Does the Lie derivative of a form $\alpha$ in the direction of $X$ at $p$ depend only on $X(p)$?

Question

Let $\alpha$ be a differential $k$-form and $X$ a smooth vector field on a Riemannian manifold $(M,\,g)$. Let $p\in M$. Then, if I'm given another smooth vector field $Y$ with $Y(p) = X(p)$, is it true that $$ (\mathcal{L}_X\alpha)_p = (\mathcal{L}_Y\alpha)_p\,? $$ The reason I ask is that from the definition of the Lie derivative $$ (\mathcal{L}_XT)_p = \frac{\text{d}}{\text{d} t}\Bigg|_{t=0}\left((\Phi^X_{t})^*T_p\right) $$ it seems like the above should be true. However, suppose that $M = \mathbb{R}^n$ with the standard metric. Then for an arbitrary vector field $X$ it is not true that $\mathcal{L}_X\star\alpha = 0$ implies $\mathcal{L}_X\alpha= 0$ because the Hodge star operator $\star$ commutes with $\mathcal{L}_X$ only if $X$ is a Killing field. But at each point $p\in M$, can't we find a Killing field $Z$ that is parallel to $X$ so that $\mathcal{L}_Z\star\alpha = 0 \implies\mathcal{L}_Z\alpha = 0\implies \mathcal{L}_X\alpha = 0$?

Answer 1

Here's a hint: You know that for a vector field $Z$, $\mathcal L_X Z = [X,Z]$. Does the value of the Lie bracket at $p$ depend only on $X(p)$? To use this to answer your question, then note that, for a $1$-form $\alpha$, $\mathcal L_X(\alpha(Z)) = (\mathcal L_X\alpha)(Z) + \alpha(\mathcal L_X Z)$.