Given a smooth variety $X$ and two smooth subvarieties $Y_{1},Y_{2} \subset X$ with smooth intersection $Z=Y_{1} \cap Y_{2}$, I'd like to compute $\mathcal{Ext}^{i}_{X}(\mathcal{O}_{Y_{1}},\mathcal{O}_{Y_{2}})$. Something that I am reading seems to suggest that the answer should be $\Omega^{i}_Z$, or at least something in terms of the normal bundle of $Z$ in $X$. But I don't see this myself.
$\begingroup$
$\endgroup$
1
-
6$\begingroup$ See Hartshorne, proof of Theorem III 7.11 - there he computes ${\scr E}xt^r(\mathcal{O}_{Y_1}, \omega_X)$ by applying $Hom(-, \Omega_X)$ to the Koszul complex of $\mathcal{O}_{Y_1}$. The Koszul complex exists only locally, but then he shows that these local calculations glue together. I think you can apply the same argument to $\mathcal{O}_{Y_2}$ in place of $\omega_X$. $\endgroup$– Piotr AchingerCommented Feb 27, 2012 at 18:26
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
If $c$ is the codimension of $Z$ in $Y_{2}$, then $$\mathcal{E}xt^{i}_{X}(\mathcal{O}_{Y_{1}},\mathcal{O}_{Y_{2}}) = \wedge^{c} N_{Z/Y_{2}}\otimes \wedge^{i -c} \left( N_{Z/X}/N_{Z/Y_{2}}\right).$$ You can see this by Hom-ing one Koszul complex into another. You can find the calculation e.g. in Appendix A.3 of this paper of Katz and Sharpe.
answered Feb 29, 2012 at 13:47