Does the Euler characteristic increase if I add an effective divisor

Question

Let $D$ be an effective divisor on a smooth projective connected complex algebraic variety $X$. Suppose that $D\leq E$. Is it true that $$\chi(X,\mathcal{O}_X(D)) \leq \chi(X,\mathcal{O}_X(E))?$$

Answer 1

This is clearly true for curves by classical Riemann-Roch thoerem which computes Euler characteristic of a line bundle in terms of it's degree.

I claim that this is already false for surfaces and will prove that using some basic theory. All of this is covered in Hartshorne.

Let $X$ be a non-singular, projective surface with trivial canonical sheaf, for example a generic surface in $\mathbb{P}^{3}$ of degree $4$ is of this type by adjunction. Choose some point $P \in X$ and let $\tilde{X}$ be the blow-up of $X$ in $P$ and $E \subseteq \tilde{X}$ be the exceptional divisor. By classical arguments, $E^{2} = -1$ and $K _{\tilde{X}} = E + K _{X} = E$. We can use this data to compute Euler characteristic of $\mathcal{O} _{\tilde{X}} (2E)$ using Riemann-Roch for surfaces, which tells us that

$\chi(\mathcal{O} _{\tilde{X}} (2E)) = \chi(\mathcal{O} _{\tilde{X}}) + \frac{1}{2} (2E).(2E - K _{\tilde{X}}) = \chi(\mathcal{O} _{\tilde{X}}) + \frac{1}{2} (2E).E = \chi(\mathcal{O} _{\tilde{X}}) - 1$.

So $\chi(\mathcal{O} _{\tilde{X}} (2E)) < \chi(\mathcal{O} _{\tilde{X}})$ even though $2E$ is an effective divisor.