# Describing the Poisson Structure on $\mathbb{R}^3$ to show $\mathbb{R}^3$ is a Poisson manifold.

We say that $$M$$ is a Poisson manifold if $$M$$ is a smooth manifold with a Lie bracket $$\{\cdot,\cdot\}$$ on $$C^{\infty}(M)$$ where $$\{f,g\}=p(df\wedge dg)$$, $$p\in\Gamma(\Lambda^2TM)$$.

I try to understand and see what are the possible Lie brackets $$\{\cdot,\cdot\}$$ for $$\mathbb{R}^3$$. So, I can show that $$(\mathbb{R}^3,\{\cdot,\cdot\})$$ is a Poisson manifold.

To show that $$\{\cdot,\cdot\}$$ is a Lie bracket, we need to show that it satisfies Jacobi identity (JI) i.e. $$\{f,\{g,h\}+\{g,\{h,f\}+\{h,\{f,g\}\}=0\text{ }(*).$$

We know that if $$p\in\Gamma(\Lambda^2T\mathbb{R}^3)$$, then $$p=a_{12}\partial_1\wedge \partial_2+a_{13}\partial_1\wedge \partial_3+a_{23}\partial_2\wedge \partial_3$$ in local coordinates where $$\partial_i=\frac{\partial}{\partial x_i}$$. So, we want to show that our bracket $$\{\cdot,\cdot\}$$ will satisfy for $$JI$$ for a good choice of $$p$$ i.e. $$\{a_{ij}\}$$. I tried to use the fact that $$\{f,g\}=p(df\wedge dg)=\sum_{i=1}^3\sum_{j=1}^3\frac{\partial f}{\partial x_i}\frac{\partial g}{\partial x_j}\{x_i,x_j\}$$ and just do straightforward computations to figure out what conditions we need to put on $$a_{ij}$$, but I failed.

Is there a nice way deriving the conditions on $$a_{ij}$$?

• @QiaochuYuan thank you for your reply. I will take a look. However, in this article: en.wikipedia.org/wiki/Poisson_manifold, they say "The condition that the ensuing {⋅,⋅} be a Poisson bracket — i.e., satisfy the Jacobi identity — can be characterized by the non-linear partial differential equation [p,p]=0." So, I am a little bit confused now. But, I think I need to try to figure out that blog post first. However, if you will elaborate on that, I will greatly appreciate. Thanks! Commented Sep 29, 2020 at 0:45
• Ah, sorry, I misremembered; it's been awhile since I wrote that post. What I said earlier doesn't include satisfying the Jacobi identity. What I think should still be true is that the Jacobi identity is satisfied iff it's satisfied for the $x_i$; that is, iff $\{ \{ x_i, x_j \}, x_k \} + \{ x_j, x_k \}, x_i \} + \{ \{ x_k, x_i \}, x_j \} = 0$. On $\mathbb{R}^3$ this is a single condition to check. Commented Sep 29, 2020 at 0:57

The coefficients $$a_{12},a_{13}, a_{23}\in C^{\infty}(\mathbb{R}^3)$$ need to satisfy the following equation \begin{align*}\sum_{l=1}^3 a_{lk}\partial_la_{ij}+a_{li}\partial_l a_{jk} +a_{lj}\partial_la_{ki} =0 \end{align*} for all $$i,j,k\in \{1,2,3\}$$. As mentinoed in the comments it is induced by evalutating the Jacobi identity on the coordinate functions $$x_i$$. Here $$a_{ji}:=-a_{ij}$$ for $$i\le j$$.