# What is the conormal sheaf?

I'm having some silly confusion about the definition so I would really appreciate it if someone can help me out with this definition.

Let $$i: X \hookrightarrow Y$$ be a closed immersion cut out by $$\mathscr{I} \subseteq \mathcal{O}_Y$$. Then, Vakil defines the conormal sheaf of this closed immersion to be $$\mathscr{I}/\mathscr{I}^2$$ as viewed as a quasi-coherent sheaf on X. The italicized part is essentially my confusion.

How are we viewing this as a quasi-coherent sheaf on $$X$$? The most obvious way to do this is to take $$i^*(\mathscr{I}/\mathscr{I}^2)$$ but I don't think this is right. In particular, this question seems to take it to be $$i^* \mathscr{I} = i^{-1}(\mathscr{I}/\mathscr{I}^2)$$. I find this unsettling though.

By definition, $$\Delta: X \to X\times_YX$$ is a locally closed immersion and we define $$\Omega_{X/Y}$$ to be the conormal sheaf of this embedding. This would be fine, although Hartshorne defines $$\Omega_{X/Y} = \Delta^*(\mathscr{I}/\mathscr{I}^2)$$ instead of what I would have expected from above, $$\Delta^*(\mathscr{I})$$. What is going on here?

To summarize, is the conormal sheaf of $$i: X \hookrightarrow Y$$ defined to be $$i^*(\mathscr{I})$$, $$i^*(\mathscr{I}/\mathscr{I}^2)$$, or both?

Thank you very much for any help.

These are all the same sheaf.

Basically, this is the isomorphism $$I/I^2 \cong I \otimes A / I$$ for an ideal $$I \subset A$$. This tells you that if $$\iota :Z \to X$$ is a closed immersion cut out by an ideal $$\mathcal{I}$$ then the following sheaves are the same,

1. $$\iota^{*} (\mathcal{I} / \mathcal{I}^2)$$
2. $$\iota^{-1} (\mathcal{I} / \mathcal{I}^2)$$
3. $$\iota^{*} \mathcal{I}$$

This is because $$\iota^* \mathcal{F} = \iota^{-1} \mathcal{F} \otimes \mathcal{O}_Z$$ but $$\mathcal{I} / \mathcal{I}^2$$ is already a $$\mathcal{O}_Z = \mathcal{O}_X / \mathcal{I}$$-module and $$\mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{O}_Z = \mathcal{I} / \mathcal{I}^2$$.

Furthermore, $$\iota_*$$ induced an equivalence of categories from $$\mathcal{O}_Z$$-modules to $$\mathcal{O}_X$$-modules $$\mathcal{F}$$ such that $$\mathcal{I} \cdot \mathcal{F} = 0$$ (in particular supported along $$Z$$). If this is confusing, think about what the difference between an $$A$$-module and an $$A / I$$-module is. Therefore, we can view $$\mathcal{O}_Z$$ and $$\iota_* \mathcal{O}_Z$$ as essentially the same sheaf as well as view, $$\iota^* \mathcal{I} = \iota^* (\mathcal{I} / \mathcal{I}^2) = \iota^{-1} (\mathcal{I} / \mathcal{I}^2)$$ as essentially the same sheaf as $$\mathcal{I} / \mathcal{I}^2$$. This is why Ravi said viewed as a sheaf on $$X$$ rather than pulled back to $$X$$ because once you quotient its already a $$\iota_* \mathcal{O}_Z$$-module and therefore by the equivalence of categories already essentially the same as $$\iota^{-1} (\mathcal{I} / \mathcal{I}^2)$$ on $$Z$$.

• @Daniel edited thanks Dec 17, 2021 at 1:18