Let $A{\buildrel F\over\rightarrow}B{\buildrel G\over\rightarrow}C$ be additive functors between abelian categories.

Hartshorne, in Proposition 5.4 of Residues and Duality, constructs the obvious natural transformation $\zeta_{G,F}:R(GF)\Rightarrow (RG)(RF)$ and notes its obvious properties. He then remarks:

This proposition shows the convenience of derived functors in the context of derived categories. What used to be a spectral sequence becomes now simply a composition of functors. (And of course one can recover the old spectral sequence from this proposition by taking cohomology and using the spectral sequence of a double complex).

I've always felt like I'm missing something here. In what sense does the composition of functors (together with the natural transformation $\zeta_{G,F}$) replace the old spectral sequence? I understand that there's a conceptual insight here, but my questions are these:

  1. What (if anything) is an example of a statement that used to be proved by invoking the spectral sequence but can now be proved more succinctly using the composition of functors?

  2. In what sense does the recovery of the spectral sequence actually use the derived formalism? Isn't the prescribed double complex exactly the same one I'd have written down if I'd never heard of derived categories?

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    $\begingroup$ One thing which becomes painful in terms of spectral sequences is what happens when you compose three functors. On the other hand, when you want to actually compute something, spectral sequences are your only friend, mostly. $\endgroup$ Mar 21, 2012 at 2:50
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    $\begingroup$ Grothendieck's spectral sequence is just a chain rule. On the one hand you have the derivatives of the composition, on the other hand you have a 'formula' (the spectral sequence) in terms of the composition of the derivatives. $\endgroup$ Mar 21, 2012 at 11:07

1 Answer 1


1 Easy

Proposition Let $f:X\to Y$ be a continuous map of topological spaces, $\mathscr F$ a sheaf of abelian groups on $X$ such that $R^jf_*\mathscr F=0$ for $j>0$. Then for all $i\geq 0$ there exists a natural isomorphism $$ H^i(Y, f_*\mathscr F)\simeq H^i(X,\mathscr F) $$

Proof Apply the composition rule for the derived functors of $G=\Gamma(Y, \_ )$ and $F=f_*({\_})$. By definition, $G\circ F = \Gamma(X, \_ )$. Then $$ R\Gamma(Y, f_*\mathscr F) \simeq R\Gamma(Y, Rf_*\mathscr F) \simeq R\Gamma (X, \mathscr F). $$ Taking cohomology shows the result. $\square$

(edit to please anon, see the comments below) This is usually exhibited as an example of how to use the Leray spectral sequence. Doing it that way is not much harder than the above, but perhaps a bit less "automatic".

Furthermore, this proof shows more: Not only the cohomologies of these sheaves are isomorphic, but they come from the same complex! That's a much stronger statement. It is easy to give examples when the cohomologies of two complexes are isomorphic, but the complexes are not. I suppose one may argue that the word "natural" in the statement means exactly this, but then I'd say that proving naturality with the Leray spectral sequence is certainly possible, but it definitely needs more care.

This last point is actually an important one regarding the derived category language. You get a higher level notion. The fact that you can work with the complex whose cohomologies give the derived functors of your original functor is very very useful.

2 Less Easy

In case you were not convinced by the above example, here is one that should do the trick:

A special case of Grothendieck duality says that if $f:X\to Y$ is a proper morphism between not too horrible schemes, let's say finite type over a field $k$ (let me not try to make a precise statement, this is in Residues and Duality that you mentioned) and $\mathscr F$ is a coherent sheaf on $X$, then $$ Rf_*R\mathscr Hom_X(\mathscr F, \omega_X^{\bullet})\simeq R\mathscr Hom_Y(Rf_*\mathscr F, \omega_Y^{\bullet}). $$ Here $\omega_{Z}^{\bullet}=\varepsilon^!k$ is "the" dualizing complex where $\varepsilon: Z\to \mathrm{Spec}\ k$ is the structure map of $Z$.

Now try to imagine how one could state this using spectral sequences. Both sides actually correspond to spectral sequences, so the statement would be something like "there is a natural map between this an this spectral sequences, such that they converge to the same thing".

I would argue that already the statement of this theorem would be tiring in the language of spectral sequences, but using it would be pure pain.

3 Even Less Easy

Here is an application of Grothendieck duality where one can see how the derived category formalism makes life easier and arguments that seemed complicated are reduced to a one liner.

Theorem (a.k.a. Kempf's Criterion) Let $Y$ be a normal variety over $\mathbb C$ with a resolution of singularities $f:X\to Y$. Then $Y$ has rational singularities (i.e., $R^if_*\mathscr O_X=0$ for $i>0$) if and only if

  1. $Y$ is Cohen-Macaulay
  2. $f_*\omega_X\simeq \omega_Y$

Proof Let $n=\dim Y=\dim X$ and suppose $Y$ has rational singularities. Then $$ \omega_Y^{\bullet}\simeq R\mathscr Hom_Y(\mathscr O_Y, \omega_Y^{\bullet})\simeq R\mathscr Hom_Y(Rf_*\mathscr O_X, \omega_Y^{\bullet})\simeq Rf_*R\mathscr Hom_X(\mathscr O_X, \omega_X^{\bullet})\simeq Rf_*\omega_X[n]\simeq f_*\omega_X[n]. $$ (The isomorphisms follow by the assumptions, Grothendieck duality, and the last one is the Grauert-Riemenschneider vanishing theorem). This implies that $\omega_Y=h^{-n}(\omega_Y^{\bullet})\simeq f_*\omega_X$, which is the second condition to prove and also that $h^i(\omega_Y^{\bullet})=0$ for $i\neq -n$ which is equivalent to $Y$ being Cohen-Macaulay.

The other direction goes essentially in the same fashion. $\square$

Now try to do this with spectral sequences.

To answer your second question, I think you are right. In order to get the spectral sequences you do not need to go through the derived category formalism. However, if you are indeed "recovering" the spectral sequence, then you start with the derived category formalism. In other words, if you've never heard of derived categories, why (or perhaps more importantly how) would you want to recover anything from a derived category statement? (Since written word lacks intonation, let me add that I'm not trying to be confrontational, but I feel that this question is somehow off target.)

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    $\begingroup$ In my country Grothendieck duality is not compulsory for High School students. I hang my head in shame. $\endgroup$ Mar 21, 2012 at 12:05
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    $\begingroup$ Cher Georges, please excuse my joke! First I wrote (as somewhat customary) "baby case" for the first one. Then I was stuck with what I should write for the second and third. My next attempt was "toddler" and "preschool" for the next two, but I thought that's a little ridiculous, I should somehow get to an advanced level. I tried various combination, but they all looked silly. At the end I just got tired and went with the latest one. As always, you have eloquently pointed to a fault without making the culprit feel too bad about it. Thank you for that! Cheers! $\endgroup$ Mar 21, 2012 at 15:16
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    $\begingroup$ @Yosemite Sam: the Grauert-Riemenschneider vanishing is probably in any book on higher dimensional geometry, especially one that discusses generalizations of the Kodaira vanishing theorem. A possible source is Lazarsfeld's Positivity in AG. $\endgroup$ Mar 21, 2012 at 15:21
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    $\begingroup$ Dear Sándor: but there is no fault and no culprit! On the contrary, I loved your joke and since I had nothing mathematical to contribute to your perfect answer, I just wrote a little tongue-in-cheek, good-humoured comment as a witness that I had read and appreciated your answer. $\endgroup$ Mar 21, 2012 at 15:29
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    $\begingroup$ Looks like I'm still in elementary school... $\endgroup$ Mar 23, 2012 at 6:35

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