# In Algebraic Compact Quantum Groups, is an Irreducible Corepresentation equivalent to its Conjugate?

A quantum group $A$ here is an algebraic compact quantum group --- a Hopf*-algebra with a Haar State. Here $\hat{A}$ is the set of linear functionals $\{\mathcal{F(a)}:a\in A\}$ of the form $\mathcal{F}(a)(b)=h(ba)$.

We have for $f\in A$ a result by Van Daele that

$$f=\hat{\psi}(\hat{S}(\cdot)\mathcal{F}(f)).\qquad(*)$$

Here $\mathcal{F}:A\rightarrow \hat{A}$ is, as before, defined by $\mathcal{F}(f)(a)=h(af)$. The antipode on $\hat{A}$ is given by $\hat{S}:\hat{A}\rightarrow \hat{A}$ and defined by

$$\hat{S}(\mathcal{F}(a))b=\mathcal{F}(a)S(b)=h(S(b)a),$$ where $S:A\rightarrow A$ is the antipode of $A$. The Haar state on $\hat{A}$ is $\hat{\psi}$ and is defined by $$\hat{\psi}(\mathcal{F}(a))=\varepsilon(a),$$ where $\varepsilon\in A'$ is the counit of $A$.

In $(\star)$, $f$ is taken to be in the second hat-dual which is isomorphic to $A$. Hence $(\star)$ is giving us

$$f(\mathcal{F}(x))=\mathcal{F}(x)(f)=\hat{\psi}(\hat{S}(\mathcal{F}(x))\mathcal{F}(f)).$$

The multiplication is that of $A'$ namely the convolution defined by $$\nu\star\mu(a)=(\nu\otimes\mu)\Delta(a).$$

I have been trying to look at $(\star)$ by expanding

$$\mathcal{F}(x)=\sum_i\beta_i\mathcal{F}(\rho_i),$$

where the sum is over the matrix elements, $\rho_i$, of the inequivalent, unitary, one dimensional corepresentations... actually I want $A$ to be cocommutative so that these guys span $A$.

I think $\mathcal{F}(\rho_i)=\delta^{\rho_i^*}$ and when I run the numbers through I end up wanting to calculate
$$\hat{S}(\mathcal{F}(x))\mathcal{F}(\rho_i)\rho_k.$$ I believe these are equal to $$\delta_{i,k}h(\rho_kf).$$

Now what I want to do is expand $f=\sum_i\alpha_i\rho_i$ and using the result $h(\rho_i^*\rho_j)=\delta_{i,j}$, get a nice answer. However I have $\rho_kf$ not $\rho_k^*f$...

OK, that is the context, now onto my question.

Now, following Timmermann's book consider the following.

Let $\kappa:V\rightarrow V\otimes A$ be a corepresentation of $A$ on a vector space $V$. The matrix elements of $\kappa$, $\{\rho_{ij}\}$ are given by $$\kappa(e_j)=\sum_i e_i\otimes\rho_{ij},$$ where $\{e_i\}$ is a basis of $V$. Denote by $\bar{V}$ the conjugate vector space of $V$ and by $v\mapsto \bar{v}$ is the canonical conjugate-linear isomorphism. Then the map $$\overline{\kappa}:\bar{V}\rightarrow \bar{V}\otimes A,\,\,\bar{e_j}\mapsto \sum_i\bar{e_i}\otimes\rho_{ij}^*,$$ is a corepresentation again, called the conjugate of $\kappa$. More details below but now my question

1. Are $\kappa$ and $\bar{\kappa}$ equivalent or inequivalent?
2. If inequivalent in general, is it the case that they are equivalent for one dimensional corepresentations?

Corepresentations $\kappa_1:V_1\rightarrow V_1\otimes A$ and $\kappa_2:V_2\rightarrow V_2\otimes$ are equivalent if there is an invertible linear map $T:V_1\rightarrow V_2$ such that

$$\kappa_2\otimes T=(T\otimes I_A)\kappa_1.$$ Maps that satisfy this are called intertwiners.

We have

$$\bar{\kappa}C=(C\otimes\text{Inv})\kappa,$$ where $C:V\rightarrow \bar{V}$ is the conjugate-linear isomorphism and $\text{Inv}(a)=a^*$.

More details by Timmermann:

• If $T$ intertwines $\kappa_1$ and $\kappa_2$, then $\bar{T}$ intertwines $\overline{\kappa_1}$ and $\overline{\kappa_2}$ where $\bar{T}\bar{v}=\overline{Tv}$.
• $\bar{\kappa}$ is irreducible if and only if $\kappa$ is (no invariant subspaces $U\subset V$ such that $U\mapsto \kappa(U)\subset U\otimes A$).
• Perhaps the downvoter might have explained why the downvote. I have asked some questions about quantum groups on math.stackexchange but have always had to put it to the MO for lack of answers. – JP McCarthy Nov 6 '14 at 8:30

Before thinking about complicated quantum groups, it's best to first understand the simplest situation. So let's just look at when G is a finite group and A is the Hopf algebra of functionals on G. In the finite dimensional setting A-comodules is the same thing as $A^*$-modules, and in this case $A^*$ is just the group ring $\mathbb{C}[G]$. Modules over the group ring are exactly the same thing as group representations. This is the simplest example that you should always have in mind when thinking about quantum groups.
• I have the matrix elements for $A=F(\mathbb{Z}_3)$ are $\rho_1=1_A$, $\rho_2=(1\,\,\,e^{2\pi i/3}\,\,\,e^{4\pi i/3})^T$ and $\rho_3=(1\,\,\,e^{4\pi i/3}\,\,\,e^{2\pi i/3})^T$. But $\kappa_2$ is not equivalent to $\kappa_3$. Is not $\overline{\kappa_1}=\kappa_2$? Is this (en.wikipedia.org/wiki/Frobenius%E2%80%93Schur_indicator) worth looking at? – JP McCarthy Nov 5 '14 at 20:18
• Even easier, you can take consider the dual of Noah's finite group example. Then the representation category is equivalent to the symmetric category $\mathrm{Vec}_G$ of $G$-graded vector spaces in which every simple object is self-dual iff $G$ is an elementary abelian 2-group. – Marcel Bischoff Dec 28 '16 at 19:00