1}+\deg(W_{1})\leq 4d+1$ and in particular $\deg(W_{1})\leq 3d-3$. Lemma 1 gives either the existence of a line $R\subset\mathbb{P}^{4}$ with $\deg(R\cap W_{1})\geq d+1$ or the existence of a conic $T^{\prime}\subset\mathbb{P}^{4}$ with $\deg(T^{\prime}\cap W_{1})\geq 2d$ or the existence of a plane cubic $F$ with $\deg(F\cap W_{1})\geq 3d-3$. The latter case is impossible, because it would give $e_{1}\geq 3d-2$ and hence $\deg(W_{1})\leq d+4$. Since $\deg(W_{1})\geq d+1$, then $e_{1}\leq 3d$.

(b3.3.1) Assume the existence of the line $R$. Since $R\cap(B\setminus B\cap H_{1})\neq\emptyset$, we have $R\nsubseteq H_{1}$ and hence $R\neq L$. Since $L$ is the tangent line of $C$ at $O$, we have $\deg(R\cap A)\leq 1$. Remark 5 gives $O\in R$, $O\notin B$ and $\sharp(B\cap R)=d$. Let $M\subset\mathbb{P}^{4}$ be a hyperplane containing $R\cup L$ and with maximal $h:=\deg(M\cap W_{0})$. Since $A\nsubseteq M$, we have $h^{1}(\mathcal{I}_{\mbox{Res}_{M}(W_{0})}(d-1))>0$. We have $h\geq 2d+3$ and hence $\deg(\mbox{Res}_{M}(W_{0}))\leq 2d-2$. Therefore there is a line $D\subset\mathbb{P}^{4}$ such that $\deg(D\cap\mbox{Res}_{M}(W_{0}))\geq d+1$. Since $D\cap(B\setminus B\cap M)\neq\emptyset$, we have $D\nsubseteq\langle R\cup L\rangle$. Since $L$ is the tangent line of $C$ at $O$, Remark 5 gives $O\in D$ and $\sharp(D\cap B)=d$. Let $N$ be the hyperplane spanned by $D\cup R\cup L$. Since $O\notin B$, we have $e_{1}\geq\deg(M\cap W_{0})\geq d+2+d+d$, contradicting the inequality $e_{1}\leq 3d$ proved in step (b3.3).

(b3.3.2) Assume the existence of the conic $T^{\prime}$ but the non-existence of any line $R$ with $\deg(R\cap W_{1})\geq d$. If $T$ is reducible, then $\deg(J\cap W_{1})=d$ for each component $J$ of $T^{\prime}$ (Remark 3). Since $A$ spans $\mathbb{P}^{4}$, we have $e_{1}>\deg(T^{\prime}\cap W_{1})$. Therefore $e_{1}=2d+1$, $\deg(T^{\prime}\cap W_{1})=2d$ and $W_{1}\subset T^{\prime}$. Since $e_{1}\leq 3d-2$, we have $L\cap\langle T^{\prime}\rangle=\emptyset$. We have $h^{0}(\mathcal{I}_{L\cup T^{\prime}}(2))\geq 7$. Let $Q$ be any quadric hypersurface such that $\deg(W_{0}\cap Q)>\deg(W_{0}\cap(L\cup T^{\prime}))$. Since $\deg(\mbox{Res}_{Q}(W_{0}))\leq 4d+1-(3d+2)$, we have $h^{1}(\mathcal{I}_{\mbox{Res}_{Q}(W_{0})}(d-2))=0$ and hence $A\subset Q$ ((bb2, , Lemma 5.1)). Since $L\cap\langle T^{\prime}\rangle=\emptyset$, then $L\cup T^{\prime}$ is the scheme-theoretic base-locus of $|\mathcal{I}_{L\cup T^{\prime}}(2)|$. Since $A$ is connected, we get $A\subset L$, a contradiction.

(c) Proposition 1 covers the case $b_{1}=2$ and $b_{2}=b_{3}=b_{4}=1$. Proposition 2 covers the case $b_{1}=b_{2}=2$ and $b_{3}=1$. Proposition 4 covers the case $b_{1}=3$ and $b_{2}=b_{3}=1$. Proposition 3 covers the case $b_{1}=3$ and $b_{2}=2$.

(d) Now assume $s=2$, $b_{1}=4$, $b_{2}=1$. Assume $\sharp(B)\leq 3d-2$ and hence $\deg(W_{0})\leq 3d+3$. Write $A=A_{1}\sqcup\{O_{2}\}$ and set $\{O_{1}\}:=(A_{1})_{red}$. We repeat the proof of Proposition 3, except that now $\{O_{2}\}$ is the unique reduced connected component of $A$ and hence we cannot freely use (bb2, , Lemma 5.1); however, $\deg(W_{0})<3d+4$. We write (++xxxx) for the set-up of step (xxx) of the proof of Proposition 3 and use the notation of that step.

(++b1.1.1) Assume $h^{1}(\mathcal{I}_{\mbox{Res}_{M}(W_{0})}(d-1))=0$. We get $O_{2}\in B$ and $A_{1}\cup B_{1}\subset M$, where $B_{1}:=B\setminus\{O_{2}\}$. Since $P\neq\nu_{d}(O_{2})$, we get that $\langle\nu_{d}(A_{1})\rangle\cap\langle\nu_{d}(B_{1})\rangle\neq\emptyset$, that $\nu_{d}(O_{2})\notin\langle\nu_{d}(A_{1})\rangle\cap\langle\nu_{d}(B_{1})\rangle$ and that $\langle\nu_{d}(A)\rangle\cap\langle\nu_{d}(B)\rangle$ is the linear span of $\nu_{d}(O_{2})$ and $\langle\nu_{d}(A_{1})\rangle\cap\langle\nu_{d}(B_{1})\rangle$. Hence there is a unique $P_{1}\in\langle\nu_{d}(A_{1})\rangle\cap\langle\nu_{d}(B_{1})\rangle$ such that $P\in\langle\{P_{1},O_{2}\}$. Since $P\notin\langle W\rangle$ for any $W\subseteq A$ and any $P\subsetneq W$, we get $P_{1}\notin\langle A^{\prime}\rangle$ for any $A^{\prime}\subsetneq A^{\prime}$ and any $A^{\prime}\subsetneq B$. Since $A_{1}$ evinces the scheme-rank of $P_{1}$, (bb2, , Proposition 5.19) gives $r_{m,d}(P_{1})=3d-2$.

(++b1.2) If $A\cup B\subset Q$, then that proof works. Assume $A\cup B\nsubseteq Q$. We get $O_{2}\in B$ and $A_{1}\cup B_{1}\subset Q$, where $B_{1}:=B\setminus\{O_{2}\}$. Varying $Q$ we get $A_{1}\cup B_{1}\subset F\cup L$. Since $F\cap L=\emptyset$ and $\langle F\rangle$ is a plane, we have $A_{1}\nsubseteq F\cup L$, a contradiction.

(++b2) Assume $h^{1}(\mathcal{I}_{Z_{1}}(d-1))=0$. We repeat step (++b1.1.1) with the hyperplane $N_{1}$ instead of the hyperplane $M$.

(++c) Assume $h^{1}(\mathcal{I}_{W_{1}}(d-1))=0$. We repeat step (++b1.1.1) with the hyperplane $H_{1}$ instead of the hyperplane $M$.

(++c1) Assume $h^{1}(\mathcal{I}_{\mbox{Res}_{M}(W_{0})}(d-1))>0$. We repeat step (++b1.1.1).

Proposition 1

Fix integers $d,s,m$ such that $0\leq s\leq m-1$, and $d\geq m$. Fix a degree $2$ connected zero-dimensional scheme $A_{1}\subset\mathbb{P}^{m}$ and a set $S\subset\mathbb{P}^{m}$ such that $S\cap A_{1}=\emptyset$, $\sharp(S)=s$ and the scheme $A:=A_{1}\cup S$ is linearly independent. Set $\{O\}:=(A_{1})_{red}$. Fix any $P\in\langle\nu_{d}(A)\rangle$ such that $P\notin\langle\nu_{d}(A^{\prime})\rangle$ for any $A^{\prime}\subsetneq A$. Then $r_{m,d}(P)=s+d$ and every $B\in\mathcal{S}(P)$ is the disjoint union of $S$ and $d$ points of $\langle A_{1}\rangle\setminus\{O\}$.

Proof

By concision ((l, , Exercise 3.2.2.2)) it is sufficient to do the case $\langle A\rangle=\mathbb{P}^{m}$, i.e. the case $m=s+1$. Since the case $m=1$ is true by Sylvester’s theorem (cs ), we may assume $m\geq 2$ and use induction on $m$. Since $h^{1}(\mathcal{I}_{A}(d))=0$ and $A$ has only finitely many proper subschemes, $P$ exists and there is a unique point $Q\in\langle\nu_{d}(A_{1})\rangle\setminus\{O\}$ such that $P\in\langle\{Q\}\cup\nu_{d}(S)\rangle$. Since $r_{1,d}(Q)=d$ by a theorem of Sylvester, we get $r_{m,d}(P)\leq s+d$. Fix $B\in\mathcal{S}(P)$. Set $W_{0}:=A\cup B$. We have $\deg(W_{0})\leq d+2s+2=d+2m$. Since $A$ is not reduced, we have $A\neq B$ and hence $h^{1}(\mathcal{I}_{W_{0}}(d))>0$. Take $H_{i},e_{i},W_{i},g$ as in the proof of Theorem 1.1. Since $A$ spans $\mathbb{P}^{m}$, if $e_{i}\leq m-1$, then $e_{i+1}=0$ and $W_{i}=\emptyset$. Hence $g\leq d$. If $2\leq g<d$ we also get $e_{i}\geq d+m+2-g$ for $i<g$ and hence $d+2m\geq g(d+m+2-g)-m+1$ and hence $g\leq 2$. If $g=d$, then $e_{d}\geq 3$, $e_{i}\geq m$ if $i<d$ and hence $\deg(W_{0})\geq m(d-1)+3$, a contradiction.

(a) Assume $g=2$. Set $k:=\dim(\langle W_{1}\rangle)$. Remark 4 for the integer $d-1$ gives $e_{2}\geq k+d$. Since $A$ spans $\mathbb{P}^{4}$ we have $e_{1}\geq e_{2}+m-k\geq m+d$. Hence $2d+m+k\leq d+2m$, contradicting the inequality $d\geq m$.

(b) Assume $g=1$. Since $A$ spans $\mathbb{P}^{m}$, Remark 4 gives $e_{1}\geq d+m+1$.

(b1) Assume $h^{1}(\mathcal{I}_{W_{1}}(d-1))=0$. We get $A_{1}\subset H_{1}$ and $S\setminus S\cap H_{1}=B\setminus B\cap H_{1}$ ((bb2, , Lemma 5.1)). Since $A$ is in linearly general position and spans $\mathbb{P}^{m}$, we get that $S\setminus S\cap H_{1}$ is a point; call it $Q$. Since $h^{1}(\mathcal{I}_{W_{1}}(d-1))=0$, Grassmann’s formula gives that $\langle\nu_{d}(A)\rangle\cap\langle\nu_{d}(B)\rangle$ is the linear span of $\nu_{d}(Q)$ and $\langle\nu_{d}(A\setminus\{Q\})\rangle\cap\langle\nu_{d}(B\setminus\{Q\})\rangle$. Fix $P_{1}\in\langle\nu_{d}(A\setminus\{Q\})\rangle\cap\langle\nu_{d}(B\setminus\{Q\})\rangle$ such that $P\in\langle\{\nu_{d}(Q),P_{1}\}\rangle$. Since $P\notin\langle A^{\prime}\rangle$ for any $A^{\prime}\subsetneq A$, we get $P_{1}\notin\langle\nu_{d}(A\setminus\{Q\})\rangle$. In the same way we get $P_{1}\notin\langle\nu_{d}(B^{\prime})\rangle$ for any $B^{\prime}\subsetneq B\setminus\{Q\}$. We may use the inductive assumption in $m$ and get that $B$ is as claimed in the statement of Proposition 1.

(b2) Assume $h^{1}(\mathcal{I}_{W_{1}}(d-1))>0$ and set $k:=\dim(\langle W_{1}\rangle)$. Remark 4 gives $\deg(W_{1})\geq d+k$ and $e_{1}\geq d+m$. and hence $d+k\leq m$, a contradiction.

Proposition 2

Assume $d\geq 7$. Fix $O_{3}\in\mathbb{P}^{m}$, $m\geq 4$ and let $A_{1},A_{2}$ be connected degree 2 subschemes of $\mathbb{P}^{m}$ such that $A:=A_{1}\cup A_{2}\cup\{O_{3}\}$ spans a 4-dimensional linear subspace. Set $\{O_{i}\}:=(A_{i})_{red}$. Fix $P\in\langle\nu_{d}(A)\rangle$ such that $P\notin\langle\nu_{d}(A^{\prime})\rangle$ for any $A^{\prime}\subsetneq A$. Then $r_{m,d}(P)=2d+1$. There are uniquely determined $P_{i}\in\langle\nu_{d}(A_{i})\rangle$, $i=1,2$, such that $A_{i}$ evinces the cactus rank of $P_{i}$ and every $B\in\mathcal{S}(P)$ has a decomposition $B=B_{1}\sqcup B_{2}\sqcup\{O_{3}\}$ with $B_{i}\subset\langle A_{i}\rangle\setminus\{O_{i}\}$ $\sharp(B_{i})=d$ and $B_{i}\in\mathcal{S}(P_{i})$.

Proof

By concision ((bb2, , Exercise 3.2.2.2)) we may assume $m=4$. Since $\nu_{d}(A)\rangle$ is linearly independent, there are unique $P_{i}\in\langle\nu_{d}(A_{i})\rangle$, $i=1,2$, such that $P\in\langle\{P_{1},P_{2},\nu_{d}(O_{3})\}\rangle$. Since $P\notin\langle\nu_{d}(A^{\prime})\rangle$ for any $A^{\prime}\subsetneq A$, we have $P_{i}\neq\nu_{d}(A_{i})$. Hence $P_{i}$ has border rank 2 with respect to the degree $d$ rational normal curve $\nu_{d}(\langle A_{i}\rangle$. Since $P\in\langle\nu_{d}(B_{1}\cup B_{2}\cup\{O_{3}\})\rangle$ for all $B_{i}\in\mathcal{S}(P_{i})$, we get $r_{m,d}(P)\leq 2d+1$. Fix any $B\in\mathcal{S}(P)$ and set $W_{0}:=A\cup B$. We have $\deg(W_{0})\leq 2d+6$. Define $H_{i},e_{i},W_{i},g$ as in the proof of Theorem 1.1. Since $\deg(W_{0})\leq 2d+6$, we get $g\leq 2$.

(a) Assume $g=2$. Since $e_{1}\geq e_{2}$ and $e_{1}+e_{2}\leq 2d+6$, there is a line $L\subset H_{2}$ such that $\deg(L\cap W_{1})\geq d+1$. Hence $e_{1}\leq d+5$. Let $N_{1}\subset\mathbb{P}^{4}$ be a hyperplane containing $L$ and with maximal $f_{1}:=\deg(N_{1}\cap W_{0})$. Define $N_{i},f_{i},Z_{i},g^{\prime}$ as in step (b1) of the proof of Theorem 1.1. We get $g^{\prime}\leq 2$.

(a1) Assume $g^{\prime}=2$. Since $A$ spans $\mathbb{P}^{4}$ we have $f_{1}\geq\deg(L\cap W_{1})+2\geq d+3$ and hence $f_{2}\leq d+3$. Therefore there is a line $R\subset N_{2}$ such that $\deg(Z_{1}\cap R)\geq d+1$. Since $\deg(R\cap A)\leq 2$ and $Z_{1}\cap B\subseteq B\setminus B\cap L$, we get $L\neq R$. Any hyperplane $M$ containing $L\cup R$ shows that $e_{1}\geq(d+1)+(d+1)-1$, a contradiction.

(a2) Assume $g^{\prime}=1$.

(a2.1) Assume $\deg(L\cap W_{0})\geq d+2$. The case $t=1$ of Remark 5 gives $\deg(L\cap A)=2$, $\sharp(B\cap L)=d$ and $L\cap A\cap B=\emptyset$. Since $A$ spans $\mathbb{P}^{4}$, we get $f_{1}\geq d+4$.

(a2.1.1) Assume $h^{1}(\mathcal{I}_{Z_{1}}(d-1))>0$. Since $\deg(Z_{1})\leq 2d+6-f_{1}\leq 2(d-1)+1$, there is a line $R\subset\mathbb{P}^{4}$ such that $\deg(Z_{1}\cap R)\geq d+1$ and hence with $R\setminus R\cap L\neq\emptyset$ Hence $R\neq L$. Any hyperplane $M$ containing $L\cup R$ has $\deg(M\cap W_{0})\geq d+2+d+1-1$, contradicting the inequality $e_{1}\leq d+5$.

(a2.1.2) Assume $h^{1}(\mathcal{I}_{Z_{1}}(d-1))=0$. Since $A\nsubseteq N_{1}$, (bb2, , Lemma 5.1) gives $O_{3}\in B$ and $A_{1}\cup A_{2}\cup(B\setminus\{O_{3}\}\subset N_{1}$. Since $e_{1}\leq d+5$, we get $\sharp(B)\leq d+2$. Since $\deg(W_{0})\leq d+7$, we immediately get $g=1$, a contradiction.

(a2.2) Assume $\deg(L\cap W_{0})=d+1$. Since $e_{1}\leq d+5\leq 2d+1$, there is a line $R\subset N_{1}$ such that $\deg(N_{1}\cap R)\geq d+2$ ((bgi, , Lemma 34)). Since $R\neq L$, we get $d+5\geq e_{1}\geq\deg(N\cap(L\cup R)\geq d+1+d+2-1$, a contradiction.

(b) Assume $g=1$.

(b1) Assume $h^{1}(\mathcal{I}_{W_{1}}(d-1))>0$. We get $\deg(W_{1})\geq d+1$ ((bgi, , Lemma 34)) and hence $e_{1}\leq d+5$. Since $h^{1}(H_{1},\mathcal{I}_{H_{1}\cap W_{0}}(d))>0$ and $e_{1}\leq 2d+1$, there is a line $L\subset H_{1}$ such that $\deg(L\cap W_{0})\geq d+2$. In particular $e_{1}\geq d+2$ and hence $\deg(W_{1})\leq d+4$. Since $2(d-1)+1\leq d+4$, we get the existence of a line $R\subset\mathbb{P}^{4}$ such that $\deg(W_{1}\cap R)\geq d+1$. Since $\deg(A\cap R)\leq 2$, we obtain $R\cap(B\setminus B\cap L)\neq\emptyset$ and hence $R\neq L$. Any hyperplane containing $L\cup R$ gives $e_{1}\geq(d+2)+(d+1)-1>d+5$, a contradiction.

(b2) Assume $h^{1}(\mathcal{I}_{W_{1}}(d-1))=0$. Since $A\nsubseteq H_{1}$, we get $O_{3}\in B$ and $A_{1}\cup A_{2}\cup(B\setminus\{O_{3}\})\subset N_{1}$ ((bb2, , Lemma 5.1)). Since $\nu_{d}(A)$ is linearly independent, there is a unique $Q\in\langle\nu_{d}(A_{1}\cup A_{2})\rangle$ such that $P\in\langle\{Q\cup\nu_{d}(O_{3})\}\rangle$ (we have $\{Q\}=\langle\nu_{d}(A_{1}\cup A_{2})\rangle\cap\langle\{P,\nu_{d}(O_{3})\}\rangle$). We get that $A_{1}\cup A_{2}$ evinces the cactus rank and the border rank of $Q$ and that $Q\in\langle\nu_{d}(B\setminus\{O_{3})\rangle$. Since $B\in\mathcal{S}(P)$, we get $B\setminus\{O\}\in\mathcal{S}(Q)$. Now we repeat the construction in $H_{1}$ with $A^{\prime}:=A_{1}\cup A_{2}$ and $B^{\prime}:=B\setminus\{O_{3}\}$. From now on we work inside the 3-dimensional projective space $H_{1}$ and we use planes contained in $H_{1}$ as hyperplane. We use the notations $W_{i},H_{i},e_{i},g,Z_{i},f_{i},g_{i}$ in this new set-up. For instance $W_{0}=(A_{1}\cup A_{2})\cup(B\setminus\{O\})$ and with respect to the point $Q$ instead of the point $P$. We need to check that $B\setminus\{O_{3}\}\in\mathcal{S}(Q)$ and describe $B\setminus\{O_{3}\}$. For the first part it is sufficient to apply (bb2, , Proposition 5.17) (which gives $r_{m,d}(Q)\leq 2d$), because we have $\sharp(B\setminus\{O_{3}\})\leq 2d$. In particular we checked that $r_{m,d}(P)=2d+1$, i.e. the part of Proposition 2 used in Theorem 1.1. The only difference with respect to the proof of Theorem 1.1 is that (since we are working with planes) we only now that if $e_{i}\leq 2$, then $e_{i+1}=0$ and $W_{i}=\emptyset$ and that if $f_{i}\leq 2$, then $f_{i+1}=0$ and $Z_{i}=0$. We have $w_{0}:=\deg(A_{1}\cup A_{2}\cup(B\setminus\{O_{3}\}))\leq 2d+4$. First assume $g\geq d+2$ (resp. $g^{\prime}\geq d+1$). We get $w_{0}\geq 1+3(d+1)$, a contradiction. Now assume $g=d+1$ (resp. $g^{\prime}=d+1$). We find $w_{0}>3d$, a contradiction. Now assume $g<d$ and $e_{g}\geq 2(d+1-g)+2$ (resp. $g^{\prime}<d$ and $f_{g^{\prime}}\geq 2(d+1-g)+2$). We get $w_{0}\geq 2g(d+1-g)+4$ (resp. $w_{0}\geq 2g^{\prime}(d+1-g^{\prime})+4$) and hence $g=1$ (resp. $g^{\prime}=1$). In particular if $H_{1}$ is a plane of $\mathbb{P}^{3}$ with $\deg(H_{1}\cap W_{0})$ maximal, then $h^{1}(H_{1},\mathcal{I}_{H_{1}\cap W_{0}}(d))>0$ and hence (since $H_{1}\cap W_{0}$ spans $H_{1}$) $e_{1}\geq d+3$ (Remark 4). Since $A_{1}\cup A_{2}$ spans $\mathbb{P}^{3}$ and no connected component of $A_{1}\cup A_{2}$ is reduced, (bb2, , Lemma 5.1) gives $h^{1}(\mathcal{I}_{W_{1}}(d-1))>0$ and in particular $\deg(W_{1})\geq d+1$. Hence $e_{1}\leq d+3$, which implies $\deg(W_{1})=d+1$ and the existence of a line $R$ such that $W_{1}\subset R$. Since $e_{1}\leq d+3$, there is a line $L\subset H_{1}$ such that $\deg(L\cap W_{0})\geq d+2$. Remark 5 gives that either $L=\langle A_{1}\rangle$ or that $L=\langle A_{2}\rangle$, $\sharp(B\cap L)=d$ and $B\cap(A_{1}\cup A_{2})\cap L)=\emptyset$. Up to renaming $A_{1}$ and $A_{2}$ we may assume $L=\langle A_{1}\rangle$. Remark 5 gives $O_{1}\notin B$ and $\sharp(B\cap L)=d$, i.e. $B\cap\langle A_{1}\rangle$ is as described in the statement of Proposition 2. Remark 5 gives $O_{2}\in R$ and that either $O_{2}\notin H_{1}$, $R=\langle A_{2}\rangle$ and $\sharp(B\cap W_{1})=d-1$ or $O_{2}\in H_{1}\cap R$, $R=\langle A_{2}\rangle$ and $\sharp(B\cap(R\setminus\{O_{2},O_{3}\}))=d+1$. In the latter case we get $O_{2}\notin B$, because we know that $\sharp(B)\leq 2d+1$. Hence in the latter case $B\setminus\{O_{3}\}\subset\langle A_{1}\rangle\cup\langle A_{2}\rangle$ and each $B\cap\langle A_{i}\rangle$ is as described in Proposition 2. Now assume $O_{2}\notin H_{1}$. Recall that $R=\langle A_{2}\rangle$ and $L=\langle A_{1}\rangle$. Since $e_{1}=d+3$, there is a unique $o\in(B\setminus\{O_{3}\})\cap H_{1}$ such that $H_{1}=\langle L\cup\{o\}\rangle$. Fix any $Q\in(B\setminus B\cap H_{1})$ and set $M:=\langle\{Q\}\cup L\rangle$. Since $e_{1}=d+3$, we have $o\notin M$. Since $A_{1}\cup A_{2}$ has no reduced connected component, (bb2, , Lemma 5.1) gives $h^{1}(\mathcal{I}_{\mbox{Res}_{M}(W_{0})}(d-1))>0$. We first get the existence of a line $R^{\prime}\subset\mathbb{P}^{3}$ such that $\deg(R^{\prime}\cap\mbox{Res}_{M}(W_{0}))\geq d+1$ and then $\deg(\mbox{Res}_{M}(W_{0}))=d+1$, $\mbox{Res}_{M}(W_{0})\subset R^{\prime}$ and $R^{\prime}=R$. We get $o\in\langle A_{2}\rangle$. Hence $B\cap\langle A_{2}\rangle$ is as described in the statement of Proposition 2 and $B\setminus\{O_{3}\}\subset\langle A_{2}\rangle\cup\langle A_{1}\rangle$.

For a description of $\mathcal{S}(P_{1})$ and $\mathcal{S}(P_{2})$ below see (b, , part (b3) of Theorem 4 and part (f) of §4) and(b, , Proposition 1 and Theorem 3), respectively. The case in which the conic $C\supset A_{1}\cup B_{1}$ is reducible is described in (b, , Proposition 7).

Proposition 3

Fix an integer $d\geq 9$. Let $A_{1}\subset\mathbb{P}^{4}$ be a degree $3$ connected curvilinear scheme spanning a plane and $A_{2}\subset\mathbb{P}^{4}$ be a degree two connected curvilinear scheme. Assume $\langle A_{1}\rangle\cap\langle A_{2}\rangle=\emptyset$, i.e. assume $A_{1}\cap A_{2}=\emptyset$ and $\dim(\langle A_{1}\cup A_{2}\rangle)=4$. Then $r_{m,d}(P)=3d-1$. Fix any $B\in\mathcal{S}(P)$. Then there is a unique decomposition $B=B_{1}\sqcup B_{2}$ with $B\cap A_{red}=\emptyset$, $\sharp(B_{1})=2d-1$, $\sharp(B_{2})=d$, $B_{1}\subset\langle A_{1}\rangle$, $B_{2}\subset\langle A_{2}\rangle$, each $B_{i}$ evincing the symmetric tensor rank of a uniquely determined point of $P_{i}\in\langle\nu_{d}(A_{i})\rangle$.The scheme $A_{1}\cup B_{1}$ is contained in a reduced conic $C\subset\langle A_{1}\rangle$.

Conversely, for any $P_{i}\in\langle\nu_{d}(A_{i})\rangle$, $i=1,2$, such that $P_{i}\notin\langle\nu_{d}(A^{\prime})\rangle$ for any $A^{\prime}\subsetneq A_{i}$, any $B_{i}\in\mathcal{S}(P_{i})$ and any $P\in\langle\{P_{1},P_{2}\}\rangle\setminus\{P_{1},P_{2}\}$ we have $B_{1}\cap B_{2}=\emptyset$ and $B_{1}\cup B_{2}\in\mathcal{S}(P)$.

Proof

Since $P\in\langle\nu_{d}(A)\rangle$ and $\langle\nu_{d}(A_{1})\rangle\cap\langle\nu_{d}(A_{2})\rangle=\emptyset$, there are uniquely determined $P_{i}\in\langle\nu_{d}(A_{i})\rangle$, $i=1,2$, such that $P\in\langle\{P_{1},P_{2}\}\rangle$. Since $P\notin\langle\nu_{d}(A^{\prime})\rangle$ for any $A^{\prime}\subsetneq A$, then $P\neq P_{i}$ and $P_{i}\notin\langle\nu_{d}(A^{\prime})\rangle$ for any $A^{\prime}\subsetneq A$. Every element of $\mathcal{S}(P_{i})$ is contained in $\langle A_{i}\rangle$ ((l, , Exercise 3.2.2.2)) and each element of $\mathcal{S}(P_{i})$ is known ((b, , Proposition 1 and Theorem 2) for $i=1$, (b, , Theorem 3 and §4) for $i=2$). Since $\langle A_{1}\rangle\cap\langle A_{2}\rangle=\emptyset$, we have $B_{1}\cap B_{2}=\emptyset$ for all $B_{i}\in\mathcal{S}(P_{i})$.

Set $\{O_{1}\}=(A_{1})_{red}$ and $\{O_{2}\}=(A_{2})_{red}$. Fix $B\in\mathcal{S}(P)$. Since $r_{m,d}(P_{1})=2d-1$ and $r_{m,d}(P_{2})=d$, we have $r_{m,d}(P)\leq 3d-1$. Hence to prove all the assertions of the proposition it is sufficient to prove that $B=B_{1}\sqcup B_{2}$ with $B_{i}\in\mathcal{S}(P_{i})$ for all $i$. Set $W_{0}:=A\cup B$. We have $w_{0}:=\deg(W_{0})\leq 3d+4$. Define $e_{i},H_{i},W_{i},g$ as in step (b) of the Theorem 1.1. Since $w_{0}\leq 3d+4$, we get $g\leq 3$.

(a) Assume $g=3$. Since $e_{1}\geq e_{2}\geq e_{3}$ and $e_{1}+e_{2}+e_{3}\leq 3d+4$, we have $e_{3}\leq d+1\leq 2(d-2)+1$. Since $h^{1}(H_{3},\mathcal{I}_{W_{2}\cap H_{3}}(d-2))>0$, there is a line $L\subset H_{3}$ such that $\deg(L\cap W_{2})\geq d$. Since $A$ spans $\mathbb{P}^{4}$, the maximality property of the integer $e_{2}$ gives $e_{2}\geq 2+\deg(L\cap W_{1})\geq d+2$. Since $e_{1}\geq e_{2}$, we get $e_{1}=e_{2}=d+2$ and $e_{3}=d$. Set $Z_{0}:=W_{0}$ and define $Z_{i},N_{i},f_{i},g^{\prime}$ as in step (b1) of the proof of Theorem 1.1. As above we get $g^{\prime}\leq 3$ and $f_{1}=d+2$.

(a1) Assume $g^{\prime}=3$. As above we get $f_{1}=f_{2}=d+2$, $f_{3}=d$ and the existence of a line $R\subset N_{3}$ such that $Z_{3}\subset R$. Since $\deg(R\cap A)\leq 2$, we have $R\cap(B\setminus B\cap N_{1})\neq\emptyset$ and hence $R\neq L$. Hence $\deg(W_{0}\cap(L\cup R))\geq d+d-1>d+2=e_{1}$, contradicting the existence of a hyperplane containing $L\cup R$.

(a2) Assume $g^{\prime}=2$. Since $f_{2}\leq 3d+4-f_{1}$, Lemma 1 gives that either there is a line $D\subset N_{2}$ with $\deg(D\cap Z_{1})\geq d+1$ or there is a conic $F\subset N_{2}$ with $\deg(F\cap W_{1})\geq 2d$. The latter case is impossible, because $e_{1}<2d$. Hence $D$ exists. Since $\deg(D\cap A)\leq 2$, we have $D\cap(B\setminus B\cap N_{1})\neq\emptyset$ and hence $D\neq L$. Any hyperplane $M\supset D\cup L$ gives $e_{1}\geq(d+1)+d-1>d+2$, a contradiction.

(a3) Assume $g^{\prime}=1$. Since $f_{1}\leq e_{1}=d+2$, (bgi, , Lemma 34) gives the existence of a line $T\subset N_{1}$ such that $\deg(Z_{0}\cap T)\geq d+2$. Since $A$ spans $\mathbb{P}^{4}$, we get $e_{1}\geq 2+\deg(Z_{0}\cap T)$, a contradiction.

(b) Assume $g=2$. Since $e_{1}\geq e_{2}$ and $e_{1}+e_{2}\leq 3d+4$, we have $e_{2}\leq 2(d-1)+1$. Hence (bgi, , Lemma 34) gives the existence of a line $L\subset N_{2}$ such that $\deg(L\cap W_{1})\geq d+1$. We define $Z_{i},f_{i},N_{i},g^{\prime}$ with respect to the line $L$ met in this step. We have $f_{1}\geq\deg(L\cap W_{1})+2\geq d+3$. Hence $\sum_{i\geq 2}f_{i}\leq 2d+1$, excluding the case $g^{\prime}=3$. Therefore $1\leq g^{\prime}\leq 2$.

(b1) Assume $g^{\prime}=2$. Since $f_{2}\leq 2d+1$, either there is a line $R\subset N_{2}$ with $\deg(R\cap Z_{1})\geq d+1$ or there is a reduced conic $F\subset N_{2}$ with $\deg(F\cap W_{1})\geq 2d$ (Lemma 1 and Remark 3).

(b1.1) Assume the existence of the line $R\subset N_{2}$ with $\deg(R\cap Z_{1})\geq d+1$. Since $L\subset N_{1}$ and $R\cap(B\setminus B\cap N_{1})\neq\emptyset$, we have $R\neq L$. Let $M$ be a hyperplane containing $R\cup L$ and such $h_{1}:=\deg(M\cap W_{0})$ is maximal. If $R\cap L=\emptyset$, then $h_{1}\geq 2d+2$. If $R\cap L\neq\emptyset$, then $h_{1}\geq 1+\deg(W_{0}\cap(L\cup R))\geq 2d+2$.

(b1.1.1) Assume $h^{1}(\mathcal{I}_{\mbox{Res}_{M}(W_{0})}(d-1))=0$. Since no connected component of $A$ is reduced, (bb2, , Lemma 5.1) gives $A\cup B\subset M$, a contradiction.

(b1.1.2) Assume $h^{1}(\mathcal{I}_{\mbox{Res}_{M}(W_{0})}(d-1))>0$. Since $h_{1}\geq 2d+2$, we have $\deg(\mbox{Res}_{M}(W_{0}))\leq d+2\leq 2(d-1)+1$. Hence there is a line $D\subset\mathbb{P}^{4}$ such that $\deg(D\cap\mbox{Res}_{M}(W_{0}))\geq d+1$. Since $R\cup L\subset M$, the lines $D,L,R$ are different. We have $h^{0}(\mathcal{I}_{D\cup L\cup R}(2))\geq 6$. Fix a general $Q\in|\mathcal{I}_{D\cup L\cup R}(2)|$. We have $\deg(Q\cap W_{0})\geq\deg(W_{0}\cap R)+\deg(W_{0}\cap L)-1+\deg(W_{0}\cap D)-2$ and hence $h^{1}(\mathcal{I}_{\mbox{Res}_{Q}(W_{0})}(d-2))=0$. Since no connected component of $A$ is reduced, (bb2, , Lemma 5.1) applied to the degree two hypersurface $Q$ instead of a hyperplane gives $W_{0}\subset Q$. Hence $W_{0}$ is contained in the scheme-theoretic base locus $\mathcal{B}$ of $|\mathcal{I}_{D\cup L\cup R}(2)|$. The scheme $\mathcal{B}$ depends from the mutual position of the 3 lines, $D,L,R$, but just knowing that $A\subseteq\mathcal{B}$ gives that $\mathcal{B}$ spans $\mathbb{P}^{4}$. Hence either the 3 lines $L,D,R$ are disjoint and $\langle L\cup D\cup R\rangle=\mathbb{P}^{4}$ or two of these lines, say $J_{1},J_{2}$ meets, while the other one, $J_{3}$, is disjoint from the plane $\langle J_{1}\cup J_{2}\rangle$. The former case cannot occur, because $A_{1}$ is contained in no line. Assume that $J_{1}\cup J_{2}$ is a conic and that $J_{3}\cap\langle J_{1}\cup J_{2}\rangle=\emptyset$. We have $\mathcal{B}=J_{1}\cup J_{2}\cup J_{3}$. Since $A_{1}$ is not contained in a line, we get $A_{1}\subset J_{1}\cup J_{2}$ (and hence $A_{2}\subset J_{3}$) and $\{O_{1}\}=J_{1}\cap J_{2}$. Set $B_{1}:=B\cap(J_{1}\cup J_{2})$ and $B_{2}:=B\cap J_{3}$. Since $h^{1}(\mathcal{I}_{\langle J_{1}\cup J_{2}\rangle\cup J_{3}}(d))=0$ and $J_{3}\cap\langle J_{1}\cup J_{2}\langle=\emptyset$, the projective space $\langle\nu_{d}(J_{1}\cup J_{2}\cup J_{3})\rangle$ is spanned by its two subspaces $\langle\nu_{d}(J_{1}\cup L_{2})\rangle$ and $\langle\nu_{d}(J_{3})\rangle$. Hence there are uniquely determined points $U_{1}\in\langle\nu_{d}(J_{1}\cup J_{2})\rangle$ and $U_{2}\in\langle\nu_{d}(J_{3})\rangle$ such that $P\in\langle\{U_{1},U_{2}\}\rangle$. For the same reason there are uniquely determined points $Q_{1}\in\langle\nu_{d}(A_{1})\rangle$, $Q_{2}\in\langle\nu_{d}(A_{2})\rangle$, $Q^{\prime}_{1}\in\langle\nu_{d}(B_{1})\rangle$, $Q^{\prime}_{2}\in\langle\nu_{d}(B_{2})\rangle$ such that $P\in\langle\{Q_{1},Q_{2}\}\rangle$ and $P\in\langle\{Q^{\prime}_{1},Q^{\prime}_{2}\}\rangle$. The uniqueness of $U_{i}$, $i=1,2$, gives $U_{i}=Q_{i}$ and $U_{i}=Q^{\prime}_{i}$. Hence $Q_{i}=Q^{\prime}_{i}$, $i=1,2$. Obviously $A_{i}$ evinces the scheme-rank and the border-rank of $Q_{i}$. Since $B$ evinces the rank of $P$, $B_{i}$ evinces the rank of $Q_{i}$. Since $(J_{1}\cup J_{2})\cap J_{3}=\emptyset$, we also get $B_{1}\cap B_{2}=\emptyset$. Hence $B$ is as claimed in the statement of Proposition 3.

(b1.2) Assume the existence of the conic $F$ with $\deg(F\cap W_{1})\geq 2d$, but that there is no line $R$ as in (b1.1). Since $L\subset N_{1}$, $W_{1}\cap B\cap N_{1}=\emptyset$ and $\deg(L\cap A)\leq 2$, we get $\deg(W_{0}\cap(L\cup F))\geq 3d-2$. Since $e_{2}\geq d+1$, we have $e_{1}\leq 3d+4-e_{2}<3d-2$. Hence $\langle L\cup F\rangle=\mathbb{P}^{4}$, i.e. $L\cap\langle F\rangle=\emptyset$. Therefore $L\cup F$ is the scheme-theoretic intersection of the linear system $|\mathcal{I}_{F\cup L}(2)|$. Fix a general $Q\in|\mathcal{I}_{F\cup L}(2)|$. Since $F\cap L=\emptyset$, we have $\deg(Q\cap W_{0})\geq 3d+1$ and hence $\deg(\mbox{Res}_{Q}(W_{0}))\leq 3\leq d-1$. Therefore $h^{1}(\mathcal{I}_{\mbox{Res}_{Q}(F\cup L}(d-2))=0$. Since no connected component of $A$ is reduced, (bb2, , Lemma 5.1) gives $A\cup B\subset Q$. Since $L\cup F$ is the scheme-theoretic intersection of the linear system $|\mathcal{I}_{F\cup L}(2)|$, we get $A\cup B\subset L\cup F$. Since $L\cap\langle F\rangle=\emptyset$ and $A_{1}\nsubseteq F$, we get $A_{2}\subset L$ and $A_{1}\subset F$. We repeat the last part of the proof of step (b1.1.1) with $F$ instead of $J_{1}\cup J_{2}$ and $L$ instead of $J_{3}$.

(b2) Assume $g^{\prime}=1$. If $h^{1}(\mathcal{I}_{Z_{1}}(d-1))=0$, then (bb2, , Lemma 5.1) gives $A\subset N_{1}$ (because no connected component of $A$ is reduced), a contradiction. Hence we may assume $h^{1}(\mathcal{I}_{Z_{1}}(d-1))>0$. Since $f_{1}\geq d+3$, we have $\deg(Z_{1})\leq 2d+1$ and hence either there is a line $R\subset\mathbb{P}^{4}$ such that $\deg(R\cap Z_{1})\geq d+1$ or there is a reduced conic $F\subset\mathbb{P}^{4}$ such that $\deg(F\cap Z_{1})\geq 2d$. In the second case we repeat the proof of step (b1.2.1). In the first case we repeat all steps of (b1.2).

(c) Assume $g=1$. Since $A\nsubseteq H_{1}$ and no connected component of $A$ is reduced, (bb2, , Lemma 5.1) gives $h^{1}(\mathcal{I}_{W_{1}}(d-1))>0$. Remark 4 gives $e_{1}\geq d+4$ and hence $\deg(W_{1})\leq 2d$. Hence either there is a line $R\subset\mathbb{P}^{4}$ with $\deg(R\cap W_{1})\geq d+1$ or $\deg(W_{1})=2d$, $e_{1}=d+4$ and there is a reduced conic $F\subset\mathbb{P}^{4}$ such that $W_{1}\subset F$. The second case is impossible, because it would give $e_{1}\geq 2d+1$. Therefore there is a line $R\subset\mathbb{P}^{4}$ with $\deg(R\cap W_{1})\geq d+1$. We also have $e_{1}\leq 2d+3$. Since $h^{1}(H_{1},\mathcal{I}_{N_{1}\cap W_{0}}(d))>0$ and $e_{1}\leq 2d+3$, either there is a line $L\subset H_{1}$ with $\deg(L\cap W_{0})\geq d+2$ or there is a conic $F\subset H_{1}$ such that $\deg(F\cap W_{0})\geq 2d+2$ (Lemma 1).

(c1) Assume the existence of a line $L\subset H_{1}$ with $\deg(L\cap W_{0})\geq d+2$. Since $\deg(R\cap A)\leq 2$, we have $R\cap(B\setminus B\cap H_{1}))\neq\emptyset$ and hence $R\neq L$. Let $M\subset\mathbb{P}^{4}$ be a hyperplane containing $\langle L\cup R\rangle$ and with maximal $\deg(M\cap W_{0})$. If $L\cap R\neq\emptyset$, then $\deg(W_{0}\cap(L\cup R))\geq 2d+2$ and hence $\deg(M\cap W_{0})\geq 2d+3$. If $L\cap R=\emptyset$, then $\deg(M\cap W_{0})\geq 2d+3$. Since $h^{1}(\mathcal{I}_{\mbox{Res}_{M}(W_{0})}(d-1))>0$ by (bb2, , Lemma 5.1), we first get $\deg(\mbox{Res}_{M}(W_{0}))\geq d+1$ and then $e_{1}=\deg(W_{0}\cap M)=2d+3$, $\deg(\mbox{Res}_{M}(W_{0}))=d+1$ and the existence of a line $D\subset\mathbb{P}^{4}$ such that $\mbox{Res}_{M}(W_{0})\subset D$. See step (b1.1.1); in this case $J_{3}=L$, $\{J_{1},J_{2}\}=\{R,D\}$. A posteriori we get $R\cap D=\{O_{1}\}$ and $A_{2}\subset L$; $B\cap(R\cup D)$ is as described in (b, , Proposition 7).

(c2) Assume the existence of the conic $F$, but the non-existence of any line $L\subset H_{1}$ with $\deg(L\cap W_{0})\geq d+2$. Remark 4 gives that $F$ is reduced and that if $F=J_{1}\cup J_{2}$ with each $J_{i}$ is a line, then $\deg(J_{i}\cap W_{0})=d+1$, $i=0,1$. Since $e_{1}\leq 2d+3$ and $A$ spans $\mathbb{P}^{4}$, we get $e_{1}=2d+3$, $\deg(F\cap W_{0})=2d+2$ and $\deg(W_{0}\cap\langle F\rangle)=2d+2$. Since $h^{1}(\mathcal{I}_{W_{1}}(d-1))>0$ by (bb2, , Lemma 5.1) and $\deg(W_{1})\leq d+1$, we get $\deg(W_{1})=d+1$ and the existence of a line $J\subset\mathbb{P}^{4}$ such that $W_{1}\subset J$. Since $\deg(J\cap A)\leq 2$, we have $J\cap(B\setminus B\cap H_{1})\neq\emptyset$ and hence $J\nsubseteq F$. If $J\cap\langle F\rangle\neq\emptyset$, then $e_{1}\geq 2d+2+d+1-2$, a contradiction. Therefore $J\cap\langle F\rangle=\emptyset$. We go as in step (b1.2), i.e. as in the last part of step (b.1.1.1).

Proposition 4

Fix integers $s\geq 0$, $m\geq s+2$ and $d\geq 2m+1$. Fix a degree $3$ curvilinear zero-dimensional scheme $A_{1}\subset\mathbb{P}^{m}$ with $\dim(\langle A_{1}\rangle)=2$ and a set $S\subset\mathbb{P}^{m}$ such that $S\cap A_{1}=\emptyset$, $\sharp(S)=s$, $S$ is linearly independent and $\langle S\rangle\cap\langle A_{1}\rangle=\emptyset$. Set $A:=A_{1}\cup S$. Fix any $P\in\langle\nu_{d}(A)\rangle$ such that $P\notin\langle\nu_{d}(A^{\prime})\rangle$ for any $A^{\prime}\subsetneq A$. There is a unique $P_{1}\in\langle\nu_{d}(A_{1})\rangle$ such that $P\in\langle\{P_{1}\}\cup\nu_{d}(S)\rangle$. We have $P_{1}\notin\langle\nu_{d}(A^{\prime})\rangle$ for any $A^{\prime}\subsetneq A$. Then $r_{m,d}(P)=s+2d-1$ and every $B\in\mathcal{S}(P)$ is of the form $B=B_{1}\sqcup S$ with $B_{1}\in\mathcal{S}(P_{1})$ (and the converse holds). The elements of $\mathcal{S}(P_{1})$ are described in the case $m=2$ of (b, , part (b3) of Theorem 4 and part (f) of §4).

Proof

The last sentence follows from concision ((l, , Exercise 3.2.2.2)).

The scheme $A$ is linearly independent and $\deg(A)=3+s$. Hence $\nu_{d}(A)$ is linearly independent. Every element of $\mathcal{S}(P)$ is contained in $\langle A\rangle$ ((l, , Exercise 3.2.2.2.)). Hence it is sufficient to do the case $s=m-2$. Fix any $B\in\mathcal{S}(P)$. Since $P\in\langle\nu_{d}(E\cup S)\rangle$ for any $E\in\mathcal{S}(P_{1})$, we have $r_{m,d}(P)\leq m+2d-3$ and it is sufficient to prove that $B=S\sqcup B_{1}$ with $B_{1}\in\mathcal{S}(P_{1})$. Set $\{O\}:=(A_{1})_{red}$ and let ${\bf{w}}$ the degree two subscheme of $A_{1}$. If $m=2$, then this proposition is the case $m=2$ of (b, , part (b3) of Theorem 4). Hence we may assume $m>2$ and use induction on $m$. Set $W_{0}:=A\cup B$ and define $H_{i}$, $e_{i}$ and $g$ as in step (b) of the proof of Theorem 1.1. If $e_{i}\leq m-1$, then $e_{i+1}=0$ and $W_{i}=\emptyset$. We have $w_{0}:=\deg(W_{0})\leq 2d+2m-2$. Assume for the moment $g=d+1$. We get $h^{1}(\mathcal{I}_{W_{d}\cap H_{d+1}})>0$, i.e. $e_{d+1}\geq 2$. Since $e_{d+1}>0$, we have $e_{i}\geq m$ for $i\leq d$ and hence $2d+2m-2\leq md+2$, which is absurd if $m>2$; for $m=2$ we have $s=0$, $P_{1}=P$ and the statement is a tautology. Assume $g\geq d+2$. We get $W_{d+1}\neq\emptyset$ and hence $2d+2m-2\geq 1+m(d+1)$, a contradiction. Now assume $2\leq g\leq d$. Using Remark 4 we get $2d+2m-2\geq g(d+1+m-g)-m+2$. Since $d\geq 2m+1$, we get $g\leq 2$. Hence $1\leq g\leq 2$.

(a) Assume $g=2$. Since $e_{1}\geq e_{2}$ and $4d>2d+m-2$, we have $e_{1}\leq 2d-1$. By (bgi, , Lemma 34) there is a line $L\subset H_{2}$ such that $\deg(L\cap W_{1})\geq d+1$. Since $e_{2}\geq d+1$, we have $e_{1}\leq d+2m-3$. Set $Z_{0}:=W_{0}$. Set $k:=\dim(\langle H_{2}\cap W_{1}\rangle)$. If $k\leq m-2$, then the maximality property of the integer $e_{2}$ gives $W_{2}=\emptyset$. Remark 1 for the integer $d-1$ gives $e_{2}\geq d+k$. Since $e_{1}\geq e_{2}$, if $k=m-1$ we have $e_{1}+e_{2}=2d+2m-2$. Hence in all cases we have $W_{2}=\emptyset$. Let $M\subset\mathbb{P}^{m}$ be a hyperplane containing $\langle H_{2}\cap W_{1}\rangle$ and with maximal $\deg(W_{0}\cap M)$ (we have $M=H_{2}$ if $k=m-1$). Since $A$ spans $\mathbb{P}^{m}$, we have $\deg(W_{0}\cap M)\geq d+m-1$ and hence $\deg(\mbox{Res}_{M}(W_{0}))\leq d+m-1$.

(a1) Assume $h^{1}(M,\mathcal{I}_{M\cap W_{0}}(d))=0$. Set $Z_{0}:=W_{0}$. A residual exact sequence gives $h^{1}(\mathcal{I}_{\mbox{Res}_{M}(W_{0})}(d-1))>0$. Since $\deg(\mbox{Res}_{M}(W_{0}))\leq d+m-1\leq 2d-1$, there is a line $L\subset\mathbb{P}^{m}$ such that $\deg(L\cap\mbox{Res}_{M}(W_{0}))\geq d+1$. Let $N_{1}\subset\mathbb{P}^{m}$ be a hyperplane such that $N_{1}\subset L$ and $f_{1}:=\deg(N_{1}\cap W_{0})$ is maximal and set $Z_{1}:=\mbox{Res}_{N_{1}}(Z_{0})$. Since $A$ spans $\mathbb{P}^{m}$, then $f_{1}\leq d+m-1$ and hence $\deg(Z_{1})\leq d+m-2$. For all integers $i\geq 2$ define $f_{i},Z_{i},N_{i}$ as in step (b1) of the proof of Theorem 1.1. We have $f_{i}\geq f_{i+1}$ for all $i\geq 2$ and $f_{i+1}=0$ if $f_{i}\leq m-1$. By residual exact sequence like (2) we get the existence of a minimal integer $g^{\prime}>0$ such that $h^{1}(\mathcal{I}_{W_{g^{\prime}-1}\cap N_{g^{\prime}}}(d+1-g^{\prime}))>0$. As in step (b1) of the proof of Theorem 1.1 we get $g^{\prime}\leq 2$.

(a1.1) Assume $g^{\prime}=2$. Since $d+m-2\leq 2(d-2)+1$, there is a line $R\subset\mathbb{P}^{m}$ such that $\deg(R\cap Z_{1})\geq d+1$. Since $\deg(R\cap A)\leq 2$, we have $\deg(R\cap Z_{1}\cap B)>0$. Since $Z_{1}\cap B=B\setminus B\cap N_{1}$ and $L\subset N_{1}$, we have $R\neq L$. First assume that either $L\cap R\neq\emptyset$ or $m\geq 4$ (i.e. assume that $L\cup R$ is contained in a hyperplane of $\mathbb{P}^{m}$). Since $\deg(L\cap R)=1$ as schemes, we get $e_{1}\geq d+1+d+1-1$. Hence $2d+1\leq d+2m-3$, contradicting our assumption $d\geq 2m+1$. Now assume $m=3$ and $L\cap R=\emptyset$. Since $A_{1}$ is connected and spans a plane, we have $A_{1}\nsubseteq L\cup R$. Fix a general $Q\in|\mathcal{I}_{L\cup R}(2)|$. Since $A_{1}$ is curvilinear and $L\cup R$ is the scheme-theoretic base locus of the linear system $|\mathcal{I}_{L\cup R}(2)|$, we have $A_{1}\nsubseteq Q$. Since $\deg(\mbox{Res}_{Q}(W_{0}))\leq 2d+2m-2-(d+1)-(d+1)\leq d-1$ (by our assumption $d\geq 2m+1$), we have $h^{1}(\mathcal{I}_{\mbox{Res}_{Q}(W_{0})}(d-2))=0$. By (bb2, , Lemma 5.1) applied to the degree two surface $Q$ instead of a plane we get $A_{1}\subset Q$, a contradiction.

(a1.2) Assume $g^{\prime}=1$.

(a1.2.1) Assume $h^{1}(M,\mathcal{I}_{M\cap W_{0}}(d))=0$. By (bb2, , Lemma 5.1) we have $A_{1}\subset M$ and $S\setminus S\cap M=B\setminus B\cap M$. Since $A$ spans $\mathbb{P}^{m}$, we have $S\neq S\cap M$. Set $E:=S\cap B$. We just saw that $E\neq\emptyset$ and that $(W_{0}\setminus E)\subset M$. Since $E\subset S$, we have $h^{1}(\mathcal{I}_{E}(d-1))=0$. Hence a residual exact sequence and Grassmann’s formula gives that $\langle\nu_{d}(A)\rangle\cap\langle\nu_{d}(B)\rangle$ is spanned by its subspaces $\langle\nu_{d}(A\setminus E)\rangle\cap\langle\nu_{d}(B\setminus E)\rangle$ and $\langle\nu_{d}(E)\rangle$ and that these subspaces are disjoint. Take any $Q_{1}\in\langle\nu_{d}(A\setminus E)\rangle\cap\langle\nu_{d}(B\setminus E)\rangle$ such that $P\in\langle\{Q_{1}\}\cup\nu_{d}(E)\rangle$. Since $P\notin\langle\nu_{d}(A^{\prime})\rangle$ for any $A^{\prime}\subsetneq A$, we have $Q_{1}\notin\langle\nu_{d}(A^{\prime})\rangle$ for any $A^{\prime}\subsetneq A$. $Q_{1}$ is the only point of $\langle\nu_{d}(A\setminus E)\rangle$ such that $P\in\langle\nu_{d}(E)\cup\{Q_{1}\}\rangle$ and $P_{1}$ is associated also to $Q_{1}$. By induction on $m$ (or b if $m-\sharp(E)=2$) we get $r_{m,d}(Q_{1})=2d+m-3-\sharp(E)$ and that every $U\in\mathcal{S}(Q_{1})$ is of the form $U\sqcup S\setminus E$ with $\sharp(U)=2d-1$ and $U\in\mathcal{S}(P_{1})$. See Step (b2) below.

(a1.2.1) Assume $h^{1}(M,\mathcal{I}_{M\cap W_{0}}(d))>0$. Since $\deg(M\cap W_{0})\leq e_{1}\leq 2d-1$, there is a line $R\subset M$ such that $\deg(R\cap W_{0})\geq d+2$.

(a1.2.1.1) Assume $R=L$. Take $N_{i},Z_{i},f_{i}$ as in step (a1). Now it is easier, because $f_{1}\geq d+m$.

(a1.2.1.2) Assume $R\neq L$. First assume that either $m\geq 4$ or $m=3$ and $R\cap L\neq\emptyset$, i.e. assume $\langle R\cup L\rangle\neq\mathbb{P}^{m}$. Since $\deg(R\cap L)=1$, we get $e_{1}\geq d+2+d+1-1$, a contradiction. Now assume $m=3$ (and hence $w_{0}\leq 2d+4$) and $L\cap R=\emptyset$. We have $\deg(W_{0}\cap(L\cup R))\geq 2d+3$ and $A_{1}\nsubseteq L\cup R$. Hence one of the lines $L,R$ (call it $D$ and call $J$ the other one) is spanned by the only degree two subscheme of $A$ and $B\cup S\subset L\cup R$. Since $A_{1}$ is curvilinear and $L\cup R$ is the scheme-theoretic base locus of $|\mathcal{I}_{L\cup R}(2)|$, there is a quadric surface $Q\supset L\cup R$ such that $A_{1}\nsubseteq Q$. Since $\deg(\mbox{Res}_{Q}(W_{0}))=1$, we have $h^{1}(\mathcal{I}_{\mbox{Res}_{Q}(W_{0})}(d-2))=0$. By (bb2, , Lemma 5.1) applied to the degree two surface $Q$ instead of a plane we get $A_{1}\subset Q$, a contradiction.

(b) Assume $g=1$. Since $W_{0}\cap H_{1}$ spans $H_{1}$, we have $e_{1}\geq d+m+1$ (Remark 4).

(b1) Assume $h^{1}(\mathcal{I}_{W_{1}}(d-1))>0$. Therefore $\deg(W_{1})\geq d+1$ and hence $e_{1}\leq d+2m-3$. Since $(d+m+1)+2d>2d+2m-2$, we have $\deg(W_{1})<2d$. Hence there is a line $L\subset\mathbb{P}^{m}$ such that $\deg(W_{1}\cap L)\geq d+1$. Hence $e_{1}\leq d+2m-3\leq 2d+1$. Therefore there is a line $R\subset H_{1}$ such that $\deg(R\cap W_{0})\leq d+2$. Since $\deg(A\cap L)\leq 2$, we have $L\cap(B\setminus B\cap H_{1})\neq\emptyset$. Since $R\subset H_{1}$, we have $R\neq L$. As in step (a1.1) we get a contradiction.

(b2) Assume $h^{1}(\mathcal{I}_{W_{1}}(d-1))=0$. Copy step (a1.2.1.2) with $H_{1}$ instead of $M$.

Remark 6

Fix integers $b>0$ and $m\geq b-1$. Let $E\subset\mathbb{P}^{m}$ be a curvilinear connected scheme of degree $b$. Set $\{O\}:=E_{red}$. The local Hilbert scheme of the regular local ring $\mathcal{O}_{\mathbb{P}^{m},O}$ is smooth and irreducible at $E$ and of dimension $(m-1)(b-1)$ (g ). Therefore, if we don’t prescribe $O$, but only the integer $b$ we get that the set of all possible $E$ is parametrized by an irreducible variety of dimension $m+(m-1)(b-1)$. Alternatively, we may consider the equisingular deformations of $E$ in $\mathbb{P}^{m}$ and get (in the curvilinear case) that it is parametrized by a smooth variety of dimension $m+(m-1)(b-1)$. If $d\geq b-1$, then $\dim(\langle\nu_{d}(E)\rangle)=b-1$. If we take a type $(s;b_{1},\dots,b_{s})$ with $\sum_{i}b_{i}=5$ and general $E_{1},\dots,E_{s}$ with $E_{i}$ curvilinear and $\deg(E_{i})=b_{i}$, then in the computation of Remark 8 below we only need to use that $\dim(\langle\nu_{d}(E_{1}\cup\cdots\cup E_{s})\rangle)=4$ and get the dimensions claimed in Remark 8 and Proposition 5 below.

Remark 7

Fix a combinatorial type $(s;b_{1},\dots,b_{s})$ with $s\geq 1$ and $b_{1}+\cdots+b_{s}=5$. All $A$ evincing the the border rank of some degree $d\geq 9$ polynomial and with combinatorial type $(s;b_{1},\dots,s)$ is obtained in the following way. Fix a $4$-dimensional linear subspace $M\subseteq\mathbb{P}^{m}$ and $s$ linearly independent points $O_{1},\dots,O_{s}\in M$. For $i=1,\ldots,s$ let $M_{i}\subseteq M$ be a $(b_{i}-1)$-dimensional linear subspace such that $O_{i}\in M_{i}$, $\dim(M_{i})=b_{i}$ and $M_{1}\cup\cdots\cup M_{s}$ spans $M$, i.e. the subspaces $M_{1},\dots,M_{s}$ are linearly independent. Let $D_{i}\subseteq M_{i}$ be the rational normal curve of $M_{i}$ with the convention that $D_{i}=\{O_{i}\}$ if $b_{i}=1$ (i.e. if $M_{i}=\{O_{i}\}$) and $D_{i}=M_{i}$ if $b_{i}=2$ (i.e. if $M_{i}$ is a line). Let $A_{i}\subset D_{i}$ be the effective degree $b_{i}$ divisor of $D_{i}$ with $O_{i}$ as its support with the convention $A_{i}=\{O_{i}\}$ if $b_{i}=1$. Set $A:=A_{1}\cup\cdots\cup A_{s}$. Since each $A_{i}$ spans $M_{i}$, $A$ is a curvilinear degree $5$ scheme of combinatorial type $(s;b_{1},\dots,b_{s})$ spanning $M$. Since $d\geq 4$, we have $\dim\langle\nu_{d}(A)\rangle=4$. Take any $P\in\langle\nu_{d}(A)\rangle$ such that $P\notin\langle\nu_{d}(A^{\prime})\rangle$ for any $A^{\prime}\subsetneq A$. Since $d\geq 5$, $P$ has border rank $5$ and cactus rank $5$ and $A$ is the only scheme evincing the cactus rank of $P$ ((bgl, , Corollary 2.7)). All possible $M,O_{1},\dots,O_{s},D_{1},\dots,D_{s}$ are projectively equivalent and so all schemes with combinational type $(s;b_{1},\dots,b_{s})$ are projectively equivalent.

Remark 8

Since $d\geq 5$, a theorem of Alexander and Hirschowitz gives that $\sigma_{5}(X_{m,d})$ and $\sigma_{5}(X_{m,d})\setminus\sigma_{4}(X_{m,d})$ have dimension $5m+4$. Now assume $d\geq 9$ so that each point of $\sigma_{5}(X_{m,d})\setminus\sigma_{4}(X_{m,d})$ has a unique type. For any type $(s;b_{1},\dots,b_{s})$ let $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})$ denote the set of all $P\in\sigma_{5}(X_{m,d})\setminus\sigma_{4}(X_{m,d})$ with type $(s;b_{1},\dots,b_{s})$. Let $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime}$ denote the subset of the algebraic set $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})$ formed by all $P$ whose cactus rank is evinced by a linearly independent scheme. Let $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime\prime}$ denote the subset of $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})$ formed by all $P$ whose cactus rank is evinced by a curvilinear scheme. In the proof of Theorem 1.1 we checked that $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime\prime}\subseteq\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime}$. By a theorem of Chevalley on the image of an algebraic variety by an algebraic map ((h, , Ex. I.3.18)) all sets $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})$, $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime\prime}$, $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime}$ are constructible. Thus we may speak about the irreducible components of the sets $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})$, $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime\prime}$, $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime}$ and the dimension of each of this irreducible component; indeed, they are the irreducible components of the closure (in the Zariski topology) of the subsets $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})$ or $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime}$ of the projective variety $\sigma_{5}(X_{m,d})$. Take two schemes $A_{1},A_{2}$ with the same type. Since $\deg(A_{1}\cup A_{2})\leq 10\leq d-1$, we have $h^{1}(\mathcal{I}_{A_{1}\cup A_{2}}(d))=0$. Hence Grassmann’s formula gives $\langle\nu_{d}(A_{1})\rangle\cap\langle\nu_{d}(A_{2})\rangle=\langle\nu_{d}(A_{1}\cap A_{2})\rangle$. Using the description of all possible schemes of type $(s;b_{1},\dots,b_{s})$ given in the proof of Theorem 1.1 and Remark 6 we get that each set $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime\prime}$ is irreducible and with the following dimension:

• (i)

$\sigma_{5}(X_{m,d})(1;5)^{\prime\prime}$ has dimension $5m$.

• (ii)

$\sigma_{5}(X_{m,d})(2;3,2)^{\prime\prime}$ has dimension $5m+1$.

• (iii)

$\sigma_{5}(X_{m,d})(2;4,1)^{\prime\prime}$ has dimension $5m+1$.

• (iv)

$\sigma_{5}(X_{m,d})(3;3,1,1)^{\prime\prime}$ has dimension $5m+2$.

• (v)

$\sigma_{5}(X_{m,d})(3;2,2,1)^{\prime\prime}$ has dimension $5m+2$.

• (vi)

$\sigma_{5}(X_{m,d})(4;2,1,1,1)^{\prime\prime}$ has dimension $5m+3$.

• (vii)

$\sigma_{5}(X_{m,d})(5;1,1,1,1)^{\prime\prime}$ has dimension $5m+4$.

We get that $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime\prime}$ has codimension $5-s$ in $\sigma_{5}(X_{m,d})$ (this is the expected codimension). Since each $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime\prime}$ is irreducible, we get that $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime}$ is irreducible of dimension $5m+s-1$ and that it contains a non-empty Zariski open subset of $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime\prime}$, proving the following result.

Proposition 5

For each $(s;b_{1},\dots,b_{s})$ the sets $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime}$ and $\sigma_{5}(X_{m,d})(s;b_{1},\dots,b_{s})^{\prime\prime}$ are irreducible and of dimension $5m+s-1$.

To get the dimension of the the set of all associated degree $d$ homogeneous polynomials, add $+1$ to each dimension, because proportional polynomials, say $f$ and $cf$ with $c\in\mathbb{C}\setminus\{O\}$, have the same associated point $P\in\mathbb{P}^{r}$

4 Remarks on possible further works

Remark 9

We briefly explain what is missing to complete the stratification by rank of $\sigma_{5}(X_{m,d})\setminus\sigma_{4}(X_{m,d})$. The case with very low $d$ seems to be difficult, full of nasty cases and not very enlightening, but cases like $d=7,8$ may be easy (when $A$ is not unique, just take the one which gives a lower rank). It may be easy to check which integers $r_{m,d}(P)$ appears for some $P\in\sigma_{5}(X_{m,d})\setminus\sigma_{4}(X_{m,d})$ just because only very few integers are possible. For a fixed $P$ the proof of Theorem 1.1 may be not enough to compute the integer $r_{m,d}(P)$, but probably all a priori possible integers arise for some $P$ for which the proof of Theorem 1.1 works.

Fix any integer $d\geq 5$ and any $A$ evincing the cactus rank of some $P$. If $A$ is curvilinear, but not connected, i.e. if $s\geq 2$, then we know its single pieces $A_{i}$ and in each case by bgi or bb2 we know the rank, $\rho_{i}$, of any $Q_{i}\in\langle\nu_{d}(A_{i}))$ such that $Q_{i}\notin\langle\nu_{d}(A^{\prime})\rangle$ for any $A^{\prime}\subsetneq A_{i}$. We always have $r_{m,d}(P)\leq\sum_{i=1}^{s}\rho_{i}$. This is a good upper bound (e.g. we always have $r_{m,d}(P)\leq 3d+1$ if $s=2$ and $r_{m,d}(P)\leq 2d+1$ if $s=3$), but in some cases we know that it is not the exact value. The case $s=5$ is trivial and hence we assume $s\leq 4$. If $\dim(\langle A\rangle)=1$, then we have $r_{m,d}(P)=d-3$ by a theorem of Sylvester (cs ). If $\dim(\langle A\rangle)=2$ (and in particular always if $m=2$), then an upper bound is obtained looking at the minimal degree of a reduced curve containing $A$ (see Proposition 6). When $s=1$ and $\dim(\langle A\rangle)=3$ there is a unique non-curvilinear case ((eh, , Case III of Theorem 1.3)). For this case we have a lower bound, $3d-3$, and an upper bound, $4d$, but we don’t even know if the integer $r_{m,d}(P)$ is the same for all non-curvilinear $A$ and all $P$ with $A$ as its associated scheme.

Proposition 6

Assume $d\geq 9$. Fix $P\in\mathbb{P}^{r}$ with $b_{r,m}(P)=5$ and assume the existence of a plane $H\subset\mathbb{P}^{m}$ such that $P\in\langle\nu_{d}(H)\rangle$. Then $r_{m,d}(P)\leq 3d$. Let $A$ be the only scheme evincing the border rank of $A$. If either $h^{0}(H,\mathcal{I}_{A}(2))\geq 2$ or $A$ is contained in a reduced conic, then $r_{m,d}(P)\leq 2d$.

Proof

By concision ((l, , Exercise 3.2.2.2)) we may assume $m=2$. Since $d\geq 9$, there is a unique zero-dimensional scheme $A\subset\mathbb{P}^{2}$ with $\deg(A)=5$ and evincing the scheme rank and the border rank of $P$. Since $h^{0}(\mathcal{O}_{\mathbb{P}^{2}}(2))=6$, there is a conic $C\supset A$. If $C$ is reduced, then $r_{m,d}(P)\leq 2d$, because $C$ is connected, $\dim(\langle\nu_{d}(C)\rangle)=2d$ and (lt, , Proposition 5.1) holds (with the same proof) for reduced and connected non-degenerate curves. Hence we may assume that $C$ is a double line, say $C=2L$. It is easy to check that $\deg(L\cap A)\geq 3$. If $\deg(L\cap A)\geq 4$, then $\mbox{Res}_{L}(A)$ is a single point, $O$, and hence $A$ is contained in any conic $L\cup D$ with $D$ a line containing $O$ (we use that $L$ and $D$ are Cartier divisors of $\mathbb{P}^{m}$). Hence we may assume $\deg(L\cap A)=3$. In this case we have $A\subset L\cup D_{1}\cup D_{2}$ with $D_{1}$ and $D_{2}$ any two distinct lines, $\neq L$, and containing the scheme $\mbox{Res}_{L}(A)$. We get $r_{m,d}(P)\leq 3d$ quoting again the same modified version of (lt, , Proposition 5.1). We have $h^{0}(\mathcal{I}_{A}(2))\geq 2$ if and only if there is a line $L$ with $\deg(L\cap A)\geq 4$.

5 Conclusions

We give the stratification by the symmetric tensor rank of all degree $d\geq 9$ homogeneous polynomials with border rank $5$ and which depend essentially on at least 5 variables, extending two previous works (bgi for border ranks 2 and 3, bb2 for border rank 4). For the polynomials which depend on at least 5 variables only 5 ranks are possible: $5$, $d+3$, $2d+1$, $3d-1$, $4d-3$, but each of the ranks $3d-1$ and $2d+1$ is achieved in two geometrically different situations. These ranks are uniquely determined by a certain degree 5 zero-dimensional scheme $A$ associated to the polynomial. The polynomial depends essentially on at least 5 variables if and only if $A$ is linearly independent. The polynomial has rank $4d-3$ (resp $3d-1$, resp. $2d+1$, resp. $d+3$, resp. $5$) if $A$ has $1$ (resp. $2$, resp. $3$, resp. $4$, resp. $5$) connected components. The assumption $d\geq 9$ guarantees that each polynomial has a uniquely determined associated scheme $A$.

In each case we describe the dimension of the families of the polynomials with prescribed rank, each irreducible family being determined by the degrees of the connected components of the associated scheme $A$. Each family of polynomials has dimension $5m+s$, where $s=1,\dots,5$ is the number of connected components of $A$ (Remark 8).

When $A$ has at least $3$ connected components we also describe all linear forms evincing the rank $\rho$ of the polynomial $f$, i.e. all linear forms $\ell_{i}$ such that $f=\sum_{i=1}^{\rho}\lambda_{i}\ell_{i}^{d}$, $\lambda_{i}\in\mathbb{C}\setminus\{0\}$, up to a non-zero multiple of each $\ell_{i}$ and a permutation of $\ell_{1},\dots,\ell_{c}$.

The proofs require projective geometry and some algebraic geometry.

Acknowledgements.

We thanks the referee for useful comments.