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If you derive a right exact functor $F$ you get a functor normally denoted by $RF$ on the derived category. Similarly, if you start with a left exact functor $G$ you get a functor normally denoted by $LG$. These are simply two triangulated functors on a triangulated categories. Suppose they are both defined in the same category (for example, the derived category of bounded complexes in some abelian category). Is there any way to distinguish the right-ness or left-ness of $RF$ and $LG$ from inside the triangulated category? If there isn't, then the $R$ and $L$ don't seem to be a very good choice of notation...

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  • $\begingroup$ $F=H^0RF$ and $G=H^0LG$. $\endgroup$
    – Fernando Muro
    Commented Nov 3, 2011 at 20:07
  • $\begingroup$ 1) It does, just evaluate $H^0$ of the functor on the heart and see what happens. 2) Derived functors of left and right exact functors live in derived categories of abelian categories. 3) Yes. $\endgroup$
    – Fernando Muro
    Commented Nov 3, 2011 at 20:31
  • $\begingroup$ I think the answer is no, but I don't know much about derived categories, so I won't speak in that language. The reason for this is that, in a homology/cohomology theory with a fixed bound on complexes, the right derived functors of the first functor are the same as the left derived functors of the last functor, because they're both the original theory. I have heard that there is sometimes disagreement over whether a series of functors / single derived functor is a homology or cohomology functor, for instance with Khovanov homology. $\endgroup$
    – Will Sawin
    Commented Nov 3, 2011 at 20:34
  • $\begingroup$ Fernando, 1) I'm sorry, but I don't see how $F=H^0\Phi$ says anything about $\Phi$ being right in some sense? Take $F=H^0(X, )$ and $\Phi=L(H^n(X, ))[-n]$ for $X$ a smooth projective variety of dimension $n$. You probably had more in mind, but perhaps you should share it. :) 2) Yes, but I still think the OP is talking about triangulated categories and not derived categories. 3) Cheers! 4) Unfortunately, I accidentally deleted my first comment and I don't think I can reproduce it, especially not at that "slot". $\endgroup$
    – Sándor Kovács
    Commented Nov 3, 2011 at 21:11
  • $\begingroup$ Will, this argument would work if the question was whether you can decide that a functor on the derived category is a right or a left derived functor, but knowing the functor $F$ and the functor $RF$ tells you that it is a right derived functor. As you say under some finiteness conditions $RF$ is also a left derived functor, that is, $RF=L(R^nF)$, but in general $R^nF$ is different from $F$. It is right exact for starters, so if $F$ is not exact then they can't be the same. If $F$ is exact, then obviously its left and right derived functors are the same, they are both equal to $F$. $\endgroup$
    – Sándor Kovács
    Commented Nov 3, 2011 at 21:15

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I am not sure what your goal is with this question, but I think there is an inherent problem with your set up. $RF$ stands for a functor derived from $F$ and not just a functor on the derived category. If you consider the derived category only as a triangulated category via a forgetful functor then you are also forgetting $F$, so there is no reason to still denote your functor by $RF$. On the other hand if you remember the derived category structure or at least you still deal with complexes and so you can take cohomology of these complexes, then $RF$ induces $R^iF$ and you have the condition that $R^iF=0$ for $i<0$ and $R^0F=F$. In other words, the functor $RF$ is such that the leftmost non-zero cohomology of it is $F$ and all the interesting stuff is on the right. Homework: do the same for $LG$.

It is important that $RF$ is the right derived functor of $F$. It can happen that $RF=LG$, but then $F$ and $G$ are different. Typically $G=R^nF$ such that $R^mF=0$ for $m>n$.

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  • $\begingroup$ Dear Sandor, thanks for your answer. My question had to do with the fact that if one stays in the derived/triangulated world there is no difference between left and right exact functors: they are both triangulated. For example, a left exact functor will give you a long exact sequence that goes to the left in cohomology. A right exact functor will give you a long exact sequence going to the right in cohomology. But in the derived category they both 'just' preserve triangles. There is also a spectral sequence that relates the cohomology of any two derived functors to the cohomology of their... $\endgroup$
    – Nicolás
    Commented Nov 3, 2011 at 23:37
  • $\begingroup$ ... composition, irrespective if the functors appearing are left exact or right exact or one and one: you just use that they are triangulated. From this perspective using the letters $R$ and $L$ seem to only serve the purpose of remembering that the negative cohomologies are zero (in the case of $R$) or its positive cohomologies are zero (in the case of $L$). Rereading what I wrote, I'm not sure anyone will understand anything... $\endgroup$
    – Nicolás
    Commented Nov 3, 2011 at 23:39
  • $\begingroup$ Dear unknown, I think I understand what you are saying, but the point I am trying to make is that $RF$ is relative to $F$. It is not saying that an abstract functor is a right derived functor but that this particular one is the right derived functor of $F$ and that does make sense. You are right that this distinction would not make sense for an arbitrary functor on a triangulated category. $\endgroup$
    – Sándor Kovács
    Commented Nov 4, 2011 at 0:07

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