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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.

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Are you sure you aren't confusing the left regular action with the conjugation action? –  Jason Starr Oct 1 '13 at 16:56
    
@Jason: I don't think I am. I want left-invariant differential forms on $G$, not conjugation-invariant differential forms. –  Qiaochu Yuan Oct 1 '13 at 16:59
    
Okay, got it. So you are observing that the de Rham algebra on $G$ is (left) $G$-equivariantly isomorphic to the pullback of the Chevalley-Eilenberg algebra under projection to a point. In what sense is that a problem with quotients that needs to be "fixed"? –  Jason Starr Oct 1 '13 at 17:11
    
Well one can work with G/G as an ordinary stack and then the natural version of de Rham cohomology will be the Cartan model for equivariant cohomology, you can get back to the de Rham algebra of G by taking derived tensor over H^*(BG) with H^*(pt). Don't see any riches down that path... –  user36931 Oct 1 '13 at 17:24
    
@Jason: the nicest thing would be to have $\Omega^{\bullet}(M/G) \cong \Omega^{\bullet}(M)^G$ but this doesn't hold in the above example. The quotienting procedure is maybe not so important; what I really want is to write down some smooth object which has de Rham algebra $\Lambda^{\bullet}(\mathfrak{g}^{\ast})$, vector fields $\mathfrak{g}$, etc. –  Qiaochu Yuan Oct 1 '13 at 18:04
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3 Answers

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I fully agree with A non's answer and comments. However, here is another perspective which you might find useful.

Recall that the de Rham stack $M_{dR}$ of a manifold $M$ is obtained from $M$ by identifying infinitesimally close points. In general the space of (derived) functions on the de Rham stack is the de Rham complex, and the category of quasi-coherent sheaves on $M_{dR}$ is modules for the ring of differential operators on $M$. Thus the de Rham stack is a way of encoding he kinds of invariants you are interested in.

In the case $M=G$, we have $G_{dR} = G/\widehat{G}$, where $\widehat{G}$ is the formal group associated to $G$ (a formal neighbourhood of the identity in $G$). As you have noticed, the quotient $(G/G)_{dR} = G_{dR}/G_{dR} = pt_{dR} = pt$. So the de Rham complex of $pt = G/G$ is not the $G$-invariants of the de Rham complex of $G$.

I think what you want to look at is the stack $G_{dR}/G$. The space of functions on $G_{dR}/G$ is indeed $G$-invariant forms on $G$, i.e. the CE complex. And quasi-coherent sheaves on $G/G_{dR}$ are modules for $G$-invariant differential operators, i.e. $\mathcal U(\mathfrak{g})$-modules.

Another name for $G/G_{dR}$ would be the classifying stack $B\widehat{G}$ or $B\mathfrak{g}$.

Sorry if this is a little vague - I am thinking in an algebraic setting, but it should be possible to carry these things over to a differential geometric setting.

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My impression is that the literature is a bit sparse when it comes to doing this type of thing differentio-geometrically. I don't feel like doing the leg work myself, but if I did, I'd start with Roytenberg and collaborators. –  Theo Johnson-Freyd Oct 2 '13 at 2:04
    
Thanks! I think this captures what I was looking for conceptually, although maybe to find references I should look at papers on Lie algebroids instead as Ben suggests. –  Qiaochu Yuan Oct 2 '13 at 17:57
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The point of derived manifolds is to fix up limits, not colimits, of manifolds. A standard "fix" for what you bring up is to instead derive the notion of $G$-invariants, rather than to change the quotient:

  • Regard $\Omega^\bullet(G)$ as a commutative algebra in naive rational $G$-spectra, and take homotopy invariants. That recovers something quasi-isomorphic to $\Omega^\bullet(G/G)$.

In this viewpoint, the formation of literal $G$-invariants is providing a nice, small, model for the de Rham forms on $M$ (at least when the $G$ action is free). So you already have a manifold whose de Rham cochains "are" the Chevalley-Eilenberg complex: $G$ itself. (This is the point of the Chevalley-Eilenberg complex, read backwards.)

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Hmm. I guess I didn't ask the question I wanted to ask, then. What I really want is some sort of smooth object whose de Rham forms are the Chevalley-Eilenberg complex, whose vector fields are $\mathfrak{g}$, and whose differential operators are $U(\mathfrak{g})$. –  Qiaochu Yuan Oct 1 '13 at 19:45
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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.

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