The antipode and the compact quantum group von Neumann algebra

Question

I came across a technical question (at least I believe so) of the following nature. Let $(\mathbb{G},\Delta)$ be a compact quantum group, where $\mathbb{G}$ is a $C^*$-algebra and $\Delta$ is a co-product. Let $S$ be the antipode which is in general unbounded operator on $\mathbb{G}$. Is the operator $S$ affiliated with the group von Neumann algebra $L^{\infty}(\mathbb{G})$? The affiliation is in the sense of Segal. It would be enough to claim that the spectral projections of $S$ are in $L^{\infty}(\mathbb{G})$.

Answer 1

So the answer is straightforward. Let $(\mathbb{G},\Delta)$ be a compact group. Let $S$ be the antipode on $\mathbb{G}$. Piotr Soltan claims [2] that each intertwinning operator $F^{\pi}\in Hom(\pi,S^2\pi)$ is affiliated with the dual group $\widehat{\mathbb{G}}$ in the sense of Woronowicz [1].

It is clear that the square $S^2$ of $S$ is the direct sum of $F^{\pi}\in Hom(\pi, S^2 \pi)$ is $S^2$. Therefore, the operator $S^2$ is affiliated and hence the spectral projections $E_s(S^2)$ lie in $L^{\infty}(\mathbb{G})$. The operators $S^2$ and $S$ have the same spectral projections.

[1] S. Woronowicz. Unbounded elements affiliated with C∗-algebras and noncompact quantum group. Comm. Math. Phys., 1991.

[2] P. Soltan. Introduction to discrete quantum groups. Lecture notes, Bedlewo 2015, http://bcc.impan.pl/15TQG/uploads/pdf/discrete.pdf.