Explicit tensors of border rank at least 2n-1

For odd m, I write down tensors in C m ⊗C m ⊗C m of border rank at least 2m − 1, showing the non-triviality of the Young-flattening equations of [6] that vanish on the matrix multiplication tensor. I also study the border rank of the tensors of [1] and [3]. I show the tensors T 2 k ∈ C k ⊗C 2 k ⊗C 2 k , of [1], despite having rank equal to 2 k+1 − 1, have border rank equal to 2 k. I show the equations for border rank of [3] on C m ⊗C m ⊗C m are trivial in the case of border rank 2m − 1 and determine their precise non-vanishing on the matrix multiplication tensor. Let A, B, C be complex vector spaces of dimensions a, b, c. A tensor T ∈ A⊗B⊗C is said to have rank one if T = a⊗b⊗c for some a ∈ A, b ∈ B, c ∈ C. More generally the rank of a tensor T ∈ A⊗B⊗C is the smallest r such that T may be written as the sum of r rank one tensors. Letˆσ 0 r ⊂ A⊗B⊗C denote the set of tensors of rank at most r. This set is not closed (under taking limits or in the Zariski topology) so letˆσ r denote its closure (the closure is the same in the Euclidean or Zariski topology). The varietyˆσ r is familiar in algebraic geometry, it is cone over the r-th secant variety of the Segre variety, but we won't need that in what follows. The rank and border rank of a tensor are measures of its complexity. While rank is natural to complexity theory, border rank is more natural from the perspective of geometry, as one can obtain lower bounds on border rank via polynomials. Let R(T), R(T) respectively denote the rank and border rank of T. The maximum rank of a tensor in C m ⊗C m ⊗C m is at most m 2 , although it is not known in general if this actually occurs. The maximum border rank of a tensor in C m ⊗C m ⊗C m is ⌈ m 3 3m−2 ⌉ for all m = 3 and five when m = 3, see [9, 8]. It is an important problem to find explicit tensors of high rank and border rank, and to develop tests that …