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

$$\dim \left(X/S^{1}\right)=\dim X-1$$

$$\dim \left(X/S^{1}\right)=\dim X-1$$

is odd, and symplectic manifolds always have even dimension.

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

 

Definition 6.9

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

 

Remark 6.7

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

$$(\mu ')^{-1}(0)/G=\mu ^{-1}(c)/G.$$

$$(\mu ')^{-1}(0)/G=\mu ^{-1}(c)/G.$$

In general, different values of $c$$c$ will give different symplectic reductions. We often write

$$X//_{c}G=\mu ^{-1}(c)/G,$$

$$X//_{c}G=\mu ^{-1}(c)/G,$$

where the choice of moment map is implicit.

 

Proof (Sketch of proof of Proposition before)

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

$${\begin{aligned}T_{p}B&=T_{\tilde {p}}M/{\mathfrak {g}}\\&=T_{\tilde {p}}M/(\mathbb {R} v_{H_{1}}\oplus \cdots \oplus \mathbb {R} v_{H_{n}}),\end{aligned}}$$

$${\begin{aligned}T_{p}B&=T_{\tilde {p}}M/{\mathfrak {g}}\\&=T_{\tilde {p}}M/(\mathbb {R} v_{H_{1}}\oplus \cdots \oplus \mathbb {R} v_{H_{n}}),\end{aligned}}$$

where ${\tilde {p}}\in M$${\tilde {p}}\in M$ is any preimage of $p\in B$$p\in B$, and $H_{1},\ldots ,H_{n}$$H_{1},\ldots ,H_{n}$ are the components for the moment map on $X$$X$ corresponding to some basis for ${\mathfrak {g}}^{*}$${\mathfrak {g}}^{*}$. Define the symplectic form $\omega _{B}\in \Omega ^{2}(B)$$\omega _{B}\in \Omega ^{2}(B)$ by

$$(\omega _{B})_{p}(v,w)=\omega _{\tilde {p}}({\tilde {v}},{\tilde {w}}),$$

$$(\omega _{B})_{p}(v,w)=\omega _{\tilde {p}}({\tilde {v}},{\tilde {w}}),$$

where ${\tilde {p}}\in M$${\tilde {p}}\in M$ is a preimage of $p\in B$$p\in B$, and ${\tilde {v}},{\tilde {w}}\in T_{\tilde {p}}M$${\tilde {v}},{\tilde {w}}\in T_{\tilde {p}}M$ are preimages of $v,w\in T_{p}B$$v,w\in T_{p}B$. The form $\omega _{B}$$\omega _{B}$ is well-defined since $\omega$$\omega$ is invariant under $G$$G$ and the $H_{i}$$H_{i}$ are constant restricted to $M$$M$ (so that $\omega (v_{H_{i}},{\tilde {w}})=0$$\omega (v_{H_{i}},{\tilde {w}})=0$ for all $w\in T_{p}M$$w\in T_{p}M$). One can check that $\omega _{B}$$\omega _{B}$ is indeed a symplectic form.

 

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

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

$$e^{i\theta }(z_{0},\ldots ,z_{n})=(e^{i\theta }z_{0},\ldots ,e^{i\theta }z_{n}).$$

$$e^{i\theta }(z_{0},\ldots ,z_{n})=(e^{i\theta }z_{0},\ldots ,e^{i\theta }z_{n}).$$

The moment map (which is just a Hamiltonian in this case) is

$$\mu (z_{0},\ldots ,z_{n})={\frac {1}{2}}|z_{0}|^{2}+\cdots +{\frac {1}{2}}|z_{n}|^{2}.$$

$$\mu (z_{0},\ldots ,z_{n})={\frac {1}{2}}|z_{0}|^{2}+\cdots +{\frac {1}{2}}|z_{n}|^{2}.$$

So for every $r>0$$r>0$, we get a symplectic structure on

$$\mu ^{-1}(r^{2}/2)/S^{1}=S^{2n+1}/S^{1}=\mathbb {CP} ^{n}.$$

$$\mu ^{-1}(r^{2}/2)/S^{1}=S^{2n+1}/S^{1}=\mathbb {CP} ^{n}.$$

The associated symplectic form is the unique form $\omega _{FS}$$\omega _{FS}$ such that

$$\pi ^{*}\omega _{FS}=\omega _{\mathbb {C} ^{n+1}}|_{S^{2n+1}},$$

$$\pi ^{*}\omega _{FS}=\omega _{\mathbb {C} ^{n+1}}|_{S^{2n+1}},$$

where $\pi :S^{2n+1}\to \mathbb {CP} ^{n}$$\pi :S^{2n+1}\to \mathbb {CP} ^{n}$ is the quotient map.

 

Exercise 6.7

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

 

Example 6.10

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

$$(e^{i\theta _{0}},\ldots ,e^{i\theta _{n}})(z_{0},\ldots ,z_{n})=(e^{i\theta _{0}}z_{0},\ldots ,e^{i\theta _{n}}z_{n}),$$

$$(e^{i\theta _{0}},\ldots ,e^{i\theta _{n}})(z_{0},\ldots ,z_{n})=(e^{i\theta _{0}}z_{0},\ldots ,e^{i\theta _{n}}z_{n}),$$

via the diagonal map

$${\begin{aligned}S^{1}&\longrightarrow T^{n+1}\\e^{i\theta }&\longmapsto (e^{i\theta },\cdots ,e^{i\theta }).\end{aligned}}$$

$${\begin{aligned}S^{1}&\longrightarrow T^{n+1}\\e^{i\theta }&\longmapsto (e^{i\theta },\cdots ,e^{i\theta }).\end{aligned}}$$

So we get a residual action of $T=T^{n+1}/S^{1}$$T=T^{n+1}/S^{1}$ on the symplectic reduction $\mathbb {CP} ^{n}=\mathbb {C} ^{n+1}//_{r^{2}/2}S^{1}$$\mathbb {CP} ^{n}=\mathbb {C} ^{n+1}//_{r^{2}/2}S^{1}$ with moment map

$$\mu ':\mathbb {CP} ^{n}=H^{-1}(r^{2}/2)/S^{1}\longrightarrow \mathbb {R} ^{n+1},$$

$$\mu ':\mathbb {CP} ^{n}=H^{-1}(r^{2}/2)/S^{1}\longrightarrow \mathbb {R} ^{n+1},$$

induced by the moment map

$${\begin{aligned}\mu :\mathbb {C} ^{n+1}&\longrightarrow \mathbb {R} ^{n+1}\\(z_{0},\ldots ,z_{n})&\longmapsto \left({\frac {1}{2}}|z_{0}|^{2},\ldots ,{\frac {1}{2}}|z_{n}|^{2}\right)\end{aligned}}$$

$${\begin{aligned}\mu :\mathbb {C} ^{n+1}&\longrightarrow \mathbb {R} ^{n+1}\\(z_{0},\ldots ,z_{n})&\longmapsto \left({\frac {1}{2}}|z_{0}|^{2},\ldots ,{\frac {1}{2}}|z_{n}|^{2}\right)\end{aligned}}$$

for the $T^{n+1}$$T^{n+1}$-action. Note that the image of $\mu '$$\mu '$ is contained in

$$\{(x_{0},\ldots ,x_{n})\in \mathbb {R} ^{n+1}\mid x_{0}+\cdots +x_{n}=r^{2}/2\},$$

$$\{(x_{0},\ldots ,x_{n})\in \mathbb {R} ^{n+1}\mid x_{0}+\cdots +x_{n}=r^{2}/2\},$$

which, up to translation, is the same as

$$\operatorname {Lie} (T)^{*}\subseteq \operatorname {Lie} (T^{n+1})^{*}=\mathbb {R} ^{m+1}.$$

$$\operatorname {Lie} (T)^{*}\subseteq \operatorname {Lie} (T^{n+1})^{*}=\mathbb {R} ^{m+1}.$$

So $\mu '$$\mu '$ does make sense as a moment map. For example, the moment image of $\mathbb {CP} ^{1}$$\mathbb {CP} ^{1}$ is the interval shown below.

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

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

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

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