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`ISRN GeometryVolume 2013, Article ID 704072, 4 pageshttp://dx.doi.org/10.1155/2013/704072`
Research Article

## Rank 1 Decompositions of Symmetric Tensors Outside a Fixed Support

Department of Mathematics, University of Trento, 38123 Povo (TN), Italy

Received 18 October 2013; Accepted 12 December 2013

Academic Editors: A. M. Cegarra, R. Farnsteiner, F. B. Gallego, S. K. Kar, and P. Schenzel

Copyright © 2013 E. Ballico. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### 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).

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