Lie's third theorem via differential graded algebras?

2010-10-22T19:00:32Z

Dennis Sullivan, "Infinitesimal computations in topology", Publ. IHES: At the end of section 8, he writes, among other things, roughly the following. Let $\mathfrak{g}$ be a (finite-dimensional, real) Lie algebra and let $\Lambda \mathfrak{g}^{\ast}$ be the Chevalley-Eilenberg complex (i.e. the exterior algebra, with the differential dual to the Lie bracket). He considers the spatial realization $\langle \Lambda \mathfrak{g}^{\ast}\rangle $ with respect to the simplicial d.g.a. of $C^{\infty}$-forms on the standard simplices. This is a simplicial set, the $p$-simplices are the d.g.a.-homomorphisms $\Lambda \mathfrak{g}^{\ast} \to \mathcal{A}(\Delta^p)$ to the de Rham forms on the simplex. Let $G$ be the simply-connected Lie group with Lie algebra $\mathfrak{g}$. "Theorem" (8.1)' (the quotation marks are due to Sullivan) says that the fundamental group of $\langle \Lambda \mathfrak{g}^{\ast} \rangle $ is isomorphic to $G$. Now the set of $p$-simplices of $\langle \Lambda \mathfrak{g}^{\ast} \rangle $ has a topology, induced from the $C^{\infty}$-topology) on the space of $\mathfrak{g}$-valued forms on the simplices and so $\pi_1 (\langle \Lambda \mathfrak{g}^{\ast} \rangle )$ is a topological group. Sullivan also says that the above isomorphism is a homeomorphism. It is not difficult to verify these assertions. Since it is rather clear how to describe the exponential map in this construction, we can also recover the differentiable structure on $G$ from this method. The upshot of this discussion is: once the existence of the simply-connected Lie group is known (this is Lie's third theorem, proven only decades after Lie by Cartan), it has the given abstract description. It is well-known that Lie's third theorem is a pretty hard result (the standard proof goes via Ado's theorem).

It seems possible to reverse the logic of that argument and give a proof of Lie's 3rd theorem.

Given $\mathfrak{g}$, \emph{define} $G:=\pi_1 (\langle \Lambda \mathfrak{g}^{\ast} \rangle)$ as a topological group. The exponential map $\mathfrak{g} \to G$ is given by the following formula: any $x \in \mathfrak{g}$ defines a constant $1$-form on $\Delta^1$, hence a 1-simplex of $\pi_1 (\langle \Lambda \mathfrak{g}^{\ast} \rangle)$. That was the easy part; here are the nontrivial parts:

Show that $G$ is Hausdorff (probably difficult)
Put a smooth stucture on it, such that the exponential map is a local chart (probably the hardest part)
Once this is done, the simple connectivity of $G$ and the fact that $Lie (G)=\mathfrak{g}$ are probably both obvious.

After all these preliminaries, I can ask my question: has this approach been written down properly? I am aware that a generalization of this argument has been used by Getzler http://arxiv.org/abs/math/0404003 and Henriques http://arxiv.org/abs/math/0603563, but in these papers, I do not find the details. It is of course also possible (maybe desirable) to banish all the fancy language from the discussion, leaving a definition of $G$ as the quotient of the space of $\mathfrak{g}$-valued 1-forms on the interval.

Answer by Urs Schreiber for Lie's third theorem via differential graded algebras?

2010-10-23T12:19:36Z

The details are here:

Marius Crainic, Rui Fernandes, Integrability of Lie brackets http://arxiv.org/abs/math/0105033

(To connect this to your question, notice that a morphism $T X \to \mathfrak{g}$ of Lie algebroids, which is the language they use, is dually the same as a morphism $\Omega^\bullet(X) \leftarrow CE(\mathfrak{g})$ of dg-algebras. For more see here)

Lie's third theorem via differential graded algebras? - MathOverflow

Lie's third theorem via differential graded algebras?

Answer by Urs Schreiber for Lie's third theorem via differential graded algebras?