Corollary 20.1.35DF4E1BF96D34

Proof. We only handle the nontrivial implication. Assume that $\mathbb {D}$ is nef in the sense of definition 20.1.12. We want to show that $\mathbb {D}$ is nef in the sense of definition 20.1.11. After replacing $X$ by a realization of $\mathbb {D}$, we may assume that $\mathbb {D}$ is realized by $(X,\alpha )$ for some cohomology class $\alpha \in \mathrm{H}^{1,1}(X,\mathbb {R})$.

We first assume that $\mathbb {D}$ is big. Choose a non-divisorial closed positive $(1,1)$-current $T$ on $X$ such that $\mathbb {D}=\mathbb {D}(T)$. Now $\mathbb {D}=\mathbb {D}(T)$ means that for each modification $\pi \colon Y\rightarrow X$, the current $\pi ^*T$ is non-divisorial. In particular, $T$ has vanishing generic Lelong number along each prime divisor over $X$, see (20.1.30). Equivalently, $T$ has vanishing Lelong number everywhere. By Demailly's regularization theorem, a pseudo-effective class carrying a positive current with vanishing Lelong number at every point is nef. Hence $\alpha =\{ T\} $ is nef.

In the general case, fix a Kähler form $\omega $ on the realization of $\mathbb {D}$. For every $\epsilon >0$, the Cartier b-divisor $\mathbb {D}+\epsilon \mathbb {D}(\omega )$ is big and nef in the sense of definition 20.1.12, hence is nef in the sense of definition 20.1.11 by the big case. Thus $\alpha +\epsilon \{ \omega \} $ is nef for all $\epsilon >0$, and the closedness of the nef cone shows that $\alpha $ is nef. $\square$