Let $X$ be a smooth projective surface (over complex numbers). Let $D$ be a divisor on $X$. Then we know that its volume is defined as $$\text{vol}_X(D):= \lim \sup_{m \rightarrow \infty} \frac{h^0(X, \mathcal O_X(mD))}{{m^2}/2}.$$

Suppose that, for a divisor $D$ on $X$, it is known that $\text{vol}_X(D)=D^2$, where $D^2$ stands for its self-intersection number.

Question. What does this signify algebro-geometrically? And what are some interesting properties of $D$ that one can deduce once the volume is explicitly known?

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    $\begingroup$ Well, $D$ nef surely implies your condition (by asymptotic Riemann-Roch). $\endgroup$ Commented Apr 11, 2021 at 13:34
  • $\begingroup$ @FrancescoPolizzi, yes that's true. But here the situation is the converse question. And more importantly what can we say about the divisor by looking at the value of it's volume? I mean is there any classification result in the literature in terms of the volume ( or anything partially close to that?) $\endgroup$
    – HARRY
    Commented Apr 11, 2021 at 13:39

1 Answer 1


At least for effective divisors, the answer is strongly related to Zariski decomposition.

If $D$ is an effective divisor on a smooth surface $X$, Zariski proved in [Z62] that there exists a unique decomposition $D=P + N$, where

  • $P$ is a nef $\mathbb{Q}$-divisor
  • $N$ is an effective $\mathbb{Q}$-divisor
  • $PC=0$ for every curve $C$ appearing in $\operatorname{Supp}(N)$
  • if $N \neq 0$ and $\operatorname{Supp}(N)=C_1 \cup \ldots \cup C_k$, then the intersection form $I(C_1, \ldots, C_k)$ is negative defined.

Furthermore, one also shows that, for all positive integers $m$, the natural map $$H^0(X, \left \lfloor{mP}\right \rfloor) \to H^0(X, \, mD) $$ is an isomorphism. This means that "$P$ carries all sections of $D$" and so $$\operatorname{vol}_X(D)=\operatorname{vol}_X(P)= P^2,$$ where the last equality is a consequence of the asymptotic form of Riemann-Roch theorem, because $P$ is nef.

Thus, $\operatorname{vol}_X(D)=D^2$ is equivalent to $D^2=P^2$; since $PN=0$, this happens if and only if $N^2=0$. But the intersection form must be negative defined on $N$ when $N \neq 0$, hence the only possibility is $N=0$, namely, $D=P$.

Summing up:

given an effective divisor $D$ on a smooth surface $X$, we have $\operatorname{vol}_X(D)=D^2$ if and only if $D$ is nef.

More generally, the difference $$D^2-\operatorname{vol}_X(D)$$ equals the self-intersection $N^2$ of the negative part of the Zariski decomposition of $D$, so it can be seen as a measure of how much "$D$ fails to be nef".


[Z62] O. Zariski The theorem of Riemann-Roch for high multiples of an effective divisor on an algebraic surface, Ann. Math. (2) 76, 560-615 (1962). ZBL0124.37001.

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    $\begingroup$ nef is a numerical property, so I would not use the expression "$\mathbb{Q}$-nef". If you start with an effective $\mathbb{Q}$-divisor $D$, then the argument above simply shows that $D^2=\text{vol}(D)$ if and only if $D$ is a nef $\mathbb{Q}$-divisor. $\endgroup$ Commented Apr 11, 2021 at 14:41
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    $\begingroup$ Yes, of course. If you start with an integral effective divisor $D$, the argument above shows that $D^2=\mathrm{vol}(D)$ if and only if $D=P$, so $D$ is an integral nef divisor. $\endgroup$ Commented Apr 11, 2021 at 14:46
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    $\begingroup$ thank you very much for the answer and the clarifications. $\endgroup$
    – HARRY
    Commented Apr 11, 2021 at 14:49
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    $\begingroup$ I am not sure that I understand your last question. Zariski's decomposition holds for $\mathbb{Q}$-divisors, so the general statement is (as I said before) "If an effective $\mathbb{Q}$-divisor $D$ is such that $\text{vol}(D)=D^2$, then $D$ is nef". The case where $D$ is effective and integral is a particular case of this. $\endgroup$ Commented Apr 11, 2021 at 19:24
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    $\begingroup$ Oh yes, by "integral" divisor, in this context, I meant a divisor with coefficients in $\mathbb{Z}$. Sorry for the ambiguity. $\endgroup$ Commented Apr 12, 2021 at 5:54

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