Invariant forms on a manifold

Question

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.

Answer 1

$\omega=g^\ast\omega$ 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.

Answer 2

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 $(g^\ast\omega)(X)=(\omega(g_\ast X))\circ g$.