Group action induced on the cotangent bundle.

Question

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?

Answer 1

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. :(

Answer 2

I believe there is an error in this formula.

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 $p \mapsto g\cdot p$.

Answer 3

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 $f$.

In the same way that the tangent functor takes manifolds to its tangent bundles and a smooth map to its global derivative (this is a covariant functor), the cotangent functor takes manifolds to its cotangent bundles and smooth maps to the dual maps of their global derivatives (this is a contravariant functor).

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: $g\cdot (x,p) = \hat{g}_x(p)= p\circ {\rm d}(g^{-1})_{gx}$.