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.

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

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