The monotone closure of a $C^*$-algebra - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:01:44Z http://mathoverflow.net/feeds/question/103860 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103860/the-monotone-closure-of-a-c-algebra The monotone closure of a $C^*$-algebra Masayoshi Kaneda 2012-08-03T12:56:59Z 2012-08-04T15:20:25Z <p>Related to <a href="http://mathoverflow.net/questions/103306/strong-monotone-limits-and-dense-subalgebras-of-von-neumann-algebras-again" rel="nofollow">Jon's question</a>, I have two questions. Let $\mathcal{A}$ be a concrete $C^*$-algebra on a Hilbert space $\mathcal{H}$. For any selfadjoint subset $S$ of $\mathbb{B}(\mathcal{H})$, let $S^m$ denote the set of elements of $\mathbb{B}(\mathcal{H})_{sa}$ that can be obtained as the strong limits of monotone increasing nets from $S$.</p> <blockquote> <p><strong>Question 1.</strong> Is $((\mathcal{A}_{sa})^m)^m=(\mathcal{A}_{sa})^m$? (Maybe this is very basic.)<br/><strong>Question 2.</strong> Does the $C^*$-algebra $C^*((\mathcal{A}_{sa})^m)$ generated by $(\mathcal{A}_{sa})^m$ in $\mathbb{B}(\mathcal{H})$ coincide with the strong closure of $\mathcal{A}$ in $\mathbb{B}(\mathcal{H})$?</p> </blockquote> <p>For Question 2, I have been thinking that Pedersen's up-down-up theorem [Theorem 2 in <em>American Journal of Mathematics</em> <strong>94</strong> (1972), 955-962] might be useful, but I couldn't figure out.</p> http://mathoverflow.net/questions/103860/the-monotone-closure-of-a-c-algebra/103876#103876 Answer by Nik Weaver for The monotone closure of a $C^*$-algebra Nik Weaver 2012-08-03T15:36:50Z 2012-08-04T15:20:25Z <p>Edited. As Masayoshi points out, my reading of Hamana's paper was incorrect. I'm quite sure question 1 is false in general but I don't have a reference. (Masayoshi, did you look in Pedersen's book? I feel the answer may be there but I don't have access to it right now.)</p> <p>I guess I'd better be more explicit about question 2. For example, take $A = C[0,1]$ acting by multiplication on $l^2[0,1]$. Then the strong closure of $A$ equals $l^\infty[0,1]$, but $A_{sa}^m$ is contained in e.g. the set of bounded Borel measurable functions, so the C*-algebra it generates is also contained in that class.</p>