## Are almost commuting hermitian matrices close to commuting matrices (in the 2-norm)?

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I consider on $M_n(\mathbb C)$ the normalized $2$-norm, i.e. the norm given by $\|A\|_2 = \sqrt{Tr(A^* A)/n}$.

My question is whether a $k$-uple of hermitian matrices that are almost commuting (with respect to the $2$-norm) is close to a $k$-uple of commuting matrices (again with respect to the $2$-norm). More precisely, for an integer $k$, is the following statement true?

For every $\varepsilon>0$, there exists $\delta>0$ such that for any $n$ and any matrices $A_1,\dots, A_k\in M_n(\mathbb C)$ satisfying $0\leq A_i\leq 1$ and $\|A_iA_j - A_j A_i\|_2 \leq \delta$, there are commuting matrices $\tilde A_1,\dots,\tilde A_k$ satisfying $0\leq \tilde A_i\leq 1$ and such that $\|A_i - \tilde A_i\|_2 \leq \varepsilon$.

The important point is that $\delta$ does not depend on $n$.

I could not find a reference to this problem in the litterature. However, this question with the $2$-norm replaced by the operator norm is well-studied. And the answer is known to be true if $k=2$ (a result due to Lin) and false for $k=3$, and hence $k\geq 3$ (a result of Voiculescu).

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## 3 Answers

There is a recent paper by Glebsky titled "Almost commuting matrices with respect to normalized Hilbert-Schmidt norm" which shows that this is indeed true for any $k$ for Hermitian matrices (and in fact also unitary and normal matrices).

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Thanks a lot for your answer. I did not yet have time to read the whole proof, but this seems to completely answer my question. I just wonder why this paper was posted in the algebraic geometry section of arXiv.  – Mikael de la Salle Jun 21 2011 at 14:31

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The answer is yes, and much more is true. Any hyperfinite von Neumann algebra (with separable predual) has a unique embedding (up to conjugacy) into the ultra-product of the hyperfinite $II_1$-factor.

This implies in particular, that almost commuting matrices in Hilbert-Schmidt are close to commuting matrices. The proofs goes by contradiction; assume that there is a sequence of counterexamples and construct non-conjugate embeddings. Since any abelian von Neumann algebra is hyperfinite, this yields a contradiction.

Kenley Jung showed that uniqueness of the embedding also implies that the algebra is hyperfinite.

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Andreas, is your argument presupposing that the matrices almost commute in HS-norm and are uniformly bounded in operator norm? – Yemon Choi Jun 21 2011 at 21:16
Thanks Andreas. I already accepted Ashley's answer, but your answer is more what I was hoping for (Glebsky's proof has however the advantage of giving an explicit bound on $\delta$ depending on $k$ and $\varepsilon$). – Mikael de la Salle Jun 22 2011 at 6:42
I guess that the precise statement is :"Given a non principal ultrafilter $\mathcal U$, any two embeddiing of a hyperfinite von Neumann algebra into $\prod_{\mathcal U} M_n$ (or $R^{\mathcal U}$) are conjugate". (unless I missed something, the ultrapower might depend on $\mathcal U$). By the way, do you have a reference for this statement? – Mikael de la Salle Jun 22 2011 at 6:57
Yemon: yes, just as in the question. – Andreas Thom Jun 22 2011 at 7:32

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Mikael: you are right, this might depend on the ultrafilter. I would try looking at the paper by Kenley Jung (Math. Ann. 2007 vol. 338 (1) pp. 241-248) and see whether he gives a reference. This is the easy part anyway, it is based on the fact that a full matrix-algebras embeds (unitally) in a unique way (up to conjugacy) into each other. The same holds if you take block-sums of matrix-algebras and remember their relative multiplicities. – Andreas Thom Jun 22 2011 at 7:36
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I just found the discussion. In the paper there is the better estimates then the ones of mine and it contains citations on proofs using von Neumann algebras.

(It was a surprise for mi too why my paper is in Algebraic Geometry. Probably it is my error. I have not found an easy way to fix it.)

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