TITLE: For G a Lie group, can I make sense of G/G as a derived manifold in a nice way?
QUESTION [9 upvotes]: The functor sending a smooth manifold $M$ to its de Rham algebra $\Omega^{\bullet}(M)$ does not send quotients by actions of Lie groups to invariant subalgebras. The example I have in mind is a connected Lie group $G$ acting on itself by left multiplication. The quotient is a point, which has de Rham algebra $\mathbb{R}$ concentrated in degree $0$, but the $G$-invariant subalgebra of $\Omega^{\bullet}(G)$ is much more interesting: it's the Chevalley-Eilenberg algebra $\Lambda^{\bullet}(\mathfrak{g}^{\ast})$ of the Lie algebra of $G$. 
If I wanted to fix this, it seems like I ought to derive something, maybe the operation of taking quotients. Is there a nice category of derived manifolds in which the derived quotient $G/G$ has de Rham algebra the Chevalley-Eilenberg algebra (and vector fields given by $\mathfrak{g}$, and differential operators given by $U(\mathfrak{g})$, and so forth)? Or should I be thinking in terms of Lie algebroids, or what? 
(I think I know what the category should be based on a talk I attended recently, but I know very little about it and would appreciate references. It should be the opposite of a category whose objects are something like commutative dg-algebras with degree $0$ part a smooth algebra.) 
Edit: If what I said about quotients above is silly, let me ask a slightly different question: what I really want to know is what kind of smooth object, in some category of generalized smooth spaces, has de Rham algebra the Chevalley-Eilenberg algebra, vector fields $\mathfrak{g}$, differential operators $U(\mathfrak{g})$, etc.

REPLY [5 votes]: The standard answer to: what has de Rham algebra the Chevalley-Eilenberg algebra, vector fields g, differential operators U(g), etc. is "the point with the Lie algebroid structure given by the Lie algebra $\mathfrak{g}$."  All the notions you listed make sense for Lie algebroids, and you've listed the special cases for the one-point guy and the tangent Lie algebroid. (Funnily enough, I think usually I have to give this speech to explain in what sense differential operators are a universal enveloping algebra.)
I think Sam's answer is that in the modern world, we can think of Lie algebroids in terms of their quotients, which is surely true, but what you'll find good literature for is Lie algebroids.