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Assume that $A$ is a $C^{*}$-algebra such that $\forall a,b \in A, ab=0 \iff ba=0$.

Is $A$ necessarily a commutative algebra?

In particular does "$\forall a,b \in A, ab=0 \iff ba=0$" imply that $\parallel ab \parallel$ is uniformly dominated by $\parallel ba \parallel$? In the other word $\parallel ab \parallel \leq k \parallel ba \parallel$, for a uniform constant $k$. Of course the later imply commutativity.

Note added: As an example we look at the Cuntz algebra $\mathcal {O}_{2}$. There are two elements $a,b$ with $ab=0$ but $ba\neq 0$. This algebra is generated by $x,y $ with $$\begin{cases}xx^{*}+yy^{*}=1\\x^{*}x=y^{*}y=1\end{cases}$$ This implies $x^{*}(yy^{*})=0$ but $(yy^{*})x^{*} \neq 0$.

This shows that for every properly infinite $C^{*}$ algebra, there are two elements $a,b$ with $ab=0$ but $ba\neq 0$

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    $\begingroup$ Who is voting to close this? What is going on? $\endgroup$ Commented Mar 8, 2016 at 3:19
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    $\begingroup$ @Arthur: who is "we"? $\endgroup$ Commented Mar 8, 2016 at 9:57
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    $\begingroup$ @Arthur: I don't think there's anything "official" about that meta thread. But, in any case, this question satisfies the vast majority of the recommendations in that thread. $\endgroup$ Commented Mar 8, 2016 at 12:10
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    $\begingroup$ @Arthur Characterization of commutative rings and algebra is a self motivating general problem. I think there is no need to explain how such type of problem arise to mind of an asker. Regarding my attempt to solve the problem: I have no idea how to start on this question.I think MSE and MO and some of their participants who easily "vote to close"could be more flexible on questions which are interesting but the asker did not spend a lot of time to think about it. Posting his question is a way to find some people who are interested in the problem for a possible on line discussion. $\endgroup$ Commented Mar 8, 2016 at 14:38
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    $\begingroup$ I personally find the usual "I tried to rewrite $ab = ab + a -a$ but don't know where to go from here; maybe the fundamental theorem of calculus will help" not very enlightening. This question is refreshingly elegant. $\endgroup$
    – user138530
    Commented Mar 9, 2016 at 3:01

1 Answer 1

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The property in the question is equivalent to non existence of non trivial nilpotent element, see the elegant answer of Leonel Robert to this question, but the later is equivalent to commutativity

So $A$ is commutative if and only if $$\forall a,b \in A, ab=0 \iff ba=0 $$

P.S: I asked the moderators to consider this answer as a community wiki.

The following related property is proven here:

$A$ is commutative if and only if $$\forall a,b \in A,\;\; ab\in A_{sa}\iff ba \in A_{sa}$$

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