# How do I show $X_{\omega(Y,Z)}=-[Y,Z]$?

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.