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Let $\mathcal{G}$ be compact quantum group in the sense of S. L. Woronowicz. As is well-known, every compact quantum group contains a dense Hopf algebra, called the polynomial Hopf algebra Pol$(\mathcal{G})$. For example, consider the famous $C(SU_q(2))$, the $q$-deformation of $SU(2)$, with generators $\alpha$ and $\beta$ :

https://en.wikipedia.org/wiki/Compact_quantum_group

The dense Hopf algebra is now the polynomial $*$-algebra generated by $\alpha$ and $\beta$. A well-known fact about Pol($SU_q(2)$) is that it has no zero-divisors, that is, it is a domain. What is a good example of a compact quantum group $\mathcal{G}$ such that Pol$(\mathcal{G})$ has zero divisors? On the other hand, is there any abstract characterization of the compact quantum groups such that the polynomial Hopf algebra is a domain?

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Concerning the first question, consider the quantum permutation group $S_N^+$. Its polynomial algebra is generated by an $N\times N$ matrix $u=(u_{ij})$ of self-adjoint projections which sum up to $1$ on each row and column. As a consequence, it has lots of zero divisors.

As for a general criterion, note that for any discrete group $\Gamma$, there is a "dual" compact quantum group with polynomial algebra $\mathbb{C}[\Gamma]$. Characterizing when this is a domain is an open problem. For instance, Kaplansky's zero divisor conjecture asserts that if $\Gamma$ is torsion-free then $k[\Gamma]$ is a domain for any field $k$ and even for $k = \mathbb{C}$ this is, as far as I know, still open in full generality.

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  • $\begingroup$ It is worth pointing out (I'm sure you know this, but the phrasing above is potentially ambiguous) that the "open problem" is about characterising when the group ring is a domain. If one merely wants to get examples of ${\bf C}[\Gamma]$ with non-trivial idempotents then every non-trivial torsion element of $\Gamma$ gives rise to such an idempotent $\endgroup$
    – Yemon Choi
    May 4 '19 at 15:17
  • $\begingroup$ Thanks for your comment. I edited the answer to make remove that ambiguity. $\endgroup$
    – user131654
    May 4 '19 at 20:23

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