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For a compact Kaehler manifold $M$, a basic structural result for its de Rham cohomology is the hard Lefschetz theorem. See here or here for an overview of the result.

What happens in the non-compact case? Does the theorem fail, or is it just not understood?

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Hard Lefschetz implies Poincaré duality with real coefficients, in particular $H^0\cong H^{2n}$ and for a non-compact connected manifold that never happens ($H^{2n}=0$).

For example: $M:=\mathbb{C}P^2\setminus\mathbb{C}P^1$ is contractible, but hard Lefschetz theorem (as stated) should have given an isomorphism between $H^0(M)=\mathbb{R}$ and $H^4(M)=0$.

UPD. Poincaré duality modification for non-compact manifolds (pointed out by BS.) suggests a Lefschetz-type map $\mu: H_c^{n-k}\to H^{n+k}$ defined by multiplication by $\omega^k$: $H_c^{n-k}\to H_c^{n+k}$ and subsequent composition with the canonical map from $H_c$ to $H$. $\mu$, hovewer, needs not be an isomorphism:

for $\mathbb{C}^2\setminus \{0\}$ Lefschetz map is identically zero, but $H^{3}=H_c^1$ are one-dimensional.

So, all in all, hard Lefschetz theorem fails for non-compact Kahler manifolds.

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  • $\begingroup$ But are these examples K\"ahler manifolds? $\endgroup$ Commented Oct 4, 2019 at 10:45
  • $\begingroup$ @PierreDubois yes, an open subset of a Kahler manifold is Kahler $\endgroup$ Commented Oct 4, 2019 at 11:16
  • $\begingroup$ For non-compact manifolds, there is a duality between (de Rham) cohomologies $H^k$ and $H^{n-k}_c$ --- compactly supported. $\endgroup$
    – BS.
    Commented Oct 4, 2019 at 11:27
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    $\begingroup$ @DmitryK Cohomology with compact support works fine in the algebraic category. Although it doesn't have the same functoriality properties as usual cohomology, it has others. $\endgroup$
    – Will Sawin
    Commented Oct 5, 2019 at 3:57
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    $\begingroup$ The answers of mathoverflow.net/q/47852/6451 (about mixed Hodge structures) could perhaps help, at least in the algebraic case. $\endgroup$
    – BS.
    Commented Oct 5, 2019 at 8:15

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