Let $f: X \rightarrow X'$ be a morphism of schemes, and let $\mathcal{I}^{\bullet}$ be a complex of $\mathcal{O}_X$-modules. There are two spectral sequences (well, more than that, but these are the two I care about) abutting to the hypercohomology $\mathbb{H}^n(X, \mathcal{I}^{\bullet})$. The first is the second spectral sequence of hypercohomology, which has $E_2$-term

$E_2^{p,q} = H^p(X, H^q(\mathcal{I}^{\bullet}))$

(where $H^q(\mathcal{I}^{\bullet})$ just means the $q^{th}$ cohomology object of the complex $\mathcal{I}^{\bullet}$)

and the second is the Leray spectral sequence associated with $f$, which has $E_2$-term

$E_2^{p,q} = H^p(X', R^q f_* \mathcal{I}^{\bullet})$

(Here, since the terms in the spectral sequence are $\Gamma(X', \mathcal{O}_{X'})$-modules, the abutment $\mathbb{H}^n(X, \mathcal{I}^{\bullet})$ must be viewed as a $\Gamma(X', \mathcal{O}_{X'})$-module by restricting scalars.)

Are there conditions on $X, X', f$, and/or $\mathcal{I}^{\bullet}$ under which we can say that these $E_2$-terms are the same, that is, that $H^p(X, H^q(\mathcal{I}^{\bullet})) \simeq H^p(X', R^q f_* \mathcal{I}^{\bullet})$ as $\Gamma(X', \mathcal{O}_{X'})$-modules.

In this (non-degenerate) case, it's not really a spectral sequence question at all, but a question of somehow comparing two iterated cohomology objects. In the case I care about, in which I suspect for other reasons that there will be such an isomorphism but the spectral sequences will almost surely not degenerate at $E_2$, there are lots of nice properties that can be assumed: $X$ and $X'$ are smooth and projective over a field, $f$ is smooth and projective (hence, in particular, proper and flat), and the complex $\mathcal{I}^{\bullet}$ is a complex of injective $\mathcal{O}_X$-modules (but not, however, an injective resolution of any single $\mathcal{O}_X$-module).

(Edited 7/2: as Karl Schwede points out, even if both degenerate at the $E_2$-term there's no reason to assume all the terms are the same.)

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    $\begingroup$ Shouldn't some significant hypothesis be put on the dynamics of $f$?: One s.s. does not depend on $f$ while the other does. $\endgroup$ Jul 2, 2013 at 0:00
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    $\begingroup$ Even if they both degenerate at the E2 page (meaning the maps of the spectral sequence are zero), I don't see why two spectral sequences have the same terms. $\endgroup$ Jul 2, 2013 at 1:58
  • $\begingroup$ @Mariano: indeed, appropriately restrictive hypotheses on $f$ making these $E_2$-terms isomorphic are exactly what I'm trying to find. $\endgroup$ Jul 2, 2013 at 20:34
  • $\begingroup$ @Karl: thanks, I've edited the question. $\endgroup$ Jul 2, 2013 at 20:35
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    $\begingroup$ Ok, as Dan Peterson points out below, there probably isn't much hope unless one knows more. Is there any chance that the complex $I^{\bullet}$ is made up of sheaves of differentials, or pushed down from some other scheme made up of sheaves of differentials from some other scheme (ie, a dualizing complex, or something similar even from a simplicial scheme). That's one general case where there are lots of vanishing theorems like the kind you need. Alternately, if you know a lot more about the fibers of your morphism $f$ (are they all Fano?), there might be other things you can try. $\endgroup$ Jul 3, 2013 at 15:56

1 Answer 1


In the second spectral sequence you write $R^q f_\ast I^\bullet$. This is correct but $R^qf_\ast$ must be considered as the $q$th hyper-derived functor applied to the complex $I^\bullet$, it is not the ordinary derived functor $R^qf_\ast$ applied termwise. See my answer to Construction of the spectral sequence of Katz/Oda for more details.

Once you observe this, then you find that the hypercohomology spectral sequence is just the special case of the Leray spectral sequence when $f = \mathrm{id}$. Indeed the hyper-derived functor $R^q \mathrm{id}_\ast$ is just the functor $\mathcal H^q$.

Addendum: One situation where what you want is actually true (and which is more general than $f=\mathrm{id}$) is if all cohomology sheaves $\mathcal H^q(I^\bullet)$ are $f_\ast$-acyclic, i.e. if they vanish under $R^qf_\ast$ for $q> 0$. This implies that $$ R^qf_\ast I^\bullet \cong f_\ast \mathcal H^q I^\bullet$$ (where again on the left hand side I mean the hyper-derived functor), and also that $$ H^p(X,\mathcal H^q I^\bullet) \cong H^p(X',f_\ast\mathcal H^q I^\bullet).$$It's not hard to show that there is even an isomorphism between the two spectral sequences.

But this is a very restrictive condition. What you're asking for is not very natural -- in general there is not even a map between the two spectral sequences.

  • $\begingroup$ But if $f \neq id$, we'll have two a priori different $E_2$-terms, since we're looking at two different compositions of functors; will there be any conditions (on $f$, say) under which we can say these are isomorphic? $\endgroup$ Jul 2, 2013 at 20:37

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