Research Article | Open Access
E. Ballico, "An Upper Bound for the Tensor Rank", International Scholarly Research Notices, vol. 2013, Article ID 241835, 3 pages, 2013. https://doi.org/10.1155/2013/241835
An Upper Bound for the Tensor Rank
Let be a tensor of format , , over . We prove that has tensor rank at most .
Fix integers and , , and an algebraically closed base field . Let be a tensor of format over . The tensor rank of is the minimal integer such that with (see [1–6]). Classical papers (e.g., ) continue to suggest new results (see ). Let be the maximum of all integers , . In this paper we prove the following result.
Theorem 1. For all integers and one has .
This result is not optimal. It is not sharp when , since by elementary linear algebra. For large the bound should be even worse. In our opinion to get stronger results one should split the set of all into subregions. For instance, we think that for large the cases with and the cases with are quite different.
We make the definitions in the general setting of the Segre-Veronese embeddings of projective spaces (i.e., of partially symmetric tensors), but we only use the case of the usual Segre embedding, that is, the usual tensor rank. The tensor has zero as its tensor rank. If , then the tensors and have the same rank. Hence it is sufficient to study the function “tensor rank” on the projectivisation of the vector space . We may translate the tensor rank and the integer in the following language. For each subset of a projective space, let denote the linear span of . For each integral variety and any the -rank of is the minimal cardinality of a finite set such that . Now assume . The maximal -rank is the maximum of all integers , . Fix integers , , , and , . Set . Let , be the Segre-Veronese embedding of multidegree , that is, the embedding of induced by the -vector space of all polynomials , , , whose nonzero monomials have degree with respect to the variables , . Set . The variety is the Segre embedding of . Fix , . Let be the set of all nonzero tensors of format associated with . We have . Hence . To prove Theorem 1 we refine the notion of -rank in the following way.
Definition 2. Fix positive integers , , , and , . A small box of is a closed set with being a hyperplane of for all . A large box of is a product such that there is with being a hyperplane, while for all . A small polybox (resp., large polybox) of is a finite union of small (resp., large) boxes of . A small box (resp., small polybox, resp., large box, resp., large polybox) is the image by of a small box (resp., small polybox, resp., large box, resp., large polybox) of .
Definition 3. Fix positive integers , , , and , , and set . Fix . The rank of is the minimal cardinality of a finite set such that . The unboxed rank (resp., small unboxed rank) (resp., ) of is the minimal integer such that for each large polybox (resp., small polybox) there is a finite set with and . Let (resp., , resp., ) be the maximum of all integers (resp., , resp., ), .
Notice that .
Since for all , we have . Hence Theorem 1 is an immediate corollary of the following result.
Theorem 4. For all integers and one has .
We hope that the definitions of unboxed rank and small unboxed rank are interesting in themselves, not just as a tool. As far as we know the best upper bound for the symmetric tensor rank is due to Biaynicki-Birula and Schinzel ([9, 10]). In  Białynicki-Birula and Schinzel used the corresponding notion in the case .
2. Proof of Theorem 4
Remark 5. Fix integers and , . Fix and let be the projection. For any small polybox the set is a small polybox of the Segre variety .
In the case we also need the following notation. Fix integers . For each , , let be the minimal integer with the following property: for each finite union of hyperplanes there is a set such that and . Let be the maximum of all integers , . Obviously . Linear algebra gives .
Lemma 6. For all integers one has .
Proof. It is sufficient to prove the inequality . Without losing generality we may assume . Set and . Fix , , and a union of finitely many hyperplanes. Fix inducing and inducing . Fix a basis of such that for all . We may write for some . Hence .
Proof of Theorem 4. Lemma 6 gives the case . Hence we may assume and use induction on . Fix and a small polybox . For a fixed integer we also use induction on , starting from the case (in which we use instead of ).
Take a general hyperplane . Set , , , and . We have . Let denote the linear projection from . Notice that . If , then we have and hence we use induction on to apply Theorem 4 to . If , then we use induction on to apply Theorem 4 to . Set . Notice that induces a surjection (projection onto the first factors). Let denote the closure of in . Since is a small polybox, is a small polybox (Remark 5). For general we may also assume that is a small polybox of . First assume . Since is a small polybox of , the inductive assumption gives the existence of a set such that and . Hence . Now assume . Hence is defined. Since is a small polybox, there is such that and . Since is surjective, there is such that . Since and , we have . Hence is defined at each point of . Since and , there is such that . Since is a small polybox, there is such that and . We have and .
The author was partially supported by MIUR and GNSAGA of INdAM (Italy).
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