# Invariance of Finite Dimensional Woronowicz $\mathrm{C}^*$-ideals under the Antipode

Let $$(A,\Delta)=:F(G)$$ be a finite dimensional $$\mathrm{C}^*$$-Hopf algebra, and so the algebra of functions on a quantum group $$G$$.

Let $$J$$ be a closed ideal in $$F(G)$$ and $$\pi:F(G)\rightarrow F(G)/J$$ the quotient map. Suppose that $$J$$ is such that $$\Delta(J)\subset \ker(\pi\otimes\pi).$$ Such ideals are called Woronowicz $$\mathrm{C}^*$$-ideals.

When $$A$$ is commutative, so that $$G$$ is a finite group, Woronowicz $$\mathrm{C}^*$$-ideals correspond to functions that vanish on a distinguished subgroup $$H$$, and the quotient map, via $$F(G)/J\cong F(H)$$, $$\pi_H$$ given by: $$\pi_H\left(\sum_{t\in G}a_t\delta_t\right)=\sum_{t\in H}a_t\delta_t,$$ and $$J=\{f\in F(G)\,:\,f(H)=\{0\}\}$$. In this commutative case, if $$f\in J$$, then so is $$S(f)$$. If $$J$$ is proper, then $$\varepsilon(f)=0$$.

In Remark 2.10 of this paper (p.667), Wang says that for noncommutative $$F(G)$$ we still have $$S(J)\subset J$$ and $$\varepsilon(J)=\{0\}$$ for $$J$$ a Woronwicz $$\mathrm{C}^*$$-ideal ($$J$$ must be proper for this second condition to hold).

I can show using the antipodal property that if $$S(J)\subset J$$ then $$\varepsilon(J)=\{0\}$$... but $$S(J)\subset J$$ is eluding me.

Question: How does $$S(J)\subset J$$ follow?

Some Efforts:

We have that $$F(G)\otimes \ker \pi+\ker\pi\otimes F(G)\subset \ker(\pi\otimes \pi),$$ and I believe that due to finite dimensionality

$$\ker(\pi\otimes\pi)=F(G)\otimes \ker \pi+\ker\pi\otimes F(G).$$

This means that if $$j\in J$$, and $$\Delta(j)=\sum_i j_{(1),i}\otimes j_{(2),i},$$ that for each $$i$$, either $$j_{(1),i}$$ or $$j_{(2),i}\in J$$.

If we write $$j\in J$$ in terms of the matrix elements of irreducible unitary representations: $$j=\sum_\underset{\alpha\in\text{Irr}(G)}{i,j=1}^{d_\alpha}a_{ij}^{\alpha}\rho_{ij}^\alpha,$$ that $$j^*=\sum_\underset{\alpha\in\text{Irr}(G)}{i,j=1}^{d_\alpha}\overline{a_{ij}^{\alpha}}S(\rho_{ji}^\alpha)\in J,$$ as $$J$$ is a $$\mathrm{C}^*$$-algebra and $$S(\rho_{ij}^\alpha)=(\rho_{ji}^\alpha)^*$$.

Any help would be appreciated... I just fear there is something simple here that I am missing.

If you continue reading Wang's paper (project Euclid link), it becomes a little clearer I think. His Theorem 2.11 says that if $$J$$ is a Woronowicz $$C^*$$-ideal is a Woronowicz $$C^∗$$-algebra $$A$$, then $$A/J$$ has canonically the structure of a Woronowicz $$C^∗$$-algebra. The proof only verifies this for the compact matrix quantum group case (here using modern language) but the proof also works for the compact quantum group case (verify the "cancellation laws", that is, the axioms in definition 2.1'' in Wang's paper).
So, we reduce the problem to having two compact quantum group algebras $$A,B$$ and a (surjective) $$*$$-homomorphism between them $$\theta:A\rightarrow B$$ which intertwines the coproducts. In generality (e.g. this still holds for locally compact quantum groups, in an appropriate sense, see section 12 of Kustermans's paper (arXiv)) this means that $$\theta S \subseteq S \theta$$.
For you, $$A$$ is finite dimensional, and so $$B$$ is, and so $$S$$ is everywhere defined, and so $$\theta S = S\theta$$. In particular, if $$a\in J$$ then $$\theta(a)=0$$ so $$\theta S(a)=0$$ so $$S(a)\in J$$. Further, as $$A$$ and $$B$$ are actually Hopf algebras, I'm sure one can give a Hopf-algebraic proof; I don't know how to do this quickly, so I'll again refer to some references. Van Daele's paper (AMS Pdf) is good to read anyway, and we find Lemma 5.5 which says:
1. $$a\otimes S^{-1}b = \sum_i (a_i\otimes 1)\Delta(b_i)$$
2. $$\Delta(a)(1\otimes b) = \sum_i \Delta(a_i)(b_i\otimes 1)$$.
Notice that the Woronowicz cancellation axioms fit into this idea, showing in particular that we can write any element as a sum as in (1) (this is always true in a Hopf algebra). Thus given $$a\otimes S^{-1}b = \sum_i (a_i\otimes 1)\Delta(b_i)$$ we have that (2) holds, so apply $$\theta\otimes\theta$$ to see that $$\Delta(\theta(a))(1\otimes\theta(b)) = \sum \Delta(\theta(a_i))(\theta(b_i)\otimes 1)$$ giving (2) in the algebra $$B$$, so also (1) holds, then use intertwing again to see $$\theta(a)\otimes S^{-1}\theta(b) = (\theta\otimes\theta)(\sum (a_i\otimes 1)\Delta(b_i))$$ but this equals $$\theta(a)\otimes \theta S^{-1}(b)$$. Hence $$S^{-1}\theta = \theta S^{-1}$$.