If I have an action of a Lie group on a configuration space。
$G\to \text{Diff}(M)$, $g \mapsto \rho_g$, $\rho_g : q \mapsto \rho_g(q)$ (for example a rotation).
Then when we consider the phase spaces $T^*M$, we provide it with the action :
$G\to \text{Diff}(T^*M)$, $g \mapsto \rho^*_{g^{-1}}$, $\rho_{g^{-1}}^* \colon (q,p) \mapsto (\rho_g(q),\rho^*_{g^{-1}}(p))$.
Now I understand that the point $q$ is send to its image under the action, but I don't understand why the moment $p$ is transformed as $\rho^*_{g^{-1}}(p)$ under the action?
Why is the reason we consider this weird action with a pullback on the moment?
Why do we want the moment to transform in this way?
You want the lifted action to be by exact symplectomorphisms (actually it turns out even better, the lift gives a Hamiltonian action of $G$ on $T^*M$, but that is a bonus).
Suppose you demand that the lifted action commutes with projection and is linear on the fibers. Then on each fiber it is given by a linear map $A_q: T^*_q M\to T^*_{\rho_q} M$. If we demand that the lifted action preserves the tautological 1-form, then since the position part of the tangent vector is transformed by $D\rho_g|_q$, the momentum needs to transform in a way that "undoes" this, i.e. $A_q$ needs to be such that $<p, v>=<A_q p, D\rho_g|_q(v) >$, i.e. $A_q$ is the inverse of the adjoint of $D\rho_g|_q$; since $\rho$ is an action, this is the adjoint of $D\rho_{g^{-1}}$, aka $\rho^*_{g^{-1}}$.