How to show that Jacobi identity for $\{,\}$ is equivalent to $\omega$ being closed?

Question

I am reading the book a guide to quantum groups. I have a question on page 18. How to show that Jacobi identity for $\{,\}$ is equivalent to $\omega$ being closed? Any help will be greatly appreciated!

Answer 1

When you have a symplectic manifold $(M,\omega)$, the Poisson bracket is given by $\{f,g\} = \omega(X_f,X_g)$, where $X_f$ is the Hamiltonian vector field. So the triple bracket expressions from the Jacobi identity of the form $\{f,\{g,h\}\}$ can be expressed as $\omega (X_f, [X_h,X_g])$, where we are using that $X_{\{g,h\}} = -[X_g,X_h]$. So now compare:

The Jacobi identity for the bracket says

$$ \begin {eqnarray} 0 &=& \{f,\{g,h\}\} + \{h,\{f,g\}\} + \{g,\{h,f\}\} \\ &=& \omega(X_f, [X_h,X_g]) + \omega(X_h, [X_g,X_f]) + \omega(X_g, [X_f,X_h]) \end {eqnarray} $$

Now compare with the "coordinate-free" expression for the exterior derivative (see the wikipedia page, under the heading "in terms of invariant formula"):

$$ \begin{eqnarray} d\omega(X_f,X_g,X_h) &=& X_f \omega(X_g,X_h) - X_g \omega(X_f,X_h) + X_h \omega(X_f,X_g) \\ &\phantom{=}& - \omega([X_f,X_g],X_h) + \omega([X_f,X_h],X_g) - \omega([X_g,X_h],X_f) \end{eqnarray} $$

Using one more identity, that $X_f \omega(X_g,X_h) = \omega(X_f, [X_g,X_h])$, you can see that the top and bottom rows on the right hand side are the same, and so the right-hand side is twice the expression you get from the Jacobi identity. Since that is zero, so is $d \omega$.