The Jacobi Identity for the Poisson Bracket

The Jacobi identity for the Poisson bracket does indeed follow from the fact that $d\Omega =0$.

I claim that (twice) the Jacobi identity for functions $f,g,h$ is precisely $$d\Omega(X_f,X_g,X_h) = 0.$$

To see this, simply expand $d\Omega$.

You will find six terms of two kinds:

• three terms of the form $$X_f \Omega(X_g,X_h) = X_f \lbrace g,h \rbrace$$

• and three terms of the form $$\Omega([X_f,X_g],X_h).$$

To deal with the first kind of terms, notice that from the definition of $X_f$, for any function $g$, $$X_f g = \lbrace g, f \rbrace.$$ This means that $$X_f \Omega(X_g,X_h) = \lbrace \lbrace g,h \rbrace, f \rbrace.$$

To deal with the second kind of terms, notice that $$\iota_{[X_f,X_g]}\Omega = [L_{X_f},\iota_{X_g}]\Omega,$$ but since $d\Omega=0$, $$L_{X_f}\Omega = d \iota_{X_f}\Omega = 0,$$ and hence $$\iota_{[X_f,X_g]}\Omega = d \iota_{X_f}\iota_{X_g}\Omega = d\lbrace g,f\rbrace,$$ whence $$\Omega([X_f,X_g],X_h) = d\lbrace g,f\rbrace (X_h) = \lbrace\lbrace g,f\rbrace, h \rbrace.$$

Adding it all up you get twice the Jacobi identity.

I wanted to add this as a comment to Jose's answer but it seems that I cannot do that as a new user.

For any bivector field $\sigma$, you can define a bracket on smooth functions by $\{f,g\} = \sigma(df, dg)$. This bracket is skew and will automatically satisfy the Liebniz rule. It will satisfy the Jacobi identity precisely when $[\sigma, \sigma] = 0$, where $[\cdot,\cdot]$ is the Schouten bracket. This point of view is important, for example, in defining Poisson cohomology.

Now suppose you take $\sigma = \omega^{-1}$, where $\omega$ is a nondegenerate 2-form. Jose's calcuation shows that $[\sigma, \sigma] = 0$ iff $d\omega = 0$.

Here's the way it's done in John Lee's Smooth Manifolds book:

$${\small \iota_{X_{\lbrace f,g\rbrace}}\omega = d\lbrace f,g \rbrace=d(X_gf) = d(\mathcal{L}_{X_g}f) =\mathcal{L}_{X_g}df=\mathcal{L}_{X_g}(\iota_{X_f}\omega)=\iota_{[X_g,X_f]}\omega + \iota_{X_f}\mathcal{L}_{X_g}\omega=\iota_{[X_g,X_f]}\omega} $$ which by nondegeneracy of $\omega$ implies the desired result.