@johncarlosbaez Thanks! That makes sense. As you pointed out, if (G) has more than one component then, for example, the degree (0) de Rham cohomology will be isomorphic to (k)-copies of (\mathbb{R}) where (k) is the number of components of (G). On the other hand, the degree (0) cohomology of its Lie algebra is always isomorphic to (\mathbb{R}). So there's no way (H^\bullet_{dR}(G)\simeq H^\bullet(\mathfrak{g})) could be true without the assumption that (G) is connected.