The n-Category Café

Symplectic reduction is about forming quotients by group actions of symplectic manifolds.

So consider $(X,\omega)$ be a symplectic manifold with symplectic 2-form $\omega$ and let $G$ be a compact Lie group acting on $X$ by symplectomorphisms:

$$R : X \times G \to X \,.$$

Instead of trying to directly form the quotient $X/G$, one forms the quotient of subsets of $X$ obtained as follows:

Write $g$ for the Lie algebra of $G$. For each $\xi \in g$ we have the corresponding vector field $\xi_X \in \Gamma(T X)$ along $X$ $$\xi_X(x) = R(x,\cdot)_* \xi \,.$$

With respect to $\omega$ this vector field has a Hamiltonian generating function $J(\xi) \in C^\infty(X)$, which means that $$\iota_{\xi_X} \omega = d (J(\xi)) \,.$$

This construction is well behaved in $\xi$ so that $J$ is indeed a map $$J : X \to g^*$$ from $X$ to the linear dual space of $g$. This is the moment map of the $G$-action (since it generalizes the concept of angular momentum). This map is $G$-equivariant with respect to the coadjoint action of $G$ on its dual Lie algebra $g^*$.

The preimage $J^{-1}(0)$ of $0 \in g^*$ under $J$ is a submanifold of $X$

$$\array{ J^{-1}(0) &\stackrel{i}{\hookrightarrow}& X &\stackrel{J}{\to}& g^* } \,,$$ but not a symplectic one. However, it still has $G$ acting on it. Under some conditions ($0$ being a regular value of $J$ and $G$ acting freely and properly on $J^{-1}(0)$) the quotient of $J^{-1}(0)$ by $G$ is a symplectic manifold $$\array{ J^{-1}(0) &\stackrel{i}{\hookrightarrow}& X &\stackrel{J}{\to}& g^* \\ \downarrow \\ (J^{-1}(0)/G, \tilde \omega) }$$ called the symplectic quotient or symplectic reduction or Marsden-Weinstein quotient.

Notice how it involves in a way quotienting by $G$ twice. Accordingly, the dimension of the quotient is that of $X$ minus twice that of $G$.

The same story can be told in terms of function algebras in a way that generalizes also to Poisson reduction.

So let now $(P,\{\cdot,\cdot\})$ be a manifold $P$ with Poisson bracket $\{\cdot,\cdot\}$ on its algebra of functions such that $(C^\infty(X), \{\cdot,\cdot\})$ is a Poisson algebra.

For $I \subset C^\infty(X)$ an algebra ideal (not necessarily a Poisson ideal), for instance the collection of functions vanishing on some submanifold $$C \hookrightarrow X\,,$$ we say that $I$ is first class if it is closed under the Poisson bracket $$\{I,I\} \subset I \,.$$ For a Poisson ideal we would even have $\{I, C^\infty(X)\} \subset I$.

The quotient algebra $$C^\infty(X)/I \,,$$ which plays the role of functions on $C$ if $I$ is functions that vanish on $C$, is not in general itself a Poisson manifold. But if we further restrict to the algebra $$\left( C^\infty(X) / I \right)^{I}$$ of those elements in there which are invariant under the Poisson action $\{I, \cdot\}$ by $I$, then this does inherit an induced Poisson structure, simply because for those elements $$\{ f + I , g + I\} = \{g,f\} + I \,,$$ by assumption. This is the reduced Poisson algebra.

Notice how it is again not just a single quotient but a two-step quotient one performs. In a way $I$ is divided out twice.

To make the connection with symplectic reduction, assume that the Poisson structure is actually symplectic and identify the submanifold $C$ with the level set of the moment map $$C := J^{-1}(0)$$ and notice that then $$\{J_\xi , \cdot\} = \xi_X \,,$$ where the left hand side is a derivation which we identify with the action of the vector field $\xi_X$. Assume that the symplectic quotient exists as a symplectic manifold.

Then the reduced Poisson algebra is the algebra of functions on the symplectic quotient:

$$\left( C^\infty(X)/I \right)^I \simeq C^\infty(J^{-1}(0)/G) \,.$$

Next time: the BV complex giving a cohomological realization of this quotient.