# Convolution of functionals on compact quantum group

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 $$A$$.

• 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. – Jamie Gabe Oct 31 at 13:57
• @JamieGabe Thanks a lot! That makes sense! If you want, you can make that an answer! – user839372 Oct 31 at 14:17

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.
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 $$\beta$$.