Let $M$ be a smooth manifold on which acts a compact Lie group $G$.
Let's suppose we have a diffeomorphism on $M$ $$f : M \rightarrow M $$
If $\Phi : M \rightarrow \mathfrak{g}^* $ is a map between $M$ and the dual of the Lie algebra of $G$. Then what does the notation $f^*(\Phi)$ mean ?
I Know that if $\Phi$ were a function or a differential form, then $f^*(\Phi)$ is the the pullback of the the fonction or of the differential form. but if $\Phi$ is as above, I don't what is its pullback by $f$.
(I've come across this in symplectic geometry when $M$ is symplectic and the action of$G$ on $M$ is Hamiltonian, and $\Phi$ denotes the moment map).
Edit: (Example of $\Phi$ which confuses me): let $\lambda$ be a 1-form, we define $\Phi=\Phi_\lambda $ as follows:
$$\langle\Phi_\lambda(m), X \rangle = \lambda_m(X_M(m)) , \quad m \in M, X \in \mathfrak{g}.$$ Where $X_M $ is the fundamental vector field defined by
$X_M(m)= \frac{d}{dt}\vert_{t=0} e^{-tX}.m$
What is $ f^*(\Phi_\lambda)$ ?