Let $M$ be an $S^1$-manifold and let $\omega$ be a $k$ form on $M$. Let $X\in\mathfrak{X}(M)$ a vector field on $M$. Now my question is,
If $\mathcal{L}_X(\omega)=0$, then $(\Phi_X^t)^*\omega=\omega.$
By definition $$ \frac{d}{dt}\bigg|_{t=0}(\Phi_X^t)^*\omega =0 $$
Now, I want to say something like, $\frac{d}{dt}(\Phi_X^t)^*\omega=0$ for all $t$, so $(\Phi_X^t)^*\omega=(\Phi_X^0)^*=\omega$. How can I do this?
Use the fact that $\Phi_X^{s+t} = \Phi_X^s\circ\Phi_X^t$ to deduce that $$ \frac{d}{dt}\bigg\rvert_{t=t_0} (\Phi_X^t)^*\omega = (\Phi_X^{t_0})^* \frac{d}{dt}\bigg\rvert_{t=0} (\Phi_X^t)^*\omega = 0. $$