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.

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.

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 deRham cohomology groups $H^k_{dR}(M)$ 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 deRham 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.

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.