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Unformatted text preview: 18.966 – Homework 1 – Solutions. 1. Let E be a Lagrangian subspace of a symplectic vector space ( V, Ω), and let e 1 , . . . , e n be a basis of E . We proceed by induction, assuming we have constructed f 1 , . . . , f k − 1 ∈ V such that the family ( e 1 , . . . , e n , f 1 , . . . , f k − 1 ) is free and Ω( e i , f i ) = 1, Ω( e i , f j ) = 0 for i negationslash = j , and Ω( f i , f j ) = 0. Because ( e 1 , . . . , e n , f 1 , . . . , f k − 1 ) is free, there exists a (nonunique) linear form τ ∈ V ∗ such that τ ( e i ) = 0 for i negationslash k , τ ( f i ) = 0 for i < k , and τ ( e k ) = 1. Using the fact that Ω is = nondegenerate (induces an isomorphism between V and V ∗ ), there exists f k ∈ V such that Ω( · , f k ) = τ . Let us check that the family ( e 1 , . . . , e n , f 1 , . . . , f k ) is free. Indeed, if v = n i =1 λ i e i + k i =1 µ i f i = 0, then Ω( e i , v ) = µ i = 0 for all 1 ≤ i ≤ k , and v = λ i e i = 0; since the ( e i ) form a basis of E , we also have λ i = 0 for all i . Moreover, Ω( e i , f k ) and Ω( f i , f k ) are as prescribed. Therefore, by induction we can construct f 1 , . . . , f n such that ( e 1 , . . . , e n , f 1 , . . . , f n ) is a basis of V (it’s a free family and dim V = 2 n ) and the expression of Ω in this basis is the standard one. 2. S 2 is an orientable surface and hence carries a symplectic structure (its standard area form, for example); however, for n ≥ 2, the compact manifold S 2 n has H 2 ( S 2 n , R ) = 0, so it cannot be symplectic (for any closed 2form, S 2 n ω n = [ ω ] ∪ n · [ S 2 n ] = 0). ] = 0)....
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 Spring '10
 DenisAuroux
 Geometry, Vector Space, Ω, Symplectic manifold, Symplectic geometry, ψT, Np Σ0

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