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Let $\Phi:G\times M\ni(g,m)\mapsto\Phi_g(m)\in M$ be a smooth left action of a Lie group $G$ on a manifold $M$, and let us denote by $\Psi:G\times T^\ast M\ni(g,\alpha)\mapsto(T^\ast\Phi_g)\alpha\in T^\ast M$ its cotangent lifting.

It's easy to see that $\Phi$ is free (resp. proper) if and only if $\Psi$ is free (resp. proper).
So if $\Phi$ is free and proper then both the orbit spaces $M/G$ and $(T^\ast M)/G$ get a uniquely determined smooth manifold struscture such that the respective projection maps are submersions.

I was wondering if there exists some kind of relation between the smooth manifolds $T^\ast(M/G)$ and $(T^\ast M)/G$.

Until now I have realized the existence of a diffeomorphism between $T^\ast(M/G)$ and $J/G$, where $J$ is the $G$-invariant vector sub-bundle of $\tau_M^\ast:T^\ast M\to M$ which is the annihilator of the involutive distribution on $M$ generated by its fundamental vector fields w.r.t. the action $\Phi$ of $G$ on $M$.

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If you considered the tangent bundle, I could imagine there being an isomorphism between $T(M/G)$ and $TM/TG$. Maybe through this you can say something about the dual side of the story. –  Olivier Bégassat Oct 30 '11 at 18:31

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Your last paragraph desribes symplectic reduction. Your J is the zero set of the canonical momentum map $\mu:T^*M\rightarrow \mathfrak{g}^*$ associated to the canonical lift of the action of $G$ on $M$ to $T^*M.$ One then has a canonical symplectomorphism $T^*(M/G)\cong \mu^{-1}(0)/G=:T^*M//G.$ Notice that $T^*M/G$ is a vector bundle over $\mu^{-1}(0)/G$ with fiber $\mathfrak{g}$. In particular both have the same homotopy type.

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what does the double slash stand for in $T^*M//G$? –  Olivier Bégassat Oct 30 '11 at 18:48
    
Dear Orbicular, infact I was reading about the realization of the symplectic reduction in the context of cotangent bundles. –  Giuseppe Tortorella Oct 30 '11 at 18:53
    
@Olivier: The double slash stands for the symplectic reduction, being defined by $\mu^{-1}(0)/G,$ see the answer. –  OrbiculaR Oct 30 '11 at 18:56
    
@Giuseppe: The way you formulated the question did not make it apparent to me that you knew that. Sorry! In general $T^*M/G$ should be the union of the symplectic reductions $\mu^{-1}(\mathcal{O})/G$ for all the different $G$-orbits $\mathcal{O}$ in $\mathfrak{g}.$ This can be made more precise in terms of Poisson reduction I think. (Notice that $T^*M/G$ is not necessarily even-dimensional, in particular you have no hope of getting a symplectic structure on it!) –  OrbiculaR Oct 30 '11 at 19:00
    
Dear Orbicular, Sorry? for what? I was positevily surprised that even if I haven't written about the motivation of my question, nevertheless you have given an answer to my unexpressed questions. –  Giuseppe Tortorella Oct 30 '11 at 19:37

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