Skip to main content
1 of 2
Qiaochu Yuan
  • 118.2k
  • 40
  • 447
  • 741

For G a Lie group, can I make sense of G/G as a derived manifold in a nice way?

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.)

Qiaochu Yuan
  • 118.2k
  • 40
  • 447
  • 741