Constructing a Lie group from a Lie algebra.

 November 30, 2006 mathematics

Cartan proved that every finite-dimensional real Lie algebra Error:LaTeX failed:
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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 [1] 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}&39; \to 0 . But this central extension of \mathfrak{g}&39; by Z = \mathbb{R} corresponds to a 2-cocycle \omega \in H^2(\mathfrak{g}; \mathbb{R}) .

Lemma. Let D : H^2(G;\mathbb{R}) \to H^2(\mathfrak{g}; \mathbb{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;\mathbb{R}) with Df = \omega . Since \dim Z(\mathfrak{g}&39;) < \dim Z(\mathfrak{g}) , by induction there is a Lie group G&39; having \mathfrak{g}&39; as its Lie algebra. We build the central extension of G&39; by \mathbb{R} using the cocycle f , namely, 0 \to \mathbb{R}\to G \to G&39; \to 0 , where G \cong G&39; \times \mathbb{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} .

[1] V.V. Gorbatsevich, Construction of a simply connected group with a given lie algebra, Uspekhi Mat. Nauk. 41 (1986) 177–178.