Poisson manifold

Overview

A Poisson manifold is a generalization of symplectic manifolds. It is a way to describe families of phase spaces of physical systems.

There are two ways to define a Poisson manifold. One uses the Poisson bracket on the algebra of smooth functions $(C^{\infty }(M))$. The other uses bivector fields.

Poisson bracket definition

A Poisson manifold $(M,\{-,-\})$ is a smooth manifold $M$ and a Poisson bracket on the algebra of smooth functions $C^{\infty }(M)$

$\{-,-\}:C^{\infty }(M)\times C^{\infty }(M)⟶C^{\infty }(M).$

A morphism in the category of Poisson manifolds is a Poisson map.

Bivector field definition

A Poisson manifold $(M,\sigma )$ is a smooth manifold $M$ with a bivector fields $\sigma \in {\mathfrak{X}}^2(M)$ satisfying $[\sigma ,\sigma ]=0$ where $[-,-]$ denotes the Schouten bracket.

Intuition behind conditions

For a bivector field $\sigma$, we can look at $\sigma (\text{d}f,\text{d}g)=\{f,g\}$

Since $\{f,g\}={\mathcal{L}}_{\sigma }(f,g)$, then we know that the bracket is an anti-symmetric derivation. That takes care of the antisymmetric and Liebneiz rule condition on the Poisson bracket given the definition above.

Why do we need the condition $[\sigma ,\sigma ]=0$?

Let $\pi \in {\mathfrak{X}}^2(M)$, using definition of the Schouten bracket we can look at ${\mathcal{L}}_{[\pi ,\pi ]}(f,g,h)$ for functions $f,g,h\in C^{\infty }(M)$

${\mathcal{L}}_{[\pi ,\pi ]}(f,g,h)=({\mathcal{L}}_{\pi }\circ {\mathcal{L}}_{\pi }-(-1{)}^1{\mathcal{L}}_{\pi }\circ {\mathcal{L}}_{\pi })(f,g,h)=2({\mathcal{L}}_{\pi }\circ {\mathcal{L}}_{\pi })(f,g,h)$

In order to compute ${\mathcal{L}}_{\pi }\circ {\mathcal{L}}_{\pi }$ we can see that there are only 3 (1,2)-shuffles.

$(1,2,3)(2,1,3)(3,1,2)$

Therefore, letting $(f_1,f_2,f_3)=(f,g,h)$, using the second part of the explicit construction of the Schouten bracket we have

$$\begin{array}{rl}{\mathcal{L}}_{\pi }\circ {\mathcal{L}}_{\pi }& =(1){\mathcal{L}}_{\pi }(f,{\mathcal{L}}_{\pi }(g,h))+(-1){\mathcal{L}}_{\pi }(g,{\mathcal{L}}_{\pi }(f,h))+(1){\mathcal{L}}_{\pi }(h,{\mathcal{L}}_{\pi }(f,g))\\ & =\pi (\text{d}f,\pi (\text{d}g,\text{d}h))-\pi (\text{d}g,\pi (\text{d}f,\text{d}h))+\pi (\text{d}h,\pi (\text{d}f,\text{d}g))\\ & =\pi (\text{d}f,\pi (\text{d}g,\text{d}h))+\pi (\text{d}g,\pi (\text{d}h,\text{d}f))+\pi (\text{d}h,\pi (\text{d}f,\text{d}g))\\ & =\{f,\{g,h\}\}+\{g,\{f,h\}\}+\{h,\{f,g\}\}\end{array}$$

with the Poisson bracket defined above.

Therefore, $\frac{1}{2}{\mathcal{L}}_{[\pi ,\pi ]}(f,g,h)=\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}.$

So the only way a derivation induced from a bivector field will satisfy the Jacobi identity is if and only if $[\pi ,\pi ]=0$.