Timeline for Can one bypass the geometric realization in the definition of algebraic $K$-theory?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 5, 2023 at 8:52 | comment | added | Stabilo | The reference of Grayson is helpful, thanks @DanRamras ! | |
| Aug 5, 2023 at 8:46 | vote | accept | Stabilo | ||
| Aug 5, 2023 at 2:24 | comment | added | Dan Ramras | One answer to the question in the title is provided by work of Grayson (ams.org/journals/jams/2012-25-04/S0894-0347-2012-00743-7/…) where he gives presentations of the K-theory groups in terms of combinatorial data about $\mathcal{E}$. See also more recent work of Kasprowski and Winges (arXiv:1705.09116, arXiv:1805.06175) | |
| Aug 5, 2023 at 2:22 | comment | added | Dan Ramras | To be clear, Quillen's definition is that the K-theory groups of $\mathcal{E}$ are the homotopy groups of $NQ(\mathcal{E})$, not $N(\mathcal{E})$ (so this is a typo in the question). | |
| Aug 4, 2023 at 16:25 | answer | added | Dmitri Pavlov | timeline score: 6 | |
| Aug 4, 2023 at 16:14 | comment | added | Connor Malin | It can't be the case that K-theory factors as $\pi_* \circ \mathrm{N} \circ F$ where $F:\mathrm{ExactCat} \rightarrow \mathrm{Grpd}$ because the higher homotopy groups of the nerve of a groupoid are all trivial, and in general higher $K$-theory is nontrivial. | |
| Aug 4, 2023 at 15:03 | comment | added | Stabilo | @ConnorMalin For sure, it is. Yet, I had the impression that the construction of $Q$ was some kind of localization at admissible epi (not exactly though). I was expecting that localizing further would lead to a groupoid, then to a fibrant simplicial set, with no need of any fibrant replacement. | |
| Aug 4, 2023 at 14:23 | comment | added | Connor Malin | In what way is $\mathcal{Q}(\mathcal{E})$ intermediate? That is the central construction in this method of defining $K$-theory. | |
| Aug 4, 2023 at 13:48 | comment | added | Stabilo | @PhilTosteson Thanks! Hmm... a very naive interrogation: if one is going to take a fibrant replacement after all, why bother with the intermediate step $Q(\mathcal{E})$? | |
| Aug 4, 2023 at 13:30 | comment | added | Phil Tosteson | You don't need topological spaces to construct a fibrant replacement. For instance you can use Kan's Ex infinity functor: ncatlab.org/nlab/show/Kan+fibrant+replacement | |
| Aug 4, 2023 at 13:23 | history | asked | Stabilo | CC BY-SA 4.0 |