Assume that $A$ is a $C^{*}$ algebra with self adjoint elements $A_{sa}$. Assume that for all $a,b\in A$ we have $$ab\in A_{sa} \iff ba \in A_{sa}$$
Is $A$ necessarily a commutative algebra?
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asked Mar 22, 2016 at 22:06
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Yes. I will show that any two positive elements of $A$ commute. Since every element is a linear combination of positive elements, this suffices.
Say $a$ and $b$ are positive. Then $a^{1/2}ba^{1/2} \in A_{sa}$, so by hypothesis $ba^{1/2}a^{1/2} = ba \in A_{sa}$. That is, $ba = (ba)^* = a^*b^* = ab$. QED
answered Mar 22, 2016 at 22:34
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$\begingroup$ Isn't it quicker to observe that any element is a linear combination of (two) self-adjoint ones? $\endgroup$ Commented Mar 23, 2016 at 9:33
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2$\begingroup$ @ChrisHeunen: I guess positivity is needed to take the square root that makes the proof work. (Notice that OP's assumption is not that self-adjoint elements commute but that $ab$ is self-adjoint iff $ba$ is.) $\endgroup$– RasmusCommented Mar 23, 2016 at 11:47
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$\begingroup$ Ah, thanks Rasmus, I had indeed wrongly read the assumption as saying that self-adjoints commute. $\endgroup$ Commented Mar 23, 2016 at 13:32