For an infinite dimensional Hopf algebra $H$, a non-degenerate dually pairing Hopf algebra $H'$, and a choice of basis $e\_i$ of $H$, is the dual basis $e^i$ (defined of course by $e^i(e\_j) = \delta\_{ij}$) contained in $H'$?


I am interested in the specific case of $SU\_q(N)$ and the dually paired Hopf algebra $\mathfrak{sl}\_N$.



Dual of a Basis for a Hopf Algebra Conatined in all Dually Paired Hopf Algebras
No. Let $k$ be a field of characteristic $0$. Consider the symmetric algebra on one generator $k[x]$, with comultiplication $x \mapsto x\otimes 1 + 1\otimes x$. It has a Hopf pairing with itself, given by $\langle x^m,x^n\rangle = n! \hspace{.5ex} \delta\_{m=n}$. Then consider the bases of $k[x]$ given by expansion around $1$, i.e. $e\_n = (x-1)^n$. The dual basis, if it exists, includes $e^0$ such that $\langle e^0, (x-1)^n\rangle = \delta\_{0,n}$. So suppose that $e^0 = \sum a\_n x^n$; then:
$$ \begin{aligned}
 1 & = a\_0 \\
 0 & = a\_1 - a\_0 \\
 0 & = a\_2 - 2a\_1 + a\_0 \\ 
 \dots & \phantom= \dots \\
 0 & = \sum\_{k=0}^n (-1)^{n-k} \hspace{.5ex} n!\hspace{.5ex} a\_n \hspace{.5ex} \binom{n}{k} \\
 \dots & \phantom= \dots \\
\end{aligned} $$
The solution is that all $a\_n = 1/n!$, i.e. $e^0 = \sum x^n/n! = \exp(x)$. But this is not a polynomial.


In the case you ask about, you will similarly have some bases with dual bases, and some without.


