5
$\begingroup$

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?

$\endgroup$
2
  • 3
    $\begingroup$ Well, $D$ nef surely implies your condition (by asymptotic Riemann-Roch). $\endgroup$ – Francesco Polizzi Apr 11 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 Apr 11 at 13:39
8
$\begingroup$

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".

References.

[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.

$\endgroup$
9
  • 1
    $\begingroup$ 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. $\endgroup$ – Francesco Polizzi Apr 11 at 14:41
  • 1
    $\begingroup$ 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. $\endgroup$ – Francesco Polizzi Apr 11 at 14:46
  • 1
    $\begingroup$ thank you very much for the answer and the clarifications. $\endgroup$ – HARRY Apr 11 at 14:49
  • 1
    $\begingroup$ 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. $\endgroup$ – Francesco Polizzi Apr 11 at 19:24
  • 1
    $\begingroup$ Oh yes, by "integral" divisor, in this context, I meant a divisor with coefficients in $\mathbb{Z}$. Sorry for the ambiguity. $\endgroup$ – Francesco Polizzi Apr 12 at 5:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.