Let $\mathcal{A}, \mathcal{B}$ be two abelian categories with sufficiently many injective objects (in my case these are categories of sheaves of vector spaces on a manifold). Let $f_*\colon \mathcal{A}\to\mathcal{B}$ be a left exact functor which commutes with direct sums (in my case it is push-forward on sheaves). Let $Rf_*\colon D^+\mathcal{A}\to D^+\mathcal{B}$ be the derived functor between the derived categories. Let $F^\bullet$ be a bounded complex of objects of $\mathcal{A}$.

Question 1. Assume that $R^qf_*(F^p)=0$ for every $q\ne i$ where $i$ is a fixed number, and for every $p$. Is it true that $Rf_*F^\bullet$ is isomorphic in $D^+\mathcal{B}$ to the complex $$\dots\to R^if_*(F^k)\to R^if_*(F^{k+1})\to R^if_*(F^{k+2})\to\dots \,?$$ (For $i=0$ this is well known to be true. I believe that it should be true in general. A reference would also be helpful.)

Question 2. Assume now that for a fixed integer $a$ one has $$\begin{eqnarray*} R^qf_*(F^a)=0 \mbox{ for every } q;\\ R^qf_*(F^p)=0 \mbox{ for every } q\ne 0, p>a;\\ R^qf_*(F^p)=0 \mbox{ for every } q\ne 1, p<a. \end{eqnarray*}$$ Is it true that $Rf_*(F^\bullet)$ is isomorphic to the complex (with an appropriate cohomological shift) $$\to R^1f_*F^{a-2}\to R^1f_*F^{a-1} \overset{\Delta}{\to}R^0f_*F^{a+1}\to R^0f_*F^{a+1}\to \dots$$ where $\Delta$ is the only non-trivial differential of second order in the well known spectral sequence $E^{p.q}_r$ which converges to $R^{p+q}f_*(F^\bullet)$ and whose first term is equal to $E_1^{p,q}=R^qf_*(F^p)$? (The other arrows in the complex are induced in the obvious way by arrows in the complex $F^\bullet$.)

Remark. The second question is motivated by the fact that the two complexes have the same cohomology, as it follows from the spectral sequence considerations.

  • $\begingroup$ I'm assuming that in question 1 it should be $$ \to R^if_*(F^{k+1}) \to R^if_*(F^{k+2}) \to? $$ $\endgroup$ – Simon Rose Oct 12 '14 at 10:50
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    $\begingroup$ If by "any q" in question 1 you mean "every q", this is true by the same (simple) argument as the i=0 case. You should understand the argument yourself. $\endgroup$ – anon Oct 12 '14 at 13:09
  • $\begingroup$ @anon: Corrected to "every q". Thanks. $\endgroup$ – makt Oct 12 '14 at 14:44

First of all, the functor $Rf_\ast$ is a red herring. Take an injective resolution $I^{\bullet,\bullet}$ of $F^\bullet$ and apply $f_\ast$; then you just have two questions regarding double complexes. E.g. in the first case you're asking about double complexes whose columns have cohomology concentrated in a single degree.

Let $C^\bullet$ be a cochain complex. Recall that there are wise truncation functors $\tau_{\geq i}$, $\tau_{\leq i}$ with canonical maps $C^\bullet \to \tau_{\geq i}C^\bullet$ and $\tau_{\leq i}C^\bullet \to C^\bullet$, which are quasi-isomorphisms if $C^\bullet$ has no cohomology below (resp. above) degree $i$.

If $A^{\bullet,\bullet}$ is a double complex, let $\tau^V_{\geq i}$ be the double complex obtained by applying $\tau_{\geq i}$ to each column separately (V is for "vertical"). Then there is a zig-zag $$ A^{\bullet,\bullet} \to \tau_{\geq i}^V A^{\bullet,\bullet} \leftarrow \tau_{\leq i}^V\tau^V_{\geq i} A^{\bullet,\bullet}.$$ If all columns have cohomology concentrated in degree $i$ then both these maps are quasi-isomorphisms of double complexes. This answers your first question affirmatively, since $\tau_{\leq i}^V\tau^V_{\geq i} A^{\bullet,\bullet}$ can be identified with the cochain complex you wrote down. (Remark: I slightly disagree with anon's comment that the case $i>0$ is identical to the case $i=0$, since in the latter case we don't need to apply $\tau_{\geq i}$. So in that case we don't even need a zig-zag, there is just a quasi-isomorphism staring us in the face.)

The second requires a bit more fiddling, I think, but maybe I'm missing an easy argument. But the following should work. First do a truncation argument similar to the one above to reduce to the case when $A^{p,q} = 0$ unless $p\leq a$ and $q=1$, or $p \geq a$ and $q=0$; moreover, $A^{a,0} \to A^{a,1}$ is an isomorphism. Now let $B^{\bullet,\bullet}$ be the double complex with $B^{p,q} = A^{p,q}$ for $p \neq a$, but $B^{a,0} = B^{a,1} = A^{a-1,1}$. The differentials in $B$ are given by letting $B^{a-1,1} \to B^{a,1}$ and $B^{a,0} \to B^{a,1}$ be the obvious isomorphisms and $B^{a,0} \to B^{a+1,0}$ the map given by the $E_2$ differential in the spectral sequence for $A^{\bullet,\bullet}$. There is a natural map $B^{\bullet,\bullet} \to A^{\bullet,\bullet}$, and also a natural map from $\mathrm{Tot}(B^{\bullet,\bullet})$ to the cochain complex which you wrote down, both of which are quasi-isomorphisms.

  • $\begingroup$ Thank you very much for the nice answers! I think I understood the argument. I would need one more property, which seems to follow from your constructions, but I would like to make sure. It would be helpful for me to have in both questions the constructed isomorphisms to be functorial on the category of complexes satisfying the corresponding vanishing conditions. Is it the case? $\endgroup$ – makt Oct 14 '14 at 11:15
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    $\begingroup$ I agree that both constructions should be functorial. $\endgroup$ – Dan Petersen Oct 14 '14 at 14:39

The previous answer may be all you need, but, say, in question one one can also proceed by induction in q, to make the step of induction embed your complex into an acyclic complex of injective objects.


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