Let $A$ be a selfadjoint bounded operator on a Hilbert space. Let $M$ be another bounded selfadjoint operator. Let me assume the commutation property $$ [A, e^{iM}]=0. \tag 1$$ Does (1) imply that $$ [A, M]=0? \tag 2$$ Is there a version of this for $M$ unbounded? I guess that something should be said on the domain of the unbounded $M$. Are the above properties obvious (positively or negatively) in finite dimension?
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$\begingroup$ In finite dimension, two diagonalizable operators commute iff there is a basis in which they are both diagonal. So in finite dimensuion your (1) and (2) are equivalent. $\endgroup$– Alexandre EremenkoCommented Dec 21, 2019 at 14:56
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3$\begingroup$ @AlexandreEremenko: Not quite: Let $M$ be the diagonal matrix with diagonal entries $2\pi$ und $-2\pi$, and let $A$ be any matrix that does not commute with $M$. $\endgroup$– Jochen GlueckCommented Dec 21, 2019 at 15:03
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$\begingroup$ As is so often the case, the counterexample suggests what is going wrong here. The general problem would be the question: if $A$ and $f(M)$ commute, is the same true of $A$ and $M$? So what you need is that $f$ be invertible in a suitable sense. It is easy to formulate a precise version which I will leave as an exercise to the OP since I am typing this on a pad. $\endgroup$– user131781Commented Dec 21, 2019 at 16:09
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2$\begingroup$ To sum up the discussion so far, this will work (for arbitrary $A$) if and only if $e^{ix}$ is injective on $x\in\sigma(M)$ (because then $M$ is a function of $e^{iM}$, and if you don't have this condition, then there's Jochen's counterexample). $\endgroup$– Christian RemlingCommented Dec 21, 2019 at 18:50
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