User:LSGNT/Advanced Topics In Geometry/Symplectic reduction/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 S¹ on a symplectic manifold X. Naively, we might hope to
find a symplectic structure on the topological quotient X/S¹. 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 S¹ 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, ω) be a symplectic manifold with a Hamiltonian action of a compact
Lie group G and associated equivariant moment map μ: X →\mathfrak{g} ^* =
\mathrm{Lie} (G)^*. If 0 \in \mathfrak{g} ^* is a regular value of μ such
that G acts freely on M = μ^{-1}(0), then M/G is a symplectic manifold with
symplectic structure induced by ω.

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 μ: X →\mathfrak{g} ^* is any equivariant moment map and c \in
\mathfrak{g} ^* is invariant under the coadjoint action of G, then μ'(x) =
μ(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
μ such that the G-action on μ^{-1}(c) is free, then we can form another
symplectic reduction

(μ')^{-1}(0)/G = μ^{-1}(c)/G.

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

X//_c G = μ^{-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

T_p B = T_{\tilde{p} } M/\mathfrak{g} &= T_{\tilde{p} } M / (\mathbb{R}
v_{H_1} \oplus ⋯\oplus \mathbb{R} v_{Hₙ}),

where \tilde{p} \in M is any preimage of p \in B, and H_1, …, Hₙ are the
components for the moment map on X corresponding to some basis for
\mathfrak{g} ^*. Define the symplectic form ω_B \in Ω²⁽B) by

(ω_B)_p(v, w) = ω_{\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 TₚB. The form ω_B is
well-defined since ω is invariant under G and the H_i are constant
restricted to M (so that ω(v_{H_i}, \tilde{w} ) = 0 for all w \in TₚM). One
can check that ω_B is indeed a symplectic form.

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

Consider S¹ acting on \mathbb{C} ^{n + 1} diagonally by

e^{iθ} (z_0, …, z_n) = (e^{iθ}z_0, …,e^{iθ}zₙ₎.

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

μ(z_0, …, z_n) = \frac{1}{2} |z_0|^2 + ⋯+ \frac{1}{2} |zₙ|².

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

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

The associated symplectic form is the unique form ω_{FS} such that

π^* ω_{FS} = ω_{\mathbb{C} ^{n + 1}}|_{S^{2n + 1}},

where π:S^{2n + 1} →\mathbb{CP} ⁿ is the quotient map.

Exercise 6.7

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

Example 6.10

The S¹-action of Example before factors through the T^{n + 1}-action

(e^{iθ_0}, …, e^{iθ_n})(z_0, …, z_n) = (e^{iθ_0}z_0, …, e^{iθₙ}zₙ₎,

via the diagonal map

S^1 \longrightarrow T^{n + 1} e^{iθ} &\longmapsto (e^{iθ}, ⋯, e^{iθ}).

So we get a residual action of T = T^{n + 1}/S¹ on the symplectic reduction
\mathbb{CP} ^n = \mathbb{C} ^{n + 1}//_{r²/2} S¹ with moment map

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

induced by the moment map

μ: \mathbb{C} ^{n + 1} \longrightarrow \mathbb{R} ^{n + 1} (z_0, …, z_n)
&\longmapsto \left (\frac{1}{2} |z_0|^2, …, \frac{1}{2} |z_n|^2\right )

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

{(x_0, …, x_n) \in \mathbb{R} ^{n + 1} \mid x_0 + ⋯+ 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 μ' 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, …, x_n) \in \mathbb{R} ^{n + 1} \mid x_i \geq 0, x_0 + x_1 + ⋯+
x_n = 1}.