# 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 . 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 , there is a simply-connected, connected Lie group having as its Lie algebra.

First, if , then the exponential map gives , and we define . It turns out is a Lie group, and is its Lie algebra.

If has no center, then is injective, so we have realized as a Lie subalgebra of endomorphisms of a vector space, and by the above, there is a Lie group with as its Lie algebra. Taking its universal cover proves the theorem in this case.

Now we induct on the dimension of the center . Let be a one-dimensional central subspace of , and construct a short exact sequence . But this central extension of by corresponds to a 2-cocycle .

Lemma. Let be the map which differentiates a (smooth!) -cocycle of the group cohomology of . The map is injective.

Consequently, we can find with . Since , by induction there is a Lie group having as its Lie algebra. We build the central extension of by using the cocycle , namely, , where and the operation is . Since , it turns out that the Lie algebra corresponding to is . We finish the proof by taking the universal cover .

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