# Constructing a Lie group from a Lie algebra.

##### 2006-11-30 04:56:41 +0000mathematics

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 > MR0854249 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 $\omega \in H^2(\mathfrak{g}; \R)$. \begin{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. \end{lemma} 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 $\tilde{G}$.