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

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

The set $W(X)q$ is the main actor of this paper. We often write $Wq$ if $X$ is clear from the context.

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

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

• (a)

  We have ${q}=∩S∈S(X,q)〈S〉$ for all $q∈U$.

• (b)

  For all $(q,q′)∈U×ℙr$ if $S(X,q′)=S(X,q)$, then $q′=q$.

Take a non-degenerate $X⊂ℙr$ and $q∈ℙr$⁠. For any integer $t>0$ the $t$-secant variety $σt(X)$ of $X$ is the closure in $ℙr$ of the union of all linear spaces $〈S〉$ with $S⊂X$ and $|S|=t$⁠. The border rank or border $X$-rank $bX(q)$ of $q∈ℙr$ is the minimal integer $b≥1$ such that $q∈σb(X)$⁠. We say that a finite set $A⊂ℙr$ irredundantly spans $q$ if $q∈〈A〉$ and $q∉〈A′〉$ for any $A′⊊A$⁠. We use Theorem 2 to prove the following result for the order $d$ Veronese embedding of $ℙn$⁠.

• (1)

  $rX(q)=k$ and $S(X,q)⊇{E∪U}E∈S(Y,q′)$.

• (2)

  If $k≤2d−3$, then $S(X,q)={E∪U}E∈S(Y,q′)$ and $Wq=〈U∪{q′}〉$.

In Section 4 we consider the following problem. For any positive integer $t$ let $S(X,q,t)$ be the set of all $S⊂X$ such that $|S|=t$ and $S$ irredundantly spans $q$⁠. We have $S(X,q,t)=ø$ for all $t<rX(q)$ and $S(X,q,rX)=S(X,q)≠ø$⁠. By the definition of irredundantly spanning set we have $S(X,q,t)=ø$ for all $t≥r+2$⁠. Since $X$ is integral and non-degenerate, for all $(X,q)$ we have $S(X,q,r+1)≠ø$ and $S(X,q,r+1)$ contains a general subset of $X$ with cardinality $r+1$⁠. There are easy examples of triples $(X,q,t)$ such that $r>t>rX(q)$ and $S(X,q,t)=ø$ (Remark 3). It easy to check that $S(X,q,t)≠ø$ for all $t$ such that $r+1−dimX≤t≤r$ (Lemma 2). Set $W(X)q,t:=∩S∈S(X,q,t)〈S〉$⁠, with the convention $W(X)q,t:=ℙr$ if $S(X,q,t)≠ø$⁠. We often write $Wq,t$ instead of $W(X)q,t$⁠.

In Section 4 we prove the following result.

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

We thanks a referee for useful comments.

The cactus rank or cactus $X$-rank $cX(q)$ of $q∈ℙr$ is the minimal degree of a zero-dimensional scheme $Z⊂X$ such that $q∈〈Z〉$⁠. Let $Z(X,q)$ denote the set of all zero-dimensional schemes $Z⊂X$ such that $deg(Z)=cX(q)$ and $q∈〈Z〉$⁠.

(i) We have $bX(q)=cX(q)$ ([18, Lemma 1.38]) and $|Z(X,q)|=1$ ([18, Part (i) of Theorem 1.43]).

(ii) If $cX(q)<rX(q)$⁠, then $cX(q)+rX(q)=d+2$ and $S(X,q)$ is infinite. Let $Z$ be the only element of $Z(X,q)$⁠, $d+2−cX(q)$ is the minimal degree of a scheme $A⊂X$ such that $q∈〈A〉$ and $A⊉Z$⁠.

(iii) If $rX(q)>cX(q)$⁠, then ${q}=〈Z〉∩〈S〉,$ where ${Z}=Z(X,q)$ and $S$ is any element of $S(X,q)$ (this also follows from the fact that $h1(ℙ1,L)=0$ for any line bundle $L$ on $ℙ1$ with $deg(L)≥−1$⁠, as in the proof of Claim 1).

(iv) then $IfrX(q)>bX(q),then dimS(X,q)=d+3−2b$ ([14, eq. (9)]).

(v) If $d$ is odd and $rX(q)=(d+1)/2$ (i.e. $rX(q)=bX(q)$ is the generic rank), then $S(X,q)=Z(X,q)$ and $|S(X,q)|=1$ ([18, Theorem 1.43]).

(vi) Assume $d$ even and $rX(q)=d/2+1$ and so $q$ has the generic rank and $bX(q)=rX(q)$⁠, but we do not assume that $q$ is general in $ℙd$⁠. Fix $S,S′∈S(X,q)$ such that $S≠S′.$

Proof of Claim 1. Since $S≠S′$ and $S∈S(X,q)$⁠, we have $q∉〈S∩S′〉$⁠. The Grassmann’s formula gives $h1(ℙ1,IS∪S′(d))>0$⁠. Since $h1(ℙ1,L)=0$ for any line bundle $L$ on $ℙ1$ with $deg(L)≥−1$ and $q∉X$⁠, we have $S∩S′=ø$ and $h1(ℙ1,IS∪S′(d))=1$⁠. Thus the Grassmann’s formula implies $dim(〈S〉∩​〈S′〉)=0$⁠, proving Claim 1.

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

Note that (iii) implies that each $q∈ℙr$ with $cX(q)≠rX(q)$ is uniquely determined by the zero-dimensional scheme evincing its cactus rank and by one single set evincing its rank (any $S∈S(X,q)$ would do the job). Obviously part (i) implies that most $q∈ℙr$ (the ones with $rX(q)=bX(q)$ ) are not uniquely determined by $S(X,q)$⁠. By parts (i) and (ii) for each $q∈ℙr$ such that $rX(q)=bX(q)$ there are exactly $∞t$⁠, $t:=rX(q)−1$⁠, points $o∈ℙr$ with $S(X,o)=S(X,q)$⁠.

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

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

2. $|F|=2d+2$ and $F$ is contained in a reduced conic $D$ ; if $D=L1∪L2$ with each $Li$ a line we have $L1∩L2∉F$ and $|F∩L1|=|F∩L2|=d+1$;

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

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

Since part (a) is trivial in the case $r=2$⁠, we assume $r≥4$⁠. Since no non-degenerate curve is defective ([23, Corollary 1.5 and Remark 1.6]), there is a non-empty Zariski open subset $V⊂ℙr$ such that $rX(q)=r/2+1$ and $dimS(X,q)=1$ for all $q∈V$⁠.

For each set $S⊂X$ such that $|S|=r/2+1$ and $dim〈S〉=r/2$ let $ℓS:ℙr\〈S〉→ℙr/2−1$ denote the linear projection from $〈S〉$⁠. For a general $S$ we have $〈S〉∩X=S$ (scheme-theoretically) by Bertini’s theorem and the trisecant lemma ([24, Corollary 2.2]) and $ℓS|X\S$ is birational onto its image, again by the trisecant lemma and the assumption $r≥4$⁠.

Fix a general $S⊂X$ such that $|S|=r/2+1$⁠. Let $XS⊂ℙr/2−1$ be the closure of $ℓS(X\S)$ in $ℙr/2−1$⁠. There is a finite set $E⊂XS$ containing $XS\ℓS(X\S)$ and such that for each $p∈XS\E$ there is a unique $o∈X\S$ such that $ℓS(o)=p$⁠. For any set $A⊂XS\E$ let $AS⊂X\S$ denote the only set such that $ℓS(AS)=A$⁠. Any general $A⊂XS\E$ such that $|A|=r/2+1$ is linearly dependent, but each proper subset of $A$ is linearly independent. Thus $〈S〉∩〈AS〉$ is a single point, $qS,A$⁠, and $qS,A∉〈B〉$ for any $B⊊AS$⁠. For a general $A$ we get as $AS$ a general subset of $X$ with cardinality $r/2+1$⁠. Thus for a general $A$ we have $qS,A∉S′$ for any $S′⊊S$⁠. We start with $S∈S(X,o)$ for a general $o∈ℙr$⁠. Thus $rX(q)=r/2+1$ for a general $q∈〈S〉$⁠. Thus for a general $A$ we get $S∈S(qS,A)$ and $AS∈S(qS,A)$⁠. By construction we have ${qS,A}=〈S〉∩〈AS〉$⁠. For a general $A$ the point $qS,A$ is general in $〈S〉$⁠. By the generality of $S$ we get that the points $qS,A$ ’s (with $(S,A)$ varying, but general), cover a non-empty Zariski open subset of $ℙr$⁠.□

(a) First assume $b=2$⁠. We use the proof of [20, Proposition 5.1]. Fix $a∈ℙd\{q}$⁠. Let $H⊂ℙd$ be a general hyperplane containing $q$⁠. Since $q∉X$⁠, Bertini’s and Bezout’s theorems give that $X∩H$ is formed by $d$ distinct points. Since $X$ is connected, the exact sequence

gives that $X∩H$ spans $H$⁠. Thus $q∈〈X∩H〉$⁠. Since $rX(q)=d$⁠, we get $X∩H∈S(X,q)$⁠. The generality of $H$ gives $a∉H$⁠, concluding the proof that $Wq={q}$⁠.

(b) Step (a) and part (i) (resp. part (vi)) of Remark 2 for the case $d$ odd (resp. $d$ even) and $q$ with generic rank cover all cases with $d≤4$⁠. Thus we may assume $d≥5$ and use induction on $d$⁠. Fix a general $o∈X$⁠. Let $ℓo:ℙd\{o}→ℙd−1$ denote the linear projection from $o$⁠. Let $Y⊂ℙd−1$ denote the closure of $ℓo(X\{o})$ in $ℙd−1$⁠. $Y$ is a rational normal curve of $ℙd−1$⁠. Set $q′:=ℓo(q)$ and $Z′:=ℓo(Z)$ (by the generality of $o$ we have $o∉〈Z〉$ and hence $Z′$ is well-defined, $deg(Z′)=b$ and $dim〈Z′〉=b−1$ ). The generality of $o$ also implies that $q∉〈Z″∪{o}〉$ for any $Z″⊊Z$ (here we use that $X$ is a smooth curve and hence $Z$ has only finitely many subschemes). Thus $q′∈〈Z′〉$ and $q′∉〈Z″〉$ for all $Z″⊊Z′$⁠. Since $Y$ is a degree $d−1$ rational normal curve and $b≤d/2$⁠, parts (i) and (ii) of Remark 2 imply $bY(q′)=b$ and $Z(Y,q′)={Z′}$⁠. Fix an irreducible family $Γ⊆S(X,q)$ such that $dimΓ=d+3−2b$ (it exists by part (v) of Remark 2). Let $B$ denote the set of all $A∈Γ$ such that $o∈A$⁠. By part (v) of Remark 2 and the generality of $o$ we have $B≠0$ and $dimB=d+2−2b$⁠. Set $A:={ℓo(B\{o})}B∈B$⁠. Since $Y$ is a rational normal curve, parts (i) and (ii) of Remark 2 imply $A⊆S(Y,q′)$⁠. We have $dimA=(d−1)+3−2b$⁠. The inductive assumption gives ${q′}=∩A∈A〈A〉$⁠. Thus $〈{o,q}〉=∩B∈B〈B〉$⁠. Since $dimΓ=dimS(X,q)=d+3−2b$ (part (v) of Remark 2) and $o$ is general in $X$⁠, there is $S∈Γ$ such that $o∉S$⁠. Thus $∩A∈Γ〈A〉={q}$⁠.□

Note that we have $Wq=〈νd(U)∪{q′}〉$ for any $q$ such that $S(X,q)={E∪U}E∈S(Y,q′)$ by Theorem 2. Assume either $rX(q)<k$ or $k≤2d−3$ and the existence of $B∈S(X,q)\{E∪U}E∈S(Y,q′)$⁠. In the former case take $B∈S(X,q)$⁠. Set $S:=A∪B$⁠. In both cases we have $|B|≤|A|$ and $|A|+|B|≤4d−5$⁠. Since $h1(IS(d))>0$ ([6, Lemma 1]) there is $F⊆S$ in one of the cases listed in Lemma 1.

(a) Assume the existence of a plane cubic $T⊂ℙn$ such that $|T∩S|≥3d$⁠.

(a1) Assume $n=2$⁠. Thus $T$ is an effective divisor of $ℙ2$⁠. Consider the residual exact sequence of $T$ in $ℙ2$⁠:

Since $|S\S∩T|≤4d−5−3d=d−5$⁠, we have $h1(IS\S∩T(d−3))=0$⁠. Thus either [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] give $A\A∩T=B\B∩T$⁠. Assume for the moment $L⊈T$⁠. Bezout gives $|L∩T|≤3$⁠. Since $U$ is general and $h0(Oℙ2(3))=10$⁠, we have $|U∩T|≤9$⁠. Thus $|B∩T|≥3d−12>12≥|T∩A|$ and hence $|B|>|A|$⁠, a contradiction. Now assume $L⊂T$⁠. Since $h0(Oℙ2(2))=6$ and $U∩L=ø$⁠, we get $|A∩T|≤d+8−b$⁠. Thus $|B∩T|≥2d−5+b$ and again $|B|>|A|$⁠, a contradiction.

(a2) Assume $n>2$⁠. Let $M⊂ℙn$ be a general hyperplane containing the plane $〈T〉$ (so $M=〈T〉$ if $n=3$ ). Since $S$ is a finite set and $M$ is a general hyperplane containing $〈T〉$⁠, we have $S∩M=S∩〈T〉$⁠. Consider the residual exact sequence of $M$ in $ℙn$⁠:

Since $|S\A∩M|≤4d−5−3d=d−5$⁠, we have $h1(IS\S∩M(d−1))=0$⁠. Thus either [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] give $A\A∩M=B\B∩M$⁠. Since no $4$ points of $U$ are coplanar, we have $|A∩M|≤d+5−b<3d−d−2+b$⁠. Thus $|B|>|A|$⁠, a contradiction.

(b) Assume the existence of a plane conic $D$ such that $|S∩D|≥2d+2$⁠.

(b1) Assume $n=2$⁠. Consider the residual exact sequence of $D$ in $ℙ2$⁠:

First assume $h1(IS\S∩D(d−2))>0$⁠. Since $|S\S∩D|≤4d−5−2d−2=2(d−3)−1$⁠, there is a line $R⊂ℙ2$ such that $|R∩(S\S∩D)|≥d−1$ ([9, Lemma 34]). Thus $|S∩(D∪R)|≥3d+1$⁠. Step (a1) gives a contradiction. Now assume $h1(IS\S∩D(d−2))=0$⁠. Either [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] give $A\A∩D=B\B∩D$⁠. Assume for the moment $L⊈D$⁠. Thus $|L∩D|≤2$⁠. Since $U$ is general and $h0(Oℙ2(2))=6$⁠, we have $|U∩D|≤5$⁠. Thus $|B∩D|>|A∩D|$ and so $|B|>|A|$⁠, a contradiction. Now assume $L⊂D$⁠. Write $D=L∪R$ with $R$ a line. Set ${o}:=L∩R$⁠. Since $U∩L=ø$⁠, $|L∩R|=1$ and $|U∩R|′≤2$ for each line $R′$⁠, we have $|A∩D|≤d+4−b$⁠. Let $Z′⊂L$ be the only degree $b$ zero-dimensional scheme evincing the cactus rank of $q′$ with respect to the rational normal curve $νd(L)$ (part (i) of Remark 2). Set $Z″:=Z′∪U$ and $Z:=Z″∪B$⁠. Since $A\A∩D=B\B∩D$⁠, we have $Z=Z′∪(U∩R)∪(B∩D)∪(U\U∩D)$⁠. Since $q′∈〈νd(Z′)〉$⁠, we have $q∈〈νd(Z″)〉∩〈νd(B)〉$⁠. Thus $h1(IZ(d))>0$⁠. Since $h1(IU\S∩D(d−2))=h1(IS\S∩D(d−2))=0$⁠, the residual sequence (3) of $D$ in $ℙ2$ with $Z$ instead of $S$ gives $h1(D,IZ∩D,D(d))>0$⁠. Since $D=R∪L$⁠, using either [16, Corollaire 2] or the residual exact sequences of $R$ and $L$ in $ℙ2$ we get that we are in one of the following cases:

1. $deg(Z∩L)≥d+2$⁠;

2. $deg(Z∩R)≥d+2$⁠;

3. $deg(Z∩R)=deg(Z∩L)=d+1$ and $o∉Zred$⁠.

Recall that $A\A∩B=B\B∩D$ (and hence) $|B∩D|≤|A∩D|$ and that $|A∩D|≤d+4−b$⁠.

(b1.1) Assume $deg(Z∩L)≥d+2$⁠. Since $Z∩L=Z′∪(B∩L)$⁠, we get $|B∩L|≥d+b−2$⁠. Consider the residual exact sequence of $L$ in $ℙ2$⁠:

First assume $h1(IS\S∩L(d−1))>0$⁠. Since $h1(IS\S∩(R∪L)(d−2))=0$⁠, the residual exact sequence of $R$ gives $h1(R,IS∩R\S∩R∩L(d−1))>0$⁠. Thus $|S∩R\S∩{o}|≥d+1$⁠. Since $|U∩R|≤2$⁠, we get $|B∩R\B∩{o}|≥d−1$ and hence $|B|>|A|$ (because $d≥5$⁠), a contradiction.

Now assume $h1(IS\S∩L(d−1))=0$⁠. By [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] we get $A\A∩L=B\B∩L$⁠. Since $|B∩L|≥d+2−b=|A∩L|$⁠, we get $|B∩L|=d+2−b$⁠. In this case part (1) of Theorem 3 is proved. To prove part (2) we need to prove that $B∩L∈S(Y,q′)$⁠. Since $|B∩L|=d+2−b$ and $deg(Z∩L)≥d+2$⁠, we get $Z1∩B=ø$ and $deg(Z∩B)=d+2$⁠. Thus $〈νd(B∩L)〉∩〈νd(Z′)〉$ is a single point, $q″$⁠, and $B∩L∈S(Y,q″)$⁠. Since $U$ is general, and $|U|≤(d+22)−d$⁠, we have $〈νd(U)〉∩〈νd(L)〉=ø$⁠. Since $B\B∩L=A\A∩L=U$⁠, we get $q″=q′$⁠, proving part (2) in this case.

(b1.2) Assume $deg(Z∩R)=deg(Z∩L)=d+1$ and $o∉Zred$ (i.e. $o∉Zred′$⁠). Since $deg(Z′)=b$ and $deg(Z″∩D)=b+|U∩R|≤b+2$⁠, we get $|L∩B|≥d+1−b$⁠, $|R∩B|≥d−1$ and $o∉B$⁠. Thus $|B∩D|≥2d+2−b$⁠. Since $|B∩D|≤|A∩D|≤d+4−b$⁠, we obtain a contradiction.

(b1.3) Assume $deg(Z∩R)≥d+2$⁠. Since $|U∩R|≤2$ with strict inequality if $o∈Zred′$ and every point of $〈νd(R)〉$ has rank $≤d$ by Sylvester’s theorem, we get $|U∩R|+deg(Z′∩R)=2$⁠, $|B∩R|=d$ and $U∩B∩R=ø$⁠. If $h1(IS\S∩R(d−1))=0$⁠, we have $A\A∩R=B\B∩R$ by [7, Lemma 5.1] or [8, Lemmas 2.4 and 2.5] and hence $|B|>|A|$⁠, a contradiction. Now assume $h1(IS\S∩R(d−1))>0$⁠. Since $h1(IS\S∩D(d−2))=0$ in this part of the proof, the residual exact sequence of $L$ gives $h1(L,IL∩S\L∩S∩R(d−1))>0$ and hence $|S∩(L\L∩R)|≥d+1$⁠. Thus $|B∩(L\L∩R)|≥b−1$⁠. We get $|B∩D|≥d+b−1$⁠. Since $A\A∩D=B\B∩D=U\U∩D$⁠, we get $d+b−1≤d+4−b$ and hence $b=2$ and $|B∩L|≤2$⁠. Since $|U∩R|+deg(Z′∩R)≥2$ and $q′$ uniquely determines $Z′$⁠, $R$ is uniquely determined by $Z′$ and the set $|R∩U|$⁠. If $o∈Zreg′$$R$ is uniquely determined by $q′$ and one point of $R\{o}$⁠. Since we took a general $U$ after fixing $q′$⁠, we have $|U∩R|≤1$ if $R∩L∈Zreg′$⁠. Hence (varying the points of $U\U∩R$ (if $U⊈R$ ) we may (after fixing $q′$ ) assume that $U\U∩R$ is general in $ℙn\D$⁠. Since $A\A∩D=B\B∩D=U\U∩D$⁠, $|U\U∩D|≤(d+22)−2d−1$ and $U\U∩D$ is general, we have $〈νd(U\U∩D)〉∩〈νd(D)〉$⁠. Thus there is a unique $q″∈〈νd(D)〉$ such that $q∈〈{νd(U\U∩R),q″}〉$⁠. We have $q″∈〈νd(A∩D)〉∩〈νd(B∩D)〉∩〈{q′,νd(U∩R)}〉$⁠. Thus is it sufficient to prove parts (1) and (2) of the theorem for $q″$⁠, $A∩D$ and $B∩D$ instead of $q$⁠, $A$ and $B$⁠, i.e. in the rest of this step we assume $U=U∩R$⁠. Since $|B|≤|A|$⁠, we have $|B∩L|≤2$⁠. Thus $h1(IZ′∪(L∩B)(d−1))=0$⁠. if $o∉Zred′∩B$⁠, [7, Lemma 5.1] gives a contradiction, because $Z′⊈B$⁠. Now assume $o∈Zred′∩B$⁠. Since $o∈Zred′$ and $Z′$ is uniquely determined by $q′$⁠, we observed that $|U∩R|≤1$⁠. Thus (under the assumption $U⊂D$⁠), we have $|U|≤1$⁠. Since $|B∩R|=d,$$o=L∩R∈B$ by assumption and $Z′⊈R$⁠, we get $deg(Z∩R)≤d+1$⁠, a contradiction.