# Zero divisors in compact quantum groups

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?

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.
• 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 May 4 '19 at 15:17