I have recently begun to study quasi-triangular structures and have come across a problem I can't resolve. Let ${\cal U}\_q({\mathfrak sl}\_N)$ denote the quantised enveloping algebra of ${\mathfrak sl}\_N$, and let $R$ be a universal R-matrix for ${\cal U}\_q({\mathfrak sl}\_N)$ . If we denote the usual dual pairing of ${\cal U}\_q({\mathfrak sl\_N})$ with 
$SU\_q(N)$ by $\langle \cdot , \cdot \rangle$, then it is well known that $$R^{ir}\_{js} = \langle R, u^i\_j \otimes u^r\_s \rangle = q^{\delta\_{ir}}\delta\_{ij}\delta\_{rs} + (q-q^{-1})\theta (i-r)\delta\_{is}\delta\_{jr}.$$ A natural question to ask is whether such a formula exists for $(R^{-1})\_{js}^{ir}=\langle R^{-1}, u^i\_j \otimes u^r\_s \rangle$. An obvious guess would be to take the inverse of the matrix $[R\_{js}^{ir}]\_{i,r,j,s}$. 


That is, to guess that $$[(R^{-1})\_{js}^{ir}] \_{i,r,j,s}$$ 


is equal to $$([R\_{js}^{ir}]\_{i,r,j,s})^{-1}.$$ This guess is confirmed by the fact that


$$
\delta\_{ij}\delta\_{rs} = \langle R R^{-1},u^i\_j \otimes u^r\_s\rangle = \sum\_{k,l} \langle R,u^i\_k \otimes u^r\_l \rangle \langle R^{-1},u^k\_j \otimes u^l\_s \rangle= \sum\_{k,l} R\_{kl}^{ir} (R^{-1})^{kl}\_{js}.
$$
The matrix inverse is easy to calculate and gives us the formula
$$\langle R, u^i\_j \otimes u^r\_s \rangle = q^{-\delta\_{ir}}\delta\_{ij}\delta\_{rs} - (q-q^{-1})\theta (i-r)\delta\_{is}\delta\_{jr}.
$$


Now let's try and test this result: As is very well known $(S \otimes $id)$R = R^{-1}$. Thus, $$\langle R^{-1}, u^i\_j \otimes u^r\_s \rangle = \langle R, S(u^i\_j) \otimes u^r\_s \rangle.$$ In the case of $N=2$, $i=j=r=s=1$, we have $S(u^1\_1) = u^2\_2$, and so,
$$
(R^{-1})^{11}\_{11}=\langle R^{-1},u^1\_1 \otimes u^1\_1 \rangle = \langle R,S(u^1\_1) \otimes u^1\_1 \rangle = \langle R,u^2\_2 \otimes u^1\_1 \rangle = R^{21}\_{21} = 1.
$$ 


But the formula above gives us that $(R^{-1})^{11}\_{11}$ is equal to


$$ q^{-\delta\_{11}}\delta\_{11}\delta\_{11} - (q-q^{-1})\theta (1-1)\delta\_{11}\delta\_{11} = q^{-1},$$ 


Moreover, performing the analogous calculations for the other possible values of $i,j,r,s$ we get two different matrices: Using the general formula we get
$$
\left(
\begin{array} {cccc}
 q^{-1} & 0 & 0 & 0 \\\
 0 & 1 & 0 & 0 \\\
 0 & -\lambda & 1 & 0 \\\
 0 & 0 & 0 & q^{-1} \\\
\end{array}
\right);
$$
and using the equality $(S \otimes$id$)(R) = R^{-1}$ we get
$$
\left(
\begin{array} {cccc}
 1 & 0 & 0 & -\lambda \\\
 0 & q & 0 & 0 \\\
 0 & 0 & q & 0 \\\
 0 & 0 & 0 & 1 \\\
\end{array}
\right),
$$
where $\lambda = (q-q^{-1})$.


I can't see why these two results don't agree and am guessing I have made some basic beginner's mistake. Can someone please tell me where I have gone wrong? It's driving me a little crazy!strong text



The Inverse of a Universal R-Matrix for Quantized Universal Enveloping Algebra of sl2 and the Dual Pairing with SUq(2)
I think the only issue here is a harmless error in your calculation and that there is a normalization of the $R$-matrix for $U\_q(sl\_N)$ by a factor of $q^{1/2}$ which you have omitted (See 8.4.2 of Klymik Schmudgen).


First, I get $(R^{-1})^{21}\_{12} = -q^{-1} R^{21}\_{12}$,
because $\langle(S\otimes id)(R),a^2\_1\otimes a^1\_2\rangle = \langle R,S(a^2\_1)\otimes a^1\_2 \rangle = -q^{-1} \langle R,a^2\_1\otimes a^1\_2\rangle$, using that $S(b)=-q^{-1} b$ from Proposition 4.1.2.3 of Klymik Schmudgen. So the actual matrix you should get should just be $q^{-1}$ times what you had expected to get.


Now the factor of q^{-1} here is because you had multiplied the actual universal R matrix by $q^{1/2}$ and so $(\lambda A)^{-1} = \lambda^{-1} A^{-1}$, so there's a factor of $\lambda^2$ as a discrepancy.


I hope this helps! 


