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I don't know if I'm actually using the right terminology here, to be clear I'm going to state explicitly what I'm trying to figure out to see if I can be pointed in the right direction:

Let $F: \mathcal{A} \rightarrow \mathcal{B}$ be a left exact functor between 2 abelian categories $\mathcal{A}$ and $\mathcal{B}$, let $X^\bullet$ be the complex $0 \rightarrow X^0 \rightarrow X^1 \rightarrow X^2 \rightarrow X^3 \rightarrow \cdots$ of objects in $\mathcal{A}$, let $T^\bullet$ be a complex obtained by removing the FIRST TERM (or maybe the first $q$ terms from the left?), I mean let $T^\bullet$ be $X^0 \rightarrow X^1 \rightarrow X^2 \rightarrow X^3 \rightarrow \cdots$, the complex obtained by removing the $0$ from $X^\bullet$, NO 0's to the left, so $T$ becomes "unbounded". Let's say I have a way of obtaining the derived functor $RF(T^\bullet)$ using the machinery developed for unbounded complexes. Is there any way I can compute some of the terms in $RF(T^\bullet)$ from knowing this?

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  • $\begingroup$ As an extreme case, imagine the initial complex has exactly one term! :-) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Sep 14, 2012 at 20:16
  • $\begingroup$ If the domain category has finite global dimension $d$, for example, perturbations in a complex only travel $d$ (or maybe $d+1$?) places. If you go through the steps of constructing a projective resolution of your initial complex and then applying the functor to that, you can see this at once. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Sep 14, 2012 at 20:18
  • $\begingroup$ Sorry I'm a bit lost here, what's a perturbation in this context? Thanks for answering $\endgroup$
    – Louis A
    Commented Sep 14, 2012 at 20:40
  • $\begingroup$ By perturbation I simply mean a change one of the terms in the complex (for example, replacing it with zero) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Sep 14, 2012 at 20:44
  • $\begingroup$ This is the first time I've heard the term perturbation in this context, so I don't know what it means for a perturbartion to travel $d$ places, apply the functor to what? To the perturbation? To see what at once? Sorry if I sound clueless but I'm totally unfamiliar with what you're saying $\endgroup$
    – Louis A
    Commented Sep 14, 2012 at 21:56

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Suppose for simplicity $T$ and $S$ are two bounded complexes supported in negative degrees which differ only in their $0$th term. Construct Cartan-Eilenberg projective resolutions $P_T$ and $P_S$ for them, so that $RF(T)$ and $RF(S)$ are in fact $F(\operatorname{Tot} P_T)$ and $F(\operatorname{Tot} P_S)$. Now, in view of the way one constructs C-E resolutions, $P_T$ and $P_S$ can be chosen in such a way that they only differ in specific places (the terms appearing in the $0$th column and in the differentials which have those terms as codomains, I think). If the global dimension $d$ of $\mathcal A$ is finite, and if one uses finite resolutions for objects of length at most $d$, the complexes $F(\operatorname{Tot} P_T)$ and $F(\operatorname{Tot} P_S)$ therefore differ in at most $d+1$ places.

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    $\begingroup$ More generally, if the complex $T$ differs from $S$ only in the terms with degree in $\{-r,\dots,0\}$ and the global dimension of the domain category is $d$, then $RF(T)$ and $RF(S)$ will coincide in all degrees smaller that $-r-d-1$ (plus or minus 1...) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Sep 14, 2012 at 22:19
  • $\begingroup$ Thank you! Are there any books or references for this? What's the global dimension of a category? I know what's the global dimension of a module, I found this: bit.ly/O4phq7 $\endgroup$
    – Louis A
    Commented Sep 15, 2012 at 0:18
  • $\begingroup$ Do you have a reference for this, I can't find it on Google, please, there has to be a reference, I'm in a hurry, I'm gonna get thrown out of school if I don't hurry up $\endgroup$
    – Louis A
    Commented Sep 15, 2012 at 1:25
  • $\begingroup$ For left exact functors one usually uses injective resolutions. Projective resolutions work for right exact functors. $\endgroup$
    – Sasha
    Commented Sep 15, 2012 at 2:52
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$T$ is called the stupid truncation of $X$ (in a contrast with the canonical or smart truncation).

In the derived category $D(A)$ there is a distinguished triangle $$ T \to X \to R, $$ where $R$ is the complex consisted of the ONE term (or $q$ terms) of $X$. Applying the derived functor $RF$ you get a distinguished triangle $$ RF(T) \to RF(X) \to RF(R) $$ in $D(B)$. So, $RF(R)$ measures the difference between $RF(T)$ and $RF(X)$. In particular there is a long exact sequence $$ \dots \to R^nF(T) \to R^nF(X) \to R^nF(R) \to R^{n+1}F(T) \to \dots $$ and if you know $R^nF(R)$ you can compare $R^nF(X)$ and $R^nF(T)$.

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  • $\begingroup$ Thank you so much! One thing though: The thing is this, complex X is $0 \rightarrow X^1 \rightarrow X^2 \rightarrow X^3 \rightarrow \cdots$ and complex T is $X^1 \rightarrow X^2 \rightarrow X^3 \rightarrow X^4 \rightarrow \cdots$, the same complex starting at $X^1$ so I'm treating it as an unbounded complex, I did it that way because removing the $0$'s from $X$ makes the complex exact and easier to work with $\endgroup$
    – Louis A
    Commented Sep 15, 2012 at 4:34
  • $\begingroup$ Sorry to bother you again, but isn't a stupid truncation the one where you add 0's to the left? I don't know if I'm not understanding this right but my complex $T$ is "unbounded", it doesn't have 0's to the left, $T$ is $X^0 \rightarrow X^1 \rightarrow X^2 \rightarrow X^3 \rightarrow \cdots$, NOT $0 \rightarrow 0 \rightarrow X^0 \rightarrow X^1 \rightarrow X^2 \rightarrow X^3 \rightarrow \cdots$, don't know if I'm confused there $\endgroup$
    – Louis A
    Commented Sep 21, 2012 at 17:34

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