7 Symplectic Quotients
... Theorem 7.6 Suppose G acts freely on Φ−1 (0). Then 0 is a regular value of Φ and M0 = Φ−1 (0)/G is a smooth manifold. Theorem 7.7 In the notation of the previous theorem, M0 is symplectic. Proof: Define Tm̄ M0 = Tm Φ−1 (0)/(g · m = Tm (G ◦ m)). Let η¯1 , η¯2 ∈ Tm Φ−1 (0). The quotient by Tm (G · m). ...
... Theorem 7.6 Suppose G acts freely on Φ−1 (0). Then 0 is a regular value of Φ and M0 = Φ−1 (0)/G is a smooth manifold. Theorem 7.7 In the notation of the previous theorem, M0 is symplectic. Proof: Define Tm̄ M0 = Tm Φ−1 (0)/(g · m = Tm (G ◦ m)). Let η¯1 , η¯2 ∈ Tm Φ−1 (0). The quotient by Tm (G · m). ...
L6: Almost complex structures To study general symplectic
... The tangent spaces to M are naturally complex vector spaces, which carry multiplication by i. More generally, for almost complex (M, J), J extends complex-linearly to T M ⊗ C , and splits this space into ±i-eigenspaces T M ⊗ C = ∼ T M , and T 1,0(M ) ⊕ T 0,1(M ). So T 1,0(M ) = R in the complex case ...
... The tangent spaces to M are naturally complex vector spaces, which carry multiplication by i. More generally, for almost complex (M, J), J extends complex-linearly to T M ⊗ C , and splits this space into ±i-eigenspaces T M ⊗ C = ∼ T M , and T 1,0(M ) ⊕ T 0,1(M ). So T 1,0(M ) = R in the complex case ...
Some Notes on Compact Lie Groups
... In particular, the embeddings SU (2) ,→ SU (n) and Sp(1) ,→ Sp(n) induce isomorphisms at the level of π3 . For SO(n), we use π3 (Sp(2)) ∼ = π3 (SO(5)) that follows from the relation (iii) and the homotopy exact sequence (for G = Sp(2), H = {±12 }). By this we find a map Sp(1) → SO(n) for n = 5, 6, 7 ...
... In particular, the embeddings SU (2) ,→ SU (n) and Sp(1) ,→ Sp(n) induce isomorphisms at the level of π3 . For SO(n), we use π3 (Sp(2)) ∼ = π3 (SO(5)) that follows from the relation (iii) and the homotopy exact sequence (for G = Sp(2), H = {±12 }). By this we find a map Sp(1) → SO(n) for n = 5, 6, 7 ...