Timeline for Why not $\mathit{KSO}$, $\mathit{KSpin}$, etc.?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 23, 2017 at 4:49 | comment | added | Arun Debray | This is really nifty. Thanks for writing it down! Time to play with some examples. | |
| Apr 23, 2017 at 4:48 | vote | accept | Arun Debray | ||
| Apr 18, 2017 at 14:40 | history | edited | Denis Nardin | CC BY-SA 3.0 |
Improved the condition for $K\theta$ to be a cohomology theory
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| Apr 18, 2017 at 14:35 | comment | added | Denis Nardin | @MarcHoyois Hmm.. That is a good point. We can't expect to get algebraic K-theory spectra that easily. That's probably the best we can get in general. | |
| Apr 18, 2017 at 14:17 | comment | added | Marc Hoyois | @DenisNardin I think that to express the group completion $X^+$ using filtered colimits you need to know that for every $x\in X$, there exists $m\geq 2$ such that the cyclic permutation of $x^m$ is homotopic to the identity (at least, I only know how to prove it in this case). These permutations live in $\pi_1$'s of $X$. In the $U_n$ and $Sp_n$ cases these are trivial, in the $O_n$ case we have $Z/2$ so any odd $m$ works. But this fails completely in other cases, eg when $X=\coprod_n BGL_n(R)$ for some nonzero commutative ring $R$. | |
| Apr 18, 2017 at 11:35 | comment | added | Denis Nardin | @JamesCameron It follows from the fact that in the ∞-category of spaces finite limits commute with filtered colimits, and that in those cases the group completion can be given by a filtered colimit (since $\mathrm{Map}(B,-)$ is a finite limit). Now that I write it, I have the feeling that it should work for pretty much any $E_∞$-space $X$, let me see if I can straighten out the details.... | |
| Apr 18, 2017 at 5:16 | comment | added | J Cameron | This is nice! How do you see that you can pull the group completion inside when X is BO, BU and BSp? | |
| Apr 18, 2017 at 1:17 | history | answered | Denis Nardin | CC BY-SA 3.0 |