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Composing left and right derived functors


probably this question is obvious. I apologize for this.

Given functors $F$ and $G$ left exact, with as good properties as you want, on the "correct" category, we have a spectral sequence $R^p F\circ R^q G$ abutting to $R^{p+q}F\circ G$. I am looking for an analogous for a "mixed version" in the following case: $F$ left exact and $G$ right exact. What appens to $L^pG\circ R^q F$?


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Such questions are probably best thought about in terms of composing total derived functors. $L^{p}G(Y)$ is computed by taking a $G$-acyclic resolution of your object $Y$, applying $G$ to each term of that resolution, and then taking cohomology of the resulting complex at $p$. But you could also stop after applying $G$ to each term to get the total derived functor $LG(Y)$, which is a complex. Likewise, you can get the total derived functor $RF(X)$. Now you could try to compose the two total derived functors: given $X$, resolve it with $F$-acyclics, apply $F$ term by term to a complex $Y$, and –  Chris Brav Jul 26 '11 at 11:22
now resolve the complex $Y$ by a complex of $G$-acyclics (meaning replace it with a quasi-isomorphic complex of $G$-acyclics) and apply $G$ term by term. This will give you a composition $LG \circ RF$. Now you could ask how to compute cohomology of the resulting complex $LG \circ RF (X)$. This should be doable in terms of a spectral sequence beginning with $L^{p}G \circ R^{q}F$. –  Chris Brav Jul 26 '11 at 11:25
I retagged with spectral-sequences and derived-functor tags. Hope you don't mind! –  David White Jul 26 '11 at 12:37
For the Grothendieck spectral sequence, it doesn't matter that your triangulated functors are derived functors of some other functors -- it just gives the spectral sequence relating the cohomologies of the composition to the cohomologies of the second functor applied to cohomologies of the first. This is clear e.g. from the construction is "Theorie de Hodge II." In your setting, a person needs to worry about things like convergence of the spectral sequence, but that's another question! –  Moosbrugger Jul 26 '11 at 13:30
The question was discussed here: mathoverflow.net/questions/24143/… –  Sasha Jul 26 '11 at 18:41
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marked as duplicate by Anton Geraschenko Jul 26 '11 at 20:16

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