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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 $\|S\| = 2?$

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  • $\begingroup$ 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. $\endgroup$ Dec 9, 2021 at 20:34

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$S$ 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.

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  • $\begingroup$ So either $S$ is an isometry in the Kac-case, or $S$ is unbounded. A bit funny behaviour! $\endgroup$
    – Andromeda
    Dec 9, 2021 at 21:03

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