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?



cotangent bundle symplectic reduction and fibre bundles
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.


