Timeline for derived functors and triangulated categories
Current License: CC BY-SA 3.0
9 events
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| Nov 5, 2011 at 17:40 | comment | added | Greg Friedman | I don't think anyone else has made this point yet: it's also not right, for example, that it's right exact functors that give us right derived functors. For example, tensor product with a fixed module is a right exact functor for which we usually form its left derived functor. Similarly Hom is left exact and we right derive it. | |
| Nov 3, 2011 at 23:39 | vote | accept | Nicolás | ||
| Nov 3, 2011 at 21:15 | comment | added | Sándor Kovács | 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 $F$. | |
| Nov 3, 2011 at 21:11 | comment | added | Sándor Kovács |
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".
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| Nov 3, 2011 at 20:34 | comment | added | Will Sawin | 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. | |
| Nov 3, 2011 at 20:31 | comment | added | Fernando Muro | 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. | |
| Nov 3, 2011 at 20:07 | comment | added | Fernando Muro | $F=H^0RF$ and $G=H^0LG$. | |
| Nov 3, 2011 at 20:04 | answer | added | Sándor Kovács | timeline score: 9 | |
| Nov 3, 2011 at 19:50 | history | asked | Nicolás | CC BY-SA 3.0 |