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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 $ba\neq 0$

Ali Taghavi
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    Who is voting to close this? What is going on? – Martin Argerami Mar 08 '16 at 03:19
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    @MartinArgerami It is being closed because "This question is missing context or other details." Basically, while it's a good and interesting question (which is evident by four favourites, one of which is mine), it is not the kind of question we are really looking for here. There is no background on how the OP came across it, and no own thoughts on how to go about solving it. Some don't need more reason than that to vote to close. – Arthur Mar 08 '16 at 08:09
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    @Arthur: who is "we"? – Martin Argerami Mar 08 '16 at 09:57
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    @MartinArgerami "We" is the official stance on how to ask a well-recieved question on this site. I have not voted to close (as I said, I really want to know the answer too), but I can see why the votes are there. – Arthur Mar 08 '16 at 10:40
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    @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. – Martin Argerami Mar 08 '16 at 12:10
  • @MartinArgerami But it does not satisfy the one that is specified as a reason to close in the vote-to-close menu. Also, as for "officialness", you may find the same kind of tips in the very much official help center, with quotes like "Tell us what you found and why it didn’t meet your needs" and "Make it clear how your question is relevant to more people than just you." If this is closed, then it will not the first time, nor the last time that a question is closed on those grounds, though most of the time the problem is a lot less interesting. – Arthur Mar 08 '16 at 12:16
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    @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. – Ali Taghavi Mar 08 '16 at 14:38
  • BTW the following fact was one of motivations for this question:In a Banach algebra $A$, if $\parallel ab \parallel \leq k \parallel ba \parallel$ the $A$ is commutative. – Ali Taghavi Mar 08 '16 at 17:02
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    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. –  Mar 09 '16 at 03:01
  • @ChristianRemling Thank you very much for your supportive words on this question. – Ali Taghavi Mar 09 '16 at 09:36
  • @MartinArgerami thank you very much for your attention to my question. – Ali Taghavi Mar 09 '16 at 09:38
  • Any non-commutative C*-algebra that is also a domain should provide a counterexample, no ? – Hmm. Mar 09 '16 at 20:42
  • @SoumyaSinhaBabu but every C* algebra has non trivial zero divsor:For a self adloint a write $a=a_{+}-a_{-}$ the difference of two zero divisor. – Ali Taghavi Mar 09 '16 at 20:48
  • @Ali Taghavi, $\mathbb{C}$ is a C* algebra that does not have any non-trivial zero divisor. – Hmm. Mar 09 '16 at 20:54
  • @Ali Taghavi, disk algebras also constitute counterexamples to what you said. – Hmm. Mar 09 '16 at 20:57
  • @SoumyaSinhaBabu Yes. But this is the only case. I correct my statment:"a C* algebra witrh dim at least 2 has non trivial zero divisor – Ali Taghavi Mar 09 '16 at 20:58
  • @SoumyaSinhaBabu disk algebra is a Banach algebra which does not admit any compatible involution. It is not a C* algebra. – Ali Taghavi Mar 09 '16 at 20:59
  • My bad, I apologize. – Hmm. Mar 09 '16 at 21:00
  • @SoumyaSinhaBabu you are well come. – Ali Taghavi Mar 09 '16 at 21:00
  • Oh okay, you are right. Thanks. – Hmm. Mar 09 '16 at 21:05

1 Answers1

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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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Ali Taghavi
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