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

Question

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 $a_{ij}$?

Answer 1

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 $i\le j$.