2
$\begingroup$

Let $V$ be a vector bundle on $X$, and $Y = \text{Tot}(V)$ be the total space of this bundle; we have a closed embedding $i: X \rightarrow Y$. Why is the following result true?

$$ \text{Ext}^k(i_* \mathcal{O}_X,i_* \mathcal{O}_X) = \bigoplus_{i+j=k} H^i(X, \Lambda^j V)$$

$\endgroup$
2
  • $\begingroup$ Why do you believe it's true? $\endgroup$ Nov 2, 2013 at 15:55
  • $\begingroup$ Does this work if X is a point? $\endgroup$ Nov 2, 2013 at 16:48

1 Answer 1

6
$\begingroup$

Let $\pi :Y\rightarrow X$ be the projection. The bundle $\pi ^*V$ has a canonical section $s$ (the diagonal, if you think of $\pi ^*V$ as $V\times _XV$), which vanishes exactly along $i(X)$. Thus $i_*\mathcal{O}_X$ has a Koszul resolution $$ \ldots \rightarrow \wedge^2\pi ^*V^*\rightarrow \pi ^*V^* {\buildrel {s}\over {\longrightarrow}}\ \mathcal{O}_Y\rightarrow i_*\mathcal{O}_X$$ by locally free $\mathcal{O}_Y$-modules. Applying $\ \underline{\mathrm{Hom}}_{\mathcal{O}_Y}(-,i_*\mathcal{O}_X)$ kills the differentials, so we get $R\underline{\mathrm{Hom}}_{\mathcal{O}_Y}(i_*\mathcal{O}_X,i_*\mathcal{O}_X)\cong \oplus \ \wedge^pV[p]$, then $R\mathrm{Hom}_{\mathcal{O}_Y}(i_*\mathcal{O}_X,i_*\mathcal{O}_X)\cong \oplus R\Gamma (\wedge^pV[p])$, hence the result by applying $H^k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.