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Consider the following fragment from Takesaki's second volume "Theory of operator algebras" in chapter VIII "Modular automorphism groups" p122-123.

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Why is it possible to choose an increasing net in $\mathfrak{m}_0^+$ converging $\sigma$-strongly to $1$?

Note that in this post, it is established that the following is not true in general:

If $A$ is a $\sigma$-weakly dense $*$-subalgebra of a von Neumann algebra $\mathcal{M}$, then it can be $\sigma$-weakly approximated with a net of positive elements in $A$.

Is this maybe a mistake in Takesaki's book?

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    $\begingroup$ Although this is not really answering your question, note that in the proof of this Proposition 3.15, you could also just take a bounded net a_j that converges strongly to 1 (using the Kaplansky density theorem). Later on in the proof, you can use the dominated convergence theorem instead of Dini's theorem. $\endgroup$
    – Stefaan Vaes
    Commented Mar 28, 2023 at 2:24
  • $\begingroup$ @StefaanVaes Thanks for your comment! I am not sure dominated convergence can be applied here, since we are dealing with nets: math.stackexchange.com/questions/141198/… Hopefully, I'm wrong though. $\endgroup$
    – Andromeda
    Commented Mar 28, 2023 at 7:03
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    $\begingroup$ You are right. In the fully general, potentially non-separable setting, one has to be more careful. But still: taking a bounded net $a_j$ that converges strongly to 1, we get that $\|a_j \xi - \xi\| \to 0$ uniformly on compact subsets of $\xi \in H$. Therefore, given a vector $\xi$, the net $\|a_j \Delta^{-it} \xi - \Delta^{-it}\xi\|$ converges to $0$ uniformly on compact subsets of $t \in \mathbb{R}$. Thus, $\|\sigma_t(a_j) \xi - \xi\| \to 0$ uniformly on compact subsets of $t \in \mathbb{R}$. If I'm not mistaken, this suffices for the proof of Proposition 3.15. $\endgroup$
    – Stefaan Vaes
    Commented Mar 28, 2023 at 14:24
  • $\begingroup$ @StefaanVaes Thank you. I will look into it! $\endgroup$
    – Andromeda
    Commented Apr 1, 2023 at 22:02

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The key to answering this is to realise that approximating the identity is a special case: any $C^*$-algebra has an approximate identity!

Indeed, consider first $A = \overline{\mathfrak{m}_0}$ the norm closure, which is a $C^*$-algebra $\sigma$-strongly in $\mathcal M$. Then $A$ has an increasing net of positive elements forming a bai (indeed, the positive elements in the unit ball of $A$ form such a thing) see Takesaki I, Chapter 1, Theorem 7.4. This net converges $\sigma$-strongly to $1$ in $\mathcal M$.

Given $a\in A^+$ and $\epsilon>0$ there is $m\in\mathfrak m_0$ with $\|m-a^{1/2}\| < \epsilon$. Then $\|m^*m - a\| = \| (m-a^{1/2})^*m + a^{1/2}(m - a^{1/2}) \| < \epsilon (\|m\| + \|a^{1/2}\|)$ and so we have well-approximated $a$ by some element of $\mathfrak{m}_0^+$. It is hence possible to find a bai in $\mathfrak{m}_0^+$, but this may not be increasing.

I had originally claimed that $\{ m\in \mathfrak{m}_0^+ : \|m\|< 1 \}$, with the natural order, forms an approximate identity for $A$. However, it is not obvious (to me) that this set is upward directed (and so it seems to not necessarily be a net). So this currently remains a partial answer.

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  • $\begingroup$ Thanks for your answer! Some remarks: (1) Your answer shows that if $A$ is a $\sigma$-weakly dense $*$-subalgebra of a VNA $\mathcal{M}$, then there exists an increasing net of positive elements in $A$ that converges to $1_\mathcal{M}$ $\sigma$-strongly. Doesn't this contradict this post: mathoverflow.net/questions/103240/… ? $\endgroup$
    – Andromeda
    Commented Mar 25, 2023 at 23:04
  • $\begingroup$ (2) Why is $\Lambda:=\{m\in \mathfrak{m}_0^+: \|m\|< 1\}$ an approximate unit for $A$? I don't see why it is (i) upwards directed (ii) it satisfies $\lim_{\lambda \in \Lambda} u_{\lambda} a= a$ for $a\in A$ where $u_\lambda = \lambda$. $\endgroup$
    – Andromeda
    Commented Mar 25, 2023 at 23:09
  • $\begingroup$ For (1): I see no counter-example in the link you give: you certainly cannot approximate an arbitrary positive in $M$ but you can, I believe, approximate the unit. For (2): Ah, I think I agree that $\Lambda$ is not obviously upwards directed (and so part (ii) doesn't even make sense... if $\Lambda$ is a net, then the approximation argument I gave shows that it's a bai). $\endgroup$
    – Matthew Daws
    Commented Mar 26, 2023 at 9:17
  • $\begingroup$ Alright. Thanks. Hopefully someone can come up with a fix. $\endgroup$
    – Andromeda
    Commented Mar 26, 2023 at 9:32

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