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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 $\wedge^n V$?

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:$$ T=\sum_{\sigma \in S_n}(-1)^{sign(\sigma)} e_{\sigma(1)}\otimes e_{\sigma(2)} \otimes \ldots \otimes e_{\sigma(n)}.$$

The trivial upper bound on the tensor rank of this form is $n!$. Does it know any better uper bound?

As far as I know without $(-1)^{sign(\sigma)}$(i.e. for a symmetric form) it know upper bound of $2^n$.

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What is the tensor rank? –  Sasha Dec 20 '12 at 5:37
    
@Sasha, see here for a definition: its.caltech.edu/~matilde/WeitzMa10Abstract.pdf –  Qfwfq Dec 20 '12 at 10:01
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I don't know the answer to your question, but I know that there is quite a lot of work on computing and bounding tensor ranks in the algebraic geometry community. You might try writing to any of M. Catalisano, A.V. Geramita, A. Gimigliano, J.M. Landsberg, and/or Jerzy Weyman and asking them your question. (I don't know that they read MO, so they might not otherwise know about it.) –  Robert Bryant Dec 21 '12 at 16:59
    
For symmetric tensors, I think your problem is called 'Waring Problem for polynomials'. Specifically, identifying symmetric tensors with polynomials, the Waring problem asks- given a homogeneous polynomial of degree d, what is the minimum number of d-th powers of a linear polynomial that are needed to write the given polynomial. The generic number has been known for a while and is called (i hope i'm remembering correctly) the Alexander-Hirshowitz theorem. The problem of given a monomial, how many dth forms are needed to write it was just solved and is on the arxiv. –  aginensky Jan 17 '13 at 17:15
    
Here is a link - arxiv.org/abs/1110.0745 . I think the rank of 'detrminant' considered as a symmetric tensor must be known, but I do't know it ! –  aginensky Jan 17 '13 at 17:17
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