Let $G$ be a compact connected Lie group with Lie algebra $\mathfrak{g}$. Let $M$ Be a smooth manifold on which $G$ acts . Let $p \in M$ and $X \in \mathfrak{g}$ such that the vector field $X_M$ defined by $$X_M(m) = \frac{d}{dt}{\vert_{t=0}} e^{tX}.m , \quad m \in M, $$ vanishes at $p$, which implies that $e^{tX}.p=p.$
In page 213 of the book Heat kernels and Dirac operators, the authors say that the Lie action $\mathcal{L}(X)\xi= [X_M, \xi]$ on $\Gamma (M, TM)$ gives rise to an invertible transformation $L_p$ of $T_pM$ (which is defined by $L_p : T_pM \rightarrow T_pM, Y \longmapsto L_p(Y) = [X_M, \tilde{Y}](p)$, where $\tilde{Y}$ is a vector field which at the point $p$ is equal to $Y$ .)
Next, the authors claim that the map $L_p$ has only imaginary eigenvalues since it is the Lie derivative of an action of a compact Lie group.
Why is $L_p$ the Lie derivative of an action of $G$ ?
I think what the authors mean is that if $a(t) : M \rightarrow M, m \longmapsto e^{tX}.m,$ then
$$L_p (Y)f = (da(t)(Y))f, \quad \forall Y \in T_pM , \forall f \in C^\infty(M),$$
but $L_p (Y)f = \frac{d}{dt}\vert_{t=0} Y_{a(t)(p)}(f \circ a(t))$ and $(da(t)(Y))f = Y(f \circ a(t)),$
however, I don't see why they are equal ?