# Weak center is same as center for $C^{\ast}$-Algebra? [closed]

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?

• 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).
– YCor
Jun 13 '20 at 14:37
• @YCor: I asked this question as I was trying to figure out definition of center of TRO. Please see my comment below. Jun 13 '20 at 14:45
• 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.
– YCor
Jun 13 '20 at 14:46
• You don't have to, I'm not offended.
– YCor
Jun 13 '20 at 14:49
• I’m voting to close this question for the reasons mentioned by @YCor Jun 13 '20 at 15:44

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.
• 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 $a,b,c\}$? Jun 13 '20 at 17:04