Invertible antisymmetric matrix and identities A link to the page is available here. The relevant bit is on P. 15 of the book. I would really appreciate it if somebody could help!
It is probably something quite obvious, hence left out by the author, but I don't seem to see it! 
Could someone please explain the following I've read in Solitons, Instantons and Twistors.
(I have changed the notation a bit -- I am more used to $"i,j,k"$)
$\xi_i$ are coordinates, with $i=1,...,2n$
Suppose $w^{ij}$ is an invertible, antisymmetric matrix 
Define $$\{f,g\}:=\sum_{i,j=1}^{2n} w^{ij}(\xi){\partial f\over \partial \xi_i}{\partial g\over \partial \xi_j}$$
and it satisfies $$\{a,\{b,c\}\}+\{b,\{c,a\}\}+\{c,\{a,b\}\}=0$$
Let $W_{ij}:=(w^{-1})_{ij}$
Why is it that it follows that $${\partial W_{jk}\over \partial \xi_i}+{\partial W_{ki}\over \partial \xi_j}+{\partial W_{ij}\over \partial \xi_k}=0,\,\,\,\,\,\forall i,j,k=1,...,2n$$?
It is also said that $$w^{ij}(\xi)=\{\xi^i,\xi^j\}$$
Thank you.
Please help!
 A: That $w^{ij}(\xi)=\{\xi^i,\xi^j\}$ follows directly from taking $f=\xi^i$ and $g=\xi^j$ in the definition of the Poisson bracket.
The other fact is not as obvious, although it's a rather standard calculation.
By taking $a$, $b$ and $c$ to be $\xi^i$, $\xi^j$ and $x^k$ in the Jacobi identity, you get
$$
\sum_r w^{ir} (\partial_r w^{jk}) + (\text{two similar terms obtain by cyclic permutation of $i$, $j$, $k$}) = 0
.
$$
The formula for the derivative of the inverse of a matrix says that $\partial_r w^{jk} = - \sum_{s,t} w^{js} (\partial_r W_{st}) w^{tk}$, hence
$$
- \sum_{r,s,t} w^{ir} w^{js} (\partial_r W_{st}) w^{tk} + \text{cyclic} = 0
.
$$
Using $w^{tk} = -w^{kt}$ and writing everything out, we have
$$
\sum_{r,s,t} w^{ir} w^{js} w^{kt} (\partial_r W_{st}) +
\sum_{r,s,t} w^{jr} w^{ks} w^{it} (\partial_r W_{st}) +
\sum_{r,s,t} w^{kr} w^{is} w^{jt} (\partial_r W_{st})
= 0
.
$$
Now in the second sum we relabel the summation indices: write $s$ for what previously was called $r$, write $t$ for $s$, and $r$ for $t$. Similarly for the third sum:
$$
\sum_{r,s,t} w^{ir} w^{js} w^{kt} (\partial_r W_{st}) +
\sum_{s,t,r} w^{js} w^{kt} w^{ir} (\partial_s W_{tr}) +
\sum_{t,r,s} w^{kt} w^{ir} w^{js} (\partial_t W_{rs})
= 0
.
$$
In other words,
$$
\sum_{r,s,t} w^{ir} w^{js} w^{kt} (\partial_r W_{st} + \partial_s W_{tr} + \partial_t W_{rs}) = 0
.
$$
Now multiply by $W_{ai} W_{bj} W_{ck}$ and sum over $i$, $j$, $k$. This will cancel the $w$ factors and leave you with
$$
\partial_a W_{bc} + \partial_b W_{ca} + \partial_c W_{ab} = 0
.
$$
(And it should be pretty clear that you can run this argument backwards too.)
