6
$\begingroup$

I am using this definition:

An algebra of functions on a finite quantum group $\mathbb{G}$ is a finite dimensional $C^\ast$-Hopf algebra $A=:F(\mathbb{G})$.

I have the following (very well known --- folklore --- result)

(Classification Theorem)

Let $A$ be the algebra of functions on a finite quantum group $\mathbb{G}$:

  1. if $A$ is commutative then $\mathbb{G}\cong \Phi(A)$.
  2. if $A$ is cocommutative then $A=F(\mathbb{G})\cong \mathbb{C} \Phi(A)=:F(\widehat{\Phi(A)})$.

Here $\Phi(A)$ is the set of characters on $A$.

I want to reference these results but am struggling somewhat to find good, old, authoritative references. Any help would be appreciated.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ When you speak of a "finite quantum group", do you mean "a finite dimensional Hopf algebra" ? $\endgroup$
    – Konstantinos Kanakoglou
    Commented Dec 19, 2016 at 22:19
  • $\begingroup$ @Konstantinos the definition I am using is in the question. $\endgroup$
    – JP McCarthy
    Commented Dec 20, 2016 at 10:54

3 Answers 3

Reset to default
2
+50
$\begingroup$

I took a quick look into Timmermann's book "An invitation to Quantum Groups".

It refers to:

Saad Baaj; Georges Skandalis Unitaires multiplicatifs et dualité pour les produits croisés de C*-algèbres Annales scientifiques de l'École Normale Supérieure (1993) Volume: 26, Issue: 4, page 425-488 ISSN: 0012-9593

They describe finite dimensional C${}^*$-Hopf algebras in terms of multiplicative unitaries. Theorem 2.2 shows that if you have commutative multiplicative unitaries you get a locally compact group and therefore in the finite case, a finite group.

This gives the first statement. The second statement is equivalent to the first by duality.

*edit*

Such a result is already stated as Theorem 3.3 in

L. I. Vaĭnerman and G. I. Kac, Nonunimodular Ring Groups and Hopf-Von Neumann algebras, Mathematics of the USSR-Sbornik, Volume 23, Number 2

Improve this answer
$\endgroup$
1
  • $\begingroup$ These sound like ideal references thank you. $\endgroup$
    – JP McCarthy
    Commented Dec 28, 2016 at 18:53
1
$\begingroup$

I do not work in this area, but take a look at this survey article in case it gives some directive to your question.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you very much for this. I see on p.18 the article asserts "it is not very hard to see [part 1]" and on p.21 "[part 2] is evident from the duality result below". I could use these but am possibly more interested in an older reference with fuller proofs. $\endgroup$
    – JP McCarthy
    Commented Dec 20, 2016 at 12:23
1
$\begingroup$

If the definition of a finite quantum group, you use, is a pair $(A,\Phi)$ of a finite dimensional $C^*$-algebra $A$, with a comultiplication $\Phi$, such that $(A,\Phi)$ is a Hopf $*$-algebra, then this paper may be helpful.

.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you very much for this. I see on p.16 the paper asserts that "we have the property that a finite quantum group is a group if and only if the underlying algebra $A$ is abelian". I could use this but am possibly more interested in an older reference with fuller proofs. I see no reference to part 2. of my question. $\endgroup$
    – JP McCarthy
    Commented Dec 20, 2016 at 12:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .