This is taken from Timmermann's [Invitation to Quantum Groups and Duality](http://www.ems-ph.org/books/book.php?proj_nr=72).


Hi folks I am struggling a little with a small calculation in the above text. 


I will just get right into it.


**Lemma 3.2.5**
*Let $(A,\Delta)$ be a Hopf $\*$-algebra with a normalised integral $h$, let $\chi\_V$ and $\chi\_W$ be corepresentations on finite-dimensional vector spaces $V$ and $W$, respectively, and let $Q\in \hom(V,W)$. Denote by $X$ and $Y$ the corepresentation operators corresponding to $\chi\_V$ and $\chi\_W$, respectively, and define $R,T\in\hom(V,W)$ by*
$$R:=(I\otimes h)(Y^{-1}(Q\otimes 1)X)\,\,\,\textit{ and }\,\,\, T:=(I\otimes h)(Y(Q\otimes 1)X^{-1}).$$
*Then $R,T\in\hom(\chi\_V,\chi\_W)$. If $Q\in\hom(\chi\_V,\chi\_W)$, then $R=T=Q$.*


**Proof** Since $Y$ and $X$ are corepresentations, the composition


$$\begin{align}
Y^{-1}(R\otimes 1)X&=Y^{-1}(I\otimes h)(Y^{-1}(Q\otimes 1)X)X
\\&=(I\otimes h\otimes I)\left(Y\_{[13]}^{-1}Y\_{[12]}^{-1}(Q\otimes 1\otimes 1)X\_{[12]}X\_{[13]}\right)
\end{align}$$ 



> 
> **Question**: Where does the second equality come from?
> 
> 
> 


**The Rest of the Proof**


This second expression can be rewritten in the form 
$$(I\otimes h\otimes I)((I\otimes \Delta)\left(Y^{-1}\right)(Q\otimes 1\otimes 1)(I\otimes\Delta)(X)).$$


By right-invariance of $h$, this composition is equal to 


$$(I\otimes h)\left(Y^{-1}(Q\otimes 1)X\right)\otimes 1=R\otimes 1.$$


Thus $(R\otimes 1)X=Y(R\otimes 1)$. A similar calculation shows that $(T\otimes 1)X=Y(T\otimes 1)$ (using left-invariance of $h$). A previous result shows that $R$ and $T$ are then elements of $\hom(\chi\_V,\chi\_W)$. 


If $Q\in\hom(\chi\_V,\chi\_W)$, then $Y^{-1}(Q\otimes 1)X=Q\otimes 1=Y(Q\otimes 1)X^{-1}$ by that previous result and so $R=Q=T$.


**Further Details**


*Normalised Integral*: A linear functional $h:A\rightarrow \mathbb{C}$ is a normalised integral if it is both left-and right-invariant:


$$(I\otimes h)\Delta(a)=h(a)1\_{\mathcal{M}(A)}\,\,\,\text{ and }\,\,\,(h\otimes I)\Delta(a)=h(a)1\_{\mathcal{M}(A)},$$


(where $\mathcal{M}(A)$ is the multiplier algebra of the quantum group $A$), as well as normalised


$$h(1\_A)=1.$$


*Corepresentation Operator*: Where $\chi:V\rightarrow V\otimes A$ is a *corepresentation* of $A$ on a vector space $V$, the corepresentation operator associated with $\chi$ is an $X\in\hom(V)\otimes A$ such that
$$X=\sum\_{i,j}e\_{ij}\otimes a\_{ij},$$


where the $a\_{ij}$ are elements of $A$ given by


$$\chi(e\_j)=\sum\_i e\_i\otimes a\_{ij},$$


so that $\chi(v)=X(v\otimes 1\_A)$.


The inverse of $X$ is given by $X^{-1}=(I\otimes S)(X)$ where $S$ is the antipode. 


*Leg Notation*: The leg notation is given by $X\_{[12]}=X\_0\otimes 1\_A$ and $X\_{[13]}=(I\otimes\tau)(X\otimes 1\_A)$ where $\tau:A\otimes B\rightarrow B\otimes A$ is the flip map and we can show that
$$(I\otimes\Delta )X=X\_{[12]}X\_{[13]}.$$


*Intertwiners*: An intertwiner is a map $\varphi:V\rightarrow W$ such that
$$\chi\_W\varphi=(\varphi\otimes 1\_A)\chi\_V.$$
The space of intertwiners from a corepresentation $\chi\_V:V\rightarrow V\otimes A$ to a corepresentation $\chi\_W:W\rightarrow W\otimes A$ is denoted by $\hom(\chi\_V,\chi\_W)$.


**My Thoughts**


...are few and far between. If anything I think it should be
$$Y^{-1}(R\otimes 1)X=Y^{-1}((I\otimes h)(Y^{-1}(Q\otimes 1)X)\otimes 1)X$$


I had previously tried to unravel this lemma but Timmermann called $R$ by $S$ which I confused with the antipode eventually to experience "death by a million linear maps".


Further information available of course on request.



Towards a quantum version of Schur's orthogonality relations
The formula should be


$$\begin{align}
Y^{-1}(R\otimes 1)X&=Y^{-1}(I\otimes h)(Y^{-1}(Q\otimes 1)X)X
\\&=(I\otimes h\otimes I)\left(Y\_{[13]}^{-1}Y\_{[12]}^{-1}(Q\otimes 1\otimes 1)X\_{[12]}X\_{[13]}\right)
\end{align}$$


i.e. the $Q$ in the first term should be an $R$.


The first equality is just the substitution of the definition of $R$ into the expression $Y^{-1}(R\otimes 1)X$. In the second equality the Haar state $h$ gets pulled to the outside of the expression and the resulting triple tensor product expression is re-written in leg notation (since the $Y^{-1}$ and the $X$ on the outside of the previous expression now act on the first and third factor in the tensor product).


I hope this helps.


