Are almost commuting hermitian matrices close to commuting matrices (in the 2-norm)? I consider on $M_n(\mathbb C)$ the normalized $2$-norm, i.e. the norm given by $\|A\|_2 = \sqrt{\mathrm{Tr}(A^* A)/n}$.
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). 
 A: I just found the discussion. In this paper by Filonov and Kachkovskiy there are better estimates than mines and it contains citations of proofs using von Neumann algebras. 
(It was a surprise for me too why my paper is in Algebraic Geometry. Probably it is my error. I have not found an easy way to fix it.)
A: As part of my dissertation, "Almost Commuting Operators on von Neumann Algebras," 
I have extended Glebsky's result to the normalized Schatten class for $1 \leq p < \infty$.  Moreover, for $p=2$ we recover Filonov and Kachkovskiy's theorem with the same estimate.  In our work, however, we use different techniques as the normalized Schatten p-norm does not arise from an inner product for $p \neq 2$. 
A: 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).
A: 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.
