Extension of a von Neumann algebra by a von Neumann algebra

I asked this question at MSE now I repeat it at MO:

Let $$A,B,C$$ be $$3$$ unital $$C^*$$ algebras. Assume that we have the following short exact sequence of $$C^*$$-algebras:

$$0\to A\to C\to B\to 0$$

Assume that $$A,B$$ are generated by their projections. Is $$C$$ necessarily generated by its projections, too?

Assume that $$A,B$$ are von Neumann algebras, is $$C$$ necessarily a von Neumann algebra, too?

Does the last question has an obvious answer when $$A,B$$ (hence $$C$$) are commutative algebras?

Yes, it is. Let $$C$$ be a C*-algebra and let $$A \subseteq C$$ be an ideal which is intrinsically a von Neumann algebra. Then the positive part of the unit ball of $$A$$ has a least upper bound in $$A$$ which must be a projection. (Its norm cannot be greater than $$1$$, so if it is not a projection then its square root also belongs to the unit ball and is larger.) It follows that $$C \cong A \oplus B$$, so if $$B$$ is also intrinsically a von Neumann algebra then so is $$C$$.

• Thank you and +1 for your interesting answer. i confess that I did not pay attention to the fact that the extension is trivial if the first object, $A$ , is unital. I am quite beginner in this area and also in von Neumann algebra. – Ali Taghavi Dec 1 '18 at 20:13
• You're welcome! See the edit history of my post for an answer to a similar question where we assume $A$ and $C$ are com Neumann algebras. – Nik Weaver Dec 1 '18 at 23:03

This is an extended comment on Nik Weaver's nice answer. (Unless I've made a mistake, this argument shows that $$A$$ being unital does all the work).

What exactly do we mean by $$0 \rightarrow A \rightarrow C \rightarrow B \rightarrow 0$$ is an exact sequence of $$C^*$$-algebras? I think what is meant is that we have $$*$$-homomorphisms $$\phi : A\rightarrow C, \qquad \psi :C\rightarrow B,$$ with $$\phi$$ injective, $$\psi$$ surjective, and $$\ker\psi = \operatorname{im}\phi$$. We do not assume that $$\phi$$ or $$\psi$$ is unital.

However, in the original question, we do assume that $$A,B,C$$ are unital. Let $$p=\phi(1)\in C$$ so that $$p=p^2=p^*$$ as $$\phi$$ is a $$*$$-homomorphism. So $$p$$ is a projection, so also $$1-p$$ is a projection. Also $$p\phi(a) = \phi(a)p$$ for all $$a\in A$$.

As $$\phi(A) = \ker\psi$$ is an ideal in $$C$$, if $$c\in C$$ with $$pc=c$$ (or $$cp=c$$) then $$c\in \phi(A)$$. So $$\phi(A) = \{ c\in C: cp=c \}$$. Further, for any $$c\in C$$, we see that $$pc=c$$ if and only if $$c=cp$$. So, for any $$c\in C$$, we have that $$pc=pcp$$ and $$cp=pcp$$. For $$c\in C$$ let $$d=(1-p)c$$ so $$0 = pd = dp$$ so $$d(1-p)=d$$ so $$(1-p)c=(1-p)c(1-p)$$.

Define $$B' = \{ c\in C : cp=pc=0 \} = \{c\in C: c(1-p)=(1-p)c=c\}.$$ Then $$B'$$ is a $$C^*$$-subalgebra of $$C$$. If $$c\in B'$$ with $$\psi(c)=0$$ then $$c=\phi(a)$$ for some $$a\in A$$ so $$cp=c=0$$. So $$\psi$$ is injective on $$B'$$. For any $$c\in C$$, $$c = pc + (1-p)c = pcp + (1-p)c(1-p),$$ from the discussion above. Then $$pcp\in \phi(A)$$ and $$c'=(1-p)c(1-p)\in B'$$ so $$\psi(c) = \psi(c')$$. So $$\psi$$ restricts to a $$*$$-isomorphism between $$B'$$ and $$B$$.

We have hence carefully shown that $$C$$ is isomorphic to $$A\oplus B$$.

It now immediately follows that if $$A,B$$ are generated by their projections, then so is $$C$$; if $$A,B$$ are von Neumann algebras, then so is $$C$$.

• Good answer. I wasn't assuming they had units, but of course you're right, if $A$ is intrinsically a von Neumann algebra then it has to have one. – Nik Weaver Nov 29 '18 at 13:55
• Thank you and +1 for your interesting answer. i confess that I did not pay attention to the fact that the extension is trivial if the first object, $A$ , is unital. I am quite beginner in this area and also in von Neumann algebra. – Ali Taghavi Dec 1 '18 at 20:13