Is $\mathbb{C}[G]$ dual to $U(\mathfrak{g})$? Let $G$ be a Lie group and $\mathfrak{g}$ its Lie algebra. Is $\mathbb{C}[G]$ dual to $U(\mathfrak{g})$? In the case of $G = SL_2$, we have $\mathbb{C}[SL_2] = \langle a,b,c,d\rangle / (ad-bc-1 )$ and  $U(\mathfrak{sl}_2) = \langle E, F, H \rangle$. Is it possible to define a pairing between $\mathbb{C}[SL_2]$ and $U(\mathfrak{sl}_2)$ explicitly? For example, define $(a, E) = $, $(a, F)=$ and so on. Thank you very much.
 A: A long time passed, but I'll try to give an answer anyway, mimicking Chari, Pressley, A Guide to Quantum Groups, Example 4.1.17.
First of all, from the example you gave it seems to me that you are considering polynomial functions from $G$ to $\mathbb{C}$. If this is what you meant, Tobias already mentioned that such a duality exists and you may find it in the above mentioned reference.
Alternatively, you may be interested in smooth functions from $G$ to $\mathbb{C}$. In such a case, one may proceed almost in the same way.
Denote by $A$ the $\mathbb{C}$-algebra of smooth functions from $G$ to $\mathbb{C}$, i.e. $A=\mathcal{C}^{\infty}(G)$. We have that $A$ is an augmented $\mathbb{C}$-algebra with augmentation $\varepsilon:A\to \mathbb{C}, f\mapsto f(e)$ ($e$ being the neutral element of $G$). I assume $\mathfrak{g}=\mathrm{Der}(A,\mathbb{C}_{\varepsilon})$, i.e.
$$\mathfrak{g}=\left\{\delta\in A^*\mid\delta(fg)=\delta(f)g(e)+f(e)\delta(g)\right\}.$$
First of all, let us see that $\mathfrak{g}$ is isomorphic to the vector space $\mathcal{L}(G)$ of left invariant derivations in
$$\mathrm{Der}_\Bbbk(A,A)=\left\{X\in\mathrm{Hom}_\Bbbk(A,A)\mid X(fg)=X(f)g+fX(g)\quad\mathrm{and}\quad X(1)=0\right\}.$$
By a left invariant derivation I mean the following: $G$ acts on $A$ by left translation, i.e. for every $x\in G$ and $f\in A$, $\lambda_xf:G\mapsto\mathbb{C},y\mapsto f(x^{-1}y)$. With this convention, $X\in \mathrm{Der}_\Bbbk(A,A)$ is left invariant if $\lambda_x\circ X = X\circ \lambda_x$ for every $x\in G$.
One way is easy: to every $X\in\mathcal{L}(G)$ one associate $X_e:A\to\mathbb{C},f\mapsto (X(f))(e)$. For the other way around, to every $\delta\in\mathfrak{g}$ we associate
$$*\delta:A\to A,f\mapsto \left[f*\delta:x\mapsto\delta\left(\lambda_{x^{-1}}f\right)\right].$$
This construction comes from Humphreys, Linear Algebraic Groups, §9.2.
This allows us to work with $\mathcal{L}(G)$ instead of $\mathfrak{g}$, thus I will identify them. To see why this is helpful, notice that now $\mathfrak{g}$ acts as a Lie algebra of derivations on $A$, which extends to an action of $U(\mathfrak{g})$ on $A$ by the universal property of the universal enveloping algebra.
The duality is now given simply by
$$U(\mathfrak{g})\times \mathcal{C}^{\infty}(G)\to \mathbb{C}, (u,f)\mapsto (u\cdot f)(e).$$
I hope this answers your question or at least it helps you to find the answer you were looking for.
