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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?

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Yes, it is much easier than doing pages of calculus:

In coordinates it always suffices to show the Jacobi identity for coordinate functions $(f,g,h)=(x_i, x_j, x_k)$ with $i<j<k$. 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 $\mathbb{R}$.

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    $\begingroup$ Thank you! Really useful. Can you state a proper argument for which it is sufficient to test on the coordinate functions? $\endgroup$ – Dac0 Oct 13 '18 at 10:22
  • $\begingroup$ 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 $\tfrac{1}{2}[\![P,P]\!]^{ijk} = \{\{x_i, x_j\},x_k\} + \{\{x_j, x_k\},x_i\} + \{\{x_k, x_i\},x_j\}$. $\endgroup$ – Ricardo Buring Oct 13 '18 at 10:41
  • $\begingroup$ but I don't know yet that these are Poisson Bracket, is that fine anyway? $\endgroup$ – Dac0 Oct 13 '18 at 10:45
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    $\begingroup$ 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 $\tfrac{1}{2}[\![P,P]\!]^{ijk} = 0$. $\endgroup$ – Ricardo Buring Oct 13 '18 at 11:23

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