Let $L$ be a pseudo-effective divisor, we may define its numerical fixed part $N_{\sigma}(L)$. How to prove it is a divisor? I know there is a proof in Nakayama's book, but I can't find this book.
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By definition, you first define $\sigma_\Gamma(L)$ for big divisors and then you take the limit. In other words, if $L$ is big, then clearly $\sigma_\Gamma(L)$ is non zero for only finitely many divisors. Indeed, $L=A+B$ with $A$ ample $\mathbb Q$-divisor, and $B\ge 0$. Thus $\sigma_\Gamma(L)>0$ implies $\Gamma$ is contained in the support of $B$. If $L$ is pseudo-effective, then you define $$\sigma_\Gamma(L)=\lim_{\epsilon\to 0^+} \sigma_\Gamma(L+\epsilon A)$$ where $A$ is ample (it is easy to check that the definition does not depend on the choice of $A$). It follows that if $\sigma_\Gamma(D)=\alpha>0$ then $D-\alpha\Gamma$ is pseudo-effective. Finite generation of $N^1(X)$ implies that there can only be finitely many $\Gamma$ with such a property. |
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