Differential forms of a Lie group giving cohomology of the Lie group

Question

Consider a manifold $M$. Then, we have the notion of differential forms on $M$ and complex associated to that, denoted by $$\cdots\rightarrow \Omega^{k-1}(M)\rightarrow \Omega^k(M)\rightarrow \Omega^{k+1}(M)\rightarrow \cdots$$ giving de Rham cohomology groups $H^k_{\mathrm{dR}}(M)$ for $k\in \mathbb{N}$.

Now consider a Lie group $G$. There is a notion of cohomology of this Lie group. Ignoring the group structure, we can talk about de Rham cohomology of the underlying manifold.

Question : Is there a notion of restricted complex of differential forms, that is a sub complex $\{\widetilde{\Omega^k(G)}\}$, of the complex of differential forms $\{\Omega^k(G)\}$ of $G$, whose cohomology groups gives cohomology of the Lie group?

All that (of great importance) extra structure coming in Lie group $G$ is the multiplication (and inverse) map $G\times G\rightarrow G$. So, I am expecting this restricted differential forms to show the action of $G$ on itself.

By cohomology of the Lie group, I mean the cohomology of the underlying manifold. I do not think there is any notion of cohomology of Lie group. Even Google search does not give anything.

Answer 1

For every Lie group $G$, you have in your $\Omega_G$ the subcomplex of the left-invariant differential forms. Moreover, $G$ acts on the right on this subcomplex. It has the virtue of being finite-dimensional and isomorphic to a complex defined purely in terms of the Lie algebra, thus the computation of its cohomology is reduced to linear algebra.

On the other hand, considering a maximal compact subgroup $K\subseteq G$, since $G/K$ is contractible, by Poincare's lemma, $\Omega_G$ and $\Omega_K$ have the same cohomology.