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Let $X$ be an n–dimensional compact algebraic manifold, $i \colon Z \to X$ a closed submanifold, and let $\mathcal{N}_{Z|X}$ denote the normal sheaf. I have trouble understanding the isomorphism $$ H^0(Z,\mathcal{N}_{Z|X}) \xrightarrow{\cong} \mathrm{Ext}^1_X(i_*\mathcal{O}_Z, i_*\mathcal{O}_Z) $$ stated in the paper https://arxiv.org/pdf/math/9912245.pdf in the statement of Proposition 8.7. I think I understand how the map is defined, i.e., by applying the functor $\operatorname{Hom}_X(-, i_*\mathcal{O}_Z)$ to the short exact sequence $$ 0 \to \mathcal{I} \to \mathcal{O}_X \to i_* \mathcal{O}_Z \to 0$$ and considering the associated long exact sequence.

I believe the map so obtained is injective, however I don't understand how to prove it is surjective.

Christa Wolf
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1 Answers1

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It is not surjective in general. Indeed, the associated long exact sequence looks like $$ 0 \to H^0(Z,\mathcal{N}_{Z/X}) \to \mathrm{Ext}^1(i_*\mathcal{O}_Z,i_*\mathcal{O}_Z) \to H^1(Z,\mathcal{O}_Z) \to \mathrm{Ext}^1(\mathcal{I}_Z,i_*\mathcal{O}_Z) \to \dots $$ Now assume, for instance, $Z$ is a plane cubic curve. Then $\mathcal{I}_Z \cong \mathcal{O}(-3)$, hence $$ \mathrm{Ext}^1(\mathcal{I}_Z,i_*\mathcal{O}_Z) \cong H^1(Z,\mathcal{O}_Z(3)) = 0, $$ while $H^1(Z,\mathcal{O}_Z) \ne 0$.

Sasha
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