# 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 $\mathrm{im} c_1 =H^{(1,1)}(X,\mathbb{Z})$ ? 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. – Ted Shifrin 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? – user739133 Jan 4 at 15:19