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Constructing a Lie group from a Lie algebra.

Cartan proved that every finite-dimensional real Lie algebra \germ g comes from a connected, simply-connected Lie group G. I hadn’t known the proof of this result (and apparently it is rather uglier than one might hope), but

Gorbatsevich, V. V. Construction of a simply connected group with a given Lie algebra.(Russian) Uspekhi Mat. Nauk 41 (1986), no. 3(249), 177—178.

gives a short proof of it, which I presented to the undergraduates in my Lie group seminar. I’ll sketch the proof now.

Theorem. For every Lie algebra \mathfrak{g}, there is a simply-connected, connected Lie group G having \mathfrak{g} as its Lie algebra.

First, if \mathfrak{g} \subset \mathfrak{gl}(V), then the exponential map gives U = \exp \mathfrak{g}, and we define G = \bigcup_{k=1}^\infty U^k \subset GL(V). It turns out G is a Lie group, and \mathfrak{g} is its Lie algebra.

If \mathfrak{g} has no center, then \rm{ad} : \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g}) is injective, so we have realized \mathfrak{g} as a Lie subalgebra of endomorphisms of a vector space, and by the above, there is a Lie group G \subset GL(\mathfrak{g}) with \mathfrak{g} as its Lie algebra. Taking its universal cover \tilde{G} proves the theorem in this case.

Now we induct on the dimension of the center Z(\mathfrak{g}). Let Z \subset Z(\mathfrak{g}) be a one-dimensional central subspace of \mathfrak{g}, and construct a short exact sequence 0 \to Z \to \mathfrak{g} \to \mathfrak{g}' \to 0. But this central extension of \mathfrak{g}' by Z = \R corresponds to a 2-cocycle \omega \in H^2(\mathfrak{g}; \R).

Lemma. Let D : H^2(G;\R) \to H^2(\mathfrak{g}; \R) be the map which differentiates a (smooth!) 2-cocycle of the group cohomology of G. The map D is injective.

Consequently, we can find f \in H^2(G;\R) with Df = \omega. Since \dim Z(\mathfrak{g}') < \dim Z(\mathfrak{g}), by induction there is a Lie group G' having \mathfrak{g}' as its Lie algebra. We build the central extension of G' by \R using the cocycle f, namely, 0 \to \R \to G \to G' \to 0, where G \cong G' \times \R and the operation is (g_1, t_1) \cdot (g_2, t_2) = (g_1 g_2, t_1 + t_2 + f(g_1,g_2)). Since Df = \omega, it turns out that the Lie algebra corresponding to G is \mathfrak{g}. We finish the proof by taking the universal cover \tilde{G}.

Posted: November 29th, 2006 under Mathematics.
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