Suppose a compact Lie group $G$ acts on a manifold $M$ with only one orbit type $G/H$ ($H$ denotes the stabiliser group). Then the manifold $M$ becomes a fibre bundle over the quotient manifold $X:=M/G$ with typical fibre $G/H$ and structure group $G$.
On the one hand one could look at the cotangent bundle $T^* X$ of the quotient (which carries a natural symplectic structure).
On the other hand consider the lifted action of $G$ on the cotangent bundle $T^* M$ with moment map $\mu: T^* M\to \mathfrak{g}^* .$ The symplectic quotient $T^* M//G:=\mu^{-1}(0)/G$ inherits the structure of a symplectic manifold. Here comes the question: Are $T^* X$ and $T^* M//G$ (canonically) symplectomorphic?
[Math] cotangent bundle symplectic reduction and fibre bundles
dg.differential-geometrysg.symplectic-geometry
Best Answer
These two symplectic manifolds are canonically symplectomorphic.
Notice first, that the map $\mu$ vanishes on the sub-bundle of $T^* M$ of 1-forms vanishing on the fibers of the fibration $M\to X$. Let us call this sub-bundle by $T_h ^* M$ (h- for horizontal).
To construct the symplectomorphism notice that there is an obvious projection $\pi: T_h^* M \to T^* X$. The restriction of the symplectic form of $T^* M$ to $ T_h^* M$ equals to the pullback of the symplectic form of $T^* X$ under $\pi$. The projection $\pi$ commutes with the action of $G$ and $G$ preserves the symplectic form on $T^* M$. Since the projection $\pi$ just produces the quotient of $T_h^*M$ by the action of $G$, now everything follows from definitions.