How do I show $X_{\omega(Y,Z)}=-[Y,Z]$, where $\omega$ is a symplectic 2 form (in particular non-degenerate) and $Y,Z$ are vector fields and $X_f$ is the vector field correspond to the 1 form $df$ under the pairing $\omega$.
When applied to a 1-form $\alpha$, LHS is $\omega(di_Yi_Z\omega,\alpha)$, and RHS is $d(\alpha(Y))Z-i_Yi_Zd\alpha-i_Ydi_Z\alpha$.
I must be missing something trivial...
The Poisson bracket between the differentiable functions $F$ and $G$ on a symplectic manifold $M$ is defined by the formula $$ \{F,G\} = \omega(Y, Z)= dG(Y) = \mathcal{L}_{Y} G, $$ where $\mathcal{L}$ is the Lie derivative and the symplectic vector fields $Y$ and $Z$ are defined by $$ dF=-\iota_Y\omega,\qquad dG=-\iota_Z \omega. $$ Recalling that $$ \iota_{[Y,Z]} = \mathcal{L}_Y\circ\iota_Z-\iota_Z\circ \mathcal{L}_Y, $$ we have finally that $$ \iota_{[Y,Z]} \omega = -\mathcal{L}_Y dG = -d\mathcal{L}_Y G = -d\{F,G\} = -d(\omega(Y,Z)), $$ where the fact that $\mathcal{L}_Y\omega = 0$ was used.