3
$\begingroup$

I am going through the proof of the Sherman-Takeda theorem and Fillmore's book "A User's Guide on Operator Algebras" seems to have a nice approach, but something seems off to me:

We need to prove that if $A$ is a $C^*$-algebra then $A^{**}$ is isometrically isomorphic to $\pi_u(A)''$, where $(H_u,\pi_u)$ is the universal representation.

The idea is finding an isometric isomorphism $e:A^*\to(\pi_u(A)'')_*$ taking the adjoint map and employing the uniqueness of the predual in von Neumann algebras. To that end, Fillmore takes a state $\rho\in S(A)$ and says that $\rho$ extends uniquely to a vector state on $\pi_u(A)''$. I agree with this and I can see why this is true. The map $e:A^*\to(\pi_u(A)'')_*$ is then constructed as follows: an arbitrary functional of $A^*$ is written as a linear combination of (four) states, so the above observation allows us to extend any functional of $A^*$ to a linear combination of (four) vector states on $\pi_u(A)''$, so $e$ takes our functional to that extension. I understand why this map $e:A^*\to(\pi_u(A)'')_*$ is a linear isometry but I find it weird that it is onto. Fillmore's proof says that, if $\tau\in(\pi_u(A)'')_*$ is a normal linear functional, then the restriction $\tau\vert_{\pi_u(A)}$ is a functional of $\pi_u(A)^*\cong A^*$ so $e(\tau\vert_{\pi_u(A)})=\tau$. Even though this makes sense to me, there's this awkward part:

Doesn't this imply that all normal functionals on $\pi_u(A)''$ are linear combinations of vector states, hence (SOT) continuous? Is this true or is it a hint that something has gone wrong?

I feel awkward about this because I have the impression that it is very rare for SOT continuous functionals to be the same as the ultraweakly continuous ones.

$\endgroup$
10
$\begingroup$

Doesn't this imply that all normal functionals on $\pi_u(A)''$ are linear combinations of vector states, hence (SOT) continuous?

It sounds like you understand the proof, but are suspicious of this consequence. Not to worry, the proof is correct, and yes, in general ultraweakly continuous linear functionals need not be SOT continuous, but they are in this case.

Ultraweak continuity on a von Neumann algebra $M$ is continuity for the unique dual space topology on $M$, and this is independent of any representation of $M$. In contrast, SOT continuity depends on the representation, and roughly speaking, if you pass to a bigger representation then there are more vectors, hence it is harder for a net of operators to SOT converge, hence it becomes easier to be a SOT-continuous linear functional.

If $\rho$ is any normal state on a von Neumann algebra $M$, then in the GNS representation it generates it becomes a vector state, right? So if you take the direct sum of these representations over all normal states, you get a representation in which every normal state is a vector state, and hence every normal state is SOT continuous.

$\endgroup$
2
  • $\begingroup$ Thank you very much, this truly helps me. $\endgroup$ – JustDroppedIn Sep 14 '20 at 15:36
  • $\begingroup$ No problem at all! $\endgroup$ – Nik Weaver Sep 14 '20 at 15:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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