Are Almost Commuting Hermitian Matrices Close to Commuting Matrices in 2-Norm? – fa.functional-analysis,gr.group-theory,matrices,oa.operator-algebras,von-neumann-algebras

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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).

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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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[Math] Finite subgroups of unitary groups

This might help: If $u$ and $v$ are both close enough to scalars in the operator norm, and $\varepsilon$ is small enough, then no matter how large $k$ is, if the group generated by $u^{\prime}$ and $v^{\prime}$ is finite, then it still has to be Abelian. The reasoning is something like this: I use the operator norm : given a unitary matrix $M$, there is a unique scalar $\lambda$ ( in the interior of the unit disk unless $M$ itself is a scalar matrix) with $\|M- \lambda I\|$ minimal, and this only depends on the spectrum of $M$. In particular, we can only have $\|M -\lambda I\| < \frac{1}{2}$ for a scalar $\lambda$ if the eigenvalues of $M$ all lie on an arc of length less than $\frac{\pi}{3}$ on the unit circle.It is essentially a Theorem of Frobenius that if $G$ is a finite group of unitary matrices, then the subgroup $H$ generated by the matrices at distance less than $\frac{1}{2}$ from a scalar is an Abelian normal subgroup. If $u$ and $v$ are close enough to scalars, and $\varepsilon$ is small enough then we can make both $\|u^{\prime} - \lambda I\|$ and $\| v^{\prime} - \mu I\|$ less than $\frac{1}{2}$ for suitable scalars $\lambda,\mu$. So if they generate a finite group, it must be Abelian. Note added: I will add a little more detail. One proof of Jordan's theorem uses a compactness argument and a kind of contraction Lemma. Let $[a,b]$ denote $a^{-1}b^{-1}ab$. If $a$ and $b$ are unitary, then $$\| I-[a,b] \| = \|I - a^{-1}b^{-1}ab \| = \|ab-ba \|$$ $$ = \| (a-I)(b-I) - (b-I)(a-I)\| \leq 2 \|a-I\| \|b-I\|$$. Hence if $\| I-a\| < \frac{1}{2}$, then $\|I- [a,b] \| < \|I-b\|$ for all $b.$ Hence in a finite group $G$ of unitary matrices, let $H$ be the subgroup generated by those elements of $G$ with $\|I-h\|< \frac{1}{2}$. For any $h \in H$ with $\|I - h \| < \frac{1}{2}$, we have $[g,h,h, \ldots ,h] = 1$ for all $g \in G$. By a theorem of Baer, $H$ is nilpotent. With a little more work, you actually get $H$ Abelian. (This is a kind of hybrid of arguments of Frobenius and a little more finite group theory. More or less the same line of reasoning led to a similar theorem by Zassenhaus on discrete groups, and this led to the idea of a Zassenhaus neighbourhood in a Lie group). Note that if $x,y$ are in different cosets of $H$ in $G$, we must have $\|x-y\| \geq \frac{1}{2}$, so the compactness of the unit ball of $M_{n}(\mathbb{C})$ gives a fixed bound on $[G:H]$. But the same argument works when $\| a - \lambda I \| < \frac{1}{2}$ for some scalar $\lambda$, using $$\|ab - ba \| = \| (a-\lambda I)(b-\mu I) - (b-\mu I)(a-\lambda I)\|$$ for any scalars $\lambda,\mu$, and replacing $H$ by the group generated by elements at distance less than $\frac{1}{2}$ from any scalar. (Added in response to comment below): You do need a little argument to get from $H$ nilpotent to $H$ Abelian, but the statements above are all accurate. It is important to remember that $H$ is generated by elements within distance $\frac{1}{2}$ of a scalar. Any such element has all its eigenvalues on an arc of length less than $\frac{\pi}{3}$ on $S^{1}$. No subset of the "multiset" of eigenvalues of such an element has zero sum (it's clear eg, if you rotate so that all eigenvalues have positive real part). Let $\chi$ be the character of the given representation of $H$. Let $\mu$ be an irreducible constituent of $\chi$. By the remark above, if $\|h -\alpha I \| < \frac{1}{2}$ for some scalar $\alpha$, then $\mu(h) \neq 0.$ But if $\mu$ is not linear, it is induced from a character of a maximal subgroup $M$, which is normal as $H$ is nilpotent. But then $\mu$ vanishes on all elements of $H$ which are outside $M$. Hence the elements at distance less than $\frac{1}{2}$ from a scalar must all lie within $M$. But since $M$ is proper, and such elements generate $H$, this is a contradiction. Hence $\mu$ must be linear after all. Since $\mu$ was arbitrary, $H$ is Abelian.

[Math] commuting matrices

I think it really comes down to looking at the lattice of invariant subspaces for each $A_i$. I believe this is what Igor basically is saying.

For a single real linear transformation $V \rightarrow V$ with distinct characteristic roots, $V$ decomposes as a sum of 1 and 2-dimensional invariant subspaces, and these generate the lattice of invariant subspaces. When you perturb the matrix by a known amount, the subspaces can wander around over a certain range --- how much they shift depends on how close together are the characteristic roots, and also on the angles between the subspaces. For simplicity, think of the case with all real distinct eigenvalues. The projectivized action, on $\mathbb{RP}^{n-1}$, then has $n$ fixed points. The first derivative of the projectivized map is simple to determine, given the eigenvectors and eigenvalues --- the eigenvalues of the derivative at a point for eigenvalue $\lambda$ are the ratios of the other eigenvalues to $\lambda$. Therefore, if you change the map by adding multiples of the other eigenvectors as a linear function of the projection to the eigenspace under consideration you shift your given eigenvector by a known linear function of the coefficients.

If you have several matrices with all real distinct eigenvalues that are near enough to a commuting set of matrices, then presumably the eigenvectors are also near to each other, so you can tell which eigenvectors need to line up with each other. You have a good measurement of how hard it is to move each eigenvector in a given direction, so you can find a point that's approximately equally hard for each (do this separately, for each matching collection of eigenvectors.) Then adjust each of the eigenvectors to go to that point, and repeat on the other eigenspaces.

When there are complex roots, there are 2-dimensional invariant subspaces, which appear as invariant circles in $\RP^{n-1}$. You can do much the same thing with these.

When there are multiple real or complex roots, then it's necessary to look at larger-dimensional invariant subspaces. One strategy is to first fix up invariant subspaces associated with the product of all factors of the characteristic polynomial whose roots are in a small neighborhood, or a pair of complex conjugate neighborhoods. The goal is to get such subspaces to agree with, to contain, or to be contained in similar suspaces for the other matrices. If there are repeated roots, then first see if there's a small perturbation that makes multiple eigenvectors (or multiple 2-dimensional invariant subspaces) --- when this is possible, it gives a bigger target for matching commuting matrices. If not, look at the largest invariant subspace it contains.

I doubt if there's an easy way other than looking at these invariant subspace structures. Perhaps in the actual applications you have in mind, they're not so complicated.

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[Math] Finite subgroups of unitary groups

[Math] commuting matrices

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