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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.

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    $\begingroup$ This is not true. It fails quite generally. See papers by Li Sheng Tseng with Tsai and Yau and others. $\endgroup$
    – Dan Fox
    Commented Jan 21, 2022 at 6:04
  • $\begingroup$ @DanFox which paper? is it "Cohomology and Hodge Theory on symplectic manifolds"? $\endgroup$
    – Uncool
    Commented Jan 21, 2022 at 6:09
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    $\begingroup$ One standard keyword you should be looking at is "strong Lefschetz property". $\endgroup$ Commented Jan 21, 2022 at 6:44
  • $\begingroup$ @Uncool: The paper you mention is an overview, if I remember right. There is a series of numbered papers that have all the details. $\endgroup$
    – Dan Fox
    Commented Jan 21, 2022 at 9:18
  • $\begingroup$ @DanFox Yes its a series of paper $\endgroup$
    – Uncool
    Commented Jan 21, 2022 at 9:43

1 Answer 1

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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 and its answers.

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  • $\begingroup$ I think both injectivity and surjectivity follow from the statement mentioned by you. $\endgroup$
    – Uncool
    Commented Jun 2, 2022 at 13:09
  • $\begingroup$ @Uncool: Yes, that's correct. $\endgroup$ Commented Jun 2, 2022 at 15:41

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