I am reading A Quantum Groups Primer by Shahn Majid, and I'm having trouble filling in the details for the proof of Proposition 2.5, which states that the Hopf algebra $U_q(b_+)$ is self-dual. $ \newcommand{\la}{\langle} % shortened command for left angle bracket \newcommand{\ra}{\rangle} % shortened command for right angle bracket \newcommand{\q}[1]{[#1]_q} \newcommand{\qfac}[1]{[#1]_q!} \newcommand{\qbinom}[2]{\genfrac{[}{]}{0pt}{}{#1}{#2}_q} $
Here $U_q(b_+)$ where $q$ is some invertible element of some field $k$, is given by the generators $X, g, g^{-1}$, and the relations $g g^{-1} = g^{-1}g = 1$, $g X = q Xg$. The coproduct is given by $\Delta X = X\otimes 1 + g \otimes X $, $\Delta g = g\otimes g$, and $\Delta g^{-1} = g^{-1} \otimes g^{-1}$. The counit $\epsilon$ sends $X$ to $0$, and sends both $g$ and $g^{-1}$ to $1$. The antipode $S$ swaps $g$ and $g^{-1}$, and $S(X) = - g^{-1}X$.
In Proposition 2.5, Majid claims that we can define a pairing $\la g, g \ra = 1$, $\la X, X\ra = 1$, $\la X, g\ra = \la g, X\ra = 0$ between $U_q(b_+)$ and itself. To prove this, one of the things we need to show is that the product and coproduct are adjoint to each other under this pairing.
The book gives an outline and says that you need to first work out the pairing on basis elements of the form $X^m g^n$. I was able to work out that $\la X^m g^n, X^u g^v\ra = \delta_{m,u}q^{nv}$.
Next I tried to show that $\la X^m g^n, (X^a g^b) (X^c g^d)\ra = \la \Delta(X^m g^n), X^a g^b\otimes X^c g^d\ra$. The left hand side becomes $\la X^m g^n, q^{bc}X^{a+c} g^{b+d}\ra = q^{bc} (\delta_{m, a+c} q^{n(b+d)}) = q^{bc+n(b+d)}\delta_{m, a+c}$. Using the fact that $\Delta X^m = \sum_{r=0}^m \qbinom{m}{r} X^{m-r}g^{r}\otimes X^r$, the right hand side becomes $$ \begin{align} \la \Delta X^m(g^n\otimes g^n), X^a g^b\otimes X^c g^d\ra & = \sum_{r=0}^m \qbinom{m}{r} \la X^{m-r}g^{r+n}, X^a g^b\ra \la X^r g^n, X^c g^d\ra \\ & = \sum_{r=0}^m \qbinom{m}{r} (\delta_{m-r, a} q^{(r+n)b})(\delta_{r,c}q^{nd}) \\ & = \qbinom{m}{c} q^{bc+n(b+d)}\delta_{m, a+c}\ . \end{align} $$ Now this has an extra factor of $\qbinom{m}{c}$ compared to the left hand side, and even if I use the delta function to turn it into $\qbinom{a+c}{c}$, or $\qbinom{m}{m-a}$, there doesn't seem to be an obvious reason why it should simplify to $1$, or cancel with some factor in the expression.
Therefore, my question is, can anybody help me spot the error in my attempt? Is my result for $\la X^m g^n, X^u g^v\ra$ incorrect? Or is my simplification of the product-coproduct adjointness axiom incorrect? Or is something else wrong? Thank you in advance!
As an additional bit of context, I am only a beginner in quantum groups, so while I am aware that $U_q(b_+)$ as defined here can be probably be treated in a more general framework (in fact Majid says so himself), I would appreciate an answer that is more concrete and based on this particular object.
$\newcommand{\la}{\langle} % shortened command for left angle bracket \newcommand{\ra}{\rangle} % shortened command for right angle bracket \newcommand{\q}[1]{[#1]_q} \newcommand{\qfac}[1]{[#1]_q!} \newcommand{\qbinom}[2]{\genfrac{[}{]}{0pt}{}{#1}{#2}_q}$ After some careful calculation, I've found the issue. The problem is with the $\la X^m g^n, X^a g^b\ra$ formula. It should come out to $\qfac{m} q^{nb} \delta_{m, a}$. Putting this $\qfac{m}$ factor in, we get an extra factor of $\qfac{m-c}\qfac{c}$ on the right hand side of the counit-unit adjunction relation, and the left hand side will get a factor of $\qfac{m}$. Multiplying out $\qbinom{m}{c}\qfac{m-c}\qfac{c}$ to $\qfac{m}$ on the right, the two sides are now the same.