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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?

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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 $C$.

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  • $\begingroup$ 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. $\endgroup$ – Ali Taghavi Dec 1 '18 at 20:13
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    $\begingroup$ 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. $\endgroup$ – Nik Weaver Dec 1 '18 at 23:03
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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 $C$.

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    $\begingroup$ 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. $\endgroup$ – Nik Weaver Nov 29 '18 at 13:55
  • $\begingroup$ 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. $\endgroup$ – Ali Taghavi Dec 1 '18 at 20:13

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