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!

  • $\begingroup$ possible duplicate of Poisson bracket identities/properties $\endgroup$ Oct 14 '12 at 18:30
  • 1
    $\begingroup$ @HansLundmark I think the question you are referencing deserves closing, not this one. $\endgroup$
    – Norbert
    Oct 15 '12 at 15:44
  • $\begingroup$ @Norbert: Fair enough. This is a better formulated question. (Although the other one was asked first, of course.) $\endgroup$ 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.)

  • $\begingroup$ You're welcome. I just noticed that I had made a few mistakes, but now it should be OK. $\endgroup$ Oct 16 '12 at 8:58
  • $\begingroup$ 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) $\endgroup$
    – Daphne P
    Oct 17 '12 at 6:17
  • $\begingroup$ 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!! $\endgroup$
    – Daphne P
    Oct 17 '12 at 6:21
  • $\begingroup$ OK, I've filled in the details now. $\endgroup$ Oct 17 '12 at 9:39
  • $\begingroup$ Hans, you're a saint! Thank you! $\endgroup$
    – Daphne P
    Oct 17 '12 at 12:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.