# Cup Product with Ample Line Bundles

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 $$\alpha=0$$?

• 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. Nov 21, 2019 at 2:58
• @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 ? Nov 21, 2019 at 3:03
• Yes, my comment only answers your question if $2k<n$. Nov 21, 2019 at 3:08
• @BalazsElek I see, thanks. Let me edit my post for more details. Nov 21, 2019 at 3:12

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).
• @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. Nov 21, 2019 at 7:35
• @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.