Show that $\mathcal{L}(X_M ) \alpha_X=0$

Question

Let $M$ be a compact manifold on which act a compact lie group $G$. Let $\langle\cdot,\cdot \rangle$ be a $G$-invariant Riemannian metric on M.

Let $X \in \mathfrak{g}$, we denote $X_M$ the vector field on $M$ defined by $$X_M(m) = \frac{d}{dt} \Bigg\vert_{t=0} e^{-tX}.m, \qquad m \in M.$$

Consider the $1$-form $\alpha_X$ defined by $$\alpha_X(Y): =\langle X_M, Y\rangle , \qquad Y \in TM. $$

$\textbf{Question:}$ Prove that $\mathcal{L} (X_M)\alpha_X = 0.$

What I've tried so far is the following: Let $\phi_t: M \rightarrow M$ be the integral curve of $X_M.$ (By the definition of $X_M$, we know that $\phi_t(m)= e^{-tX}.m$. )

Applying the definition of the Lie derivative, I get for $m \in M$ and $Y \in T_mM$

\begin{align*} (\mathcal{L} (X_M)\alpha_X)_m(Y) &= \frac{d}{dt} \Bigg\vert_{t=0} (\phi^*_t(\alpha_X))_m(Y))\\ &= \frac{d}{dt} \Bigg\vert_{t=0} {(\alpha_X)}_{\phi_t(m)}({(\phi_t)}_*(Y)) \\ &= \frac{d}{dt} \Bigg\vert_{t=0} \langle X_M(\phi_t(m)), (\phi_t)_*(Y) \rangle \\ &= \left\langle {\frac{d}{dt}}_{t=0} X_M(\phi_t(m)),Y \right\rangle + \left\langle X_M(m), {\frac{d}{dt}}_{t=0} (\phi_t)_*(Y) \right\rangle. \end{align*}

From here, all what I can say is that $$\frac{d}{dt} \Bigg\vert_{t=0} X_M(\phi_t(m))= \frac{d}{dt}\Bigg\vert_{t=0} \frac{d}{ds} \Bigg\vert_{s=0} e^{-(t+s)X}.m $$ and $$ \frac{d}{dt} \Bigg\vert_{t=0} (\phi_t)_*(Y) = [X_M(m),Y].$$ But, I don't know how to continue. Any help please!

Answer 1

$$\frac{d}{dt} \Bigg\vert_{t=0} \langle X_M(\phi_t(m)), (\phi_t)_*(Y) \rangle = \left\langle {\frac{d}{dt}}_{t=0} X_M(\phi_t(m)),Y \right\rangle + \left\langle X_M(m), {\frac{d}{dt}}_{t=0} (\phi_t)_*(Y) \right\rangle $$

I don't think this step is fully justified. The reason is that the metric $\langle \cdot,\cdot\rangle$ is evaluated at the point $\phi_t(m)$, and so also implicitly depends on $t$.

I think the easiest way to approach this result is to write the metric as a tensor $g\in S^2(M)$ instead. Then $\alpha_X = i_{X_M}g$, and so $$ \mathcal{L}_{X_M}\alpha_X = \mathcal{L}_{X_M}(i_{X_M}g) = i_{\mathcal{L}_{X_M}(X_M)}g + i_{X_M}\mathcal{L}_{X_M}g. $$ But $\mathcal{L}_{X_M}(X_M) = [X_M,X_M] = 0$, and $\mathcal{L}_{X_M}g=0$ by $G$-invariance of $g$.

If you want to follow the approach you attempt above instead, you would need to use

\begin{align*} g_{\phi_t(m)}(X_M(\phi_t(m)),((\phi_t)_*(Y))_{\phi_t(m)}) &= (\phi_t^*g)_m(T_{\phi_t(m)}\phi_{-t}(X_M(\phi_t(m))), Y_m) \\&= g_m(((\phi_{-t})_*X_M)_m, Y_m), \end{align*} using the $G$-invariance $\phi_t^*g = g$ in the last step. Since $\phi_t$ is the flow of $X_M$, we have that $(\phi_{-t})_*X_M = X_M$, and so the RHS is constant.

Here I'm using the notation that for a vector $X$ and diffeomorphism $\phi$, the pushforward is defined by $$ (\phi_*X)_m = T_{\phi^{-1}(m)}\phi(X_{\phi^{-1}(m)}) $$ while for the metric $g$ $$ (\phi^*g)_m(X_m, Y_m) = g_{\phi(m)}(T_m\phi(X_m), T_m\phi(Y_m)). $$ Many authors don't make a distinction between the pushforward $\phi_*$ (acting on tensor fields) and the derivative map $T_m\phi:T_mM\to T_{\phi(m)}M$, but I find it useful to do so.

Answer 2

Hint: You have not used two pieces of information: that the metric is $G$-invariant and that the variables $s$ and $t$ commute. For the latter it might be better to bracket off the statement as a basic ODE statement:

Obs: If $X$ is a vector field on $M$ with flow $\phi_\bullet:\mathbb{R}\to \operatorname{Diff}(M)$, then

$$\forall t\in\mathbb{R},\forall p\in M: \left.\dfrac{\partial}{\partial s}\right|_{s=0} \phi_s(\phi_t(p)) = T_p\phi_t(X_p).$$