# Jacobi identity in local coordinates?

1. Apr 26, 2013

### quasar987

Jacobi identity in local coordinates?!?

Apparently (i.e. according to an article written by physicists), the Jacobi identity for the Poisson bracket associated to a Poisson bivector $\pi = \sum\pi^{ij}\partial_i\wedge\partial_j$ is equivalent to $\sum_{\text{cyclic}}\pi^{i\nu}\partial_{\nu}\pi^{jk}=0$ the sum is over all cyclic permutation of the i,j,k indices and the summation convention is used on the nu index. It is easy to see that this identity is equivalent to $\sum_{\text{cyclic}}\{x^i,\{x^j,x^k\}\}$ but why does this imply the general Jacobi identity?!?

2. Apr 27, 2013

### quasar987

Recall (or learn) that by definition, $\{f,g\}:=\pi(df,dg) = \sum\pi^{ij}\partial_if \partial_j g$.