3
$\begingroup$

Let $M$ be a von Neumann algebra and $\varphi$ be a normal weight on it, and let $A$ be its definition subalgebra. We still denote $\varphi$ the extension to $A$ as a linear positive functional. It is known that $\varphi$ is lower-ultraweakly-semicontinuous on $M^+$ (the positive elements of $M$).

Questions:

  • Is $\varphi$ ultraweakly continuous on $A$ (as a linear positive functional), where $A$ has the induced ultraweak topology of $M$ ? (Clearly, if $A=M$ this assertion is classical, right ?)

  • If we fix a Hilbert space representation of $M$ (so that $M\subseteq \mathcal{B}(H)$). Do we have that $\varphi$ is strongly or weakly continuous on A ?

(If it helps, one can suppose $\varphi$ to be a trace, semifinite and faithful).

  • A third (somewhat related) question: Suppose that $B$ is a subalgebra of a von Neumann algebra, and that $f:B\to M$ is a positive linear map, such that $f$ is normal in the following sense : for any increasing net in $B^+$ with supremum in $B^+$, the image (by $f$) of this supremum is the supremum of the image of the net (the usual notion of normality, but with the hypothesis that the supremum lies in $B^+$). Do we have that $f$ is continuous (for the ultraweak topologies)?

These questions seem natural to me, but I haven't been able to locate any reference about them.

$\endgroup$
0

1 Answer 1

5
$\begingroup$

1) if $A=M$ the assertion is indeed classical: normal states are exactly those ultraweakly continuous. But consider the case where $M=B(H)$ and $\varphi$ is the trace. Then the definition subalgebra $A$ is exactly the trace-class operators. Let $\{p_k\}\subset A$ be a maximal net of pairwise orthogonal projections of trace 1 (i.e. $\{e_{kk}\}$ for any choice of matrix units). Then $p_k\to0$ ultraweakly, but $\mbox{Tr}(p_k)=1$ for all $k$. So the trace is not ultraweakly continuous on the definition algebra, only ultraweakly lower-semicontinuous.

2) The example on 1) is already explicitly represented, so no.

3) Still thinking about it.

$\endgroup$
8
  • 1
    $\begingroup$ For (3), doesn't the same example work, considered as the map $T(H)\rightarrow B(H), x\mapsto \tr(x) 1$? I think this follows, as the question insists that the supremum of our increasing net in $B^+$ exists in $B^+$. $\endgroup$ Jan 29, 2012 at 9:38
  • $\begingroup$ I first answered that same example to 3, but then I thought that maybe the question required $B$ to be a von Neumann subalgebra and the map to be defined everywhere. So I stopped to consider cases like when $B$ is a monotone complete C$^*$-subalgebra of $B(H)$. Maybe Oliver can clarify what he expects in 3). $\endgroup$ Jan 29, 2012 at 13:28
  • $\begingroup$ I have just added a possible answer to 3). $\endgroup$ Jan 29, 2012 at 14:02
  • 1
    $\begingroup$ Never mind. Let's wait for Oliver to clarify question 3. $\endgroup$ Jan 29, 2012 at 14:46
  • 1
    $\begingroup$ Indeed, this answers also my question 3. Thanks! But I am actually also curious about the case you mentioned where $f$ is everywhere defined and $B$ is a von Neumann subalgebra. $\endgroup$
    – Oliver
    Jan 29, 2012 at 20:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.