Let $G$ be a compact Lie group and denote by $\mathcal L_X$ the Lie derivative with respect to the field $X$. I have two questions:
Suppose that for a smooth form $u$ I have that $\mathcal L_X u = 0$ for every $X$ in the Lie algebra of $G$. Does it hold that $u$ is left-invariant?
Suppose that $u$ is a left-invariant form. Does is hold that $\mathcal L_X u = 0$?
I am asking these two questions because on Raoul Bott's paper "homogeneous vector bundles", if I am not reading it wrong, he appear to be using that a form is left-invariant if and only if it is annihilated by all infinitesimal left-translations, but I can not prove this statement.
Thank you,
The two conditions are equivalent if $G$ is connected. The main step twoards seeing this is that for a differential form $u$, (and indeed any tensor field) the Lie derivative in direction of a vector field $\xi$ vanishes identically if and only if $u$ is preserved by all local flows of $\xi$. Now if you take $\xi$ to be the right invriant vector field generated by $X\in\mathfrak g$, then the flow of $\xi$ up to time $t$ is $g\mapsto \exp(tX)g$. Hence the fact that $u$ has vanishing Lie derivative along all right invariant vector fields is equivalent to the fact that $u$ is invarinat under all left translations by elements of the form $\exp(X)$. If $G$ is connected, these exponentials generate $G$.