Cohomology of $G$-invariant differential forms

Question

Let $G$ be a Lie group with a left action on a manifold $M$, $\cdot : G \times M \to M$.

Define a $G$-invariant differential form $\alpha \in \Omega^{k}(M)$ as a form satisfying $g^{*}\alpha = \alpha,$ where $g:M \to M$ is the action by $g \in G$. Since pullbacks commute with differentials, the set of all $G$-invariant forms (which I will denote by $\Omega(M)^G$) is a subcomplex of $\Omega(M)$. I found the following theorem in several places:

Theorem: If $G$ is compact and connected, then the inclusion $i: \Omega(M)^{G} \to \Omega(M)$ is induces an isomorphism in de Rham cohomology.

One of the places where I found this is https://planetmath.org/invariantdifferentialform. However, there is no proof there, and I would like to find a proof. Does anyone know a good reference on this subject matter?

Answer 1

Spivak "A comprehensive introduction to Differential Geometry" vol. 5, Look at chapter 13, section 16 pg 308,309.

Answer 2

The cochain map $j:\Omega^k_G(M)\longrightarrow\Omega^k(M)$ induces an injective morphism on the cohomological level:

$$j^*:H^k_G(M)\longrightarrow H^k(M)$$

We need to show this $j^*$ is also surjective, which is equivalent to show for any cohomologous $[\beta]\in H^k(M)$, it contains at least one $G-$ invariant form $\alpha\in[\beta]$.

Recall that since $G$ is a connected compact Lie group, it admits a left invariant Haar measure, which allows us to take integral over $G$, so we can define $$\alpha=\int_G g^*\beta$$ Where the inetgrands is taking for all $g\in G$.

(1). $\alpha$ is closed. Indeed, we have $$d\alpha=\int_G dg^*\beta=\int_G g^*d\beta=0$$

(Recall that the pull-back of a smooth map is a cochain map of de Rham complex)

(2). We then claim that $\alpha-\beta$ is exact, hence $[\alpha]=[\beta]$.

To see this, we will use the fact: if for two closed forms $\alpha,\beta$, we have $$\int_{[Z]}\alpha=\int_{[Z]}\beta$$ for all $[Z]\in H_k(M;\mathbb{R})$, then $\alpha-\beta$ is exact.

In fact, notice that $[Z]=[gZ]$ for any $g\in G$, then by Fubini theorem, we have $$\begin{aligned}\int_{[Z]}\alpha&=\int_{[Z]}\int_Gg^*\beta\\&=\int_G\int_{[Z]}g^*\beta\\&=\int_G\int_{[gZ]}\beta\\&=\int_G\int_{[Z]}\beta\\&=\int_{[Z]}\beta\end{aligned}$$ that proves the claim.

(3). $\alpha$ is $G-$invariant, in fact, for all $h\in G$, we have

$$h^*\alpha=\int_G(gh)^*\beta=\alpha$$

by the change of some variables, hence $j^*$ is also surjective, which is to say, the Cohomology of $G$-invariant forms coincides with the usual de Rham Cohomology. $\blacksquare$

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Just one comment on this little conclusion. If we let $M$ be the compact Lie group $G$ itself, let $G$ acts on itself by left multiplication, then as a corollary, we have: The de Rham complex of $G$ is equivalent to the left-invariant de Rham complex of $G$. That is the differential forms can be reduced to left-invariant forms only.

Now, notice that the left-invariant forms only determined by its value at the identity, hence the left-invariant de Rham complex of $G$ coincides with the Chevalley-Elienberg complex of its Lie algebra $\mathfrak{g}$, hence we have:

Theorem: The de Rham cohomology of a compact Lie group $G$ coincides with the Chevalley-Eilenberg cohomology of its Lie algebra $\mathfrak{g}$ , that is $$H^k(G;\mathbb{R})\cong H^k(\mathfrak{g};\mathbb{R})$$