# Norm antipode on a Kac-type compact quantum group

Let $$\mathbb{G}$$ be a $$C^*$$-algebraic compact quantum group. Consider the associated dense Hopf$$^*$$-subalgebra $$\mathcal{O}(\mathbb{G})$$ and let $$S: \mathcal{O}(\mathbb{G})\to \mathcal{O}(\mathbb{G})$$ denote its antipode. In the book "Compact quantum groups and their representation categories", it is shown that $$S$$ is unbounded when $$\mathbb{G}$$ is not of Kac-type. The converse is also true, namely if $$\mathbb{G}$$ is of Kac-type, then $$S$$ is bounded. To see this, note that the assumption implies that $$S$$ is $$*$$-preserving, and thus $$S$$ is a positive map because $$S(x^*x) = S(x)S(x^*) = S(x)S(x)^*\ge 0$$ for any $$x \in \mathcal{O}(\mathbb{G})$$. Note that $$\mathcal{O}(\mathbb{G})$$ is an operator system, and since a positive unital map on an operator system is necessarily bounded (See Paulsen's book "Completely bounded maps and operator algebras", proposition 2.1), we conclude that $$S$$ is bounded. Moreover, it also follows that $$\|S\| \le 2\|S(1)\| = 2.$$

Question: Can we say something more about the norm $$\|S\|$$? Is it possible that $$\|S\| = 2?$$

• Is the antipode a $\ast$-homomorphism to the opposite algebra? Does this imply it has norm one? That is in a completion of the Hopf* algebra but we can come back down to your question. Dec 9, 2021 at 20:34

$$S$$ is an anti-$$*$$-homomorphism, and extends by continuity to $$A = C(\mathbb G)$$ (the closure of $$\mathcal O(\mathbb G)$$ acting on the GNS space for the Haar state). Let $$A^{\operatorname{op}}$$ be the opposite $$C^*$$-algebra to $$A$$. Then we can consider $$S$$ as a map $$A\rightarrow A^{\operatorname{op}}$$, which is now a $$*$$-homomorphism, and hence contractive.
The relevant result in the book of Neshveyev and Tuset is Proposition 1.7.9. This also shows that (equivalently) $$S^2=\operatorname{id}$$. So in fact $$S$$ is an isometry.
• So either $S$ is an isometry in the Kac-case, or $S$ is unbounded. A bit funny behaviour! Dec 9, 2021 at 21:03