What kind of completion is this? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T00:45:23Z http://mathoverflow.net/feeds/question/60328 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60328/what-kind-of-completion-is-this What kind of completion is this? Chris Heunen 2011-04-01T23:19:09Z 2013-05-10T16:23:41Z <p>Let $X$ be a compact Hausdorff space, and $C(X)$ the unital commutative C*-algebra of continuous functions on it. The double Banach dual <code>$C(X)^{**}$</code> is a commutative von Neumann algebra and hence has a compact Hausdorff space <code>$X^{**}$</code> as Gelfand spectrum again. What is $X^{**}$, in terms of $X$?</p> <p>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?</p> <p>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? </p> http://mathoverflow.net/questions/60328/what-kind-of-completion-is-this/60330#60330 Answer by Yemon Choi for What kind of completion is this? Yemon Choi 2011-04-01T23:44:49Z 2011-04-01T23:44:49Z <p>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</p> <p>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.</p> <p>The adjunction you describe has I think been well studied, but off-hand I am not sure about good references.</p> <p>Spaces <code>$X^{**}$</code> (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</p> <p><a href="http://www.springerlink.com/content/pm2341h8j5513670/" rel="nofollow">Hyperstonean cover and second dual extension</a></p> <p>which might, if not directly relevant, at least have pointers to the literature.</p> http://mathoverflow.net/questions/60328/what-kind-of-completion-is-this/60333#60333 Answer by Gerald Edgar for What kind of completion is this? Gerald Edgar 2011-04-02T01:00:26Z 2011-04-02T01:00:26Z <p>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 $L^\infty(\mu_i)$.</p> http://mathoverflow.net/questions/60328/what-kind-of-completion-is-this/63393#63393 Answer by Benjamin Hayes for What kind of completion is this? Benjamin Hayes 2011-04-29T10:16:19Z 2011-04-29T10:16:19Z <p>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 $\mu &lt;&lt;\nu$ 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 $\mu&lt;&lt;\nu$) then the integral of $f_{\mu}$ against $\mu$ is well-defined for each $\mu\in M(X)$ and gives the duality between $Y$ and $M(X).$</p> http://mathoverflow.net/questions/60328/what-kind-of-completion-is-this/130275#130275 Answer by Chris Heunen for What kind of completion is this? Chris Heunen 2013-05-10T16:23:41Z 2013-05-10T16:23:41Z <p>For what it's worth, I found a lot of information in [Dales, Lau &amp; Strauss, "Second duals of measure algebras", Dissertationes Mathematicae 481:1-121, 2012]. The assignment $X \mapsto X^{\ast\ast}$ is functorial, and called the <i>hyper-Stonean cover</i>. It loses information: if $X$ is countable, then $X^{\ast\ast} \cong \beta\mathbb{N}$. </p> <p>If $X$ is metrizable and uncountable, a lot of the structure of $X^{\ast\ast}$ is known -- it is characterised as follows:</p> <ul> <li>$X^{\ast\ast}$ is hyper-Stonean;</li> <li>the set $D$ of isolated points of $X^{\ast\ast}$ has cardinality $2^{\aleph_0}$, its closure $Y$ is a clopen subspace homeomorphic to $D_d$;</li> <li>$X\setminus Y$ contains a family of $2^{\alpha_0}$ pairwise disjoint, clopen subspaces, each homeomorphic to $\mathbb{H}$;</li> <li>the union $U$ of the above sets is dense in $X \setminus Y$, and $\beta U = X \setminus Y$. </li> </ul> <p>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.</p>