Timeline for Is the ind-completion of a triangulated category triangulated?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 24 at 17:21 | comment | added | Drew Heard | There is a short argument in Remark 5.9 of math.uni-bielefeld.de/~hkrause/completion.pdf, which I believe shows that the ind-completion (in the triangulated sense) of the stable module category is not triangulated if the Sylow p-subgroup of $G$ is not cyclic. | |
| Sep 20 at 11:37 | comment | added | Dave Benson | @UlrichPennig Yes it is; no it doesn't. | |
| Sep 20 at 11:31 | comment | added | Ulrich Pennig | Now I am confused: Isn't the homotopy category of a stable $\infty$-category triangulated? I guess, the Ind-completion does not commute with taking the homotopy category? | |
| Sep 20 at 10:31 | comment | added | Dave Benson | Let me reinforce Neil's answer here with another example. If you ind complete the small stable module category of a finite group as a triangulated category, you get the big stable module category modulo phantoms, which is (probably) not triangulated, and not really what you want. However, if you regard it as a stable infinity category and ind complete, you get the big stable module category that you wanted, as a stable infinity category. An intermediate solution is to use differential graded categories, and this works well, but comes with its own problems. | |
| Sep 20 at 10:15 | comment | added | Elías Guisado Villalgordo | I think I don't understand anything from this answer xD. I don't know what spectra are (my topology knowledge is not great), and I don't know anything about $\infty$-categories either. | |
| Sep 20 at 10:02 | history | answered | Neil Strickland | CC BY-SA 4.0 |