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I am wondering if the following argument is true:

Let $X$ be a $\dim n$ compact projective complex manifold, let $\alpha\in H^{2n-2}(X,\mathbb{Q})$ be a cohomology class. If for any ample line bundle $L$, we have $c_1(L)\cup \alpha=0$, can I argue that $\alpha=0$?

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    $\begingroup$ If $\alpha\in H^{2k}(X\mathbb{Q})$ and $2k<n$, where $n=\dim_\mathbb{C}(X)$ then this is a consequence of the Hard Lefschetz Theorem, which says that if $L$ is ample, then the map $(c_1(L))^{r} \cup -: H^{n-r}(X,\mathbb{Q}) \to H^{n+r}(X,\mathbb{Q})$ is an isomorphism. $\endgroup$
    – Balazs Elek
    Commented Nov 21, 2019 at 2:58
  • $\begingroup$ @BalazsElek Dear Balaza, thanks for your reply. I am a bit confused here. Say n=dim X, and k=n-1. HL says that $H^2\to H^{2n-2}$ is isomorphism, but I am actually consider $\cup c_1(L)\colon H^{2n-2}\to H^2n}$ here ? $\endgroup$
    – Winnie_XP
    Commented Nov 21, 2019 at 3:03
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    $\begingroup$ Yes, my comment only answers your question if $2k<n$. $\endgroup$
    – Balazs Elek
    Commented Nov 21, 2019 at 3:08
  • $\begingroup$ @BalazsElek I see, thanks. Let me edit my post for more details. $\endgroup$
    – Winnie_XP
    Commented Nov 21, 2019 at 3:12

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No. If $\alpha $ is of type $(n-2,n)$, its product with any class of type $(1,1)$ is zero, but $\alpha $ is not necessarily zero (you can take for instance $\alpha = c_1(L)^{n-2}[\omega ]$, where $L$ is an ample line bundle and $\omega $ a nonzero holomorphic 2-form).

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    $\begingroup$ @Winnie_XP Yes it should be true in this case. $H^{n-1,n-1}(X)$ and $H^{1,1}(X)$ are dual under cup product. The ample cone in $H^{1,1}(X)$ is open (and spans $H^{1,1}(X)$). So if a class $\alpha\in H^{n-1,n-1}(X)$ vanishes after cupping with arbitrary ample, then it vanishes when cupping with an arbitrary (1,1)-class. By duality it is 0. $\endgroup$
    – Yosemite Stan
    Commented Nov 21, 2019 at 7:35
  • $\begingroup$ @YosemiteStan Thanks ! I just realized i was asking a stupid question. Thanks for the reply ! $\endgroup$
    – Winnie_XP
    Commented Nov 21, 2019 at 7:43
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    $\begingroup$ @Yosemite Stan: no, in general the ample cone does not span $H^{1,1}$, and is not open there. Think of a general surface of degree $\geq 4$ in $\Bbb{P}^3$: the ample cone is a half-line, while $H^{1,1}$ is quite large. $\endgroup$
    – abx
    Commented Nov 21, 2019 at 9:45
  • $\begingroup$ @abx But isn't the tensor with enough power of an ample line bundle makes arbitrary line bundle ample ? I know this is a result for Noetherian schemes, but I suppose that it also holds for cpt projective algebraic complex manifolds ? $\endgroup$
    – Winnie_XP
    Commented Nov 22, 2019 at 1:39
  • $\begingroup$ @abx Oops! I guess I should say that the real (1,1) Hodge classes are dual to the real (n-1,n-1) Hodge classes and that the ample cone is open in the space of real Hodge classes. I think that corrects the argument... $\endgroup$
    – Yosemite Stan
    Commented Nov 22, 2019 at 6:05

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