# Are singular rational curves on K3 surfaces rigid?

Let $S$ be a K3 surface over the complex numbers $\mathbb{C}$. If $C\subset S$ is a smooth rational curve, the normal bundle $N_{C/S}$ is isomorphic to $\mathbb{O}(-2)$ and thus $C$ is rigid. What about a curve $C'\subset S$ whose normalization is a rational curve? I think it may admit deformation in higher genus family, but can it admit deformation in a family of (possibly singular) rational curves?

• Not over $\mathbb{C}$, but this can occur in positive characteristic. Dec 18, 2013 at 15:02

To develop what Jason says: if your curve deforms in a family of rational curves, it means that you can find a dominant rational map from a ruled surface onto your K3. This is forbidden (over $\mathbb{C}$): e.g. because the nonzero 2-form of the K3 would lift to a nonzero 2-form on the ruled surface.

• Thank you for the answer. To clarify your answer, the map is dominant because the curve has a singularity and the pull-back 2-form extends because the singularity of the ruled surface is of codimension 2.
– Hiro
Dec 18, 2013 at 22:30

Here is another proof using deformation theory:

Let $$f:\mathbb P^1\to S$$ be the composite of inclusion $$C'\subseteq S$$ and the normalization of $$C'$$. Then there is an exact sequence

$$0\to T_{\mathbb P^1}\to f^*T_{S}\to N_f\to 0,$$

where $$N_f$$ is the normal sheaf of $$f$$ and $$H^0(\mathbb P^1,N_f)$$ is the first order deformation space of the morphism $$f:\mathbb P^1\to S$$ with the target fixed.

Since $$S$$ is a K3 surface, $$c_1(T_S)=0$$, on the other hand $$c_1(T_{\mathbb P^1})=2$$, so by exact sequence we have $$c_1(N_f)=-2$$. In particular $$H^0(\mathbb P^1,N_f)=0$$. So $$C'$$ does not deform in a family of singular rational curves.

• The sequence is not exact on the left in general: consider e.g. the normalization of $y^2 - x^3$. Apr 24 at 18:38