Higher direct images of a sheaf over the diagonal along a projection vanish

253 Views Asked by At

Let $\Delta \subset \mathbf{P}^n \times \mathbf{P}^n$ be the diagonal, $E$ be a bundle over the first $\mathbf{P}^n$ factor and $p_1, p_2$ be the projections.

I am curious about a method which allows the computation of $\mathbf{R}^kp_{2\ast}(p_1^{\ast}E|_{\Delta})$. Note that Hartshorne [III, 9.3] and its generalizations do not apply and I don't have idea how to handle that extra $\otimes \ \mathcal{O}_{\Delta}$ which appears there. In this case, I expect the result to be $E$ for $k = 0$ and zero otherwise.

This is all from the proof of the Beilinson spectral sequence in Okonek's book, and any workaround that does not stray from the consequences of the hypercohomology spectral sequence would be appreciated.

1

There are 1 best solutions below

0
On BEST ANSWER

This follows from projection formula for any variety $X$. Let me call projections $p=p_2$ and $q=p_1$ for shortness and $\Delta: X \to X\times X$ is the diagonal embedding, $\mathcal{O}_\Delta=\Delta_* \mathcal{O}_X$. All functors are derived in the computation below

$$ p_*(q^*E\otimes \mathcal{O}_\Delta) = p_*(q^*E\otimes \Delta_* \mathcal{O}_X)=p_*(\Delta_*(\Delta^* q^* E\otimes_X \mathcal{O}_X)) $$ but $\Delta^* q^* E \cong E$, because $q \circ \Delta=id$, tensor product with $\mathcal{O}_X$ does not do anything on $X$, finally $p_*\Delta_*=id_*$, again because $p \circ \Delta=id$