7
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.