0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @YCor thank you for the tag edit. $\endgroup$ – JP McCarthy Apr 27 at 12:08
  • 2
    $\begingroup$ 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 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 at 13:20
  • 1
    $\begingroup$ 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 at 14:36
2
$\begingroup$

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.

$\endgroup$

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.