# Right derived functor of diagonal morphism equals direct image on line bundles?

Let $X$ be a smooth projective variety. The map $i:X\to X\times_k X$ induced by the identity is a closed immersion. Denote its image by $\bigtriangleup$. We have $\mathcal{O}_{\bigtriangleup}=i_*\mathcal{O}$. Consider the derived category $D^b(X)$ of coherent sheaves on $X$.

How do I conclude $\mathbf{R}i_*\mathcal{O}=\mathcal{O}_{\bigtriangleup}$ ? Rather: why is $\mathbb{R}i_*(L)=i_*L$ for every line bundle on $X$?

Thanks a lot.

-
For a closed immersion of topological spaces, the pushforward map on sheaves is already exact, so deriving it doesn't have any effect. – Akhil Mathew Jan 21 '12 at 22:10

Let $i>0$. Write $T:=R^ii_*(\mathcal{O})$. Let $U$ be an affine subset of $X$. Then $p^{-1}(U)$ is an open subset of $X\times X$ and $i^{-1}p^{-1}(U)=U$. We have $T(p^{-1}(U))=H^i(U, \mathcal{O}_{|U})=0$ as $U$ is affine. We can cover $X\times X$ by the such $p^{-1}(U)$. This implies $T=0$ and hence $R^ii_*(\mathcal{O})$ vanishes for all i>0. Hence $\mathcal{O}$ is $i_*$-acyclic and thus $\mathbf{R}i_*(\mathcal{O})=i_*\mathcal{O}$.