# Neron-Severi group $= H^{(1,1)}(X,\mathbb{Z})$

Where does my confusion arise from? I am getting $$H^{(1,1)}(X,\mathbb{Z})=NS(X)$$, the Neron-Severi group, however in any reference I don't find this remarkable thing.

Let $$X$$ be a Kahler manifold. Using the exponential sequence one obtains a homomorphism $$H^1(X,\mathcal{O}_X^*)\rightarrow H^2(X,\mathbb{Z})$$. This is associating to a holomorphic line bundle its Chern class $$c_1$$.

If we take any connection $$D$$ on a holomorphic line bundle $$L$$, if we do $$D\circ D$$ and then take its class in $$H^2(X,\mathbb{C})$$ we get a curvature form independent of the choice of the connection. Multiplying by $$-i/(2\pi)$$ we run into $$c_1(L)$$ when identifying $$H^2(X,\mathbb{Z})$$ as a subspace of $$H^2(X,\mathbb{C})$$. This is a theorem. The last in particualr shows that the image of $$c_1$$ is always $$(1,1)$$ form.

On the other hand, the map $$H^1(X, O_X^*)\rightarrow H^{(1,1)}(X,\mathbb{Z})$$ is surjective by the Lefschetz theorem.

So $$\operatorname{im} c_1 =H^{(1,1)}(X,\mathbb{Z})$$ ? The group of Neron-Severi $$NS:=H^1(X,O_X^*)/\ker c_1\cong \mathrm{im} c_1$$

Of course, you should be writing $H^{(1,1)}(X)\cap H^2(X,\Bbb Z)$ when you use your shorthand. And you need $X$ compact Kähler. But ... You're almost right: The Lefschetz Theorem assumes $X$ is a projective variety, does it not?
• If you want to get the result that every line bundle corresponds to a divisor on $X$, you need $X$ projective to guarantee a meromorphic section. Dec 27 '16 at 18:31
• possibly a quite stupid question: why it is allowed to identify $H^2(X,\mathbb{Z})$ as a subgroup of $H^2(X,\mathbb{C})$ here as the OP has remarked? or in other words why is the map $H^2(X,\mathbb{Z}) \to H^2(X,\mathbb{C})$ induced by canonical inclusion $\mathbb{Z} \subset \mathbb{C}$ of local constant sheaves on $X$ an injection between $2$-th cohomology groups?