Invariant forms on a manifold Probably a silly question - still, it's been bugging me for some time now.
Say that we have an invariant $1-$form $\omega$ on a smooth manifold $M$, acted on by a group $G$. Then
\begin{equation}
\omega = g^*\omega
\end{equation}
This should mean that if I pick some vector field $X$ on $M$, and plug it on both sides, I should obtain the same number:
\begin{equation}
\omega(X)=g^*\omega(X)
\end{equation}
let me evaluate both objects in some $x\in M$:
\begin{equation}
\omega_x(X_x)=(g^*\omega(X))_x=\omega_{g\cdot x}(X_{g\cdot x})
\end{equation}
Now, this seems to me a bit strange, to say the least: it depends on the vector field I'm inserting. But then what's the meaning of $\omega = g^*\omega$? How should I interpret it pointwise?
Thanks.
 A: $\omega=g^\ast\omega$ means that for all $p\in M$, $\omega_p=g^\ast(\omega_{g(p)})$.
Now, to consider your statement that $\omega(X)=g^\ast\omega(X)$, we consider the statement pointwise.  Suppose for now that the action is transitive and fix $p\in M$.  Now consider $(\omega(X))(q)$ for any $q\in M$.  By transitivity, there is some $g\in G$ such that $g\cdot p=q$.
$$
(\omega(X))(q)=\omega_q(X_q)=((g^{-1})^\ast\omega_p)(X_q)
$$
This follows since $((g^{-1})^\ast\omega_p)=((g^{-1})^\ast\omega_{g^{-1}(q)})=\omega_q$ by the invariance.  Therefore, 
$$
(\omega(X))(q)=((g^{-1})^\ast\omega_p)(X_q)=\omega_p(g^{-1}_\ast X_q).
$$
Observe that $g^{-1}_\ast X_q$ is a vector at $p$.  If, in addition, $X$ is also $G$-invariant, so that $g_\ast X_q=X_{g\cdot q}$, then the expression above simplifies to
$$
(\omega(X))(q)=\omega_p(g^{-1}_\ast X_q)=\omega_p(X_p)=(\omega(X))(p).
$$
Hence, in this case, the value of the form on the vector field depends only on the value at the point $p$.
Now, let's add a little more, let $h$ be an arbitrary element of $G$ and consider
$$
((h^\ast\omega)(X))(q)=(h^\ast\omega)_q(X_q)=(h^\ast(\omega_{h\cdot q}))(X_q)=(\omega_{h\cdot q})(h_\ast X_q)=(\omega_{h\cdot q})((h_\ast X)_{h\cdot q})=(\omega(h_\ast X))(h\cdot q)
$$
Since $(hg)\cdot p=q$, from the argument above, we know that 
$$
(\omega(h_\ast X))(h\cdot q)=\omega_p((hg)_\ast^{-1}h_\ast X)=\omega_p(g_\ast^{-1}X)=(\omega(X))(q).
$$
This is a bit of a long way to show everything (there are slicker versions of this), but it should show how a lot of this fits together.
A: Looking at this a different way, $\omega=g^\ast\omega$ means that pointwise:
$$
(g^\ast\omega)_p=g^\ast(\omega_{g\cdot p})=\omega_p.
$$
For a vector field $X$, 
$$
(\omega(X))(p)=\omega_p(X_p)
$$
On the other hand,
$$
(\omega(X))(p)=((g^\ast\omega)(X))(p)=(g^\ast\omega)_p(X_p)=(g^\ast(\omega_{g\cdot p}))(X_p)=\omega_{g\cdot p}(g_\ast X_p)=\omega_{g\cdot p}((g_\ast X)_{g\cdot p}).
$$
Therefore, $\omega_p(X_p)=\omega_{g\cdot p}((g_\ast X)_{g\cdot p})$.  In other words, as functions, $\omega(X)=(\omega(g_\ast X))\circ g$.
Without the invariance, the last inequality is just $(g^\ast\omega)(X)=(\omega(g_\ast X))\circ g$.
