if the algebra of functions on a manifold is a Poisson ring then the manifold is symplectic
Let be a smooth manifold and let be the algebra of smooth functions from to . Suppose that there exists a bilinear operation which makes a Poisson ring.
For this proof, we shall use the fact that is the sheafification of the -module generated by the set modulo the relations
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Let us define a map by the following conditions:
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for all
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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 .
Finally, we will use the Jacobi identity to show that is closed. If then, by a general identity of differential geometry,
Since this identity is trilinear in , we can restrict attention to a generating set. Because of the non-degeneracy assumption, vector fields of the form where is a function form such a set.
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 |
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Canonical name | IfTheAlgebraOfFunctionsOnAManifoldIsAPoissonRingThenTheManifoldIsSymplectic |
Date of creation | 2013-03-22 14:46:34 |
Last modified on | 2013-03-22 14:46:34 |
Owner | rspuzio (6075) |
Last modified by | rspuzio (6075) |
Numerical id | 18 |
Author | rspuzio (6075) |
Entry type | Theorem |
Classification | msc 53D05 |