Diffgeometer1,
@Diffgeometer1@mathstodon.xyz avatar

It’s a well known result that if (G) is a connected and compact Lie group, then it’s de Rham cohomology is isomorphic to the cohomology of its Lie algebra. The compactness assumption is needed to turn arbitrary differential forms into left-invariant ones by integrating over (G). Why do we need to assume (G) is also connected?

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