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I can't understand why the positive part of the Zariski decomposition of a big class is itself big. More concretely: let $X$ be a smooth projective surface over $\mathbb{C}$. Let $N^1_{\mathbb{R}}(X)$ be the Néron-Severi space associated to $X$. Let $D$ be a big class in $N^1_{\mathbb{R}}(X)$ and let $D=P_D+N_D$ be its Zariski decomposition. Recall that $P_D$ is an effective, nef $\mathbb{R}$-divisor and $N_D=\sum_i a_i C_i$ is an effective, negative $\mathbb{R}$-divisor, where "negative" means that the Gram-matrix $(C_i \cdot C_j)_{i,j}$ is negative definite. Moreover it holds $P_D \cdot C_i=0$ for every $i$. How to show that $P_D$ is a big $\mathbb{R}$-divisor? It should be an elementary fact, but I can't see it...

Thank you in advance for your support.

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  • $\begingroup$ Write $D=\sum n_iD_i$, with $n_i>0$ and $D_i$ a big integral divisor. If $D_i=P_i+N_i$ is the Zariski decomposition of $D_i$, the $P_i$ are big, so $P_D=\sum n_iP_i$ is big. $\endgroup$
    – abx
    Commented Nov 22, 2020 at 17:55
  • $\begingroup$ Why is $P_D$ ($N_D$) the sum of the positive parts (negative parts)? $\endgroup$
    – klerk
    Commented Nov 22, 2020 at 18:03
  • $\begingroup$ Oh, right, this is not clear at all. $\endgroup$
    – abx
    Commented Nov 22, 2020 at 18:19
  • $\begingroup$ @Frant isn't it true that $P_D \geq \sum n_iP_i$ (equivalently, $N_D \leq \sum n_iN_i$)? That should be enough as big + effective = big. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 22, 2020 at 21:02
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    $\begingroup$ One way to characterise $P$ is that it is the unique largest nef divisor with $D \geq P$, from which my claim is immediate. It is clearly maximal (adding a little bit of $N$ pushes it outside the nef cone), and uniqueness follows since $\max(P,P')$ is nef and $\leq D$ when $P, P'$ are (note that the Zariski decomposition is most naturally defined on the level of divisors, before passing to numerical equivalence). See Lemma 4.4 here for the integral case. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 23, 2020 at 16:28

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My copy of Lazarsfeld is quarantined in my office, and I lack the reputation for a comment, so forgive me for a non-answer. But I think there is a statement there to the effect that $h^0(\lfloor mD \rfloor) = h^0(\lfloor mP(D) \rfloor)$ for all $m$ (this is essentially from the definition, if you construct Zariski decomposition using the approach of Nakayama and specialize to dimension 2). This would mean you are done except for the nuisance of rounding, but I think it should be true that an $\mathbb R$-divisor is big if and only if $h^0(\lfloor mD \rfloor)$ grows like $m^{\dim X}$.

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  • $\begingroup$ Just to fill in the last step: this characterization is Proposition A.14 in Fujino's "Fundamental theorems for semi log canonical pairs." $\endgroup$
    – Takumi Murayama
    Commented Nov 22, 2020 at 22:54
  • $\begingroup$ Yes, @Tim Piyim, the statement is the following: Let $D=P+N$ be a big integral divisor, then the natural map $H^0(X,\mathcal{O}_X(mD-\roundup{mN})) \to H^0(X,\mathcal{O}_X(mD))$ is bijective for every $m \geq 1$. Sorry for the "\roundup" but I don't know how to do the symbol! $\endgroup$
    – klerk
    Commented Nov 23, 2020 at 9:06

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