Let $\mathcal{A}$ be a $C^*$-algebra and $p\in\mathcal{A}^{**}$ be an open projection, that is, $p=p^*=p^2$ and $p\in\overline{(p\mathcal{A}^{**}p\cap\hat{\mathcal{A}})}^{\operatorname{w}^*}$, where $\hat{\mathcal{A}}$ is the canonical copy of $\mathcal{A}$ in $\mathcal{A}^{**}$ and the closure is taken in the weak$^*$ topology on $\mathcal{A}^{**}$.

Question: Is every orthogonal projection in $\mathcal{A}^{**}$ which is Murray-von Neumann equivalent to $p$ open?


The answer is no.

Proof (Thomas Schick). The idea of the proof is due to Thomas Schick. I thank him for allowing me to reproduce it here. Let $\mathcal{A}:=C([0,1])\otimes\mathbb{M}_2$, where $\mathbb{M}_2$ is the $W^{\star}$-algebra of $2\times2$ matrices with entries in $\mathbb{C}$. Since the set of bounded Borel functions $B^{\infty}([0,1])$ is contained in $C([0,1])^{**}$, we canonically identify $B^{\infty}([0,1])\otimes\mathbb{M}_2$ with a $C^*$-subalgebra of $\mathcal{A}^{**}$. Let $F:=\{0\}$ and $F^c:=(0,1]$, then the characteristic functions $\chi_F$ and $\chi_{F^c}$ are Borel measurable. Define a partial isometry $u:=\chi_F\otimes E_{11}+\chi_{F^c}\otimes E_{12}\in\mathcal{A}^{**}$, where $E_{ij}\in\mathbb{M}_2$ is the canonical matrix unit, that is, it has $1$ in its $(i.j)$-entry and $0$ elsewhere. Then the orthogonal projection $p:=uu^{\star}=1\otimes E_{11}$ is open, whereas $q:=u^{\star}u=\chi_F\otimes E_{11}+\chi_{F^c}\otimes E_{22}$ is not. $\square$

Cf. Here is my related question.


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