In their famous paper "On the differentiation of De Rham cohomology classes..." Katz and Oda construct the spectral sequence for de Rham cohomology for the situation of a smooth morphism

$\pi: X \rightarrow S$

of smooth $k$-schemes ($k$ a field), where $S$ is assumed affine.

There is a step which is not clear to me: in Lemma 8 of this paper they say that the equality

$(*) \qquad \mathbb R^0\Gamma_X =\Gamma_S\circ \mathbb R^0\pi_*$

yields a spectral sequence of composition

$(**) \qquad E_2^{a,b}=R^a \Gamma_S\circ \mathbb R^b\pi_* \Rightarrow \mathbb R^{a+b}\Gamma_X$.

They seem to consider $(*)$ as an equality of functors on the category of complexes of abelian sheaves on $X$ - Arguments involving quasi-coherence and affineness of $S$ only appear in the next step.

My question is a very simple one which I nevertheless can't figure out:

Why do they have $(**)$?

It is a spectral sequence of composition, hence only exists when one knows that $\mathbb R^0 \pi_* $ sends injective objects in the category of complexes of sheaves on $X$ to $\Gamma_S$-acyclic ones. I don't see why this is true: imagine a two-term complex $I^0\rightarrow I^1$ of injective abelian sheaves on $X$, then $\mathbb R^0 \pi_*$ of it is $ker(\pi_*I^0 \rightarrow \pi_*I^1)$. This has no reason to be acyclic for $\Gamma_S$, hasn't it?

A similar problem arises a bit later in Lemma 9 when they consider the equality

$(+) \qquad \mathbb R^0\Gamma_S = H^0\circ \Gamma_S$

which I see as equality of functors on the category of complexes of abelian sheaves on $S$.

Can anybody give a hint how to cleanly resolve these two problems?

  • $\begingroup$ How is this different than the Leray spectral sequence? $\endgroup$ Jun 2, 2013 at 9:18
  • $\begingroup$ Because in the case of the usual topological Leray sequence you immediately know that injective sheaves are sent by $\pi_*$ to $\Gamma_S$ acyclics (because the direct image of an injective sheaf is flasque). But note that here we are dealing with complexes. $\endgroup$
    – Veen
    Jun 2, 2013 at 9:25
  • $\begingroup$ Doesnt the Grothendieck spectral sequence give $(**)$? $\endgroup$
    – J.C. Ottem
    Jun 2, 2013 at 9:52
  • $\begingroup$ The problem is that the Grothendieck sequence only exists under the hypothesis that injectives are sent by the first functor to acyclic objects with respect to the second functor. $\endgroup$
    – Veen
    Jun 2, 2013 at 10:03
  • 2
    $\begingroup$ "It is a spectral sequence of composition, hence only exists when one knows that $\mathbb R^0\pi_\ast$ sends injective objects in the category of complexes of sheaves on $X$ to $\Gamma_S$-acyclic ones." I think you're confused here: you only need injective sheaves to be mapped to acyclic ones. $\endgroup$ Jun 2, 2013 at 10:09

2 Answers 2


Let me try to say the same thing as Will but in a different way. Let $f \colon A \to B$ be left exact between abelian categories and assume there are enough injectives. Then one can distinguish between derived functors $$\newcommand{\R}{\mathrm R} \R^i f \colon A \to B,$$ as well as hyper-derived functors $$\newcommand{\RR}{\mathbb R} \RR^i f \colon \mathrm{Kom}^+(A) \to B,$$ and the total derived functor $$ \newcommand{\RRR}{\mathbf R} \RRR f \colon D^+(A) \to D^+(B).$$ Now suppose we also have $g \colon B \to C$ with the same hypotheses and that $f$ maps injective objects of $A$ to $g$-acyclic objects of $B$. Then there is a natural isomorphism $$ \RRR(g \circ f) \cong \RRR g \circ \RRR f.$$ This isomorphism specializes not only to the usual spectral sequence $$ E_2^{p,q} = \R^p g( \R^q f(X)) \implies \R^{p+q}(g \circ f)(X),$$ for $X$ any object of $A$, but also to the spectral sequence $$ E_2^{p,q} = \R^p g( \RR^q f(X^\bullet)) \implies \RR^{p+q}(g \circ f)(X^\bullet),$$ for $X^\bullet$ any object of $\mathrm{Kom}^+(A)$, in a similar way. In particular one needs no extra hypotheses to get the latter spectral sequence; even though the domain of $\RR^i f$ is $\mathrm{Kom}^+(A)$, one should not assume that complexes of injectives are sent to acyclics.

  • $\begingroup$ @Dan: do you happen to have a reference for the second spectral sequence you write down? In the standard books I have always only seen the first one. Or are the constructions simply the same for both? $\endgroup$
    – Veen
    Jun 2, 2013 at 17:45
  • $\begingroup$ I'm at home now so I can't check, but I'm quite sure you can find it either in Dimca's "Sheaves in topology" or Gelfand--Manin's "Homological algebra". $\endgroup$ Jun 2, 2013 at 18:34
  • $\begingroup$ Found it in Dimca's book (Thm. 1.3.19), thank you! Side remark: Katz in "Nilpotent connections...", Rem. (3.3.3), indeed seems to use a statement like the one questioned above, namely that injective objects in the category of complexes of sheaves on $X$ are sent to $f_*$-acyclics by $\mathbb R^0\pi_*$, where $\pi: X \rightarrow S$ and $f: S \rightarrow T$ are maps of top. spaces. I still wonder if this is correct and would gratfully appreciate any comment. $\endgroup$
    – Veen
    Jun 2, 2013 at 18:57
  • $\begingroup$ I can't comment on this off the top of my head because I don't know off hand a simple description of what is an injective object in the category of complexes of sheaves. It is certainly false if you just consider complexes of injective objects. Then it fails already for $\pi = \mathrm{id}$, in which case $\mathbb R^0 \pi_\ast = \mathcal H^0$: indeed, since every sheaf has a resolution of injectives, and $\mathcal H^0$ of this resolution is the original sheaf, this would force every sheaf to be $f_\ast$-acyclic. $\endgroup$ Jun 2, 2013 at 21:14
  • 1
    $\begingroup$ I hardly dare say it, but it's possible that Katz was momentarily confused when writing this. $\endgroup$ Jun 3, 2013 at 18:07

You only need injective objects in the underlying abelian category to be mapped to acyclics, not in the category of complexes, to get a Grothendieck spectral sequence. This is because if you take a complex $A$, and take an injective resolution over that complex (to compute hypercohomology), then after applying $R \pi_*$ to this complex, you get an acyclic resolution of $R \pi_* A$. Then you can use that to compute hypercohomology of $\Gamma_s$, which you do by just applying $R^0 \Gamma_s$ to each object, which is the same as applying $R^0 \Gamma$ to the original injective resolution, which is just computing hypercohomology of $R^0 \Gamma$.

  • $\begingroup$ @ Will: I still don't quite see how the existence of the Grothendieck spectral sequence is implied by what you said. Does one have to go into the explicit construction of that sequence to see this? $\endgroup$
    – Veen
    Jun 2, 2013 at 13:56
  • $\begingroup$ This is how you get an equality of functors in the category of complexes. If you want a spectral sequence, from this perspective, it's the spectral sequence for the hypercohomology of $\Gamma_s$. $\endgroup$
    – Will Sawin
    Jun 2, 2013 at 14:15

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