# 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.

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possible duplicate of Poisson bracket identities/properties –  Hans Lundmark Oct 14 '12 at 18:30
@HansLundmark I think the question you are referencing deserves closing, not this one. –  Norbert Oct 15 '12 at 15:44
@Norbert: Fair enough. This is a better formulated question. (Although the other one was asked first, of course.) –  Hans Lundmark Oct 15 '12 at 16:53

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.)
Hans, is there a reason why ${\partial W_{jk}\over \partial \xi_i}+{\partial W_{ki}\over \partial \xi_j}+{\partial W_{ij}\over \partial \xi_k}=0,\,\,\,\,\,\forall i,j,k=1,...,2n\implies$ Jacobi identity? (i.e. the other way round) –  Daphne P Oct 17 '12 at 6:17
Also, I'd like to know that if there is an obvious reason why ${\partial W_{jk}\over \partial \xi_i}+{\partial W_{ki}\over \partial \xi_j}+{\partial W_{ij}\over \partial \xi_k}=0,\,\,\,\,\,\forall i,j,k=1,...,2n$ follows from the point your answer ends? I am so sorry for keep asking things which are probably quite simple. I don't seem to be able to get my head around these structures. Also thanks a lot for answering!! –  Daphne P Oct 17 '12 at 6:21