3
$\begingroup$

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?

$\endgroup$

2 Answers 2

5
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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, 2019 at 3:14
  • 1
    $\begingroup$ @quasinilpotent This is Proposition 1.11 in Takesaki Chapter III $\endgroup$ Mar 7, 2019 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, 2019 at 12:26
4
$\begingroup$

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 $L^1$.

$\endgroup$
3
  • $\begingroup$ 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, 2019 at 15:23
  • 1
    $\begingroup$ @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. $\endgroup$ Mar 6, 2019 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, 2019 at 3:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.