every symplectic manifold has even dimension
All we need to prove is that every finite dimensional vector space with an anti-symmetric non-degenerate linear form has an even dimension . This is only a linear algebra result. In the case of a symplectic manifold is just the tangent space at a point, and thus its dimension equals the manifold’s dimension.
Pick any not null vector . Since is non-degenerate is a not null linear form. Therefore there exists a not null vector such that
Now and are linearly independent because if then (by anti-symmetry).
Let . Consider a space of ”orthogonal” elements to under . That is:
We now prove :
Suppose is not null, then it can be written because it belongs to . Since it also belongs to is is ”orthogonal” to both and . That is:
So must be null.
Suppose . Let , , .
Then and .
Considering we have (by construction) and and similarly for
So , and and thus
If is not null we can repeat the procedure with the restriction. Since and is finite dimensional the procedure must stop at a finite step.
At the end we get a decomposition , where and is even.
|Title||every symplectic manifold has even dimension|
|Date of creation||2013-03-22 15:44:05|
|Last modified on||2013-03-22 15:44:05|
|Last modified by||cvalente (11260)|