From the abelian von Neumann algebra, I see the hyperstonean space as its spectrum(analogy with the C*-algebra). Now I want to see some examples of hyperstonean space.

1) Can you give me a hyperstonean space which is not a stonean space?

2) Does a Cantor set be a hyperstonean space? If not, can you show me some concrete hyperstonean space?

  • $\begingroup$ A little error,the question 1)should be changed:Can you give me a stonean space which is not a hyperstonean space? $\endgroup$ – Strongart Jul 6 '14 at 5:51

The second statement in Martin's answer is not quite correct - the Cantor space is totally disconnected/zero dimensional but NOT extremally disconnected/stonean. To see this, let us first be clear about our representation of the Cantor space $C$ as, say, real numbers in the interval $[0,1]$ that have ternary expansions without $1$'s. Take a point $x\in C$ which has infinitely many $0$'s and infinitely many $2$'s in its ternary expansion. Then the points below $x$ in $C$ and those above $x$ in $C$ will both be open and have $x$ in their closures, i.e. the closure of either of these open sets is not open and hence $C$ is not stonean.

The first example of a stonean space that is not hyperstonean was given indirectly by Dixmier. He gave an example of a commutative AW*-algebra $A$ that is not a von Neumann algebra which, as Martin alluded to, means the spectrum/pure states of $A$ are stonean but not hyperstonean (when given the weak* topology induced by $A$). I have never been able to track down the original paper but I have heard that his example goes something like this.

Consider the C*-algebra $B$ of all bounded Borel functions from $\mathbb{R}$ to $\mathbb{C}$. The $f\in B$ with meagre (i.e. countable union of nowhere dense sets) support $\mathbb{R}\setminus f^{-1}\{0\}$ form a closed ideal $I$, so $A=B/I$ is a C*-algebra. Every Borel subset $S$ of $\mathbb{R}$ is, modulo a meagre subset, the same as a unique regular open subset $RO(S)$. So projections in $A$ correspond to regular open subsets of $\mathbb{R}$, which form a complete lattice. Thus $A$ is an AW*-algebra.

However, $A$ is not a von Neumann algebra, and does not even have any non-zero normal positive functionals. To see this note that, as the $RO$ function preserves countable supremums, any normal positive functional on $A$ would induce a (positive countably additive) Borel measure $\mu$ on $\mathbb{R}$ which is zero on all meagre subsets. For each $q\in\mathbb{Q}$ (or any other countable dense subset of $\mathbb{R}$), $\mu(\{q\})=0$ so we can take an open interval $O_q$ containing $q$ such that $\mu(O_q)$ is as small as we like. In particular, if $\mu\neq0$ then we can make the intervals $O_q$ small enough so that $\mu(C)\neq0$, where $C=\mathbb{R}\setminus\bigcup_{q\in\mathbb{Q}}O_q$. But $C$ is meagre (even nowhere dense) by construction, so we have a contradiction.

And lastly, if you want an explicit example of a hyperstonean space then you can simply take any set $X$ with the discrete topology (arbitrary subsets are open), or its Stone-Čech compactification $\beta X$.

  • 1
    $\begingroup$ I probably should have mentioned, for $A$ to be an AW*-algebra, the projections need to form a complete lattice AND $A$ has to have real rank zero. But this follows from the fact $B$ has real rank zero and quotients of real rank zero C*-algebras are again real rank zero. Also, the original paper of Dixmier's can be found at dmitripavlov.org/scans/dixmier.pdf. $\endgroup$ – Tristan Bice May 31 '15 at 18:14

Every hyperstonean space is stonean. This corresponds to the fact that every von Neumann algebra is an AW$^*$-algebra.

The Cantor set is stonean, but not hyperstonean. I have read this assertion by experts, but I don't know the proof.

I don't think there are any "explicit" examples of hyperstonean spaces. Hyperstonean spaces are the compact sets such that $C(X)$ is a von Neumann algebra, and that's likely their best characterization.

  • $\begingroup$ Thanks, the question 1) should exchange stonean and hyperstonean, my mistake! For the C(X), I have another question: for example X=[0,1], how about C(x)", as an abelian von Neumann algebra, it looks like C(Y), Y is hyperstonean, how to change X to Y? Restriction to some hyperstonean subspace Y of [0,1]? $\endgroup$ – Strongart Jul 6 '14 at 5:59
  • $\begingroup$ What you want to do is not intuitive at all. The $Y$ you are looking for is called the hyperstonean cover of $X$. There are a few papers about it. $\endgroup$ – Martin Argerami Jul 6 '14 at 12:10
  • $\begingroup$ I will move to the best answer to Tristan Bice's because he point that "the Cantor space is totally disconnected/zero dimensional but NOT extremally disconnected/stonean." and show me an interesting example which is commutative AW*-algebra but not W*-algebra. $\endgroup$ – Strongart Dec 20 '14 at 13:58

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