Abstract

Let , , denote the degree Veronese embedding of . For any , let be the minimal cardinality of such that . Identifying with a homogeneous polynomial (or a symmetric tensor), corresponds to writing as a sum of powers with a linear form (or as a sum of -powers of vectors). Here we fix an integral variety and and study a similar decomposition with and minimal. For instance, if is a linear subspace, then we prove that and classify all such that .

1. Introduction

Let , , denote the degree Veronese embedding of . Set . For any , the symmetric rank or symmetric tensor rank or the rank of is the minimal cardinality of a finite set such that , where denotes the linear span. In this paper we study the following problem.

Fix an integral variety . What is the minimal cardinality of a finite set such that , for any and ?

The -rank of is the minimal cardinality of a finite set such that , with the convention if there is no such set , that is, if . The classical case is when is a proper linear subspace , , of . Fix any -dimensional linear subspace and . In this case we are looking at the symmetric rank of a symmetric tensor depending only on variables. By [1, Proposition 3.1], there is such that and ; that is, . By [2, Exercise ], we have ; that is, if , , and , then .

The general case is motivated from the following query. Assume that writing the -power of the linear form associated with has a price . Find a finite set such that and is small. What happens if the points of are cheaper than the points of ? The condition “ for any ” is necessary to get a reasonable definition for the following reason. Suppose ; that is, , and take such that and . Fix any . Obviously . Hence there is a set such that , , and . We need to exclude sets like . A similar problem is to find the minimal cardinality of a set such that . It is easy to see that is always a finite integer. Let be the minimal integer among all finite sets such that , for any and . Obviously . Roughly speaking, we use if it is forbidden (or very expensive) to use the points of .

We first prove the following result.

Proposition 1. Fix any and any .(a)One  has  .(b)If  ,   then     and  every    evincing     evinces  .   If,   then  .

Quite often equality holds in part (a) of Proposition 1 (see, for instance, Theorem 6). Part (b) shows that if , then is always small.

We recall again that if is a linear subspace, then for every ([2, Exercise ]). In this paper we prove that this is a characterization of linear subspaces of . Indeed we prove the following result.

Theorem 2. Fix an integral variety which is not a linear subspace. Then there is such that

Proposition 3. Assume . Then . If evinces , then .

Theorem 4. Assume the existence of an integer such that the sheaf is spanned by its global sections outside . Fix . Then . Fix a finite set such that , , and . Then either there is a line such that , there is a conic such that , or there is a plane cubic such that .

Theorem 5. Assume the existence of a line such that and . Then there is such that , and with the following additional property. If the scheme is reduced, then and is the only set such that evinces . If the scheme is not reduced, then .

Theorem 6. Let be a proper linear subspace. Fix . One has . Fix any such that , , and . Let be a minimal subset such that . Set . Then one of the following cases occurs:(i);(ii), is contained in a line , is a point (call it ), , and .

We work over an algebraically closed field with . 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 any , let denote the set of all finite sets such that evinces , that is, such that and .

The definition of the integer gives that if , then for any .

We recall the following elementary result ([6, Lemma 1]).

Lemma 8. Fix . Let be two zero-dimensional schemes such that . Assume the existence of such that for any and for any . Then .

The following lemma was proved (with a hyperplane) in [7, Lemma 7]. The same proof works for an arbitrary hypersurface of .

Lemma 9. Fix positive integers , , and such that and finite sets . Assume the existence of a hypersurface such that and . Set . Then is linearly independent and is the linear span of its supplementary subspaces and . If there is such that for any and for any , then and .

Lemma 10. Fix and set .(a)Fix a finite set such that and . Then . If , then there is a line such that and .(b)For each line with and any such that , one has .

Proof. Lemma 8 gives . Hence , and equality holds only if there is a line such that . Now we fix the line . Fix any set such that . Since , , the set spans the -dimensional linear space . Since , we have .

Corollary 11. Fix any integral variety and any . Set . Then .

Proof. Part (a) of Lemma 10 gives , and hence . Since , there is a line such that and . Since is finite, to get , it is sufficient to take any with by part (b) of Lemma 10.

3. The Proofs

Proof of Theorem 2. Fix any smooth point of . Let be the Zariski tangent space of at . Since is not a linear space and is a smooth point of it, there is a line such that and . Hence the set is a finite set containing . Let be the degree effective divisor of with as its reduction. Since and is the tangent space of at , we have . Since , the linear space is a line. Fix any . Since , is the tangent line of the rational normal curve at . A theorem of Sylvester gives ([3], [4, Theorem 4.1], or [5, Section 3]). Since , we have . Hence . Fix any such that evinces . Since and is a linear subspace of , we have (the symmetric case of [2, Proposition 1] or [1, Proposition 3.1]). By a theorem of Sylvester, we have ([3], [4, Theorem 4.1], and [5]). Lemma 8 gives . Hence . Since and , Grassmann's formula gives . Hence for a fixed scheme , each set is associated with a unique . Since is a finite set, it has only finitely many subsets with cardinality . Hence for a general , no set computing is contained in . Hence and for a general . Fix any . Since is finite, the proof of [4, Proposition 5.1] gives the existence of such that . Hence .

Proof of Proposition 1. If , then and any evinces , because . Hence we may assume . Fix any such that evinces . Fix any . Since , there is a line such that and . Hence is finite. Take any such that . Set . We have . Since , we have . Since , we have . Since , we have .

Proof of Proposition 3. Fix computing either or . Hence . Since , for each , there is a line such that and is finite. Fix such that . Set . We have and . Since , we have for all . Hence .

Proof of Theorem 4. Taking a smaller subset of , we may assume for any . Fix computing . Since , we have . Lemma 8 gives . Let be a general hypersurface of degree containing . Since is spanned outside and is finite, the generality of implies ; that is, . Since and , Lemma 9 implies . Hence ([5, Lemma 34]). Since this is true for any , we get . By [8, Theorem 3.8], applied to the integer , we also get the second part of Theorem 4.

Proof of Theorem 5. Set (scheme-theoretic intersection). Take a finite set such that . Since , we have . Since and , we get that is a single point. We call this point. Since for every zero-dimensional scheme such that , we have for any and for any . Hence for any and for any . Lemma 8 and [5, Lemma 34] give for any zero-dimensional scheme such that and . Since , is the unique zero-dimensional subscheme of with degree and whose linear span contains . The set gives . Since , every is contained in ([2, Exercise ]). From a theorem of Sylvester ([3], [4, Theorem 4.1], and [5, Section 3]), we get that if is reduced, then and , while if is not reduced, then and every is contained in . If is a single point, , then any finite set such that has cardinality at least [5, Lemma 34]. Hence we may assume . To conclude, it is sufficient to prove that if evinces , then . Take such that evinces . Since , we have . Lemma 8 gives . By [5, Lemma 34], we have , and (since ) there is a line such that . Since , we get and . If , then . Now assume . Hence either or the scheme theoretic intersection has degree . Since , we get , a contradiction.

Proof of Theorem 6. We may assume ; otherwise there is nothing to prove. Since is minimal, we have for any . Hence the set is linearly independent. Fix such that evinces . Hence for any . Lemma 8 gives . Fix a general hyperplane such that . Since is general and is finite, we have . Hence . We may apply the second part of Lemma 9 with respect to the degree divisor . Since , we get that either or . Since , we have . Hence . Since , there is a line such that ([5, Lemma 34]). Since is linearly independent, we have and . Let be a general hyperplane containing . Since is finite and is general, we have . Set . Notice that . Hence . Therefore . Since , Lemma 9 applied to the degree hypersurface gives . Since and , we have . Since and are linear subspaces of , the restriction map is surjective. Hence . Since and for any , the set is a unique point. Call this point. Since and , we get and , where . Thus .

Acknowledgments

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