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 algebraMathworldPlanetmathPlanetmath of smooth functions from M to . Suppose that there exists a bilinear operation [,]:A×AA which makes A a Poisson ring.

For this proof, we shall use the fact that T*(M) is the sheafificationPlanetmathPlanetmath of the A-module generated by the set {df|fA} modulo the relationsPlanetmathPlanetmath

  • d(f+g)=df+dg

  • dfg=gdf+fdg

Let us define a map ω:T*(M)T(M) by the following conditions:

  • ω(df)(g)=[f,g] for all fgA

  • ω(fX+gY)=fω(X)+gω(Y) for all f,gA and all X,YT*(M)

For this map to be well-defined, it must respect the relations:

ω(f+g)(h)=[f+g,h]=[f,h]+[g,h]=ω(f)(h)+ω(g)(h)
ω(fg)(h)=[fg,h]=f[g,h]+g[f,h]=fω(g)(h)+gω(g)(h)

These two equations show that ω is a well-defined map from the presheafPlanetmathPlanetmathPlanetmath hence, by general nonsense, a well defined map from the sheaf. The fact that ω(fdg) is a derivationMathworldPlanetmath readily follows from the fact that [,] is a derivation in each slot.

Since [,] is non-degenerate, ω is invertiblePlanetmathPlanetmath. Denote its inverse by Ω. Since our manifold is finite-dimensionalPlanetmathPlanetmath, we may naturally regard Ω as an element of T*(M)T*(M). The fact that Ω is an antisymmetric tensor field (in other words, a 2-form) follows from the fact that Ω(df)(g)=[f,g]=-[g,f]=-Ω(dg)(f).

Finally, we will use the Jacobi identityMathworldPlanetmath to show that Ω is closed. If u,v,wT(M) then, by a general identityPlanetmathPlanetmath of differential geometry,

dΩ,uvw=u,dΩ,vw+v,dΩ,wu+w,dΩ,uv

Since this identity is trilinear in u,v,w, we can restrict attention to a generating setPlanetmathPlanetmath. Because of the non-degeneracy assumption, vector fieldsMathworldPlanetmath of the form adf where f is a function form such a set.

By the definition of Ω, we have Ω,adfadg=[f,g]. Then adf,dΩ,adgadh=[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 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