Suppose that $D$ and $E$ are compactly generated triangulated categories, even algebraic (i.e. equivalent to derived categories of small dg categories) if we want, and asume that their subcategories $D^c$ and $E^c$ of compact objects are triangle equivalent. Are $D$ and $E$ triangle equivalent?. By Theorem 9.2 of Keller's 'Deriving dg categories', we know that the answer is 'yes' when either $D$ or $E$ is the derived category of (the category of modules over) a small $k$-linear category, but I do not know the general answer. Any help would be highly appreciated.
$\begingroup$
$\endgroup$
4
-
1$\begingroup$ Welcome to MathOverflow, Manuel. The site supports basic LaTeX formatting ("MathJax", technically). I've added formatting to your question to make it a little easier to read: I hope you don't mind. $\endgroup$– Jeremy RickardCommented Jun 23, 2017 at 12:38
-
$\begingroup$ Dear Jeremy, Thank you very mucy for adding TeX formatting to my question. $\endgroup$– Manuel SaorínCommented Jun 23, 2017 at 18:26
-
1$\begingroup$ I fear that you cannot really write down a functor $D \to E$ because homotopy colimits are not functorial on the level of triangulated category. If you have instead an equivalence between stable infinity categories of compact object, you obtain an equivalence between $Ind(D^c)$ and $Ind(E^c)$, which agree with $D$ and $E$ when the latter are compactly generated stable infinity categories. $\endgroup$– Lennart MeierCommented Jun 26, 2017 at 8:04
-
$\begingroup$ Thank you Lennart. Your comment is very helpful for me. I had the impression that going to the context of infinity categories or that of (Grothendieck) derivators, the answer might be 'yes'. But still, without adding further structure, I would like to find examples beyond the derived categories of rings (with several objects), for which the answer to my question is positive. $\endgroup$– Manuel SaorínCommented Jun 27, 2017 at 9:59
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
Consider the case where $D$ is the ordinary stable homotopy category of spectra, and $E$ is assumed to have all coproducts (which is clearly a necessary condition), and we also assume that $D^c\simeq E^c$ as tensor triangulated categories. It is an old conjecture of Margolis that this forces $D$ to be equivalent to $E$, and I believe that this is still open in full generality. In both $D$ and $E$ one can define a square-zero ideal of phantom maps. Hovey, Palmieri and I proved that $D/\text{Ph}_D\simeq E/\text{Ph}_E$ and that $\text{Ph}_D\simeq\text{Ph}_E$ as a module over this quotient. Schwede and Shipley proved in https://arxiv.org/abs/math/0108143 that $D\simeq E$ provided that $E\simeq\text{Ho}(E_0)$ for some Quillen model category $E_0$.
answered Jun 28, 2017 at 9:09
-
$\begingroup$ Thanks Neil. Although my interest is more on algebraic triangulated categories, your information should be helpful. Hopefully, some of the proofs on the two references that you mention admit some adaptation. $\endgroup$ Commented Jul 3, 2017 at 8:17