formula in differential-geometry

Question

$(M,\omega)$ is a symplectic manifold, $\omega=d \lambda$, then I want to prove that: $$ i_v\lambda\cdot\omega^n=n\lambda\wedge i_v\omega\wedge\omega^{n-1}. $$

Answer 1

Let $N$ be a smooth manifold and $\alpha$ and $\beta$ differential forms and $X$ a vector field on it. Claim: $\imath_X(\alpha\wedge\beta)=\imath_X(\alpha)\wedge\beta+(-1)^{\mathrm{degree}(\alpha)}\alpha\wedge\imath_X(\beta)$.

Now you just need to use this formula in your particular case.

Here is an example, suppose that $\mathrm{degree}(\alpha)=k$ and $\mathrm{degree}(\beta)=k-1$:

\begin{equation} \imath_X(\alpha\wedge\beta\wedge\beta)=\beta(X)\alpha\wedge\alpha+(-1)^{k-1}\beta\wedge\imath_X(\alpha\wedge\alpha) \\ =\beta(X)\alpha\wedge\alpha+(-1)^{k-1}\beta\wedge\imath_X(\alpha)\wedge\alpha+(-1)^{k-1}(-1)^k\beta\wedge\alpha\wedge(\imath_X\alpha) \ . \end{equation}