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Suppose I have a Hamiltonian $H$ with the phase space $\mathcal{M}$, a symplectic manifold with a symplectic 2-form $\omega.$ Now assume that the Hamiltonian system has two first integrals $C_1,C_2$. Define the restricted phase space $\mathcal{N}$ of $\mathcal{M}$ by taking $C_1$=constant,$C_2$=constant. Then under the conditions that the differentials $dC_1$ and $dC_2$ are independent it follows that $\{C_1,C_2\} \neq 0$ ( where $\{\cdot , \cdot \}$ is the Poisson bracket) if and only if $\mathcal{N}$ is a symplectic submanifold. See this post.

However you also have that the result that if $\{C_1,C_2 \} =0$ then there exists a reduction by 2 degrees of freedom (,i.e., 4 dimensions).

How do these two results relate? They seem to contradict each other. What am I missing?

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There is no contradiction here. While it is true that the Marsden-Weinstein reduced space is a symplectic manifold, it is not a submanifold of $M$ but rather a quotient of a (coisotropic) submanifold of $M$. Let me explain the usual Marsden-Weinstein reduction to make sure your definitions are the same as mine, and then I will proceed to describe a process for noncommutative reduction for a Hamiltonian system with non-involutive integrals.


Marsden-Weinstein Reduction.

I'll begin by describing the Marsden-Weinstein reduced space from the viewpoint of symplectic geometry before stating things in terms of integrals of a Hamiltonian system.

Let $(M,\omega)$ be a symplectic manifold and $G$ a Lie group (with Lie algebra $\mathfrak{g}$) acting symplectically on $(M,\omega)$ via $$G \longrightarrow \mathrm{Symp}(M,\omega),$$ $$g \mapsto \psi_g.$$ For each $\xi \in \mathfrak{g}$, there is a corresponding vector field $X_\xi$ on $M$ defined by $$X_\xi(p) = \left. \frac{d}{dt}\right|_{t=0} \psi_{\exp(t\xi)}(p)$$ for all $p \in M$, where $$\exp: \mathfrak{g} \longrightarrow G$$ is the exponential map of $G$. We call the action $g \mapsto \psi_g$ Hamiltonian if

  1. The vector field $X_\xi$ is Hamiltonian for all $\xi \in \mathfrak{g}$, i.e. there exists a function $$H_\xi: M \longrightarrow \Bbb R$$ such that $$dH_\xi = \iota(X_\xi)\omega$$ for all $\xi \in \mathfrak{g}$. (Note: The assignment $\xi \mapsto H_\xi$ can be chosen to be linear).

  2. The (linear) map $$\mathfrak{g} \longrightarrow C^\infty(M),$$ $$\xi \mapsto H_\xi$$ is a Lie algebra homomorphism, where the Lie algebra structure on $C^\infty(M)$ is given by the Poisson bracket.

Note that, for example, if $G = \Bbb R$ or $G = S^1$, the second condition is trivial and we just need a single Hamiltonian function $H$.

Now let $G$ act on $(M, \omega)$ in a Hamiltonian fashion. Then a moment map for this action is a map $$\mu: M \longrightarrow \mathfrak{g}^\ast$$ such that $$H_\xi(p) = \mu(p)(\xi)$$ for all $p \in M$ and $\xi \in \mathfrak{g}$. One can show that the moment map of a Hamiltonian action is coadjoint equivariant, i.e. $$\mu(\psi_g(p)) = \mathrm{Ad}_{g}^\ast \mu(p)$$ for all $p \in M$ and $g \in G$.

Note: Here I have used the definitions as given in McDuff and Salamon's Introduction to Symplectic Topology; some other books don't require property (2) of a Hamiltonian action but instead require that moment maps be coadjoint equivariant. However, we end up with the same results no matter what definition we use.

Now let $G$ act on $(M,\omega)$ in a Hamiltonian fashion with moment map $\mu: M \longrightarrow \mathfrak{g}^\ast$. Then since $0 \in \mathfrak{g}^\ast$ is a fixed point of the coadjoint action, it follows from the equivariance of the moment map that $\mu^{-1}(0)$ is invariant under the action of $G$.

Theorem. (Marsden-Weinstein) With the setup above, suppose that $0$ is a regular value of $\mu$ and that the action of $G$ is free and proper. Then the Marsden-Weinstein reduced space $$M /\!\!/ G = \mu^{-1}(0)/G$$ is a manifold admitting a symplectic form $\tau$ such that $$i^\ast \omega = \pi^\ast \tau,$$ where $i: \mu^{-1}(0) \longrightarrow M$ is the inclusion map and $\pi: \mu^{-1}(0) \longrightarrow M /\!\!/ G$ is the natural projection.

One may also perform reduction at other levels $\xi \in \mathfrak{g}^\ast$ to form the reduced space $\mu^{-1}(\xi)/G$ as long as $\mu^{-1}(\xi)$ is preserved by $G$, which is equivalent to $$\xi = \mathrm{Ad}_g^\ast \xi$$ for all $g \in G$.

Note: We see that the Marsden-Weinstein reduced space is a quotient of a submanifold of $M$ and not a submanifold of $M$. It is possible that $M /\!\!/ G$ doesn't even embed in $M$. For example, consider the Hamiltonian circle action on $\Bbb C^n$ (with the standard symplectic structure) defined by $$\psi_t(z_1, \dots, z_n) = (e^{2\pi it}z_1, \dots, e^{2\pi it}z_n)$$ where $t \in \Bbb R / \Bbb Z \cong S^1$. A moment map for this action is $$\mu(z) = \frac{|z|^2}{2} - \frac{1}{2}.$$ Therefore the level set $\mu^{-1}(0)$ is the unit sphere $S^{2n-1}$. The Marsden-Weinstein reduced space is $$\Bbb C^n /\!\!/ S^1 = \mu^{-1}(0)/S^1 = S^{2n-1}/S^1 = \Bbb C P^{n-1}.$$ But for $n \geq 4$, $\Bbb C P^{n-1}$ doesn't embed in $\Bbb C^n$.

So what does this mean in terms of Hamiltonian systems? Let $(M, \omega, H)$ be a Hamiltonian system with $k$ first integrals $f_1, \dots, f_k \in C^\infty(M)$ in involution, i.e. $$\{f_i, f_j\} = 0$$ for all $1 \leq i, j \leq k$. Then we can define a moment map by $$\mu: M \longrightarrow \Bbb R^k,$$ $$p \mapsto (f_1(p), \dots, f_k(p)),$$ and this corresponds to a Hamiltonian action of $\Bbb R^k$ on $(M,\omega)$. Then the Marsden-Weinstein reduction of $(M,\omega)$ with respect to the first integrals $f_1, \dots, f_k$ is just $$M /\!\!/ \Bbb R^k = \mu^{-1}(0)/\Bbb R^k.$$


Noncommutative Reduction of Hamiltonian Systems.

Now let us discuss reduction of a Hamiltonian system with non-involutive integrals. Let $(M, \omega, H)$ be a Hamiltonian system and suppose we have $k$ first integrals $f_1, \dots, f_k$, not necessarily in involution, such that the span of the $f_i$ is closed under the Poisson bracket and furthermore the $1$-forms $df_1, \dots, df_k$ are linearly independent. Then $$\mathfrak{g} = \mathrm{span} \{f_1, \dots, f_k\}$$ is a finite-dimensional Lie algebra with Lie bracket given by the Poisson bracket. Let $G$ be a simply connected Lie group corresponding to $\mathfrak{g}$. Then we have a symplectic action of $G$ on $(M, \omega)$. Now we can define a moment map for this action by $$\mu: M \longrightarrow \mathfrak{g}^\ast,$$ $$\mu(p)(f_i) = f_i(p)$$ on the $f_i$ and extending by linearity.

Suppose we want to reduce our Hamiltonian system at the level $\xi \in \mathfrak{g}^\ast$, i.e. we want to take the quotient $\mu^{-1}(\xi)/G$. Of course, this may not be possible, since $\mu^{-1}(\xi)$ may not be preserved under the action of $G$. To remedy this, consider the subalgebra $$\mathfrak{h} = \{h \in \mathfrak{g} : \mathrm{Ad}_h^\ast \xi = 0\} \subseteq \mathfrak{g}.$$ Then for the simply connected Lie group $H$ associated to $\mathfrak{h}$, the corresponding $H$-action on $(M,\omega)$ preserves $\mu^{-1}(\xi)$. Then we have the following.

Theorem. (Mishchenko-Fomenko) With the setup above, suppose that $\xi \in \mathfrak{g}^\ast$ is a regular value of $\mu$ and that the action of $H$ is free and proper. Then the reduced space $$\mu^{-1}(\xi)/H$$ is a manifold admitting a symplectic form $\tau$ such that $$i^\ast \omega = \pi^\ast \tau,$$ where $i: \mu^{-1}(\xi) \longrightarrow M$ is the inclusion map and $\pi: \mu^{-1}(\xi) \longrightarrow \mu^{-1}(\xi)/H$ is the natural projection.

The difference here from the involutive case is that our action is no longer an $\Bbb R^k$ action but something more complicated.

In terms of your toy example where you had two non-involutive integrals $f_1$ and $f_2$, your conditions $f_1 = \mathrm{constant}$ and $f_2 = \mathrm{constant}$ correspond to fixing a level $\xi \in \mathfrak{g}^\ast$ in the dual of the Lie algebra $\mathfrak{g}$ spanned by $f_1$ and $f_2$ (assuming that $\mathrm{span}\{f_1, f_2\}$ is closed under the Poisson bracket) and looking at $\mu^{-1}(\xi)$. This just gives the restricted phase space, however. To get a reduced phase space you have to quotient by the action of the simply connected Lie group associated to the annihilator subalgebra for $\xi$.

I haven't seen noncommutative symplectic reduction in any books, so here is the original reference for it:

  • A. S. Mishchenko, A. T. Fomenko, “Generalized Liouville method of integration of Hamiltonian systems”, Funkts. Anal. Prilozh., 12:2 (1978), 46–56.
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This is a wonderful answer. It's a shame so few people have upvoted this! –  Jason DeVito Jul 14 '12 at 3:37
    
Took me a great time to fully digest it. But your answer gives an excellent overview of the underlying theory for the involutive and non-involutive case. Thanks! –  Novo Jul 15 '12 at 15:29
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