Tensor rank of anti-symmetric tensor - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:19:14Z http://mathoverflow.net/feeds/question/116831 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116831/tensor-rank-of-anti-symmetric-tensor Tensor rank of anti-symmetric tensor Klim Efremenko 2012-12-20T00:31:23Z 2012-12-20T00:31:23Z <p>Let $V$ be a vector space of dimension $n$. Let us consider $V^{\otimes n}=V\otimes V \ldots \otimes V$. This vector space contains one dimentional vector space $\wedge^n V$. My question is does it something is known about the tensor rank of the vector $\wedge^n V$?</p> <p>More formally let $e_1, e_2,\ldots e_n$ be a basis for $V$ than the question is what does it known about the tensor rank of:$$T=\sum_{\sigma \in S_n}(-1)^{sign(\sigma)} e_{\sigma(1)}\otimes e_{\sigma(2)} \otimes \ldots \otimes e_{\sigma(n)}.$$</p> <p>The trivial upper bound on the tensor rank of this form is $n!$. Does it know any better uper bound? </p> <p>As far as I know without $(-1)^{sign(\sigma)}$(i.e. for a symmetric form) it know upper bound of $2^n$. </p>