For a long time I've been under the impression that the Chevalley complex $\text{CE}(\mathfrak{g})$ of a semisimple (maybe can weaken this) Lie algebra $\mathfrak{g}$ can be extracted from the geometry $\text{BG}$, where $\mathfrak{g}$ is the Lie algebra of $\text{G}$. Does anyone know a statement like this with a proof?




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To give a bit more detail, the cotangent complex of $\text{BG}$ is a complex of quasicoherent sheaves on $\text{BG}$ (= $\text{G}$ equivariant complex on a point), explicitly it is
$$\mathbf{L}\_{\text{BG}}\ =\ \mathfrak{g}^\*[-1],$$
concentrated in degree $1$, with the coadjoint action of $\text{G}$. Thus the Chevalley-Eilenberg complex can be thought of as $\text{Sym}\mathbf{L}\_{\text{BG}}=\text{Sym}(\mathfrak{g}^\*[-1])$, along with a mysterious differential (which is the interesting part).


One guess might be that $d\_{CE}$ is actually the de Rham differential, but this is wrong because the correct analogue of the de Rham complex for stacks is
$$\Omega\_{dR}\ = \ \text{Sym}(\mathbf{L}\_{\text{BG}}[-1])\ =\ \text{Sym}(\mathfrak{g}^\*[-2]).$$
However, this is the sort of thing I was hoping $\text{CE}(\mathfrak{g})$ could be interpreted as: some general construction (like the de Rham complex) you can apply to arbitrary stacks, and when you apply to $\text{BG}$ you get out $\text{CE}(\mathfrak{g})$.



Chevalley complex and $\text{BG}$
It’s the ring of functions on the formal completion at the base point:
CE is self ext of the trivial rep of the Lie algebra, equivalently of the formal group of G, ie global functions on B of the formal completion. This doesn’t seem to be a functor of BG as a stack, but only as a pointed stack..


