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?


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.

  • 1
    $\begingroup$ You're right, thanks! One can even show that the pull-back of the Liouville form on $T^* M$ to $T^* X$ is the Liouville form on $T^* X$. Thanks! $\endgroup$ – Orbicular Jan 25 '10 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 co-vectors 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 non-Abelian $G$ ($H$ 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 co-adjoint 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 fiber-linear coordinates on the co-adjoint 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 $(co-adjoint orbit bundle).


In addition to Dmitri's answer: when you're doing reduction at non-zero 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 co-adjoint 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 one-form, 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 two-form constructed from the curvature.

In the case where $M = G$, you can choose the Maurer-Cartan 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 Maurer-Cartan "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 $\endgroup$ – Gourishankar Jan 4 '12 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.

  • 3
    $\begingroup$ 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? $\endgroup$ – Orbicular Jan 24 '10 at 18:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.