Background for Lie Algebra cohomology and de Rham cohomology of compact Lie Groups I have read that the de-Rham cohomology of compact Lie Groups is isomorphic to the Lie Algebra cohomology of its Lie Algebra. Do I need the knowledge of singular cohomology to understand this?
I think that de Rham's theorem (and so singular cohomology) is needed to understand the isomorphism of the de-Rham cohomology to left invariant forms, right? But do you use that in the above isomorphism (to Lie Algebra cohomology)?
 A: I don't think you need to pass through singular cohomology, this is all about de Rham complexes, and here's how the proof goes. Let $G$ be a connected compact Lie group and $\mathfrak{g}$ the corresponding Lie algebra. There are two different steps:


*

*Easy part: note that the complex $\Omega_L^\bullet (G) \subset \Omega^\bullet (G)$ formed by the left invariant differential forms is essentially the same as the standard Chevalley-Eilenberg complex calculating the Lie algebra cohomology $H^\bullet (\mathfrak{g}, \mathbb{R})$.

*Trickier part: the natural subcomplex inclusion $i\colon \Omega_L^\bullet (G) \to \Omega^\bullet (G)$ induces isomorphisms in cohomology. For this we consider a morphism of complexes $p\colon \Omega^\bullet (G) \to \Omega_L^\bullet (G)$ that from a given differential form $\alpha$ produces a left-invariant differential form. This is known as the "averaging map", and it is defined by the "Haar integral" 
$$p\colon \alpha \mapsto \int_G L_g^* \alpha$$
---and for this we need $G$ to be compact! First of all, $p\circ i$ is identity, and as for $i\circ p$, it is possible to construct an explicit chain homotopy equivalence between $i\circ p$ and the identity map.
A possible detailed reference: Werner Greub, Stephen Halperin, Ray Vanstone, ``Connections, Curvature, and Cohomology'', vol. II, Chapter IV.
P.S. Yes, the original paper of Chevalley and Eilenberg is probably the most accessible account.
