# Rank 1 Decompositions of Symmetric Tensors Outside a Fixed Support

Authors: E. Ballico
Journal: ISRN Geometry
 (2013)

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## Abstract

Letνd:ℙm→ℙn, n:=(n+dn)-1, denote the degree d Veronese embedding of ℙm. For any P∈ℙn, let sr(P) be the minimal cardinality of S⊂νd(ℙm) such that P∈〈S〉. Identifying P with a homogeneous polynomial q (or a symmetric tensor), S corresponds to writing q as a sum of ♯(S) powers Ld with L a linear form (or as a sum of ♯(S)d-powers of vectors). Here we fix an integral variety T⊊ℙm and P∈〈νd(T)〉 and study a similar decomposition with S⊈T and ♯(S) minimal. For instance, if T is a linear subspace, then we prove that ♯(S)≥♯(S∩T)+d+1 and classify all (S,P) such that ♯(S)-♯(S∩T)≤2d-1.


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## Body

## 1. Introduction

Letνd:ℙm→ℙn, n:=(n+dn)-1, denote the degree d Veronese embedding of ℙm. Set Xm,d:=νd(ℙm). For any P∈ℙn, the symmetric rank or symmetric tensor rank or the rank sr(P) of P is the minimal cardinality of a finite set S⊂Xm,d such that P∈〈S〉, where 〈·〉 denotes the linear span. In this paper we study the following problem.Fix an integral varietyT⊊ℙm. What is the minimal cardinality sr(P,=T) of a finite set S⊂ℙm such that P∈〈νd(S)〉, P∉〈νd(S′)〉 for any S′⊊S and S⊈T?TheT-rank srT(P) of P is the minimal cardinality of a finite set S⊂T such that P∈〈νd(S)〉, with the convention srT(P)=+∞ if there is no such set S, that is, if P∉〈νd(T)〉. The classical case is when T is a proper linear subspace ℙk, k<m, of ℙm. Fix any k-dimensional linear subspace T⊂ℙm and P∈〈νd(T)〉. In this case we are looking at the symmetric rank of a symmetric tensor depending only on k variables. By [1, Proposition 3.1], there is S⊂T such that ♯(S)=sr(P) and P∈〈νd(S)〉; that is, srT(P)=sr(P). By [2, Exercise 3.2.2.2], we have sr(P,=T)>sr(P); that is, if S⊂ℙm, ♯(S)=sr(P), and P∈〈νd(S)〉, then S⊂T.The general case is motivated from the following query. Assume that writing thed-power LQd of the linear form LQ associated with Q∈νd(ℙm) has a price cQ>0. Find a finite set S⊂ℙm such that P∈〈νd(S)〉 and ∑Q∈S‍cQ is small. What happens if the points of ℙr∖T are cheaper than the points of T? The condition “P∉〈S′〉 for any S′⊊S” is necessary to get a reasonable definition for the following reason. Suppose srT(P)<+∞; that is, P∈〈νd(T)〉, and take E⊂T such that ♯(E)=srT(P) and P∈〈νd(E)〉. Fix any Q∈ℙm∖T. Obviously P∈〈νd(E∪{Q})〉. Hence there is a set S1⊂ℙm such that ♯(S1)=srT(P)+1, S1⊈T, and P∈〈νd(S1)〉. We need to exclude sets like E∪{Q}. A similar problem is to find the minimal cardinality sr(P,-T) of a set B⊂ℙm∖T such that P∈〈νd(B)〉. It is easy to see that sr(P,-T) is always a finite integer. Let sr(P,≡T) be the minimal integer ♯(S∖S∩T) among all finite sets S⊂ℙm such that P∈〈S〉, P∉〈S′〉 for any S′⊊S and S⊈T. Obviously sr(P,≡T)≤sr(P,=T)≤sr(P,-T). Roughly speaking, we use sr(P,-T) if it is forbidden (or very expensive) to use the points of T.We first prove the following result.Proposition 1.
Fix anyT⊊ℙm and any P∈〈νd(T)〉.(a)
One  hassr(P,≡T)≤d+1.(b)
IfsrT(P)>sr(P),   thensr(P,=T)=sr(P)and  everySevincingsr(P)evincessr(P,=T).   IfsrT(P)=sr(P),   thensr(P,=T)≤sr(P)+d.Quite often equality holds in part (a) of Proposition1 (see, for instance, Theorem 6). Part (b) shows that if P∈〈νd(T)〉, then sr(P)-sr(P,=T) is always small.We recall again that ifT is a linear subspace, then sr(P,=T)>sr(P) for every P∈〈νd(T)〉 ([2, Exercise 3.2.2.2]). In this paper we prove that this is a characterization of linear subspaces of ℙm. Indeed we prove the following result.Theorem 2.
Fix an integral varietyT⊊ℙm which is not a linear subspace. Then there is P∈〈νd(T)〉 such that
(1)srT(P)>sr(P)=sr(P,=T)=sr(P,-T)=d.Proposition 3.
AssumeT≠ℙm. Then sr(P,-T)≤(d+1)·sr(P). If A evinces sr(P,=T), then sr(P,-T)≤♯(A∖A∩T)+(d+1)·♯(A∩T).Theorem 4.
Assume the existence of an integerk<d such that the sheaf ℐT(k) is spanned by its global sections outside T. Fix P∈〈νd(T)〉. Then sr(P,≡T)≥d-k+1. Fix a finite set B⊂ℙm such that P∈〈νd(B)〉, P∉〈νd(B∩T)〉, and ♯(B∖B∩T)≤3(d-k). Then either there is a line L⊂ℙm such that ♯(L∩(B∖B∩T))≥d-k+2, there is a conic R⊂ℙm such that ♯(R∩(B∖B∩T))≥2d-2k+2, or there is a plane cubic C⊂ℙm such that ♯(C∩(B∖B∩T))≥3d-3k.Theorem 5.
Assume the existence of a lineL⊂ℙm such that L⊈T and deg(L∩T)=k≤⌊(d+1)/2⌋. Then there is P∈〈νd(T)〉∩〈νd(L)〉 such that sr(P,=T)=sr(P,-T)=d+2-k, and with the following additional property. If the scheme L∩T is reduced, then sr(P)=k and L∩T is the only set such that νd(L∩T) evinces sr(P). If the scheme L∩T is not reduced, then sr(P)=sr(P,=T)=d+2-k.Theorem 6.
LetT⊊ℙm be a proper linear subspace. Fix P∈〈νd(T)〉. One has sr(P,≡T)=d+1. Fix any G⊂ℙm such that P∈〈νd(G)〉, P∉〈νd(G∩T)〉, and ♯(G∖G∩T)≤2d-2. Let B⊆G be a minimal subset such that P∈〈νd(B)〉. Set U:=B∖B∩T. Then one of the following cases occurs:(i)
U=∅;(ii)
♯(U)=d+1, U is contained in a line L, L∩T is a point (call it Q), Q∉B, and P∈〈νd({Q}∪(B∩T))〉.We work over an algebraically closed field𝕂 with char(𝕂)=0. The characteristic zero assumption is used to quote a theorem of Sylvester ([3], [4, Theorem 4.1], [5, Section 3]), and [4, Proposition 5.1].


## 2. Preliminary Results

Notation 7.
For anyP∈ℙn, let 𝒮(P) denote the set of all finite sets A⊂ℙm such that νd(A) evinces sr(P), that is, such that P∈〈νd(A)〉 and ♯(A)=sr(P).The definition of the integersr(P) gives that if A∈𝒮(P), then P∉〈νd(A′)〉 for any A′⊊A.We recall the following elementary result ([6, Lemma 1]).Lemma 8.
FixP∈ℙn. Let A,B⊂ℙm be two zero-dimensional schemes such that A≠B. Assume the existence of P∈〈νd(A)〉∩〈νd(B)〉 such that P∉〈νd(A′)〉 for any A′⊊A and P∉〈νd(B′)〉 for any B′⊊B. Then h1(ℙm,ℐA∪B(d))>0.The following lemma was proved (withD a hyperplane) in [7, Lemma 7]. The same proof works for an arbitrary hypersurface D of ℙm.Lemma 9.
Fix positive integersm, d, and t such that t≤d and finite sets A,B⊂ℙm. Assume the existence of a hypersurface D⊂ℙm such that deg(D)=t and h1(ℐ(A∪B)∖(A∪B)∩D(d-t))=0. Set F:=A∩B∖D∩A∩B. Then νd(F) is linearly independent and 〈νd(A)〉∩〈νd(B)〉 is the linear span of its supplementary subspaces 〈νd(F)〉 and 〈νd(A∩D)〉∩〈νd(B∩D)〉. If there is P∈〈νd(A)〉∩〈νd(B)〉 such that P∉〈νd(A′)〉 for any A′⊊A and P∉〈νd(B′)〉 for any B′⊊B, then A=(A∩D)⊔F and B=(B∩D)⊔F.Lemma 10.
FixQ∈ℙm and set P:=νd(Q).(a)
Fix a finite setA⊂ℙm such that Q∉A and P∈〈νd(A)〉. Then ♯(A)≥d+1. If ♯(A)=d+1, then there is a line L⊂ℙm such that Q∈L and A⊂L∖{Q}.(b)
For each lineD⊂ℙm with Q∈D and any B⊂D∖{Q} such that ♯(B)=d+1, one has P∈〈νd(B)〉.Proof.
Lemma8 gives h1(ℐA∪{Q}(d))>0. Hence ♯(A)≥d+1, and equality holds only if there is a line L⊂ℙm such that A∪{Q}⊂L. Now we fix the line D⊂ℙm. Fix any set B⊂D such that ♯(B)=d+1. Since hi(D,ℐB(d))=0, i=0,1, the set νd(B) spans the d-dimensional linear space 〈νd(D)〉. Since P∈〈νd(D)〉, we have P∈〈νd(B)〉.Corollary 11.
Fix any integral varietyT⊊ℙm and any Q∈ℙm∖T. Set P:=νd(Q). Then sr(P,=T)=sr(P,-T)=d+1.Proof.
Part (a) of Lemma10 gives sr(P,=T)≥d+1, and hence sr(P,-T)≥d+1. Since T≠ℙm, there is a line D⊂ℙm such that P∈D and D⊈T. Since D∩T is finite, to get sr(P,-T)≤d+1, it is sufficient to take any B⊂D∖D∩T with ♯(B)=d+1 by part (b) of Lemma 10.


## 3. The Proofs

Proof of Theorem2.
Fix any smooth pointQ of T. Let V⊂ℙm be the Zariski tangent space of T at Q. Since T is not a linear space and Q is a smooth point of it, there is a line L⊂V such that Q∈L and L⊈T. Hence the set G:=(T∩L)red is a finite set containing Q. Let Z⊂L be the degree 2 effective divisor of L with Q as its reduction. Since L⊆V and V is the tangent space of T at Q, we have Z⊂T. Since deg(Z)=2, the linear space W:=〈νd(Z)〉 is a line. Fix any P∈W∖{νd(Q)}. Since Z⊂L, W is the tangent line of the rational normal curve νd(L) at Q. A theorem of Sylvester gives srL(P)=d ([3], [4, Theorem 4.1], or [5, Section 3]). Since Z⊂T, we have W⊂〈νd(T)〉. Hence P∈〈νd(T)〉. Fix any S⊂ℙm such that νd(S) evinces sr(P). Since P∈〈νd(L)〉 and L is a linear subspace of ℙm, we have S⊂L (the symmetric case of [2, Proposition 1] or [1, Proposition 3.1]). By a theorem of Sylvester, we have #(S)=d ([3], [4, Theorem 4.1], and [5]). Lemma 8 gives h1(ℐZ∪S(d))>0. Hence Q∉S. Since deg(Z∪S)=d+2 and Z∩S=∅, Grassmann's formula gives {P}=〈νd(Z)〉∩〈νd(S)〉. Hence for a fixed scheme Z, each set S is associated with a unique P∈W∖{νd(Q)}. Since G is a finite set, it has only finitely many subsets with cardinality d. Hence for a general P∈W∖{νd(Q)}, no set computing sr(P) is contained in T. Hence sr(P,=T)=sr(P)=d and srT(P)>d for a general P∈W∖{νd(Q)}. Fix any P∈W∖{νd(Q)}. Since G is finite, the proof of [4, Proposition 5.1] gives the existence of A∈𝒮(P) such that A∩G=∅. Hence sr(P,-T)=d.Proof of Proposition1.
IfsrT(P)>sr(P), then sr(P,=T)=sr(P) and any A∈𝒮(P) evinces sr(P,=T), because A⊈T. Hence we may assume sr(P)=srT(P). Fix any A⊂T such that νd(A) evinces srT(P). Fix any Q∈A. Since T≠ℙm, there is a line D⊂ℙm such that Q∈D and D⊈T. Hence D∩T is finite. Take any B′⊂D∖D∩T such that ♯(B′)=d+1. Set B:=(A∖{Q})∪B′. We have ♯(B∖B∩T)=d+1. Since ♯(B′)=d+1, we have 〈νd(B′)〉=〈νd(D)〉. Since Q∈D, we have νd(Q)∈〈νd(D)〉. Since A∖{Q}⊂B, we have P∈〈νd(B)〉.Proof of Proposition3.
FixA⊂ℙm computing either sr(P) or sr(P,=T). Hence P∈〈νd(A)〉. Since T≠ℙm, for each Q∈A∩T, there is a line LQ⊂ℙm such that Q∈LQ and LQ∩T is finite. Fix SQ⊂L∖L∩T such that ♯(SQ)=d+1. Set B:=(A∖A∩T)∪(∪Q∈A∩TSQ). We have B⊂ℙm∖T and ♯(B)≤♯(A∖A∩T)+(d+1)·#(A∩T). Since 〈νd(SQ)〉=〈νd(LQ)〉, we have νd(Q)∈〈νd(SQ)〉 for all Q∈A∩T. Hence P∈〈νd(B)〉.Proof of Theorem4.
Taking a smaller subset ofB, we may assume P∉〈νd(B′)〉 for any B′⊊B. Fix A⊂T computing srT(P). Since P∉〈νd(B∩T)〉, we have B≠A. Lemma 8 gives h1(ℐA∪B(d))>0. Let D be a general hypersurface of degree k containing T. Since ℐT(k) is spanned outside T and A∪B is finite, the generality of D implies D∩(A∪B)=T∩(A∪B); that is, A∪B∖(A∪B)∩T=B∖B∩T. Since A⊂T and B⊈T, Lemma 9 implies h1(ℐB∖B∩T(d-k))>0. Hence ♯(B∖B∩T)≥d-k+1 ([5, Lemma 34]). Since this is true for any B, we get sr(P,≡T)≥d-k+1. By [8, Theorem 3.8], applied to the integer d-k, we also get the second part of Theorem 4.Proof of Theorem5.
SetZ:=L∩T (scheme-theoretic intersection). Take a finite set S⊂L∖L∩T such that #(S)=d+2-k. Since S∩Z=∅, we have deg(Z∪S)=d+2. Since Z∩S=∅ and 〈νd(Z∪S)〉=〈νd(L)〉, we get that 〈νd(Z)〉∩〈νd(S)〉 is a single point. We call P this point. Since h1(L,ℐW(d))=0 for every zero-dimensional scheme W⊂L such that deg(W)≤d+1, we have 〈νd(Z′)〉∩〈νd(S)〉=∅ for any Z′⊊Z and 〈νd(Z)〉∩〈νd(S′)〉=∅ for any S′⊊S. Hence P∉〈νd(Z′)〉 for any Z′⊊Z and P∉〈νd(S′)〉=∅ for any S′⊊S. Lemma 8 and [5, Lemma 34] give deg(W)≥d+2-k for any zero-dimensional scheme W⊂ℙm such that W⊉Z and P∈〈νd(W)〉. Since d>2k, νd(Z) is the unique zero-dimensional subscheme of νd(L) with degree ≤k and whose linear span contains P. The set S⊂L∖L∩T gives sr(P,-T)≤d-k+2. Since P∈〈νd(L)〉, every A∈𝒮(P) is contained in L ([2, Exercise 3.2.2.2]). From a theorem of Sylvester ([3], [4, Theorem 4.1], and [5, Section 3]), we get that if Z is reduced, then sr(P)=k and 𝒮(P)={Z}, while if Z is not reduced, then sr(P)=d+2-k and every A∈𝒮(P) is contained in L∖L∩T. If Z is a single point, Q, then any finite set A⊂ℙm∖{Q} such that P∈〈νd(A)〉 has cardinality at least d+1 [5, Lemma 34]. Hence we may assume deg(Z)≥2. To conclude, it is sufficient to prove that if νd(B) evinces sr(P,=T), then B⊂L∖L∩T. Take B⊂ℙm such that νd(B) evinces sr(P,=T). Since B⊈T, we have Z≠B. Lemma 8 gives h1(ℐZ∪B(d))>0. By [5, Lemma 34], we have deg(Z∪B)≥d+2, and (since deg(Z)+deg(B)≤d+2≤2d+1) there is a line D⊂ℙm such that Z∪B⊂D. Since ♯(B)≤d+2-k, we get ♯(B)=d+2-k and B∩Z=∅. If D=L, then B⊂L∖L∩T. Now assume D≠L. Hence either D∩L=∅ or the scheme theoretic intersection D∩L has degree 1. Since Z⊆D∩L, we get deg(Z)≤1, a contradiction.Proof of Theorem6.
We may assumeU≠∅; otherwise there is nothing to prove. Since B is minimal, we have P∉〈νd(B′)〉 for any B′⊊B. Hence the set νd(B) is linearly independent. Fix A⊂T such that νd(A) evinces srT(P). Hence P∉〈νd(A′)〉 for any A′⊊A. Lemma 8 gives h1(ℐA∪B(d))>0. Fix a general hyperplane H⊂ℙm such that T⊆H. Since H is general and B is finite, we have B∩H=B∩T. Hence U=(A∪B)∖(A∪B)∩T. We may apply the second part of Lemma 9 with respect to the degree 1 divisor D:=H. Since A⊂T⊆H, we get that either B⊂H or h1(ℐ(A∪B)∖(A∪B)∩H(d-1))>0. Since B∩H=B∩T, we have B⊈H. Hence h1(ℐ(A∪B)∖(A∪B)∩H(d-1))>0. Since ♯(U)≤2d-1, there is a line L⊂ℙm such that ♯(U∩L)≥d+1 ([5, Lemma 34]). Since νd(B) is linearly independent, we have B∩L=U∩L and ♯(U∩L)=d+1. Let H′ be a general hyperplane containing L. Since A∪B is finite and H′ is general, we have H′∩(A∪B)=L∩(A∪B)=U∩L. Set E:=(A∪B)∖(A∪B)∩(H∪H′). Notice that E=U∖U∩L. Hence ♯(E)≤d-2≤d-1. Therefore h1(ℐE(d-2))=0. Since A⊂H⊂H∪H′, Lemma 9 applied to the degree 2 hypersurface H∪H′ gives U=U∩L. Since U∩T=∅ and U∩L≠∅, we have L⊈T. Since L and T are linear subspaces of ℙm, the restriction map ρ:H0(ℙm,𝒪ℙm(d))→H0(T∪L,𝒪T∪L(d)) is surjective. Hence 〈νd(T)〉∩〈νd(L)〉=〈νd(T∩L)〉. Since P∈〈νd(B)〉 and P∉〈νd(B′)〉 for any B′⊊B, the set 〈νd(B∩T)∪{P}〉∩〈νd(U)〉 is a unique point. Call P′ this point. Since P∈〈νd(T)〉 and 〈νd(U)〉=〈νd(L)〉, we get T∩L≠∅ and P′=νd(Q), where {Q}:=T∩L. Thus P∈〈νd({Q}∪(B∩T))〉.



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