4
$\begingroup$

Let $\mathbb{G}= (A, \Delta)$ be a ($C^*$-algebraic) compact quantum group. In a paper I'm reading, the space $A^*= B(A, \mathbb{C})$ obtains a product
$$\omega_1*\omega_2:= (\omega_1\otimes \omega_2) \circ \Delta$$ and this is used to prove the existence of the Haar functional on a compact quantum group.

Question: How is $\omega_1 \otimes \omega_2$ defined here? Clearly we have a linear mapping $$\omega_1 \odot \omega_2: A \odot A \to \mathbb{C}$$ on the algebraic tensor product, but we need continuity to extend this to the completion $A \otimes A$ (with respect to the minimal $C^*$-norm on the algebraic tensor product $A \odot A$).

In general, I believe $\omega_1 \odot \omega_2$ must not be continuous, though this result does hold when one works with states on the $C^*$-algebra $A$.

$\endgroup$
  • 4
    $\begingroup$ Every bounded linear functional on a $C^\ast$-algebra is a linear combination of states, so $\omega_1\odot \omega_2$ extends to the spatial (minimal) tensor product for all $\omega_1,\omega_2 \in A^\ast$ by a theorem of Takesaki. $\endgroup$ – Jamie Gabe Oct 31 at 13:57
  • 2
    $\begingroup$ @JamieGabe Thanks a lot! That makes sense! If you want, you can make that an answer! $\endgroup$ – user839372 Oct 31 at 14:17
5
$\begingroup$

Every bounded linear functional on a $C^\ast$-algebra is a linear combination of states, so $\omega_1\odot \omega_2$ extends to the spatial (minimal) tensor product for all $\omega_1, \omega_2 \in A^\ast$ by a theorem of Takesaki.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

A result which I find a bit surprising is this. Let $A,B$ be $C^*$-algebras. Then:

  1. For $\omega_1\in A^\ast$ and $\omega_2\in B^\ast$, the functional $\omega_1\otimes\omega_2$ is also bounded as a map $A\otimes_{\max} B \rightarrow \mathbb C$;
  2. This means that the algebraic tensor product $A^\ast\otimes B^\ast$ maps into the dual of $A\otimes_\beta B$ where $\beta$ is any $C^*$-tensor norm on $A\otimes B$;
  3. The resulting norm (the map is injective) on $A^\ast\otimes B^\ast$ is the same for any norm $\beta$.

You can find this in Chapter IV, Proposition 4.10 of Takesaki's book, for example.

| cite | improve this answer | |
$\endgroup$

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.