# 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 ${\displaystyle S^{1}}$ on a symplectic manifold ${\displaystyle X}$. Naively, we might hope to find a symplectic structure on the topological quotient ${\displaystyle X/S^{1}}$. However, this cannot possibly work, since

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

Instead, we use the following trick: if the action of ${\displaystyle S^{1}}$ is Hamiltonian, then we can cut down the dimension by ${\displaystyle 1}$ 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 ${\displaystyle (X,\omega )}$ be a symplectic manifold with a Hamiltonian action of a compact Lie group ${\displaystyle G}$ and associated equivariant moment map ${\displaystyle \mu :X\to {\mathfrak {g}}^{*}=\mathrm {Lie} (G)^{*}}$. If ${\displaystyle 0\in {\mathfrak {g}}^{*}}$ is a regular value of ${\displaystyle \mu }$ such that ${\displaystyle G}$ acts freely on ${\displaystyle M=\mu ^{-1}(0)}$, then ${\displaystyle M/G}$ is a symplectic manifold with symplectic structure induced by ${\displaystyle \omega }$.

Definition 6.9

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

Remark 6.7

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

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

Proof (Sketch of proof of Proposition before)

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

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

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

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

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

Exercise 6.7

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

Example 6.10

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

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

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

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

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