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

- Are $\kappa$ and $\bar{\kappa}$ equivalent or inequivalent?
- 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 $U\mapsto \kappa(U)\subset U\otimes A$).