Let $A$ be a $C^{\ast}$-algebra. We say $A$ is weakly commutative if $ab^*c=cb^*a$ for all $a,b,c \in A$ and define weak center of $A$ as $$Z_w(A)= \{ v \in A : av^*c=cv^*a \;\forall a,c \in A \}.$$

Are these notions of weak commutativity and weak center the same as the usual notions of commutativity and center in $C^{\ast}$-algebras?

By using approximate identity, I have managed to prove that both notions weakly commutative and commutative are same. It is also clear that center of $A$, i.e., $Z(A)$, contains $Z_w(A)$ but reverse inclusion is not clear. Any ideas?

  • $\begingroup$ By the way the star sign on $b$ and $v$ is redundant (by the obvious change of variable $b\mapsto b^*$). In particular, the definition doesn't involve the star map. My opinion in this case is that the question should be amended (despite already having an answer, which could be a comment) to make it non-trivial (currently it's clearly not research-level, namely not seriously thought). $\endgroup$
    – YCor
    Jun 13, 2020 at 14:37
  • $\begingroup$ @YCor: I asked this question as I was trying to figure out definition of center of TRO. Please see my comment below. $\endgroup$
    – Math Lover
    Jun 13, 2020 at 14:45
  • 3
    $\begingroup$ Yes but asking a question about $C^*$-algebras to which $M_2(\mathbf{C})$ is a trivial counterexample is not serious (and actually every non-abelian unital $C^*$-algebra is a trivial counterexample, for the same reason). I'd have recommended to delete it, but unfortunately this is not possible now it has an answer. $\endgroup$
    – YCor
    Jun 13, 2020 at 14:46
  • 2
    $\begingroup$ You don't have to, I'm not offended. $\endgroup$
    – YCor
    Jun 13, 2020 at 14:49
  • 1
    $\begingroup$ I’m voting to close this question for the reasons mentioned by @YCor $\endgroup$
    – Yemon Choi
    Jun 13, 2020 at 15:44

1 Answer 1


Not the same. The center of $M_2$ is $\mathbb{C}\cdot I_2$, but its weak center is $\{0\}$. E.g. $I_2$ is not in the weak center because not all $a,c \in M_2$ commute.

  • $\begingroup$ Thank you. Actually I was trying to define Center for TRO and this definition what I called weak center was suggested here: mathoverflow.net/questions/362862/…. Do you know any definition of center for TROs? $\endgroup$
    – Math Lover
    Jun 13, 2020 at 14:26
  • $\begingroup$ I know very little about TROs, but it seems reasonable that there should be a good definition of center for them. My first guess would be $\{z: abz^*c = az^*bc$ for all $a,b,c\}$? $\endgroup$
    – Nik Weaver
    Jun 13, 2020 at 17:04
  • $\begingroup$ Oh, you can't do that, never mind. Not sure how a definition could go, then. $\endgroup$
    – Nik Weaver
    Jun 13, 2020 at 17:11
  • $\begingroup$ I searched in literature, I found the definition of commutativity Of TRO but I didn’t found the definition of center anywhere. In case you come up with something, please do comment on the question I have linked in above comment. $\endgroup$
    – Math Lover
    Jun 13, 2020 at 17:18
  • $\begingroup$ Okay. I'd think that really understanding the definition of commutativity should give some clues. $\endgroup$
    – Nik Weaver
    Jun 13, 2020 at 17:38

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