8
$\begingroup$

Related to Jon's question, 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 $S$.

Question 1. Is $((\mathcal{A}_{sa})^m)^m=(\mathcal{A}_{sa})^m$? (Maybe this is very basic.)
Question 2. 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 $\mathbb{B}(\mathcal{H})$?

For Question 2, I have been thinking that Pedersen's up-down-up theorem [Theorem 2 in American Journal of Mathematics 94 (1972), 955-962] might be useful, but I couldn't figure out.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
9
$\begingroup$

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.)

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.

Improve this answer
$\endgroup$
10
  • $\begingroup$ Thank you Nik for your answer. Please allow me to ask further questions following your answer. For Question 1: Yes, I know that paper by Hamana. In fact, I previously highlighted the sentence $B_{sa}$ is itself a unique real linear subspace $S$ of $B_{sa}$ such that $A_{sa}\subset S$ and $S^m=S$'' in Introduction. I think that you are referring this part. In this statement, $S$ is assumed to be a real linear subspace of $B_{sa}$. However, in my question, $(A_{sa})^m$ is just a selfadjoint subset'' of $B(H)_{sa}$ and not a linear subspace in general. Could you please clarify this point? $\endgroup$
    – Masayoshi Kaneda
    Commented Aug 4, 2012 at 10:45
  • $\begingroup$ For Question 2: If you don't mind, could you please provide me with your counterexample in the abelian case? Thank you. P.S. My Question 1 seems exactly what Jon is asking. $\endgroup$
    – Masayoshi Kaneda
    Commented Aug 4, 2012 at 11:14
  • $\begingroup$ Nik, thank you again for your response. For Question 2, I see, the point is to consider $l^{\infty}[0,1]$ instead of $L^{\infty}[0,1]$. Thank you for a nice example. For Question 1, of course, I looked over Pedersen’s book, but I couldn’t find the answer. This is why I am placing the question on this board. $\endgroup$
    – Masayoshi Kaneda
    Commented Aug 5, 2012 at 13:05
  • $\begingroup$ (Continuation) In Pedersen’s book (Theorem 2.4.4), Kadison’s other (but related) result “A $C^*$-subalgebra $M$ of $B(H)$ is a von Neumann algebra if and only if $(M_{sa})^m= M_{sa}$” (Lemma 1 of [Kadison, Operator algebras with a faithful weakly-closed representation, Ann. of Math. (2) 64 (1956), 175-181]) is presented. But the proof in Pedersen’s book is not Kadison’s original one, but one using his own lemma which was used to prove his “up-down” theorem. $\endgroup$
    – Masayoshi Kaneda
    Commented Aug 5, 2012 at 13:07
  • $\begingroup$ (Continuation) Kadison’s result cited in Hamana’s paper can be found on pp. 316-317 of [Kadison, Unitary invariants for representations of operator algebras, Ann. of Math. (2) 66 (1957), 304-379], though it is not presented as a theorem or lemma. (Note that in this paper, $A_1^m$ is used in a different meaning from the usual one. It is the union of the monotone closures of $A_1$ taken “transfinitely recursively”. Also the monotone closures are taken in both directions, i.e., both increasing and decreasing.) $\endgroup$
    – Masayoshi Kaneda
    Commented Aug 5, 2012 at 13:11
6
$\begingroup$

I can give you a partial answer to question 1, which you may already be aware of if you're familiar with Pedersen's "C*-algebras and their automorphism groups". Specifically, if $A\subseteq A^{**}$, i.e. if we are considering the universal representation of $A$, and if $A$ is either unital or separable then you do indeed have $(A_\mathrm{sa}^m)^m=A_\mathrm{sa}^m=\overline{A_\mathrm{sa}^m}(=$ norm closure of $A_\mathrm{sa}^m$). For if $A$ is unital then, by Pedersen 3.11.7, $A_\mathrm{sa}^m=\overline{A_\mathrm{sa}^m}$, while if $A$ is separable then we again have $A_\mathrm{sa}^m=\overline{A_\mathrm{sa}^m}$, by Corollary 3.25(a) of Lawrence G. Brown's "Semicontinuity and Multipliers of C*-algebras". By Pedersen 3.11.5, $\overline{A_\mathrm{sa}^m}$ consists precisely of those $x\in A^{**}_\mathrm{sa}$ that are lower semicontinuous on $Q=A^{*1}_+$, which are immediately seen to be closed under taking supremums of bounded increasing nets.

If we consider $A$ with its atomic representation instead, then the above remarks still apply, as there is then a normal morphism $\pi$ from $A^{**}$ onto $A''$ which is faithful on $A_\mathrm{sa}^m$, by Pedersen 4.3.13 and 4.3.15. Arbitrary representations can be faithful on $A$ but not on $A^m_\mathrm{sa}$, although in general you might perhaps try to use the predual $A''_*$ of $A''$, in the topology induced by $A$, as a replacement for $A^*$ with the weak* topology. The only problem is that $A''^1_{*+}$ may not be compact in this topology so you would have to somehow adjust the proof of Pedersen 3.11.2, on which 3.11.5 relies.

Incidentally, it is a problem of Akemann and Pedersen from 1973 (still open as of 2014 according to Brown) whether $A_\mathrm{sa}^m=\overline{A_\mathrm{sa}^m}$ for arbitrary $A$ in its universal representation, so if you or Nik have a counterexample it would be quite important.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Nice answer. I didn't know that problem was open. $\endgroup$
    – Nik Weaver
    Commented Jan 3, 2015 at 16:23
  • $\begingroup$ Thanks. I was a bit surprised that problem was still open too but that's what Brown says in a recent article posted to the ArXiv - see arxiv.org/abs/1404.1383 $\endgroup$
    – Tristan Bice
    Commented Jan 4, 2015 at 2:00
  • $\begingroup$ @TristanBice Thank you so much for your answer and the reference, and I am sorry for my late response. I have not accessed this site since then. $\endgroup$
    – Masayoshi Kaneda
    Commented Jul 24, 2016 at 11:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .