The spectrum of an abelian von Neumann algebra

Question

I am trying to understand the spectrum of an abelian Von Neumann algebra $A$. Let $\Omega(A)$ denote this spectrum with the weak$^*$-topology, and $\Omega_{\mathrm{disc}}(A)$ is isolated points in $\Omega(A).$

Is there a way to calculate exactly for which $A$ we have $\Omega_{\mathrm{disc}}(A)$ is dense in $\Omega(A)$?

For example, if $A=\ell^\infty(X)$ for discrete $X$, this is the case. Here we have $X=\Omega_{\mathrm{disc}}(A),$ $\beta X=\Omega(A).$ I also want to say it is also true for $A=L^\infty(X,\mu),$ where the isolated points $X_{\mathrm{disc}}$ of $X$ are dense in $X$ and $\mu|_{X_{\mathrm{disc}}}$ is the counting measure. Can it be true for more general $A$?

Answer 1

Theorem: Let $A$ be an abelian von Neumann algebra. If $\Omega_\mathrm{disc}(A)$ is dense in $\Omega(A),$ then $A=\ell^\infty(\Omega_\mathrm{disc}(A)).$

As Onur Oktay pointed out, the key is the notion of hyperstonean space, which has been discussed in Takesaki. The spectrum of an abelian von Neumann algebra $A$ is always a hyperstonean space (see Takesaki III.1.14, III.1.18). As noted in III.1.19, a hyperstonean space $\Omega$ admits a unique partition $\Omega_d,\Omega_c$ s.t. $\Omega_d$ has an open discrete dense subspace $X,$ and $\Omega_c$ has no isolated points.

In my question, we are concerned with the case, $\Omega_c=\emptyset.$ In any case, $\Omega_d=\beta X$ and so

$A=C(\beta X)=\ell^\infty(X).$

Answer 2

In fact, to show that $\Omega_{\text{disc}}$ being dense in $\Omega$ implies $A = \ell^\infty(\Omega_{\text{disc}})$, you don’t even need $A$ is a von Neumann algebra, just that it is monotone complete (i.e., every increasing uniformly bounded net of positive elements has a least upper bound) is enough. Indeed,

Lemma: If $A$ is monotone complete and $\Omega_{\text{disc}}$ is dense in $\Omega$, then for any $f \in \ell^\infty(\Omega_{\text{disc}})$, there is a unique $\tilde{f} \in C(\Omega) = A$ s.t. $\tilde{f}(\omega) = f(\omega)$ for every $\omega \in \Omega_{\text{disc}}$.

Proof: Uniqueness follows from density of $\Omega_{\text{disc}}$. For existence, note that by taking linear combinations, we may assume $f \geq 0$. Now, for any finite subset $F \subset \Omega_{\text{disc}}$, $\tilde{f}_F$ defined by $\tilde{f}_F(\omega) = f(\omega)$ when $\omega \in F$ and $\tilde{f}_F(\omega) = 0$ when $\omega \in \Omega \setminus F$ is continuous. As $f \geq 0$, we see that $\tilde{f}_F$ is positive and increases as $F$ increases, and $\tilde{f}_F \leq \|f\|$. Hence, as $A$ is monotone complete, $\{\tilde{f}_F\}_F$ has a least upper bound $\tilde{f}$. Clearly, $\tilde{f}(\omega) \geq f(\omega)$ for every $\omega \in \Omega_{\text{disc}}$. If $\tilde{f}(\omega_0) > f(\omega_0)$ for some $\omega_0 \in \Omega_{\text{disc}}$, then $\tilde{f}’$, which coincides with $\tilde{f}$ at all points except $\omega_0$, but takes value $f(\omega_0)$ at $\omega_0$, is continuous (since it only differs from the continuous $\tilde{f}$ on the clopen set $\{\omega_0\}$), is clearly also an upper bound of $\{\tilde{f}_F\}_F$, but is strictly smaller than $\tilde{f}$, contradicting $\tilde{f}$ being the least upper bound. Thus, $\tilde{f}(\omega) = f(\omega)$ for all $\omega \in \Omega_{\text{disc}}$, as required. $\square$

From this lemma, we easily see that $\ell^\infty(\Omega_{\text{disc}}) \ni f \mapsto \tilde{f} \in C(\Omega) = A$ is a $\ast$-isomorphism.