Volume of a divisor on a smooth projective surface

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?

• Well, $D$ nef surely implies your condition (by asymptotic Riemann-Roch). – Francesco Polizzi Apr 11 at 13:34
• @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?) – HARRY Apr 11 at 13:39

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.

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