Let $X$ be a compact Hausdorff space, and $C(X)$ the unital commutative C*-algebra of continuous functions on it. The double Banach dual $C(X)^{**}$ is a commutative von Neumann algebra and hence has a compact Hausdorff space $X^{**}$ as Gelfand spectrum again. What is $X^{**}$, in terms of $X$?

This gives an (idempotent?) endofunctor (monad?) on the category of compact Hausdorff spaces, that I don't recognize as any of the usual ones like Stone-Cech. What completion is it? Is it related to the functor taking a compact Hausdorff space to the $\sigma$-algebra generated by its opens?

Accounts of enveloping von Neumann algebras of (commutative) C*-algebras in terms of double Banach duals seem hard to find in the literature, and any references are welcome. What is the von Neumann algebra $C(X)^{**}$, in the first place?

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    $\begingroup$ The bidual of the space of continuous functions on a compact space was studied in great detail by S. Kaplan in a series of long articles (about 40 years ago). He gave an overview of his results in a research monograph (all easily available). He was interested in its Banach space properties but his work might contain some material which could be of interest to you. $\endgroup$ – jbc Sep 18 '12 at 8:27
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    $\begingroup$ Just to add to jbc's comment: if memory serves right, this is Lebesgue Thery in the Bidual of C(X), Mem. AMS $\endgroup$ – Yemon Choi Nov 20 '12 at 16:37

For what it's worth, I found a lot of information in [Dales, Lau & Strauss, "Second duals of measure algebras", Dissertationes Mathematicae 481:1-121, 2012]. The assignment $X \mapsto X^{\ast\ast}$ is functorial, and called the hyper-Stonean cover. It loses information: if $X$ is countable, then $X^{\ast\ast} \cong \beta\mathbb{N}$.

If $X$ is metrizable and uncountable, a lot of the structure of $X^{\ast\ast}$ is known -- it is characterised as follows:

  • $X^{\ast\ast}$ is hyper-Stonean;
  • the set $D$ of isolated points of $X^{\ast\ast}$ has cardinality $2^{\aleph_0}$, its closure $Y$ is a clopen subspace homeomorphic to $D_d$;
  • $X\setminus Y$ contains a family of $2^{\alpha_0}$ pairwise disjoint, clopen subspaces, each homeomorphic to $\mathbb{H}$;
  • the union $U$ of the above sets is dense in $X \setminus Y$, and $\beta U = X \setminus Y$.

In general, there exist a continuous projection $p \colon X^{\ast\ast} \to X$ and a (not necessarily injective) injection $i \colon X \to X^{\ast\ast}$ with $i \circ p = 1_{X^{\ast\ast}}$. Moreover, $X$ consists of the isolated points of $X^{\ast\ast}$, and is therefore open.


This perhaps should just be a comment, but it seemed to get slightly too long. It's also a bit disjointed as I am in a rush right now; sorry for that

I don't think one has a particularly good description of $C(X)^{**}$ as a von Neumann algebra, other than "it is what it is". I mean, it's called the enveloping von Neumann algebra, and it has an appropriate universal property, but that doesn't really "say what it is" in the sense your final question seems to ask.

The adjunction you describe has I think been well studied, but off-hand I am not sure about good references.

Spaces $X^{**}$ (in your notation) are necessarily hyper-stonean, and I guess what you are getting is some kind of hyperstonean cover (dual to the idea of $C(X)^{**}$ as a vN envelope). Google turns up the following paper from 1988

Hyperstonean cover and second dual extension

which might, if not directly relevant, at least have pointers to the literature.

  • $\begingroup$ Thanks for the pointer. By "What is $C(X)^{**}$?", I was hoping for a convenient description of a von Neumann algebra isomorphic to it, such as (but this is just a guess) $\ell^\infty(X)$. $\endgroup$ – Chris Heunen Apr 1 '11 at 23:50
  • $\begingroup$ If $C(X)^{**}$ were isomorphic to $\ell^\infty(X)$ as a von Neumann algebra then their preduals would have to be isometrically isomorphic, and so $M(X)$ would be isometrically isomorphic to $\ell^1(X)$. I strongly suspect that some messing around with measure theory tells you that this can only happen when X is discrete (hence in your setting finite). $\endgroup$ – Yemon Choi Apr 1 '11 at 23:59
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    $\begingroup$ $M(X) = l^1(X)$ when $X$ is countable compact, for example. Then $C(X)^{**} = l^\infty(X) = C(\beta X)$ where we do the Stone-Cech compactification of $X$ with discrete topology. But of course the original question is much more than this case. $\endgroup$ – Gerald Edgar Apr 2 '11 at 0:54
  • $\begingroup$ Oops, of course Gerald is correct (for some reason I had groups on the brain, and then of course you can't get countably infinite compact groups). $\endgroup$ – Yemon Choi Apr 2 '11 at 1:56

How about this: Take a maximal family $(\mu_i \colon i \in I)$ of mutually singular probability measures on $X$. Then $C(X)^* = M(X)$ is isometric to the $l^1$-sum of the spaces $L^1(\mu_i)$. Even when $X = [0,1]$ this is an uncountable direct sum. So $C(X)^{**}$ is the $l^\infty$ sum (or maybe call it the product) of the spaces $L^\infty(\mu_i)$.


This is done in Conway's book on Functional Analysis, (at least as a Banach space but the proof should work as a Von Neumann algebra as well), although I don't have the book on me and don't know the exact chapter/section reference. Note that if $\mu$ and $\nu$ are measures on $X$ with $\mu <<\nu$ and $f=0$ $\nu$ almost everywhere then $f=0$ $\nu$ almost everywhere so we have a well-defined map $L^{\infty}(X,\nu)\to L^{\infty}(X,\mu).$ One can endow theinverse limit, call it Y, of $L^{\infty}(X,\mu)$ under these maps as a Banach space and show that it is isomorphic to $C(X)^{**}.$ The duality between $Y$ and $M(X)$ is somewhat clear, if $f=[f_{\mu}]$ is a compatible collection of functions (so $f_{\mu}=f_{\nu}$ $\mu$ almost everywhere if $\mu<<\nu$) then the integral of $f_{\mu}$ against $\mu$ is well-defined for each $\mu\in M(X)$ and gives the duality between $Y$ and $M(X).$

  • $\begingroup$ Thanks for your satisfyingly categorical answer. However, I'm having trouble locating this material in Conway's book. Could I press you for a more precise reference? $\endgroup$ – Chris Heunen Jul 3 '12 at 9:53
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    $\begingroup$ In the second edition it's on pages 76-77, at the end of chapter III.5. $\endgroup$ – Dave Gaebler Jul 5 '12 at 17:22

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