# Ultraweak topology in abelian von Neumann algebras

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?

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.

• 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? – condexp Mar 7 '19 at 3:14
• @quasinilpotent This is Proposition 1.11 in Takesaki Chapter III – მამუკა ჯიბლაძე Mar 7 '19 at 7:41
• @მამუკა ჯიბლაძე, Got it. I guess the if instead of if and only if caused me some confusion. Thank you. – condexp Mar 7 '19 at 12:26

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 $$L^1$$.

• 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 $X$? – condexp Mar 6 '19 at 15:23
• @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. – მამუკა ჯიბლაძე Mar 6 '19 at 18:45
• @მამუკა ჯიბლაძე, looks like I have had the wrong idea the whole time. Thanks for your explanation. – condexp Mar 7 '19 at 3:11