proof of the Jacobi Identity for certain poisson brackets

I have to prove that these are effectively Poisson bracket. Specifically that the satisfy Jacobi Identity when $$a_{ij}=-a_{ji}$$. $$\left\{ f,g\right\} =\stackrel{\scriptscriptstyle i,j=1..3}{\sum}\left(a_{ij}+\stackrel{\scriptscriptstyle k=1..3}{\sum}\epsilon_{ijk}x^{k}\right)\frac{\partial f}{\partial x^{i}}\frac{\partial g}{\partial x^{j}},$$ I tried the plain and direct way but it involves pages of calculus... so I thought: maybe is there a smart way that I didn't see to prove it?

In coordinates it always suffices to show the Jacobi identity for coordinate functions $$(f,g,h)=(x_i, x_j, x_k)$$ with $$i. Since we are on $$\mathbb{R}^3$$ we only have to show it for $$(x_1,x_2,x_3)$$: $$\{\{x_1,x_2\},x_3\} + \{\{x_2,x_3\},x_1\} + \{\{x_3,x_1\},x_2\}=0.$$ For the "inner" brackets you can leave out the constants $$a_{ij}$$ because they will be differentiated away by the "outer" brackets. Then in each term the "inner" bracket will be (up to irrelevant constants) equal to $$\pm$$ the other argument in the "outer" bracket (the sum over $$k$$ contains only one nonzero term), so each term is zero by skew-symmetry.
By the way this is the sum of two Poisson structures, a constant one and the usual one on $$\mathfrak{so}(3)^*$$ with Casimir $$\tfrac{1}{2}(x_1^2 + x_2^2 + x_3^2)$$; the Jacobi identity for the sum is equivalent to their compatibility.
Since you tagged representation theory: the constant bracket defines a linear map $$C: \mathfrak{so}(3) \wedge \mathfrak{so}(3) \to \mathbb{R}$$ by $$(x_i,x_j) \mapsto a_{ij}$$ and compatibility is equivalent to saying that $$C$$ is a $$2$$-cocycle in the cohomology of $$\mathfrak{so}(3)$$ associated with the trivial representation of $$\mathfrak{so}(3)$$ on $$\mathbb{R}$$.
• The basic idea is: just as the Poisson bracket $P$ is a bi-vector field with coefficients $P^{ij} = \{x_i, x_j\}$ (see this expression), the left-hand side of the Jacobi identity is a tri-vector field with coefficients $\tfrac{1}{2}[\![P,P]\!]^{ijk} = \{\{x_i, x_j\},x_k\} + \{\{x_j, x_k\},x_i\} + \{\{x_k, x_i\},x_j\}$. – Ricardo Buring Oct 13 '18 at 10:41
• Yes. Starting from the expression of $P$ in local coordinates which I linked (which you already have, but which also exists in general) you can show that the Jacobi identity is equivalent to $\tfrac{1}{2}[\![P,P]\!]^{ijk} = 0$. – Ricardo Buring Oct 13 '18 at 11:23