1
$\begingroup$

I was trying to understand the Mukai Fourier transform from his paper : Duality between $D(X)$ and $D(\hat X)$ with its application to Picard sheaves, Nagoya Math Journal, 1981.

I am not very familiar with Derived categories and even less familiar with $D^-(X)$ and so this question, as Mukai works with $D^-(X)$. My knowledge of derived categories has been obtained from Iversen's book Cohomology of Sheaves.

I usually think of the bounded below derived category $D^+(X)$ by an analogy with injective resolutions modulo quasi isomorphism/homotopy equivalence. For example, $f:X\to Y$ is a morphism of algebraic varieties, then we can define the higher direct images $R^if_*F$ for any sheaf $F$ on $X$. If we have a short exact sequence $0\to F\to G\to H\to 0$, then we can find injective resolutions which are term wise split and we can get long exact sequences of higher direct images. Now instead of thinking of each of the $R^if_*F$ individually we could just take the whole complex $f_*I^\bullet$ and call this $Rf_*I^\bullet$, where $0\to F\to I\bullet$ is an injective resolution.

Since I have mostly seen only injective resolutions in action, could someone explain to me the motivation for working with $D^-(X)$, where I guess one would have to work with projective resolutions, which may not always exist.

For my purpose, I convinced myself about Proposition 1.3 in the paper of Mukai by working with $D^+(X)$ as in the case of interest (abelian varieties) the projection morphisms are flat and we are tensoring with a flat sheaf.

Thanks in advance.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ Thoughtless answer: read the book "Fourier-Mukai Transforms in Algebraic Geometry", by Huybrechts. (Not to obtain an answer to your specific question, necessarily.) $\endgroup$
    – user5117
    Commented Apr 28, 2011 at 11:03
  • 3
    $\begingroup$ After Spaltenstein (1988), you have unbounded flat and injective resolutions, so you don't really need to restrict to $D^+$ or $D^-$ to compute right derived functors of Hom, or left derived functors of tensor, respectively. $\endgroup$
    – Leo Alonso
    Commented Apr 28, 2011 at 11:33
  • 2
    $\begingroup$ Even if you don't have time to really read Huybrechts' book, the first two chapters are a good reference for derived category stuff. In particular, they have some information about how to deal when you don't have enough projectives, injectives, etc. And I found this source works as well as anything is likely to as a "random access" reference for this stuff--i.e., to understand a proposition, you don't have to carefully read through everything that came before (although you may need to assume some of the statements and/or look back at a few definitions). $\endgroup$
    – Charles Staats
    Commented Apr 28, 2011 at 14:20
  • 2
    $\begingroup$ Traditionally, for injective resolutions you start with the left most nonzero point and built it to the right. Since you need a place to start the process, you work in $D^+$. For flat resolutions, its the opposite, hence the need for $D^-$. But as Leo says, it is now possible to work in derived category ubounded in both directions. $\endgroup$
    – Donu Arapura
    Commented Apr 28, 2011 at 15:54
  • $\begingroup$ ...unless one worries about some finiteness conditions, e.g. $D_c$ or $D_{coh}.$ $\endgroup$
    – shenghao
    Commented Apr 28, 2011 at 17:12

1 Answer 1

Reset to default
1
$\begingroup$

For geometry the most informative category is $D^b(coh X)$. However, not every functor preserves coherence and boundedness. For example, it is possible that $L_if^*(F) \ne 0$ for arbitrary large $i$ even if $F$ is a sheaf. This is why sometimes people have to work in $D^-(X)$ --- otherwise the derived pullback will not be defined. So, the reason is not a choice of specific resolutions, but the properties of functors which you work with.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .