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}.