# Volume form as eigenform of the Lie derivative

Suppose (M,g) is a homogeneous Lorentzian manifold and $Y$ a vector field on it. $G$ is a transitively acting Lie group. It is stated that the volume form $\omega$ is $G$-invariant. It is also stated that the Lie derivative $L_Y\omega$ is $G$-invariant so that the volume element is an eigen form: $L_Y\omega=\lambda\omega$.

How does this follow? Do I need to state more explicit information about the setting?

Since the space of volume forms is one dimensional (over $C^\infty$) and $\omega$ is non-vanishing, we know there is some function $\lambda \in C^\infty(M)$ such that $L_Y \omega = \lambda \omega$; so what needs to be shown is that $\lambda$ is constant. The idea is simple: since everything we start with is invariant under the action of the group, the resulting $\lambda$ should be too.
As it's given that both $\omega$ and $L_Y \omega = \lambda \omega$ are $G$-invariant, there's very little work to do. For any two points $p,q \in M$, transitivity gives us a $g \in G$ such that $\phi_g(p) = q$ (where $\phi_g$ denotes the action of $g$ on $M$). The $G$-invariance of the two forms means that $\omega(p) = \phi_g^* (\omega(q))$ and $(\lambda \omega)(p) = \phi_g^* (\lambda \omega(q))$; so putting these together we get $$\lambda(p)\omega(p) = (\lambda \omega)(p) = \phi_g^* (\lambda \omega(q)) = \lambda(q) \phi_g^*(\omega(q))=\lambda(q)\omega(p).$$ Since $\omega$ is non-vanishing, this implies $\lambda(p)=\lambda(q)$ for all $p,q$; i.e. $\lambda$ is constant.