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Suppose that $A_1,\dots, A_k$ are unitary matrices such that any two of them can be approximated by commuting unitary matrices. i.e. for any $i$ and $j$, there are unitary matrices $A_i'$ and $A_j'$ such that $\|A_i-A_i'\|<\varepsilon$, $\|A_j-A_j'\|<\varepsilon$ and $A_i'A_j'=A_j'A_i'$. Can we find unitary matrices $X_1,\dots,X_k$ such that any two of them commute and $\|X_i-A_i\|<O(\varepsilon)$ for all $i$? What if any three (or small number) of them can be simultaneously approximated by commuting unitary matrices?

Here the matrix norm could be any unitary invariant norm. I'm specially interested in the operator norm and Hilbert-Schmidt norm.

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  • $\begingroup$ Here's a thought. Suppose we're in a vector space $V$ over an algebraically closed field, and each of the operators $A_1, \ldots, A_k$ has distinct eigenvalues. Make $\varepsilon$ small enough that the operators in the $\varepsilon$-ball around each $A_i$ also have distinct eigenvalues. That means any operator $\varepsilon$-near one of the $A_i$ decomposes $V$ into eigenlines. $\endgroup$
    – Vectornaut
    Commented Nov 4, 2015 at 1:32
  • $\begingroup$ Now, suppose we can find commuting operators $A_i'$ and $A_j'$ within $\varepsilon$ of $A_i$ and $A_j$ respectively. Since $A_i'$ and $A_j'$ commute, they have the same eigenlines $L_1, \ldots, L_n$. If $\varepsilon$ is small enough, I think the eigenlines of $A_i$ and $A_j$ should have to be near $L_1, \ldots, L_n$, and therefore near each other. In other words, for small enough $\varepsilon$, your commutative approximation condition implies that the eigenlines of the operators $A_1, \ldots, A_k$ almost match. $\endgroup$
    – Vectornaut
    Commented Nov 4, 2015 at 1:32
  • $\begingroup$ I'm a bit confused. You're saying that you know that there exists a solution and the question is how to find it? Or am I misunderstanding (because Mikael's answer says "no" for one instance...). $\endgroup$
    – Suvrit
    Commented Nov 4, 2015 at 14:36
  • $\begingroup$ The question is that you know any two (or three or small number) of them can be approximated simultaneously by commuting unitary matrices and the question is to approximate all of them together (with a good bound such as $c\varepsilon$) with commuting unitary matrices. As I understand Miakel's answer is no for approximately commuting Hermitian matrices and doesn't apply here. $\endgroup$
    – Omid Hatami
    Commented Nov 4, 2015 at 14:48
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    $\begingroup$ Well as I understand two cases of unitary and self-adjoint matrices are different. For example there are two unitary matrices $A$ and $B$ such that $||[A,B]||_{op}$ is small but they can't be approximated with commuting unitary matrices. mtm.ufsc.br/~exel/papers/asympt.pdf But the answer to the other case is yes for any pair of almost commuting matrices and it is called Huaxin Lin's Theorem math.ku.dk/~rordam/manus/short.pdf $\endgroup$
    – Omid Hatami
    Commented Nov 4, 2015 at 14:58

1 Answer 1

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Edit Now this answers the first question for the operator norm and the normalized Hilbert-Schmidt norm.

The answer depends on the norm you are considering. The answer is no for the operator norm, but is yes for the normalized Hilbert-Schmidt norm (at least if you replace $O(\varepsilon)$ by $o(1)$, see the answers to this question).

Here are some details on the counterexample for the operator norm.

  1. By a theorem of Lin (see here), for a pair of self-adjoint matrices of norm less than $1$, they approximately commute if and only if they can be approximated by commuting matrices.
  2. Voiculescu proved that the preceding does not hold for triples of self-adjoint matrices of norm less than $1$ (see the link I gave here, or the references in the paper by Exel and Loring given in the comments).

1+2 imply that there is a sequence of triples $A_1^n,A_2^n,A_3^n$ of matrices of norm less than $1$ which are pairwise close to (self-adjoint) commuting matrices, but whose distance to the triples of commuting matrices is bounded below.

  1. By continuity of the functional calculus and the fact that $t \in [-2,2] \mapsto e^{it}$ is a homeomorphism on its image, this implies that the unitary matrices $(e^{i A_1^n}, e^{i A_1^n},e^{i A_1^n})$ are pairwise close to pairs of commuting unitaries, but are at positive distance from triples of commuting unitaries. This is what you were looking for.
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  • $\begingroup$ Is there any counterexample for $k=3$ in the case that matrices are unitary? $\endgroup$
    – Omid Hatami
    Commented Nov 4, 2015 at 15:22
  • $\begingroup$ Can I check that you are talking about the same question? Here the assumption is not that any two of the matrices approximately commute, but rather the stronger assumption that any two of the matrices can be approximated by a pair of commuting matrices. I can't seem to track down a counterexample to this in the literature. $\endgroup$
    – gowers
    Commented Nov 7, 2015 at 10:06
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    $\begingroup$ Oops, you are both right, I missed the word unitary in the question. The references give a counterexample for self-adjoint matrices (I will expand my answer to make this precise). I have no idea what the answer is for unitary matrices. $\endgroup$
    – Mikael de la Salle
    Commented Nov 7, 2015 at 15:06
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    $\begingroup$ Well, it seems that the unitary version of the question also has a negative answer by considering, for a self adjoint matrix $A $, the unitary $e^{iA/10} $. $\endgroup$
    – Mikael de la Salle
    Commented Nov 7, 2015 at 17:08

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