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Physicist here, so apologies for the imprecision.

The version of the Schur-Weyl duality I am interested in states that an operator $A$ that acts in the $k$-fold tensor product Hilbert space $\mathcal{H}^{\otimes k}$ commutes with all operators $V^{\otimes k}$ where $V$ is a unitary on $\mathcal{H}$, iff $A$ is a linear combination of permutation operators that permutes the $k$ copies of $\mathcal{H}$. (Let's assume that the dimension of $\mathcal{H}$ is finite)

My question: What precisely does "all" entail? Does the statement still hold if $V$ is not any unitary but is just dense in the set of all unitaries?

My guess: Probably yes, but how to be precise?

Motivation: in quantum computation a universal gate set allows us to get arbitrarily close (but technically not equal) to any unitary. I want to be careful this doesn't introduce pathological difficulties..

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    $\begingroup$ Do you have an example where $A$ commutes with a dense set of unitaries, but not all? $\endgroup$
    – Sam Hopkins
    Commented Jul 10, 2021 at 13:26
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    $\begingroup$ Why doesn't this follow from the continuity of "$k$-th tensor power" and "commutator"? $\endgroup$
    – Andreas Blass
    Commented Jul 10, 2021 at 14:37
  • $\begingroup$ @SamHopkins Nope. $\endgroup$
    – nervxxx
    Commented Jul 10, 2021 at 16:31
  • $\begingroup$ @AndreasBlass Great, yes that is what I was looking for, thanks! $\endgroup$
    – nervxxx
    Commented Jul 10, 2021 at 16:31

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