Let $A$ be a Hopf algebra dually paired with a quasi-triangular Hopf algebra $B$. If $x$ is some fixed element of $A$, then we can define a linear map 
$$
P\_x: A \to \mathbb{C}
$$
by setting 
$$
P\_x:a \mapsto \langle R,x \otimes a \rangle.
$$


Let us take the case $A = SL\_q(2)$, $B = U\_q({\mathfrak sl}\_2)$, and let $R$ be the standard universal $R$-matrix for $U\_q({\mathfrak sl}\_2)$, for which
$$
\langle R, u^i\_m \otimes u^j\_n \rangle = R^{ij}\_{mn} = q^{-\frac{1}{2}}.(q^{\delta\_{ij}}\delta\_{im}\delta\_{jn} + (q-q^{-1})\theta (i-j)\delta\_{in}\delta\_{jm}),
$$
where $\theta$ is the Heaviside symbol. If we take $x=u^k\_l$, then 
$$
P\_{u^k\_l}(a) = \langle R, u^k\_l \otimes a \rangle.
$$
Now since $ab = qba$, we should have 
$$
P\_{u^k\_l}(u^1\_1u^1\_2) = q P\_{u^k\_l}(u^1\_2u^1\_1), \qquad \qquad \text{ for all } \quad k,l = 1,2.
$$
However,
$$
P\_{u^2\_1}(u^1\_1u^1\_2) = \langle R, u^2\_1 \otimes u^1\_1u^1\_2 \rangle = \sum\_{z=1}^2 \langle R,u^2\_z \otimes u^1\_1 \rangle \langle R, u^z\_1 \otimes u^1\_2 \rangle = \sum\_{z=1}^2 R^{21}\_{z1}R^{z1}\_{12}.
$$
From the formula for $R^{ij}\_{mn}$, we get that
$$
P(u^1\_1u^1\_2) = \sum\_{z=1}^2 R^{21}\_{z1}R^{z1}\_{12} = R^{21}\_{11}R^{11}\_{12} + R^{21}\_{21}R^{21}\_{12} = 0.0 + q^{-\frac{1}{2}}.1.q^{-\frac{1}{2}}.(q-q^{-1}) = q^{-1}(q-q^{-1}).
$$


On the other hand, we have
$$
qP\_{u^2\_1}(u^1\_2u^1\_1) = \langle R,u^2\_1 \otimes u^1\_2u^1\_1 \rangle = \sum\_{z=1}^2 \langle R,u^2\_z \otimes u^1\_2 \rangle \langle R,u^z\_1 \otimes u^1\_1 \rangle =\sum\_{z=1}^2R^{21}\_{z2}R^{z1}\_{11}.
$$
From the formula for $R^{ij}\_{mn}$, we now get that
$$
qP\_{u^2\_1}(u^1\_2u^1\_1) = q\sum\_{z=1}^2R^{21}\_{z2}R^{z1}\_{11} = qR^{21}\_{12}R^{11}\_{11} + qR^{21}\_{22}R^{21}\_{11} = q.q^{-\frac{1}{2}}.(q-q^{-1}).q^{-\frac{1}{2}}.q + q.0.0 = q(q-q^{-1}).
$$


Thus, the two results are not equal, but instead differ by a factor of $q^2$. A similar problem arises for the action of $P\_{u^2\_1}$ on $bd - qdb$. We get 
$$
P\_{u^2\_1}(u^1\_2u^2\_2) = q^{-1}(q-q^{-1}),
$$
whereas
$$
qP\_{u^2\_1}(u^2\_2u^1\_2) = q(q-q^{-1}).
$$


I've checked and rechecked everything very carefully but can't seem to spot my error. Can anyone see what is going wrong here?



Map constructed from the coquasitriangular structure of SLq(2) which appears not to respect the standard commutation relations
Dear John,


I tried to follow your computation until the first place where I couldn't understand a step. This comes at:



> 
> However,
>  $$
> P\_{u^2\_1}(u^1\_1u^1\_2) = \langle R, u^2\_1 \otimes u^1\_1u^1\_2 \rangle = \sum\_z \langle R,u^2\_z \otimes u^1\_1 \rangle \langle R, u^z\_1 \otimes u^1\_2 \rangle,
> $$
> 
> 
> 


Rather than the RHS, I would expect
$$<(\operatorname{id}\otimes \Delta)(R), u^2\_1 \otimes u^1\_1 \otimes u^1\_2> =<R\_{13}R\_{12}, u^2\_1\otimes u^1\_1\otimes u^1\_2>
=\sum\_z<R,u^2\_z\otimes u^1\_2><R,u^z\_1\otimes u^1\_1>
$$


which seems different than what you wrote. It seems you have used the opposite comultiplication in your computations so that where I wrote $R\_{13}R\_{12}$ above, you instead had $R\_{12}R\_{13}$. I hope this helps. I am aware that pairing of Hopf algebras sometimes requires matching multiplication of $H$ with opposite co-multiplication of $H^\*$. However, you seem to be working from Klymik and Schmudgen's text, which doesn't not use opposite co-product in the definition of dual pairing of Hopf algebras.


I haven't checked the details to see if the above resolves your issue. Perhaps this is still not your source of confusion, but it confused me when I first read it in your post.


Looking again at what you wrote, this means that the two computations you did for $P\_c(ab)$ and $P\_c(ba)$ are thus switched, so that you are multiplying $P\_c(ab)$ by $q$ instead of $P\_c(ba)$, as you thought. Multiplying instead of dividing by $q$ gives the discrepancy of $q^2$


thanks,
-david


