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Let $A$ be a commutative noetherian ring.

Let $K_{parf}(A)$ be the full subcategory of the homotopy category $K(A)$ of $A$-modules whose objects are bounded complexes of finitely generated projective $A$-modules. The functor $K_{parf}(A) \longrightarrow D(A)$ is fully faithful and we denote its essential image by $D_{parf}(A)$. An object of $D_{parf}(A)$ is called a perfect complex.

I have to give a talk about the Trace formula on tuesday and the audience and me are not so comfortable with derived categories (yet). (See below for what we want to do this tuesday.)

Question 1. Is a complex of $A$-modules perfect iff it is quasi-isomorphic to a bounded complex of finitely generated projective $A$-modules? Is there some condition involving ``homotopy'' missing?

Let $k$ be an algebraically closed field and let $\mathcal{F}$ be a locally constant finite sheaf of finitely generated projective $A$-modules on a finite type separated $k$-scheme $X$.

This defines an object $$K=\ldots \longrightarrow 0\longrightarrow \mathcal{F} \longrightarrow 0 \longrightarrow \ldots $$ of the derived category $D(X,A)$ of sheaves of $A$-modules.

Question 2. Can one explicitly write down a complex representing $R\Gamma_c(X,\mathcal{F})$? (One may assume $X$ is proper over $k$ for simplicity.)

One can show that $R\Gamma_c(X,\mathcal{F})$ is a perfect complex after identifying $A$-modules with sheaves of $A$-modules on $\textrm{Spec} \ k$. If the answer to Question 2 is positive, I could avoid introducing derived categories on tuesday.

[ADDED]

To prove Grothendieck's generalized Lefschetz trace formula, it suffices to prove the following statement.

Theorem. Let $X_0$ be a smooth affine geometrically integral curve over $\mathbf{F}_q$ and let $\Lambda$ be a finite ring which gets killed by a power of a prime number $\ell$ invertible in $\mathbf{F}_q$. Suppose that $X_0(k) = \emptyset$. Then, for any locally constant finite sheaf $\mathcal{F}$ of finite projective $\Lambda$-modules, we have that $$ 0= \textrm{Tr}(\textrm{Frob}^\ast, R\Gamma_c(X,K)).$$

So one has to do three things:

  1. Understand the statement above.
  2. Show that this statement implies the Trace formula
  3. Prove the statement above.

1 and 3 don't really need the language of derived categories, I believe. That's why I asked this question. Next week we will take care of 2 after introducing derived categories.

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    $\begingroup$ For Question 2, I guess it might be simpler if one assumes that X is smooth rather than proper (when proving the trace formula one can shrink X to open subspaces), so that one can use Poincaré duality to identify $R\Gamma_c(X,F)$ with the dual complex of $R\Gamma(X,F^{\vee}(d)[2d]).$ Anyway I think it's worth mentioning derived categories in the étale cohom. workshop, if no one is assigned to discussed it in detail. For Q1, I remember SGA 4.5 proved something related. You may find some sections in chapter 2 of K-W useful, too. $\endgroup$
    – shenghao
    Commented May 28, 2011 at 16:58
  • $\begingroup$ Next week someone will take care of the ''derived categorical'' part of the proof. Please see my edit to see what our plan is. As you said, we can assume $X$ to be smooth in the proof of the Trace formula. Therefore, we can use your suggestion. The only problem is that we haven't done Poincare duality yet and I also don't know what you mean by RΓ(X,F∨(d)[2d])...What is K-W? $\endgroup$
    – Ariyan Javanpeykar
    Commented May 28, 2011 at 17:50
  • $\begingroup$ By the way, for the proof of the above statement (see Edit) one doesn't really need to know what RGamma_c(X,K) is...In fact, one uses Weil's classical trace formula for smooth projective curves and combines this with some results about traces over noncommutative rings (such as Z/l^nZ[G]). Just something funny I wanted to point out. $\endgroup$
    – Ariyan Javanpeykar
    Commented May 28, 2011 at 17:53

1 Answer 1

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For question 1 the answer is yes, in fact the question is tautological. As for question 2 (I assume that $R\Gamma_c$ stands for the sections with compact support which in the case of proper $X$ coincide with $R\Gamma$), of course you can use any standard complex, for example you can use Cech complex.

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  • $\begingroup$ So if I understand correctly, one can do the following. Let C be the Cech complex of F (associated to some cofinal open covering). Then C is quasi-isomorphic to RGamma_c(X,F)? This would be great! $\endgroup$
    – Ariyan Javanpeykar
    Commented May 28, 2011 at 18:38
  • $\begingroup$ I meant f_*F, where f:X-->Y is a compactification. More precisely, given a projective curve Y over k and a locally constant sheaf F on the Y_{et}, do we have that RGamma(Y,F) is quasi-isomorphic to the Cech complex of F (associated to a cover by two open affines in Y)? $\endgroup$
    – Ariyan Javanpeykar
    Commented May 28, 2011 at 19:12
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    $\begingroup$ I don't see why the Cech complex (attached to an etale open covering) will do, i.e. the assoc. spectral sequence won't degenerate in general. Take $X$ to be a proj. curve covered by two opens, and $F$ is the constant sheaf. Then there's no way to get a rank-$2g$ space out of the Cech complex. $\endgroup$
    – shenghao
    Commented May 28, 2011 at 22:47
  • $\begingroup$ You're right. Let me say that I don't really need to write down an explicit complex. I just want to define RGamma(X,F). I could take an injective resolution for F, say I. Then what does RGamma(X,F) become? Is it the complex you get after taking global sections? $\endgroup$
    – Ariyan Javanpeykar
    Commented May 28, 2011 at 23:18
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    $\begingroup$ Yes, $R\Gamma(X,F),$ for $F$ bounded below, can be computed as (i.e. is quasi-isom. to) the complex $\Gamma(I^{\bullet}).$ The definition of $R\Gamma(X,-)$ fits into the general framework of derived categories and derived functors (i.e. some universal 2-properties), and on $D^+(X)$ one can use injective resolutions to "compute" (not in the practical sense, I think). On the full $D(X)$ one can use K-injective resolutions (and K-flat resol. for left derived functors); see Spaltenstein's "Resolutions of unbounded complexes". $\endgroup$
    – shenghao
    Commented May 29, 2011 at 11:33

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