Let M be a compact manifold provided with an action of a compact lie group G. Let $\langle.,, \rangle$ be a G-invariant Riemannian metric on M.
Let $X \in \mathfrak{g}$, let $\theta_X \in \Omega^1(M)$ be a 1-form on M, s.t $\theta_X(Y)= \langle X_M ,Y \rangle, $ $Y \in TM$.
How to prove that $\mathcal{L} (X)\theta = 0 ?$
I've tried this:
Let $\phi_t: M \rightarrow M, $ $ \phi_t(m) = e^{-tX}.m$. let $Y \in TM$ $(\mathcal{L} (X)\theta_X)(Y) = {\frac{d}{dt}}_{|t=0} (\phi^*_t(\theta_X))(Y) = {\frac{d}{dt}}_{|t=0} \langle X_M, (\phi_t)_*(Y) \rangle = \langle {\frac{d}{dt}}_{|t=0} X_M, {\frac{d}{dt}}_{|t=0} (\phi_t)_*(Y) \rangle = 0 $,
since $X_M$ does not dependent of t.
Is this correct ?
Let $x\in M$, and $Y\in T_xM$, we have:
$\phi_t^*\theta_X(Y)=\theta(d\phi_t.Y)=\langle X_M(\phi_t(x)),d\phi_t.Y\rangle=\langle d\phi_t.X(x),d\phi_t.Y\rangle=\langle X(x),Y\rangle$.
This implies that ${d\over{dt}}_{t=0}\phi_t^*\theta= L_X\theta_X=0$.