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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, ), 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 ( e i , f i ) = 1, ( e i , f j ) = 0 for i negationslash = j , and ( f i , f j ) = 0. Because ( e 1 , . . . , e n , f 1 , . . . , f k 1 ) is free, there exists a (nonunique) linear form V such that ( e i ) = 0 for i negationslash k , ( f i ) = 0 for i < k , and ( e k ) = 1. Using the fact that is = nondegenerate (induces an isomorphism between V and V ), there exists f k V such that ( , f k ) = . Let us check that the family ( e 1 , . . . , e n , f 1 , . . . , f k ) is free. Indeed, if v = n i =1 i e i + k i =1 i f i = 0, then ( e i , v ) = i = 0 for all 1 i k , and v = i e i = 0; since the ( e i ) form a basis of E , we also have i = 0 for all i . Moreover, ( e i , f k ) and ( 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 (its a free family and dim V = 2 n ) and the expression of 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 n = [ ] n [ S 2 n ] = 0). ] = 0)....
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 Spring '10
 DenisAuroux
 Geometry, Vector Space

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