if the algebra of functions on a manifold is a Poisson ring then the manifold is symplectic
Let $M$ be a smooth manifold and let $A$ be the algebra^{} of smooth functions from $M$ to $\mathbb{R}$. Suppose that there exists a bilinear operation $[,]:A\times A\to A$ which makes $A$ a Poisson ring.
For this proof, we shall use the fact that ${T}^{*}(M)$ is the sheafification^{} of the $A$module generated by the set $\{dff\in A\}$ modulo the relations^{}

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$d(f+g)=df+dg$

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$dfg=gdf+fdg$
Let us define a map $\omega :{T}^{*}(M)\to T(M)$ by the following conditions:

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$\omega (df)(g)=[f,g]$ for all $fg\in A$

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$\omega (fX+gY)=f\omega (X)+g\omega (Y)$ for all $f,g\in A$ and all $X,Y\in {T}^{*}(M)$
For this map to be welldefined, it must respect the relations:
$$\omega (f+g)(h)=[f+g,h]=[f,h]+[g,h]=\omega (f)(h)+\omega (g)(h)$$ 
$$\omega (fg)(h)=[fg,h]=f[g,h]+g[f,h]=f\omega (g)(h)+g\omega (g)(h)$$ 
These two equations show that $\omega $ is a welldefined map from the presheaf^{} hence, by general nonsense, a well defined map from the sheaf. The fact that $\omega (fdg)$ is a derivation^{} readily follows from the fact that $[,]$ is a derivation in each slot.
Since $[,]$ is nondegenerate, $\omega $ is invertible^{}. Denote its inverse by $\mathrm{\Omega}$. Since our manifold is finitedimensional^{}, we may naturally regard $\mathrm{\Omega}$ as an element of ${T}^{*}(M)\otimes {T}^{*}(M)$. The fact that $\mathrm{\Omega}$ is an antisymmetric tensor field (in other words, a 2form) follows from the fact that $\mathrm{\Omega}(df)(g)=[f,g]=[g,f]=\mathrm{\Omega}(dg)(f)$.
Finally, we will use the Jacobi identity^{} to show that $\mathrm{\Omega}$ is closed. If $u,v,w\in T(M)$ then, by a general identity^{} of differential geometry,
$$\u27e8d\mathrm{\Omega},u\wedge v\wedge w\u27e9=\u27e8u,d\u27e8\mathrm{\Omega},v\wedge w\u27e9\u27e9+\u27e8v,d\u27e8\mathrm{\Omega},w\wedge u\u27e9\u27e9+\u27e8w,d\u27e8\mathrm{\Omega},u\wedge v\u27e9\u27e9$$ 
Since this identity is trilinear in $u,v,w$, we can restrict attention to a generating set^{}. Because of the nondegeneracy assumption, vector fields^{} of the form $a{d}_{f}$ where $f$ is a function form such a set.
By the definition of $\mathrm{\Omega}$, we have $\u27e8\mathrm{\Omega},a{d}_{f}\wedge a{d}_{g}\u27e9=[f,g]$. Then $\u27e8a{d}_{f},d\u27e8\mathrm{\Omega},a{d}_{g}\wedge a{d}_{h}\u27e9=[f,[g,h]]$ so the Jacobi identity is satisfied.
Title  if the algebra of functions on a manifold is a Poisson ring then the manifold is symplectic 

Canonical name  IfTheAlgebraOfFunctionsOnAManifoldIsAPoissonRingThenTheManifoldIsSymplectic 
Date of creation  20130322 14:46:34 
Last modified on  20130322 14:46:34 
Owner  rspuzio (6075) 
Last modified by  rspuzio (6075) 
Numerical id  18 
Author  rspuzio (6075) 
Entry type  Theorem 
Classification  msc 53D05 