I'm attempting a question under Hamiltonian Dynamics. We are given that $\omega^{ab}$ is an antisymmetric matrix such that it's components depend on coordinates $x^a$ and such that the Poisson bracket $$ \{f,g\} = \omega^{ab} \frac{\partial f}{\partial x^a} \frac{\partial g}{\partial x^b}$$ satisfies the usual Jacobi identity.
In the first part of the question we prove $\{fg, h\} = f\{g,h\} + \{f,h\}g$. This was fine though I include it in case it is relevant in what follows.
Next we are asked to prove $$\partial_a W_{bc} + \partial_b W_{ca} + \partial_c W_{ab} = 0 \tag{$*$} $$ where we denote $W = \omega^{-1}$ and $\partial_a = \frac{\partial}{\partial x^a}$. A given hint is that we note $\omega^{ab} = \{x^a, x^b\}$.
To prove $(*)$ I have tried differentiating $W_{ac}\omega^{cb} = \delta_a^b$. I have tried using the Jacobi identity, of which $(*)$ is reminiscent. I have got nowhere.
Any answers or suggestions are appreciated. Thank you.
Ok so after a long break I finally got it - frustratingly its a just doing what I'd tried more carefully:
We take $W = \omega^{-1}$, then by definition $\delta_a^b = \omega^{ac} W_{cb}$ and differentiating this yields $$ 0 = \partial_i (\omega^{ac}W_{cb}) = \omega^{ac} (\partial_i W_{cb}) + (\partial_i \omega^{ac})W_{cb} \quad \implies \quad \omega^{ac} \omega^{bd} (\partial_i W_{cb}) = - W_{cb}\omega^{bd} (\partial_i \omega^{ac}) = - \partial_i \omega^{ad}\,. $$ Next in preparation for use in the Jacobi identity we consider that $$ \{x^a, \{x^b, x^c\}\} = \{x^a, \omega^{bc}\} = \omega^{pq} (\partial_p x^a) (\partial_q \omega^{bc}) = \omega^{aq} \partial_q\omega^{bc} = - \omega^{aq} \omega^{bp}\omega^{rc}(\partial_q W_{pr}) = \omega^{ap} \omega^{bq} \omega^{cr} (\partial_p W_{qr}) $$ where the first equality is the hint, the second is a definition, the third is that $\partial_px^a = \delta_p^a$, the forth is from above, and the final uses the antisymmetry of $\omega$ and relabels $p\leftrightarrow q$.
Then from the Jacobi identity we have \begin{align*} 0 &= \{x^a, \{x^b, x^c\}\} + \{x^b, \{x^c, x^a\}\} + \{x^c, \{x^a, x^b\}\}\\ &= \omega^{ap} \omega^{bq} \omega^{cr} (\partial_p W_{qr}) + \omega^{bp} \omega^{cq} \omega^{ar} (\partial_p W_{qr}) +\omega^{cp} \omega^{aq} \omega^{br} (\partial_p W_{qr}) \\ &=\omega^{ap} \omega^{bq} \omega^{cr} (\partial_p W_{qr}) + \omega^{ap} \omega^{bq} \omega^{cr} (\partial_q W_{rp}) + \omega^{ap} \omega^{bq} \omega^{cr} (\partial_r W_{pq})\\ &=\omega^{ap} \omega^{bq} \omega^{cr} [\partial_p W_{qr} + \partial_q W_{rp} + \partial_r W_{pq} ] \end{align*} and hence applying $W_{ia} W_{jb} W_{kc}$ to each side then yields $$ 0 = \partial_i W_{jk} + \partial_j W_{ki} + \partial_k W_{ij}\,. $$