# Symplectic reduction

The aim of symplectic reduction is to find a way of taking quotients of symplectic manifolds under group actions. For example, suppose we have a free action of $S^{1}$ on a symplectic manifold $X$ . Naively, we might hope to find a symplectic structure on the topological quotient $X/S^{1}$ . However, this cannot possibly work, since

$\dim \left(X/S^{1}\right)=\dim X-1$ is odd, and symplectic manifolds always have even dimension.

Instead, we use the following trick: if the action of $S^{1}$ is Hamiltonian, then we can cut down the dimension by $1$ by restricting the action to a level set of the moment map. Taking the quotient of this new manifold, we at least get something even-dimensional. The following proposition ensures that we get a natural symplectic structure.

Proposition 6.1

Let $(X,\omega )$ be a symplectic manifold with a Hamiltonian action of a compact Lie group $G$ and associated equivariant moment map $\mu :X\to {\mathfrak {g}}^{*}=\mathrm {Lie} (G)^{*}$ . If $0\in {\mathfrak {g}}^{*}$ is a regular value of $\mu$ such that $G$ acts freely on $M=\mu ^{-1}(0)$ , then $M/G$ is a symplectic manifold with symplectic structure induced by $\omega$ .

Definition 6.9

The symplectic manifold $M/G$ of the previous Proposition is called the symplectic reduction of $X$ , and is denoted by $X//_{0}G$ if the choice moment map is understood.

Remark 6.7

If $\mu :X\to {\mathfrak {g}}^{*}$ is any equivariant moment map and $c\in {\mathfrak {g}}^{*}$ is invariant under the coadjoint action of $G$ , then $\mu '(x)=\mu (x)-c$ defines another moment map. (In particular if $G$ is abelian, then we can do this for any $c\in {\mathfrak {g}}^{*}$ .) If $c$ is a regular value for $\mu$ such that the $G$ -action on $\mu ^{-1}(c)$ is free, then we can form another symplectic reduction

$(\mu ')^{-1}(0)/G=\mu ^{-1}(c)/G.$ In general, different values of $c$ will give different symplectic reductions. We often write
$X//_{c}G=\mu ^{-1}(c)/G,$ where the choice of moment map is implicit.

Proof (Sketch of proof of Proposition before)

Write $B=M/G$ . Since the $G$ -action on $M$ is free, $B$ is a manifold with tangent space

{\begin{aligned}T_{p}B&=T_{\tilde {p}}M/{\mathfrak {g}}\\&=T_{\tilde {p}}M/(\mathbb {R} v_{H_{1}}\oplus \cdots \oplus \mathbb {R} v_{H_{n}}),\end{aligned}} where ${\tilde {p}}\in M$ is any preimage of $p\in B$ , and $H_{1},\ldots ,H_{n}$ are the components for the moment map on $X$ corresponding to some basis for ${\mathfrak {g}}^{*}$ . Define the symplectic form $\omega _{B}\in \Omega ^{2}(B)$ by
$(\omega _{B})_{p}(v,w)=\omega _{\tilde {p}}({\tilde {v}},{\tilde {w}}),$ where ${\tilde {p}}\in M$ is a preimage of $p\in B$ , and ${\tilde {v}},{\tilde {w}}\in T_{\tilde {p}}M$ are preimages of $v,w\in T_{p}B$ . The form $\omega _{B}$ is well-defined since $\omega$ is invariant under $G$ and the $H_{i}$ are constant restricted to $M$ (so that $\omega (v_{H_{i}},{\tilde {w}})=0$ for all $w\in T_{p}M$ ). One can check that $\omega _{B}$ is indeed a symplectic form.

Example 6.9 ($\mathbb{CP}^n$)

Consider $S^{1}$ acting on $\mathbb {C} ^{n+1}$ diagonally by

$e^{i\theta }(z_{0},\ldots ,z_{n})=(e^{i\theta }z_{0},\ldots ,e^{i\theta }z_{n}).$ The moment map (which is just a Hamiltonian in this case) is
$\mu (z_{0},\ldots ,z_{n})={\frac {1}{2}}|z_{0}|^{2}+\cdots +{\frac {1}{2}}|z_{n}|^{2}.$ So for every $r>0$ , we get a symplectic structure on
$\mu ^{-1}(r^{2}/2)/S^{1}=S^{2n+1}/S^{1}=\mathbb {CP} ^{n}.$ The associated symplectic form is the unique form $\omega _{FS}$ such that
$\pi ^{*}\omega _{FS}=\omega _{\mathbb {C} ^{n+1}}|_{S^{2n+1}},$ where $\pi :S^{2n+1}\to \mathbb {CP} ^{n}$ is the quotient map.

Exercise 6.7

Show that for an appropriate choice of $r$ , $\omega _{FS}$ agrees with the explicit expression for the Fubini-Study form in the lecture on Kähler geometry.

Example 6.10

The $S^{1}$ -action of Example before factors through the $T^{n+1}$ -action

$(e^{i\theta _{0}},\ldots ,e^{i\theta _{n}})(z_{0},\ldots ,z_{n})=(e^{i\theta _{0}}z_{0},\ldots ,e^{i\theta _{n}}z_{n}),$ via the diagonal map
{\begin{aligned}S^{1}&\longrightarrow T^{n+1}\\e^{i\theta }&\longmapsto (e^{i\theta },\cdots ,e^{i\theta }).\end{aligned}} So we get a residual action of $T=T^{n+1}/S^{1}$ on the symplectic reduction $\mathbb {CP} ^{n}=\mathbb {C} ^{n+1}//_{r^{2}/2}S^{1}$ with moment map
$\mu ':\mathbb {CP} ^{n}=H^{-1}(r^{2}/2)/S^{1}\longrightarrow \mathbb {R} ^{n+1},$ induced by the moment map
{\begin{aligned}\mu :\mathbb {C} ^{n+1}&\longrightarrow \mathbb {R} ^{n+1}\\(z_{0},\ldots ,z_{n})&\longmapsto \left({\frac {1}{2}}|z_{0}|^{2},\ldots ,{\frac {1}{2}}|z_{n}|^{2}\right)\end{aligned}} for the $T^{n+1}$ -action. Note that the image of $\mu '$ is contained in
$\{(x_{0},\ldots ,x_{n})\in \mathbb {R} ^{n+1}\mid x_{0}+\cdots +x_{n}=r^{2}/2\},$ which, up to translation, is the same as
$\operatorname {Lie} (T)^{*}\subseteq \operatorname {Lie} (T^{n+1})^{*}=\mathbb {R} ^{m+1}.$ So $\mu '$ does make sense as a moment map. For example, the moment image of $\mathbb {CP} ^{1}$ is the interval shown below.

The moment image of $\mathbb {CP} ^{2}$ is the triangle below.

In general, the moment image of $\mathbb {CP} ^{n}$ is the $n$ -simplex

$\{(x_{0},x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n+1}\mid x_{i}\geq 0,\,x_{0}+x_{1}+\cdots +x_{n}=1\}.$ 