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.


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.

  • 3
    $\begingroup$ 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.) $\endgroup$
    – Mohan
    Commented Jul 5, 2022 at 20:08
  • $\begingroup$ 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. $\endgroup$ Commented Jul 7, 2022 at 11:12

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Commented Jul 10, 2022 at 9:59
  • $\begingroup$ @WiktorVacca As I said in the real life: see Hartshorne - Algebraic Geometry, LemmaV.2.1. ;) $\endgroup$ Commented Jul 12, 2022 at 19:15

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