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?
4 Answers
These two symplectic manifolds are canonically symplectomorphic.
Notice first, that the map $\mu$ vanishes on the subbundle of $T^* M$ of 1forms vanishing on the fibers of the fibration $M\to X$. Let us call this subbundle 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.

1$\begingroup$ You're right, thanks! One can even show that the pullback of the Liouville form on $T^* M$ to $T^* X$ is the Liouville form on $T^* X$. Thanks! $\endgroup$ Jan 25, 2010 at 10:48
This is more a comment towards Gourishankar than an answer to the original question. It was part of my thesis, (UCB, about 1986), so, apologies, I chime in. For simplicity, I take the case $G$ Abelian, and $H = $ trivial. To map $J^{1} (\mu)$ equivariantly to $J^{1}(0)$ subtract $\mu \cdot A$ where $A$ is any $G$connection for $\pi: X \to X/G$. $J^{1} (0)/G = T^* (X/G)$ canonically, independent of connection. The map `momentum shift map of subtracing $\mu \cdot A$ from covectors is not symplectic, relative to the standard structure, but it becomes symplectic if you subtract $\mu \pi^* F_A$, where $F_A = curv(A)$, from the standard structure. So the reduced space at $\mu$ is $T^*(X/G)$ with the standard structure minus the ``magnetic term'' $\mu F_A$.
For nonAbelian $G$ ($H$ still trivial), it is easier to explain things in Poisson terms.
$T^* X/G$ is a Poisson manifold whose symplectic leaves
are the reduced spaces in question. The momentum
shift trick turns it into $T^* (X/G) \oplus Ad^* (X)$ where $Ad^* (X) \to X/G$
is the coadjoint bundle associated to $X \to X/G$  its fibers are the
dual Lie algebras for $G$. This direct sum bundle admits coordinates $s_i, p_i, \xi_a$ where $s_i, p_i$ are canonical coordinates on $T^*(X/G)$ induced by coordinates $s_i$ on $X/G$
and where $\xi_a$ are fiberlinear coordinates on the coadjoint bundle induced
by a choice of local section of $\pi$. Then the main tricky part of the bracket is that
the bracket of $p_i$ with $p_j$ is $\Sigma \xi_a F^a _{ij}$, $F$ being the curvature of the connection
relative to the choice of local section. The symplectic leaves = reduced spaces
are of the form $T^*(X/G) \oplus $(coadjoint orbit bundle).
In addition to Dmitri's answer: when you're doing reduction at nonzero momentum some interesting things happen: the symplectic reduced space $J^{1}(\mu)/G_\mu$ then becomes a fiber bundle over $T^\ast X$ with typical fiber the coadjoint orbit $\mathcal{O}_\mu$. In the case $\mu = 0$, this reduces to the case discussed previously. This realization is not canonical, though, and depends on the choice of a connection in $M \to M/G$.
The idea of the proof is to relate $J^{1}(\mu)$ with $J^{1}(0)$ using the connection oneform, and then to use the isomorphism mentioned previously. Afterwards, the curvature shows up in the reduced symplectic form on $J^{1}(\mu)/G_\mu$, which is then the sum of the canonical form on $T^\ast X$ and a twoform constructed from the curvature.
In the case where $M = G$, you can choose the MaurerCartan form as your connection, and then the isomorphism is $J^{1}(\mu)/G_\mu = \mathcal{O}_\mu$ with the KKS symplectic form, which is in some sense the "curvature" of the MaurerCartan "connection".

$\begingroup$ Hello Jvkersch Could you please elaborate on the method of choosing a connection one form in order to get an expression for the reduced symplectic form Thanks $\endgroup$ Jan 4, 2012 at 2:59
No. Take $M=G$, with action by left translations. Then $T^*G//G$ gives the coadjoint orbits with the Kirilov symplectic str (eg for G=U(n) the complex flag manifolds with the standard Kahler str), while $X$ is just a point.

4$\begingroup$ Keep in mind that the reduction takes place at 0 momentum. The coadjoint orbit of 0 in $\mathfrak{g}^* $ is 0 itself. So, what's the problem? $\endgroup$ Jan 24, 2010 at 18:22