Asymptotic Riemann-Roch formula with exceptional divisor involved

Question

Let $X$ be a projective surface, let $D$ be a curve on $X$ viewed as a Cartier divisor. Let $P$ be a point on $D$, and let $\pi: \tilde{X} \to X$ be the blow up $X$ at $P$ with exceptional divisor $E$.

We know that $\pi^*D.E=0$ and $E^2=-1$. For sufficiently large integer $N$, consider the dimensions of $H^0(X,\mathcal{O}_X(N\pi^*D)),H^0(X,\mathcal{O}_X(N(\pi^*D-E)))$ and $H^0(X,\mathcal{O}_X(N(\pi^*D+E)))$. Using asymptotic Riemann-Roch, I get:

$h^0(X,\mathcal{O}_X(N\pi^*D))=N^2\pi^*D^2/2 + O(N)$

$h^0(X,\mathcal{O}_X(N(\pi^*D-E)))=h^0(X,\mathcal{O}_X(N(\pi^*D+E)))=N^2(\pi^*D^2-1)/2 + O(N)$.

That confuses me, since $E$ being effective, I should expect $$H^0(X,\mathcal{O}_X(N(\pi^*D-E))) \subset H^0(X,\mathcal{O}_X(N\pi^*D)) \subset H^0(X,\mathcal{O}_X(N(\pi^*D+E))).$$ But this contradicts the dimension count. Where I have things wrong?

Any comment is appreciated!

Answer 1

The statement of Asymptotic Riemann Roch should be

Let $D$ be a cartier divisor on a variety $X$ of dimension $n$. Then, $$\chi(\mathcal{O}_X(mD)) = \frac{D^n}{n!}m^n + O(m^{n - 1}).$$

Using Serre vanishing, we deduce a statement about $h^0$ as a consequence.

Let $D$ be an ample cartier divisor on a variety $X$ of dimension $n$. Then, $$h^0(\mathcal{O}_X(mD)) = \frac{D^n}{n!}m^n + O(m^{n - 1}).$$

Your example shows that if $D$ is not ample, this result may no longer hold. Indeed, $\pi^*D + E$ is not ample since it intersects negatively with $E$.