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Let $A$ be an abelian von Neumann algebra acting on the (not necessarily separable) Hilbert space $\mathcal{H}$ (with identity $I$). From the Gelfand-Neumark theorem, there is a compact Hausdorff space $X$ such that $A \cong C(X)$, the $*$-algebra of complex-valued continuous functions on $X$. The space $X$ is in fact extremally disconnected.

Here goes my question(s). Just as the norm topology of $A$ is captured by the $\sup$ norm of the function algebra over $X$, what is an analogous description of the ultraweak topology on $A$? For instance, if we want to say $f_{\alpha} \rightarrow f$ in the ultraweak topology, how do we phrase such a statement in purely topological terms (in terms of $X$ and its topology)?

Thank you.

Edit: Clarified some aspects to make the question more specific. Looking around mathoverflow brought me to the following two discussions which captures the spirit of my question: 1) Reference for the Gelfand-Neumark theorem for commutative von Neumann algebras 2) What kind of completion is this?

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მამუკა ჯიბლაძე hints that "hyperstonian" is an important definition here. I have struggled to find a good internet reference, and so am following Section 1 of Chapter III of Takesaki's book.

A Stonian space is a compact Hausdorff Extremally disconnected space. A rare or nowhere dense set $M$ is such that the closure of $M$ has empty interior.

Let $X$ be a Stonian space and let $C_{\mathbb R}(X)$ be the space of real-valued continuous functions on $X$. A (positive, Radon) measure $\mu$ on $X$ is normal if whenever $(f_i)$ is an increasing bounded net in $C_{\mathbb R}(X)$ and $f$ is the least upper bound of $\{f_i\}$ in $C_{\mathbb R}(X)$ (which exists as $X$ is Stonian), then $$ \int_X f \ d\mu = \sup_i \int_X f_i \ d\mu. $$ Alternatively, $\mu$ is normal if $\mu(M)=0$ for all (closed) rare sets $M$.

Finally, $X$ is hyperstonian if it admits sufficiently many normal measures: for any non-zero $f\in C_{\mathbb R}(X)$ there is a normal $\mu$ with $\int_X f\ d\mu\not=0$.

For a compact Hausdorff $X$, we have that $C(X)$ is a von Neumann algebra exactly when $X$ is hyperstonian.

Define a real-valued Radon measure to be normal if its positive and negative parts are normal; analogously for complex Radon measures. Following the proofs through, it follows that the predual of $C(X)$ is exactly the collection of complex normal measures on $X$.

But I wonder if this answers the original question: is this in terms of "$X$ and its topology"? However, notice that the difference between Stonian and Hyperstonian is very measure theoretic in nature.

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  • $\begingroup$ Alternatively, $\mu$ is normal if $\mu(M)=0$ for all (closed) rare sets $M$. Is this statement in the sense of an alternative definition of normality or a result? $\endgroup$ – condexp Mar 7 at 3:14
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    $\begingroup$ @quasinilpotent This is Proposition 1.11 in Takesaki Chapter III $\endgroup$ – მამუკა ჯიბლაძე Mar 7 at 7:41
  • $\begingroup$ @მამუკა ჯიბლაძე, Got it. I guess the if instead of if and only if caused me some confusion. Thank you. $\endgroup$ – condexp Mar 7 at 12:26
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The fact that a $C^\ast$-algebra is a von Neumann algebra means that it is a dual space (as a Banach space). The ultraweak topology is then what functional analysts call the weak-$\ast$ topology. In the case where it is an $L^\infty$ space then it is the weak topology induced in the natural way by $L^1$.

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  • $\begingroup$ Thank you for your answer. I am curious about a description in terms of the topology of $X$. For an extremally disconnected space $X$, is there an explicit description of the pre-dual of $C(X)$ in terms of $X$? $\endgroup$ – condexp Mar 6 at 15:23
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    $\begingroup$ @quasinilpotent Not all extremally disconnected spaces are hyperstonean, i. e. have von Neumann $C$. See an answer on math.SE for an example: the Stone space of the complete Boolean algebra of regular open sets in $\mathbb R$ is extremally disconnected but not hyperstonean. $\endgroup$ – მამუკა ჯიბლაძე Mar 6 at 18:45
  • $\begingroup$ @მამუკა ჯიბლაძე, looks like I have had the wrong idea the whole time. Thanks for your explanation. $\endgroup$ – condexp Mar 7 at 3:11

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