# Group action induced on the cotangent bundle.

I was reading the wiki on moment maps.

[...]Let $$N$$ be a smooth manifold and let $$T^*N$$ be its cotangent bundle, with projection map $$\pi : T^*N \rightarrow N$$. Let $$\tau$$ denote the tautological 1-form on $$T^*N$$. Suppose $$G$$ acts on $$N$$. The induced action of $$G$$ on the symplectic manifold $$(T^*N, \mathrm{d}\tau)$$, given by $$g \cdot \eta := (T_{\pi(\eta)}g^{-1})^* \eta$$ [...]

What does $$T_{\pi(\eta)}g^{-1}$$ mean? I know some people wright $$T_x \phi$$ for the pushforward $$d_x \phi$$ but what is the map $$\phi$$ here? $$g^{-1}$$ is an element of the group $$G$$. Not a map, no?

Also, what is the intuitive meaning of this action on the cotangent bundle?

If $$G$$ acts on $$N$$, then every $$g$$ produces a smooth function $$g\cdot : N\to N$$ given by $$x\mapsto g\cdot x$$. I'm guessing is the function you are taking the pullback of. Regarding the intuitive meaning, I'm sorry. I'm not familiar with this formalism. :(
If $$\eta \in T^*_p N$$ then it is natural to ask that $$g \cdot \eta \in T^*_{g \cdot p} N$$. With this in mind, $$g \cdot \eta$$ should eat vectors in $$v \in T^\ast_{g \cdot p} N$$. The only reasonable way to define this is $$g \cdot \eta (v) = \eta( (\phi_g^{-1})_{\ast} (v)) = ((\phi_g^{-1})^\ast \eta )(v)$$ where $$\phi_g: N \to N$$ is the map $$p \mapsto g\cdot p$$.
If $$f:M\to N$$ is a diffeomorphism, then we have $$f^{-1}:N\to M$$. The global derivative is a map $${\rm d}(f^{-1}):TN\to TM$$. Dualize to get $$\hat{f}=({\rm d}(f^{-1}))^*:T^*M\to T^*N$$. This $$\hat{f}$$ is called the cotangent lift of $$f$$.
The construction of the cotangent lift is just an application of the cotangent functor to the inverse diffeomorphism $$f^{-1}$$. Now, if $$G$$ acts on $$N$$, then $$G$$ acts on the tangent bundle $$TN$$ via derivative ("tangent lift") by $$g\cdot (x,v) = {\rm d}g_x(v)$$, and acts on the cotangent bundle $$T^*N$$ via cotangent lift: $$g\cdot (x,p) = \hat{g}_x(p)= p\circ {\rm d}(g^{-1})_{gx}$$.