If a divisor $D$ on a surface has positive self-intersection, then $nD$ or $-nD$ has nontrivial sections, eventually in $n$.

Question

Let $S$ be a smooth complex projective surface. Let $D$ be a divisor on $S$, such that $D^2 = D.D> 0$. Then, at least one of the following holds:

For $ n \gg 0$, $H^0(nD) \neq 0$;
for $n \gg 0$, $H^0(-nD) \neq 0$.

Here $H^0(D)$ denotes the space of holomorphic sections of the line bundle $[D]$ associated to $D$. With the notation "n≫0" I mean: "for every n save for a finite number of them".

How do I prove this? I was given a proof in a course on algebraic surfaces, which makes use of Riemann-Roch theorem, but I'm afraid it's flawed...

EDIT: actually, with the same hypothesis as above I can prove that $H^0(nD)$ or $H^0(-nD)$ is unbounded as a sequence indexed by $n$. I don't know whether the above statement is true.

EDIT 2: The proof I have is as follows. Apply Riemann-Roch to $nD$ and $-nD$:

$$h^0(nD) + h^2(nD) \geq \chi(\mathcal O_S) + \frac{1}{2}((n^2 D^2 - nD.K_S),$$ $$h^0(-nD) + h^2(-nD) \geq \chi(\mathcal O_S) + \frac{1}{2}((n^2 D^2 + nD.K_S).$$ Then, in both cases $\mathrm{LHS}$ goes to infinity as $n \to +\infty$. Now, suppose that $H^0(nD) = 0$ for infinite $n \in \mathbb N$ (that is, the negation of "For $ n \gg 0$, $H^0(nD) \neq 0$"). Now, I would like to go on as follows: $h^2(nD) = h^0(K_S - nD) \neq 0$ for all $n \gg 0$ (but this is not assured, I think!), then for $n \gg 0$ there is $s_n \in H^0(K_S - nD)$, $s_n \neq 0$, and multiplication by $s_n$ induces an injective map $$H^0(K_S + nD) \longrightarrow H^0(K_S + nD + K_S - nD) = H^0(2K_S),$$ so that $h^0(K_S + nD) = h^2(-nD)$ is a bounded sequence, and then $h^0(-nD) \to +\infty$, in particular $h^0(-nD) > 0$ for all $n \gg 0$.

Can you tell us the proof you saw, and how it is flawed? I claim there is a non-flawed proof using R--R, but rather than go through the whole thing, it would be easier to "debug" the proof you have. — , Jul 11 '13 at 13:58
Isn't this cor 1.8 in section 5.1 of Hastehorne's algebraic geometry book? — Youngsu, Jul 11 '13 at 16:49
I don't see the problem. Since we are assuming $h^0(nD)=0$ and the LHS goes off to infinity, surely $h^2(nD) = \mathrm{LHS}$ is nonzero!? — Andrew, Jul 12 '13 at 16:42
The problem is that I want $h^2(nD) > 0$ for all $n$ sufficiently large! — Francesco Genovese, Jul 12 '13 at 17:49

score 1 · Answer 1 · answered Jul 14 '13 at 18:50

Sorry it took me a little while to get back to this. Let me complete the above proof; my argument probably isn't the optimal one.

The proof you already gave shows that either $H^0(nD) \neq 0$ or $H^0(-nD)$ for some $n$. Let's assume by renaming that $H^0(nD) \neq 0$ for some $n$. It's easy to see that then $H^0(nD) \neq 0$ for infinitely many $n$.

Now suppose that $H^0(nD) \neq 0$, but $H^0((n+1)D)=0$. In your first displayed equation above, let's write $Q(n)$ to denote the quadratic function of $n$ on the right-hand side. Then my assumption on $(n+1)D$ implies that $h^0(K_S-(n+1)D) \geq Q(n+1)$. But now multiplying sections of $K_S-(n+1)D$ by some nonzero section of $nD$, we get that $h^0(K_S-D) \geq Q(n+1)$.

Since $Q$ is quadratic with positive leading coefficient, and $H^0(K_S-D)$ is some fixed finite-dimensional vector space, this can only happen for finitely many $n$.

If a divisor $D$ on a surface has positive self-intersection, then $nD$ or $-nD$ has nontrivial sections, eventually in $n$.

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