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Aug 4, 2023 at 23:43 comment added Basil Mm I'm not sure I understood everything that's implied in your answer, may I ask some further questions, to steer my doubts away? For one, what kind of space classifies higher-order deformations? Does it represent some functor, so as to guarantee the map from $A$ is a morphism? And secondly I tried writing down the argument with the formal function theorem: it implies that the pushforward of $\mathcal{C}_a$ onto C is locally free, and thus has a global nowhere-vanishing section on an open set $U\times A$. But the set where it has trivial fibres is closed so its fibres must all be free. Thanks!
Aug 1, 2023 at 12:04 comment added Jason Starr . . . Correction: replace $\textbf{GL}$ by $\text{Hom}$.
Aug 1, 2023 at 10:14 comment added Jason Starr Welcome new contributor. The operation of $A$ on first-order deformations $\mathcal{L}$ of $\mathcal{O}_A$ sending $\mathcal{L}$ to $\mathcal{L}^\vee\otimes_{\mathcal{O}_A} t_a^*\mathcal{L}$ induces a morphism from $A$ to $\textbf{GL}(H^1(A,\mathcal{O}_A))$. The domain is proper, and the target is affine, therefore the morphism is constant. The same argument applies for higher-order deformations as well. Thus, your sheaf $\mathcal{C}_a$ is isomorphic to $\mathcal{O}_{C\times A}$ to all infinitesimal orders near $\mathcal{O}_A$. Now use the theorem on formal functions.
S Jul 30, 2023 at 16:41 review First questions
Jul 30, 2023 at 16:53
S Jul 30, 2023 at 16:41 history asked Basil CC BY-SA 4.0