4
$\begingroup$

One of the annoying things about derived categories is that they come with a host of different finiteness conditions, which are all subtlely different, and for each situation you want a particular one.

For a project I'm doing right now, I think I want to use the derived category of modules over a particular finite-dimensional algebra composed of bounded above complexes (so I have projective resolutions) with finite-dimensional cohomology modules, which are only non-zero in finitely many degrees.

Is there a name for this category I should know about?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ One usually (well always) consider the bounded derived category as being a strictly full subcategory of the full derived category and hence always closed under resolutions. I edited my answer to reflect this. $\endgroup$
    – Greg Stevenson
    Commented Nov 22, 2009 at 22:14
  • 1
    $\begingroup$ Or you could just name it! Call them Webster Derived Categories! $\endgroup$
    – Harry Gindi
    Commented Nov 22, 2009 at 23:02
  • $\begingroup$ Are these the compact objects in the derived category of modules? $\endgroup$
    – Reid Barton
    Commented Nov 23, 2009 at 3:45
  • $\begingroup$ No they aren't - Ben said in his comment to my answer below that the algebra in question does not have finite global dimension. In particular, the compact objects will be the perfect complexes not the whole of the bounded derived category of finitely presented guys. $\endgroup$
    – Greg Stevenson
    Commented Nov 23, 2009 at 3:57
  • $\begingroup$ I called them compactly supported objects in arxiv.org/abs/0804.3986 (the paper needs rewriting, and I never got around to submitting it, so make of it what you will). $\endgroup$
    – Aaron Bergman
    Commented Nov 23, 2009 at 11:33

3 Answers 3

Reset to default
4
$\begingroup$

Hopefully there is not a name for this category nor will you invent one, either. The usual «put the conditions on the homology of complexes as a subindex to the $D$ and say "derived category of complexes with finite dimensional cohomology"» approach does wonders in introducing no new notation and no new names. We have too many of those already.

If in your paper you will deal exclusively with that category, make a local convention and say something to the point that «in this paper the phrase "derived category" will mean "derived category of complexes with finite dimensional cohomology", and the notation $D(A)$ will stand for that category» somewhere at the end of the introduction (do use the explicit long name in the abstract and in the introduction itself, though)

Improve this answer
$\endgroup$
3
  • $\begingroup$ I like this answer actually (I removed the hopefully from mine but it is here now so hopefully the italics will still work!) Although there are cases where these sorts of guys do have names which are better than just a string of descriptors. $\endgroup$
    – Greg Stevenson
    Commented Nov 23, 2009 at 4:00
  • $\begingroup$ Hmm, I actually found this answer to be written with a rather peculiar concept of "name." "derived category of complexes with finite dimensional cohomology" is a name, and exactly the sort of name I was looking for (though I fear it would not get across that I only want cohomology in finitely many degrees). I certainly didn't have anything like "Webster derived categories" in mind. How gauche! $\endgroup$
    – Ben Webster ♦
    Commented Nov 23, 2009 at 21:31
  • $\begingroup$ To avoid the perils of not conveying exactly what you want, you can be two syllables more verbose and say "derived category of complexes with finite dimensional total cohomology". (By avoiding a name I had in mind the "Webster derived category" example and the like :P ) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Nov 23, 2009 at 21:53
3
$\begingroup$

I'm not sure if this category has a particular name - usually until someone cares enough to give one of these a name or nice notation they just have long unwieldy names. I can suggest some notation though - if your algebra is $A$ then I think $$D^{-,b}_{\mathrm{fd}}(A)$$ is pretty much standard - the fd for finite dimensional maybe not so much but conditions on the cohomology traditionally go there. If want the components finite dimensional as well then you could drop the fd and use $A$-$\mathrm{mod}$ instead of $A$. You probably already know all of this though...

Post Coffee Edit: Actually the above is slightly silly - it is perfectly resonable to consider the homotopy category of bounded above complexes (usually of projectives) with bounded finite dimensional cohomology but in the derived case you may as well just look at $D_{\mathrm{fd}}(A)$. Unless you have a particularly good reason for wanting bounded above complexes you may as well take all isomorphs in $D(A)$ - this is correct philosophically. You can then drop the bounded (since finite dimensional total cohomology implies boundedness) and just call it the derived category of $A$-modules with finite dimensional total cohomology.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Well, mainly I feel deep uncomfortable talking about derived functors in a category where it's not clear I have acyclic resolutions. This is an algebra which does not have finite global dimension, so in $D_{fd}(A)$ objects don't always have projective resolutions in that category, but they do in bounded above. I think the functors I use actually do have acyclic resolutions in the bounded category, but this is not immediately apparent. $\endgroup$
    – Ben Webster ♦
    Commented Nov 22, 2009 at 23:24
  • $\begingroup$ By $D_{fd}(A)$ I mean the unbounded derived category of complexes with finite dimensional total cohomology. So it has resolutions, all that removing the superscript here really does is allow one to consider unbounded complexes - one can always truncate above to produce a quasi-isomorphic complex which is bounded above. I don't think this adds anything undesirable (it is not early anymore but my brain isn't really functioning). Although even in the bounded case where you insist all complexes are literally bounded you have an equivalence with $K^{-,b}(A\text{-}proj)$ so you can derive away. $\endgroup$
    – Greg Stevenson
    Commented Nov 22, 2009 at 23:37
2
$\begingroup$

Following an analogy of Drinfeld's, this is like complaining that there are lots of different function spaces (all the Sobolev spaces, L^p spaces etc) with different growth conditions.. these kind of restrictions are the algebraic analog of functional analysis..

In algebraic geometry what you describe is basically the bounded coherent category (bounded cohomologies, each of which is a finitely presented module)..?

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .