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Suppose we have some tensor product of vector spaces. By tensor rank, I mean the minimal number of simple tensors required to write down an element of this tensor product of spaces. Is there much known about maps that preserve this tensor rank?

Also an easier (more specific) question: suppose I'm working in $U(4)$. Might there exist a $U \in U(4)$ such that for any pair of matrices $A,B \in U(2)$, $U(A \otimes B)U^{-1}$ is another simple tensor in $U(4)$, without $U$ itself being the tensor product of two matrices?

Morally I feel like this wouldn't happen and that the set of maps that preserve simple tensors in this way are themselves simple tensors, but I'm unsure. Thanks!

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    $\begingroup$ See theorem 1 here: projecteuclid.org/download/pdf_1/euclid.pjm/1103038893 $\endgroup$ – Daniel Apr 27 '15 at 20:09
  • $\begingroup$ This seems helpful! I'll have to work out what conjugation looks like as left multiplication, but thank you! $\endgroup$ – MGN Apr 27 '15 at 20:57
  • $\begingroup$ You are welcome. $\endgroup$ – Daniel Apr 27 '15 at 21:21
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    $\begingroup$ There is an example of unitary matrix $U\in U(4)$ satisfying this condition which is not a simple tensor. It is the flip operator $F=\sum_{i,j=1}^2e_ie_j^t\otimes e_je_i^t$, where $\{e_1,e_2\}$ is the canonical basis of $\mathbb{C}^2$. Notice that $F$ is real symmetric and $F^2=Id$, so it is unitary. Now check that $F(A\otimes B)F=B\otimes A$. $\endgroup$ – Daniel Apr 28 '15 at 12:27

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