Edit: I just realized that this question is related to Andreas Thom's very interesting question here. I think the question below is more crude...

Michael's question here reminded me of the first lemma of this paper of Kadison, 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

Question: 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)?

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.

(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 why this question should be tough...as I'd learn some new things from that insight!)

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 $C^{*}$-algebras and their automorphism groups but if I remember correctly this seems to use the definition of a $C^{*}$-algebra in an essential way. Is there a way around this?!)


This is not an answer. Recently, I have been thinking questions related to yours, and I posted some of them on the board, 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.

  • $\begingroup$ No in both cases. Again, check the commutative case first! $\endgroup$
    – Nik Weaver
    Aug 5 '12 at 15:44
  • 1
    $\begingroup$ (In both cases, you can find a counterexample where $S$ contains no comparable elements, hence $S = S^m = S_m$ trivially.) $\endgroup$
    – Nik Weaver
    Aug 5 '12 at 18:07

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