# Existence of rigid curves with non-negative self intersection

The question is essentialy the title: is there a smooth algebraic surface $$X$$, say over $$\mathbb{C}$$, and an irreducible algebraic rigid curve $$C\subset X$$ with non-negative self intersection?

Here by rigid I mean that $$h^0(X,C)=1$$, so the only effective divisor linearly equivalent to $$C$$ is $$C$$ itself.

The only examples of rigid curves I know have negative self intersection.

Observations

I'm pretty sure that the answer is negative for del Pezzo or K3 surfaces. Indeed, for del Pezzo we have

• $$\chi(X)= 1$$ being $$h^1(X,\mathcal{O}_X)=h^2(X,\mathcal{O}_X)= 0$$,

• $$-K_X\cdot C> 0$$ being $$-K_X$$ ample

• $$h^2(X,C)=h^0(X,K_X-C)=0$$, using Serre duality.

Therefore, by Riemann-Roch we get $$h^0(X,C)=h^1(X,C)+\chi(X)+\frac{1}{2}C^2-\frac{1}{2} K_X\cdot C> 1+\frac{1}{2} C^2$$

so $$h^0(X,C)>1$$ if $$C^2\geq 0$$.

A similar computation holds, with some differences, for K3 surfaces.

Therefore, I tried to work on hypersurfaces of high degree in $$\mathbb{P}^3$$ but couldn't find such a curve.

• Try working out $\mathbb{P}(E)$, the projective bundle over a smooth projective curve and $E$ a rank 2 vector bundle. Find conditions on $E$ so that a section has non-negative self-intersection, but is rigid. (They exist.) Commented Jul 5, 2022 at 20:08
• Thank you for the comment. I tried the computations you said but I couldn't find an example, the point is that I can't proof that something is rigid. A curve $C$ on a rule surface is linearly equivalent to $af+bh$, where $f$ is a section of the structure morphism and $h$ is the classnof the taurological bundle (as in Beauville's book). Then if $C$ is a section of the structure morphism we have $C\cdot f=1$ hence $b=1$. With this we get that $g(C)=h^1(\mathcal{O}_X)$ and that $C^2=2a+deg(E)$. So I can find sections with positive self intersection but now prooving rigidity seems hard with RR. Commented Jul 7, 2022 at 11:12

Here is a simple example. Take an elliptic curve $$C$$ so $$H^1(O_C)$$ is one dimensional. Thus we have a non-split exact sequence $$0\to O\to E\to O\to 0$$. This section gives a section $$D$$ of $$\pi:\mathbb{P}(E)\to C$$ with $$D^2=0$$. $$\pi_* O(D)=E$$ and so $$h^0(O(D))=1$$, showing that $$D$$ is rigid.
• Sorry for the late answer, I was struggling a bit with some details of your solution. Now I think it's all clear to me except one think: how do you prove $\pi_*\mathcal{O}(D)$=E? Commented Jul 10, 2022 at 9:59