Strong monotone limits and dense subalgebras of von Neumann algebras, again - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T08:02:53Z http://mathoverflow.net/feeds/question/103306 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103306/strong-monotone-limits-and-dense-subalgebras-of-von-neumann-algebras-again Strong monotone limits and dense subalgebras of von Neumann algebras, again Jon Bannon 2012-07-27T13:19:20Z 2012-08-16T18:11:44Z <p><strong>Edit:</strong> I just realized that this question is related to Andreas Thom's very interesting question <a href="http://mathoverflow.net/questions/34692/subalgebras-of-von-neumann-algeb" rel="nofollow">here</a>. I think the question below is more crude...</p> <p>Michael's question <a href="http://mathoverflow.net/questions/103240/dense-subalgebras-of-von-neumann-algebras-and-increasing-nets" rel="nofollow">here</a> reminded me of <a href="http://www.jstor.org/stable/pdfplus/1969954.pdf" rel="nofollow">the first lemma of this paper of Kadison</a>, establishing that those $C^{*}$-algebras $\mathcal{A}\subset B(\mathcal{H})$ for which the set of self-adjoint elements $\mathcal{A_{s.a.}}$ is strong-operator closed under taking monotone increasing limits are, in fact, von Neumann algebras. I've often wondered about the following</p> <blockquote> <p><strong>Question:</strong> What are necessary and sufficient conditions on a $*$-subalgebra $\mathcal{A}\subset B(\mathcal{H})$ such that strong operator closure of $\mathcal{A_{s.a.}}$ under monotone increasing nets guarantees that $\mathcal{A}$ is closed in the strong operator topology (i.e. is actually a von Neumann algebra)?</p> </blockquote> <p>I'd like to know if there are weaker conditions than $\mathcal{A}$ being a $C^{\ast}$-algebra for which Kadison's conclusion holds. For example, conditions including things like "$\mathcal{A}$ is closed under continuous functional calculus" and such, which are essentially equivalent to $\mathcal{A}$ being a $C^{*}$-algebra in the (singly-generated) abelian case, but not in general.</p> <p>(My impression is that this question is really tough, but I hope I'm wrong. It would be nice even to have an expert's digression on precisely <em>why</em> this question should be tough...as I'd learn some new things from that insight!) </p> <p>It is also possible to ask about analogues of Pedersen's "up-down" theorem, which says that any self-adjoint element in the strong closure of a $C^{\ast}$-algebra $\mathcal{A}$ on a separable Hilbert space $\mathcal{H}$ is the strong limit of a monotone decreasing sequence of self-adjoint elements each of which is a strong limit of a monotone increasing sequence of self-adjoint elements. Can one weaken the $C^{\ast}$-condition on $\mathcal{A}$ and still get this result? (I've tried to modify the argument in Pedersen's $C^{*}$-<em>algebras and their automorphism groups</em> but if I remember correctly this seems to use the <em>definition of a</em> $C^{*}$-algebra in an essential way. Is there a way around this?!) </p> http://mathoverflow.net/questions/103306/strong-monotone-limits-and-dense-subalgebras-of-von-neumann-algebras-again/104026#104026 Answer by Masayoshi Kaneda for Strong monotone limits and dense subalgebras of von Neumann algebras, again Masayoshi Kaneda 2012-08-05T14:36:12Z 2012-08-05T14:36:12Z <p>This is not an answer. Recently, I have been thinking questions related to yours, and I posted some of them on the <a href="http://mathoverflow.net/questions/103860/the-monotone-closure-of-a-c-algebra" rel="nofollow">board</a>, so you may have seen them. I also posted related references in my comments there, but I think that you have already read those papers. It’s possible to ask a more general question: “Which subsets $S\subset\mathbb{B}(\mathcal{H})_{sa}$ have the property $S^m=S_m=S$?” or “Which real linear subspaces $S\subset\mathbb{B}(\mathcal{H})_{sa}$ have the property $S^m=S$?” It would be interesting if these simple conditions implied $S$ being strongly closed.</p>