When C(K) is closed in sigma strong topology? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T20:38:51Z http://mathoverflow.net/feeds/question/65065 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65065/when-ck-is-closed-in-sigma-strong-topology When C(K) is closed in sigma strong topology? Tomek Kania 2011-05-15T20:24:43Z 2011-05-15T22:36:17Z <p>Fix a compact Hausdorff space $K$ and think about $C(K)$ as a C*-algebra acting on a Hilbert space $H$. Suppose that $C(K)$ is closed in $\mathcal{B}(H)$ in:</p> <ul> <li>$\sigma$-strong</li> <li>$\sigma$-strong*</li> </ul> <p>topology. Must $K$ be extremelly disconnected? </p> http://mathoverflow.net/questions/65065/when-ck-is-closed-in-sigma-strong-topology/65080#65080 Answer by AndrĂ© Henriques for When C(K) is closed in sigma strong topology? AndrĂ© Henriques 2011-05-15T21:16:38Z 2011-05-15T22:36:17Z <p>Recall that if two compact spaces $K_1$, $K_2$ are such that $C(K_1)\cong C(K_2)$, then $K_1\cong K_2$. The space $K$ is called the spectrum of the abelian <i>C</i>*-algebra $C(K)$.</p> <p>Since $C(K)$ is closed in the &sigma;-strong topology, it is a von Neumann algebra (that condition is equivalent to being closed in the &sigma;-strong* topology).</p> <p>Now, the spectrum of an abelian von Neumann algebra is indeed an extremely disconnected space. So yes: $K$ has to be extremely disconnected. This kind of space is also called <i>hyperstonean</i> space.<Br><br><br></p> <p><hr> By the way, here's one way to visualize the hyperstonean space associated to the von Neumann algebra $L^\infty ([0,1])$:<br> For every measurable partition of $[0,1]$ into finitely many subsets $$[0,1]=X_1\cup\ldots \cup X_n$$ (where each $X_i$ is well defined up to measure zero sets), we form the space $$\overline{X_1}\sqcup\ldots \sqcup \overline{X_n}$$ where $\overline{X_i}$ is the closure of $X_i$ (more precisely, it is the intersection of all closures of sets that are equal to $X_i$ up to a measure zero set) and $\sqcup$ denotes disjoint union. The assignment $$X_1\cup\ldots \cup X_n \mapsto \overline{X_1}\sqcup\ldots \sqcup \overline{X_n}$$ is a functor from the poset of measurable partitions of $[0,1]$ to the category of compact topological spaces. The hyperstonean space associated to $L^\infty ([0,1])$ is the inverse limit of that functor.</p>