Timeline for Explicit construction of the quotient of a category by a group action
Current License: CC BY-SA 3.0
8 events
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| Jan 25, 2012 at 18:12 | history | edited | Kevin Walker | CC BY-SA 3.0 |
added 347 characters in body
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| Jan 25, 2012 at 18:06 | comment | added | Kevin Walker | Yes, that's what I was about to comment but you beat me to it. A simpler (than you're Z/3 counterexample) counterexample (to my claim that the constructions are the same) is $G$ acting on the trivial category. | |
| Jan 25, 2012 at 17:27 | comment | added | Chris Schommer-Pries | It looks to me like you are describing the homotopy quotient, which is something like $C \times_G EG$ where $EG$ is the pair groupoid on $G$. In contrast Roman's quotient is the strict quotient, which can be quite destructive. | |
| Jan 25, 2012 at 15:47 | comment | added | Kevin Walker | I think you're right -- there's something wrong with my claim. I'll think about it a little bit more before editing the answer. | |
| Jan 25, 2012 at 15:07 | comment | added | Chris Schommer-Pries | Are you sure this is right? Consider the example where C is the group $\mathbb{Z}/3$ thought of as a one object category with $G = \mathbb{Z}/2$ acting on C by fixing the unique object and as the automorphisms of $\mathbb{Z}/3$. Then the presentation you give looks to me like it gives the dihedral group of order 6 thought of as a one object category, while Roman's quotient gives the terminal singleton category. (I'm assuming you also meant to add the relation $m_{1,p} = id_p$ otherwise your construction is even bigger). | |
| Jan 25, 2012 at 10:53 | vote | accept | Roman Bruckner | ||
| Jan 25, 2012 at 13:59 | |||||
| Jan 25, 2012 at 0:20 | comment | added | Ryan Reich | I was about to advocate this myself, so instead you get my vote. In the example of a sheaf of groups acting non-freely on a sheaf of sets, the book (for example) "Champs Algebriques" by Laumon and Moret-Bailly describes the construction of the quotient (as a stack) of an algebraic space by a sheaf of groups in this way. | |
| Jan 25, 2012 at 0:01 | history | answered | Kevin Walker | CC BY-SA 3.0 |