Pairing higher forms of a Lie group with the universal enveloping algebra

Question

For a (compact) Lie group $G$ we have a pairing between its cotangent space $T^*$ and its Lie algebra $\frak{g}$. What happens for higher forms? Do we have a pairing between $\Lambda(T^*)$, the exterior algebra of $T^*$, and some subalgebra of the universal enveloping algebra $U(\frak{g})$?

Answer 1

(Nothing I'm about to say requires $G$ to be compact.)

$\mathfrak{g}$ can be interpreted as the left-invariant vector fields on $G$, which means it acts on the space of differential forms $\Omega^{\bullet}(G)$ via the Lie derivative. Since this is a Lie algebra action, it extends to an action of $U(\mathfrak{g})$ by differential operators on $\Omega^{\bullet}(G)$. This action restricts to an action on left-invariant differential forms, which can be identified with $\wedge^{\bullet}(\mathfrak{g}^{\ast})$, so we get an induced action of $U(\mathfrak{g})$ on the exterior algebra $\wedge^{\bullet}(\mathfrak{g}^{\ast}).$

This action does not extend the pairing between $\mathfrak{g}$ and $\mathfrak{g}^{\ast}$, which makes no use of the Lie bracket; instead it comes from thinking of $\mathfrak{g}^{\ast}$ as the coadjoint representation (the dual to the adjoint representation), which in particular means that it takes an element $X \in \mathfrak{g}$ and an element $v \in \mathfrak{g}^{\ast}$ and produces an element $Xv \in \mathfrak{g}^{\ast}$, not a scalar. The global version of this is that the Lie derivative of a $1$-form is another $1$-form.