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Let $\mathcal A\subset B(H)$ be an unital $C^*$ algebra of operators on a Hilbert space $H$. Let's denote by $\mathcal P$ the set of projections in $\mathcal A$, that is $\mathcal P:=\{a\in \mathcal A \; | \; a=a^2=a^*\}$.

I'm wondering when $\mathcal P$ is closed under composition of operators. I was able to prove the following:

$\big(\mathcal P \text{ is closed under multplication}\big)\Leftrightarrow \big(\forall_{a,b\in \mathcal P}\; ab=ba\big)$

If $\mathcal A$ is a von Neumann algebra then projections spans set which is dense in $\mathcal A$ so in this case $\big(\mathcal P \text{ is closed under multiplication}\big)\Leftrightarrow \big(\mathcal A \text{ is abelian}\big)$. Can something similar be proven in general(without assuming $\mathcal A$ being a von Neumann algebra)?.

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No. If $\mathcal A=C_r(\mathbb F_2)$ (or any other projectionless unital C$^*$-algebra), then $$\mathcal P=\{0,I\}$$ is closed under multiplication.

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  • $\begingroup$ Thank You for Your comment, Martin Argerami. So, there are noncommutative $C^*$ algebras where $\mathcal P$ is closed under multiplication. Is there any useful characterisation of this property? $\endgroup$
    – Mogget
    Mar 30, 2016 at 10:42
  • $\begingroup$ I have no answer for that, but I doubt so. It just happens that for some C $^*$-algebras projections are irrelevant. On the other hand many (most?) C $^*$-algebras contain copies of the matrix algebras, so $\mathcal P $ is not multiplicative. $\endgroup$ Mar 30, 2016 at 11:48
  • $\begingroup$ Am I right that in this case center of $\mathcal A$ is non trivial? Does something change if we assume triviality of a center? $\endgroup$
    – Mogget
    Apr 2, 2016 at 10:22
  • $\begingroup$ The center of $C_r (\mathbb F_2) $ is trivial. $\endgroup$ Apr 2, 2016 at 11:51
  • $\begingroup$ I might have not stated clearly what I meant. If we assume that the center is trivial, can we say something more about the question whether $\mathcal P$ is multiplicative? $\endgroup$
    – Mogget
    Apr 2, 2016 at 17:19

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