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Let $A$ be a C*-algebra. We identify $A$ with its canonical image in the bidual $A^{**}$. Consider the following conditions:

(1) $A$ is a von Neumann algebra.

(2) There is a multiplicative conditional expectation from $A^{**}$ onto $A$, that is, a map $\pi\colon A^{**}\to A$ that is a *-homomorphism and such that $\pi(a)=a$ for all $a\in A$.

Then (1) implies (2): Consider the predual $A_*$ of $A$, and the canonical embedding $\kappa\colon A_*\to (A_*)^{**}$. Then the dual of $\kappa$ has the desired properties.

Note that $A^{**}$ is always a von Neumann algebra. One can show that (2) implies that $A$ is a monotone complete AW*-algebra. However, is it also a von Neumann algebra?

Q1: Does (2) imply (1) ?

Assuming (2), let $J$ denote the kernel of $\pi$. Then $A$ is (*-isomorphic to) the quotient $A^{**}/J$. However, a quotient of a von Neumann algebra by a closed, two-sided ideal need not be a von Neumann algebra. (For example, the Calkin algebra is such a quotient and not a von Neumann algebra.)

In the commutative case, we might also consider the following more general question:

Q2: Is the quotient of a commutative von Neumann algebra by a closed, two-sided ideal again a von Neumann algebra?

Equivalently, is every closed subset of a hyperstonean space again hyperstonean?

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  • $\begingroup$ Your proof that $A$ is complete, doesn't it also implies that $\pi$ is normal ? If yes then the Kernel of $\pi$ is weakly closed and I think you are done, isn't ? $\endgroup$
    – Simon Henry
    Commented Mar 25, 2016 at 16:45
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    $\begingroup$ This is not clear to me. To show that $A$ is monotone complete, let $(a_j)_j$ be a norm-bounded increasing net in $A_{sa}$. Then, since $A^{**}$ is monotone complete, there is $b\in A^{**}_{sa}$ that is the supremum: $a_j\leq b$ for all $j$, and $b$ is the smallest element with these properties. Now, since $\pi$ preserves order, we have $a_j=\pi(a_j)\leq\pi(b)$ for all $j$, and thus $b=\pi(b)$. However, if we start with any increasing net in $A^{**}_{sa}$, then I don't see why $\pi(\sup_j a_j) = \sup_j \pi(a_j)$. $\endgroup$
    – Hannes Thiel
    Commented Mar 25, 2016 at 17:01
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    $\begingroup$ "is every closed subset of a hyperstonean space again hyperstonean?" mathoverflow.net/questions/378199/… $\endgroup$
    – Onur Oktay
    Commented Aug 18, 2023 at 8:53
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    $\begingroup$ @HannesThiel Again, apologies for replying to a seven-year-old comment, but $b = \pi(b)$ does not need to be true, even if $A$ is a vNa. For example, let $A = L^\infty([0, 1])$. Pick a character $\phi$ on $A$ which extends evaluation at 0 on $C([0, 1])$. Pick an increasing sequence of continuous functions $(a_n)$ s.t. $a_n(0) = 0$ for all $n$ and $a_n$ converges weakly to 1. $A^{**}$ has a direct summand of $\mathbb{C}$ corresponding to $\phi$. Then $b$, the supremum of $a_n$, evaluates to 0 on that direct summand… $\endgroup$
    – David Gao
    Commented Aug 18, 2023 at 9:12
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    $\begingroup$ @HannesThiel That seems to be the case. I’m not doubting $A$ has to be monotone complete, just pointing out a minor error in the original argument. $\endgroup$
    – David Gao
    Commented Aug 25, 2023 at 14:53

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This is a seven-year-old question, so I'm not sure if an answer is still meaningful at this point, but no, $A$ needs not be a vNa. There is even a commutative counterexample. In fact, fix any monotone complete commutative $C^*$-algebra $A$ which is not a vNa. Then $A$ is injective in the category of unital commutative $C^*$-algebras and $*$-homomorphisms. (See, for example, https://arxiv.org/abs/0706.2995 .) Hence, the identity map $A \rightarrow A$ can be extended to a multiplicative conditional expectation $\pi: A^{**} \rightarrow A$.

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The answer to Q2 is no, and $l^\infty/c_0$ is already a counterexample. Its lattice of projections is $\mathcal{P}(\omega)/fin$, which is not complete.

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