Suppose to have an additive right exact functor $F: \mathcal A \rightarrow \mathcal B$ between Abelian categories and suppose that $F(A)=B$ for an object $A$ in $\mathcal A$. Denote with $D(\mathcal A)$, $D(\mathcal B)$the derived categories of chain complexes (if you want, bounded), and denote with $Q$ the localization functor.

Suppose that $F$ has a right total derived functor $\textbf R F: D(\mathcal A) \rightarrow D(\mathcal B)$. In general, I would not expect that $Q(B) \simeq (\textbf R F)(Q(A))$, when $F$ is not exact.

Then I can't understand why this fact seems to be used roundabout in some cases. To say, I am working on Examples 5.4 of Huybrecht's Fourier-Mukai transform in algebraic geometry and it seems to be used in the following argument:

Let $X$ a smooth projective variety, $p,q: X \times X \rightarrow X$ the two projections and $i: X \rightarrow X \times X$ the diagonal morphism. The structure sheaf $\mathcal O _\Delta$ of the diagonal is canonically isomorphic to $i_* \mathcal O _X$. Regarded as complexes concentrated in degree 0, $i_* \mathcal O _X$ and $ \mathcal O _\Delta$ define objects in $D^b(X)$ (the usual derived category of $X$). If $\mathcal E$ is any object in $D^b(X)$, the integral transform can be defined. Huybrechts write:

$\Phi_{\mathcal O _\Delta}(\mathcal E) = \textbf R p_* ( q^* \mathcal E \otimes^\textbf L \mathcal O _\Delta) = \textbf R p_* ( q^* \mathcal E \otimes^\textbf L \textbf R i_* \mathcal O _X)$

where $\mathcal O _\Delta = \textbf R i_* \mathcal O _X$ seems to be used ($q^*$ is not derived since $q$ is flat, hence $q^*$ is exact).

  • 4
    $\begingroup$ For a closed embedding of varieties/schemes $i\colon X\to Y$, the direct image functor $i_*$ (acting from the category of quasi-coherent sheaves on $X$ to the category of quasi-coherent sheaves on $Y$, or from the category of sheaves of $\mathcal O_X$-modules to the category of sheaves of $\mathcal O_Y$-modules) is exact. So there is no difference between $i_*$ and $\mathbf R i_*$, or in other words, the functor $\mathbf R i_*$ does actually form a commutative square with the functor $i_*$ and the localization functors $Q$. $\endgroup$ – Leonid Positselski Jun 16 '16 at 11:19

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.