TITLE: Dual Lefschetz Operator and Contraction with the Fundamental Form
QUESTION [5 upvotes]: Let $M$ be a Kahler manifold, with metric $g$, fundamental form $\omega$, and dual Lefschetz operator $\Lambda$. Now $\Lambda$, and contraction with $\omega$, both map the two forms $\Omega^2(M)$ to $0$-forms, ie smooth functions. Are they equal? I think this is almost certainly true, but I can't see a clean argument.
Do I need Kahler here? I would guess this works for all complex manifolds.

REPLY [5 votes]: The question asks whether $\boxed{ \Lambda \alpha = g(\omega,\alpha)}$ for any $\alpha \in \Omega^2(X)$.
This is indeed true, as explained by Gunnar, but here is a simple way of seeing it. Firstly, since we're just using linear operators we can work fiber-wise (hence indeed no closedness of $\omega$ is needed, and whether we work in the context of $TM$ or general vector bundles is irrelevant). Remember that by definition $g(\Lambda \eta, \mu) = g(\eta, L \mu)$ where $L$ is the wedge product with $\omega$.
Using this in our case: let $\alpha \in \Lambda^2 E_x$, then we can rewrite $\boxed{\Lambda \alpha}$ as $g(\Lambda \alpha, 1) = g(\alpha, L \; 1) = \boxed{ g(\alpha, \omega) }$.