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?

  • $\begingroup$ @YCor thank you for the tag edit. $\endgroup$ – JP McCarthy Apr 27 '20 at 12:08
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    $\begingroup$ 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. $\endgroup$ – Nik Weaver Apr 27 '20 at 13:05
  • $\begingroup$ @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. $\endgroup$ – JP McCarthy Apr 27 '20 at 13:20
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    $\begingroup$ 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. $\endgroup$ – 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 $f\in C(X)$: $$\pi(f)=f\circ \phi.$$

A reference for this fact is given by Nik's text, Mathematical Quantisation.


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