# Showing a product on a character space is continuous

Quoting from Timmermann's An invitation to quantum groups and duality:

Prop. 5.1.3 Let $$A$$ be a commutative algebra of functions on a compact quantum group. Then there exists a compact group $$G$$ and an isomorphism $$A\cong C(G)$$ of $$\mathrm{C}^*$$-algebras.

Proof: By the Gelfand theorem, there exists a compact space $$G$$, a continuous map $$m:G\times G\rightarrow G$$...

I might reduce this to:

Let $$A$$ be a commutative unital $$\mathrm{C}^*$$-algebra with a $$\mathrm{C}^*$$-morphism $$\Delta:A\rightarrow A\otimes A$$. By the Gelfand theorem, where $$\Phi(A)$$ is the Hausdorff weak*-compact character space, the map $$m:\Phi(A)\times \Phi(A)\rightarrow \Phi(A)$$, $$(\chi,\varphi)\mapsto (\chi\otimes\varphi)\Delta$$ is continuous...

My difficulties are more appropriate to MSE, although the content knowledge might be more appropriate to here.

My natural inclination is to take sequences $$(\chi_n)\rightarrow \chi \in \Phi(A)$$ and $$(\varphi_n)\rightarrow\varphi\in\Phi(A)$$ and to look at $$(\chi_n,\varphi_n)$$, and show: $$m(\chi_n,\varphi_n)\rightarrow m(\chi,\varphi),$$ and using the dense subalgebra generated by the matrix elements of unitary irreducible (co)representations $$(\rho_{ij}^\alpha)_{\alpha\in\text{Irr}(A)}$$ shows: $$(\chi_n\otimes\varphi_n)\Delta(\rho_{ij}^\alpha)\rightarrow (\chi\otimes\varphi)\Delta(\rho_{ij}^\alpha),$$

but I don't think that $$\Phi(A)$$ is in general a sequential space. I am happy that it is when $$A$$ is separable, but I understand that the opening proposition is believed to be true in the non-separable case also.

Woronowicz originally assumed that the algebra of functions on a compact quantum group was separable in order to deduce the existence of a Haar state. Van Daele removed this assumption.

How can we show that the map $$\Phi(A)\times \Phi(A)\rightarrow \Phi(A)$$ is continuous when $$A$$ is unital, commutative, non-separable?

• @YCor thank you for the tag edit. – JP McCarthy Apr 27 '20 at 12:08
• Yeah, this isn't research level. I think there are people on MSE who would have no trouble answering this. The problem has nothing to do with groups: it's the general fact that any $*$-homomorphism from $C(X)$ to $C(Y)$ (where for you $X = G$ and $Y = G\times G$) arises as composition with a continuous function from $Y$ to $X$. Separability isn't needed. Just use nets instead of sequences. – Nik Weaver Apr 27 '20 at 13:05
• @NikWeaver that answers the question perfectly, thank you. Should I flag the question for closure, or should I delete the question, or should you answer and I accept? Or should I answer community wiki? There is no need at this point to ask the question at MSE now that it is answered. – JP McCarthy Apr 27 '20 at 13:20
• I'd say just leave the question here as is. If it bothers people enough to want to close, they can vote to do so. Just probably try MSE first next time. – Nik Weaver Apr 27 '20 at 14:36

As per Nik Weaver's comment, this is a simple consequence of the fact that for compact Hausdorff spaces $$X$$, $$Y$$, for every unital *-homomorphism $$\pi:C(X)\rightarrow C(Y)$$, there exists a continuous function $$\phi:Y\rightarrow X$$ such that, for $$f\in C(X)$$: $$\pi(f)=f\circ \phi.$$