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?
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4$\begingroup$ Not over $\mathbb{C}$, but this can occur in positive characteristic. $\endgroup$– Jason StarrCommented Dec 18, 2013 at 15:02
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2 Answers
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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.
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$\begingroup$ 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. $\endgroup$– HiroCommented Dec 18, 2013 at 22:30
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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.
answered Aug 7, 2020 at 16:38
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$\begingroup$ The sequence is not exact on the left in general: consider e.g. the normalization of $y^2 - x^3$. $\endgroup$ Commented Apr 24, 2022 at 18:38