Let $(M^{2n},\omega)$ be a symplectic manifold of dimension $2n$. Let $L^k\_{\omega}:\Omega^q(M)\to \Omega^{q+2k}(M)$ be the map given by $L^k\_{\omega}(\alpha)=\alpha\wedge\omega^k$. Then is it true that $L^k\_{\omega}$ is injective for all $q\leq n-k$ and surjective for all $q\geq n-k$?


It seems that this is true, but I could not find a proof for it. Can anyone provide a proof for this, or atleast point out where can I get the proof for this.



$\omega$ is a symplectic form then $L^k_{\omega}:\Omega^q(M)\to \Omega^{q+2k}(M)$ is injective for all $q\leq n-k$ and surjective for all $q\geq n-k$
The injectivity case is well-known and follows quite easily from the statement (usually attributed to Lefschetz) that $L^k\_\omega:\Omega^{n-k}(M)\to \Omega^{n+k}(M)$ is a isomorphism for $0\le k\le n$.


This is a purely linear algebra statement and only relies on $\omega$ being nondegenerate, i.e., that $\omega^n$ be nonvanishing; $\omega$ does not need to be closed for this isomorphism to hold.


N.B.: Just for clarity's sake, let me point out that I am assuming that the OP intends $\Omega^p(M)$ to mean the module of (smooth) $p$-forms on $M$, as is standard. I'm not sure why people are bringing up comments about cohomology and hard Lefschetz, as the question (as I understand it) really has nothing to do with that.


For a discussion of the linear algebra result, see [this MO question](https://mathoverflow.net/questions/316505/the-lefschetz-operator/316527#316527) and its answers.


