if the algebra of functions on a manifold is a Poisson ring then the manifold is symplectic
Let us define a map by the following conditions:
for all and all
For this map to be well-defined, it must respect the relations:
These two equations show that is a well-defined map from the presheaf hence, by general nonsense, a well defined map from the sheaf. The fact that is a derivation readily follows from the fact that is a derivation in each slot.
Since is non-degenerate, is invertible. Denote its inverse by . Since our manifold is finite-dimensional, we may naturally regard as an element of . The fact that is an antisymmetric tensor field (in other words, a 2-form) follows from the fact that .
By the definition of , we have . Then so the Jacobi identity is satisfied.
|Title||if the algebra of functions on a manifold is a Poisson ring then the manifold is symplectic|
|Date of creation||2013-03-22 14:46:34|
|Last modified on||2013-03-22 14:46:34|
|Last modified by||rspuzio (6075)|