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Let $S$ be a compact complex surface and $L\to S$ a holomorphic line bundle. $L$ is said to be nef if $c_1(L)[C]\geq 0$ for any curve $C\subset S$. Is it true that, for nef $L$, we have $c_1^2(L)\geq0$? This is certainly true if we can represent the class $c_1(L) \in H^2(L;\Bbb Z)\cong H_2(L;\Bbb Z)$ (Poincare duality) by a complex curve $C\subset S$, but I can't see whether this is possible or not.

I know that for an arbitrary closed oriented smooth 4-manifold $X$, any homology class in $H_2(X;\Bbb Z)$ can be represented by an embedded smooth (real) surface. But is the following corresponding statement also true?: For a closed complex surface $S$, any second homology class can be represented by an embedded complex curve.

blancket
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    Not every second homology class of a complex surface can be realised by an embedded complex curve. For example, generic tori $\mathbb{C}^2/\Lambda$ have no complex curves at all, see this answer. – Michael Albanese Sep 05 '21 at 11:27

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