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 $ℝ$. 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 $\{df|f\in A\}$ modulo the relations

• $d(f+g)=df+dg$

• $dfg=gdf+fdg$

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

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

• $\omega (fX+gY)=f\omega (X)+g\omega (Y)$ for all $f,g\in A$ and all $X,Y\in T^{*}(M)$

For this map to be well-defined, 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 well-defined 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 non-degenerate, $\omega$ is invertible. Denote its inverse by $\mathrm{\Omega }$. Since our manifold is finite-dimensional, 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 2-form) 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 $\mathrm{\Omega }$ isclosed. If $u,v,w\in T(M)$ then, by a general identity ofdifferential geometry,

$$⟨d\mathrm{\Omega },u\wedge v\wedge w⟩=⟨u,d⟨\mathrm{\Omega },v\wedge w⟩⟩+⟨v,d⟨\mathrm{\Omega },w\wedge u⟩⟩+⟨w,d⟨\mathrm{\Omega },u\wedge v⟩⟩$$

Since this identity is trilinear in $u,v,w$, we can restrict attentionto a generating set. Because of the non-degeneracy assumption, vectorfields of the form $a{d}_f$ where $f$ is a function form such a set.

By the definition of $\mathrm{\Omega }$, we have $⟨\mathrm{\Omega },a{d}_f\wedge a{d}_g⟩=[f,g]$. Then $⟨a{d}_f,d⟨\mathrm{\Omega },a{d}_g\wedge a{d}_h⟩=[f,[g,h]]$so the Jacobi identity is satisfied.