Let $M$ be a smooth manifold and let $G$ be a compact, connected Lie Group that acts on $M$ by smoothly. Now suppose that the action of $G$ on $M$ is transitive. For some $m$ in $M$, let $K$ be the stabiliser of $m$. It should be clear that $K$ is a closed subgroup of $G$, so we can associate $G/K$ with $M$. Let $\mathfrak{g}$ denote Lie Algebra of $G$ and $\mathfrak{r}$ the Lie Algebra of $K$. Because of compactness of $G$, it is known that there exists some $Ad(G)$-invariant and $ad(LG)$ invariant inner-product, say $\langle \ ,\ \rangle$ on $\mathfrak{g}$, so as a $K$ - module, $\mathfrak{g} = \mathfrak{r} \oplus \mathfrak n $ where $\mathfrak{r}$ and $\mathfrak{n}$ orthogonal with respect to $\langle \ ,\ \rangle$, and we regard $\mathfrak{n}$ as tangent space $T_m M$.
My question is towards calculating the cohomology of $M$ using exterior power of $\mathfrak{n}$.
Let $\omega$ be a $G$ - invariant differential $p$ - form on $M$. So evaluating $\omega$ at point $m$ gives us an alternating, multi-linear function $\omega_m: \mathfrak{n}^p \rightarrow \mathbb{R}$. It is mentioned that because $\omega$ is $G$ - invariant hence $\omega_m$ is $Ad(K)$ invariant, which I presume, means that for any $g \in K$ one has for left invariant vector fields $X_1,\dots, X_p$: $$\omega_m(Ad(g^{-1})X_1(e),\dots, Ad(g^{-1})X_p(e)) = \omega_m(X_1(e),\dots, X_p(e))$$
I am having some trouble with the computation to convince myself to that this is indeed the case. What I have gotten so far is if we let $L_g$ denote left multiplication by $g$ and $R_g$ denotte multiplication of $g$ and $L{_g} _*$ to denote the pushforward maps, etc. then \begin{align*} \omega_m(Ad(g^{-1})X_1(e),\dots, Ad(g^{-1})X_p(e)) &= \omega_m(L_{g^-1}{_*}R_g{_*}X_1(e),\dots,L_{g^-1}{_*}R_g{_*}X_p(e))\\ &=\omega_{g^{-1}m} (R_g{_*}X_1(e),\dots, R_g{_*}X_p(e)) \end{align*} And I am unable to proceed. Same for the converse case as well i.e. obtaining a $G$-invariant differential $p$ form from a $Ad(K)$ invariant alternating, multi-linear function $\omega_m: \mathfrak{n}^p \rightarrow \mathbb{R}$. Do let me know if I have any misconception/misunderstanding. Thanks!
from Hot Weekly Questions - Mathematics Stack Exchange
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