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There are many different definitions for "normal" in literature, and I could not see the equivalence between the two following definitions:

Let $M , N$ be von-Neumann algebra, and let $\varphi: M\to N$ be a map. We say that $\varphi$ is normal if:

  1. $\varphi(\sup x_{\alpha})=\sup \varphi (x_{\alpha})$ for all norm-bounded monotone increasing nets of self adjoint elements $\{x_{\alpha}\} \subseteq M_{sa}$.

  2. $\varphi$ is $\sigma$-weakly continuous (when identifying $M$ with its predual and recall that the weak$^*$-topology on the predual coincides with the relative ultra-weak ($\sigma$-weak) topology on $M\subseteq B(H)$).

The direction $(2)\Rightarrow (1)$:

If we let $(x_{\alpha})\subseteq M_{sa}$ be a norm-bounded increasing net, by Vigier lemma, $x_{\alpha}$ converges in SOT to some $x\in M_{sa}$ and actually $x=\sup_{\alpha} x_{\alpha}$.

We know that on bounded subsets the ultra-weak topology coincides with the weak$^*$ topology, so by $\sigma-weak$ continuity of $\varphi$ we get $\varphi(x_\alpha)\to \varphi(x)$ (in norm). However, I'm not sure why $\lim \varphi(x_\alpha)=\sup \varphi (x_{\alpha})$.

Maybe if we add an assumption that $\varphi$ is positive we could get it, again by applying Vigier's lemma.

I don't know also how to show the converse direction.

Maybe I also should mention that I'm not sure the above definitions I gave are equivalent, there is an option I did some "mix", or this is true for states?

Thank you for your time.

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    $\begingroup$ This equivalence is true (at least) for completely positive linear maps such as states and $\ast$-homomorphisms, but not in general, as you almost noted ($\phi$ might reverse the order). try to prove it under this additional assumption. $\endgroup$ – MaoWao Oct 28 '16 at 12:05
  • $\begingroup$ I had to search "Vigier's Lemma". I see that the result is mentioned like that in Murphy and a couple monographs, but it is the very first time I see those words. $\endgroup$ – Martin Argerami Oct 28 '16 at 15:48
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The equivalence of course requires the map to be positive. The equality $\lim\varphi(x_j)=\sup\varphi(x_j)$ is just the fact, in the real line, that the supremum of a monotone increasing sequence is its limit.

For the reverse implication, the proof one needs is the proof of the equivalence for states, because if you have that $\varphi$ respects suprema, the so does $f\circ\varphi$ for every normal state $f$; then $f\circ\varphi$ is $\sigma$-weakly continuous (by the proof for states) and then $\varphi$ is $\sigma$-weakly continuous. Similarly for the converse.

So the key step is to prove the assertion when $N=\mathbb C$ (and, as mentioned, assuming that $\varphi$ is positive). The way I know to prove the equivalence is in Theorem 7.1.12 in Kadison-Ringrose: one proves the equivalence of the statements

  • $\varphi=\sum_{n=1}^\infty\langle \cdot y_n,y_n\rangle$ where $\sum\|y_n\|^2=1$ and $\{y_n\}$ is orthogonal

  • $\varphi=\sum_{n=1}^\infty\langle \cdot y_n,y_n\rangle$ where $\sum\|y_n\|^2=1$

  • $\varphi$ is wot-continuous on the unit ball of $M$

  • $\varphi$ is sot-continuous on the unit ball of $M$

  • $\varphi(\sup x_n)=\sup\varphi(x_n)$ for any monotone bounded net of selfadjoints

  • $\varphi$ is completely additive $\varphi(\sum p_j=\sum\varphi(p_j)$ for every family of pairwise orthogonal projections.

I think I have seen proofs that the first three are equivalent on their own, but I'm not sure if there is a shorcut to avoid the whole path for a complete proof of your implication.

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    $\begingroup$ Thank you, Martin. Yes, I have just founded this proof in Kadison-Ringrose. Maybe I'll just read the whole path... :) $\endgroup$ – Shirly Geffen Oct 28 '16 at 15:48

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