Timeline for What is the "quaternionic" super Brauer group?
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9 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Jan 10, 2016 at 15:08 | comment | added | André Henriques | The class in $KO^4(S^4)$ comes from the relative group $KO^4(S^4,pt)$. The Lusternik-Schnirelman category of $S^4$ is two, and so every product of two elements in $KO^*(S^4,pt)$ is zero. In general, if the Lusternik-Schnirelman category of some space $X$ is $n$, and $h$ is a multiplicative cohomology theory, then the product of any $n$ elements in $h^*(X,pt)$ is necessarily zero. See en.wikipedia.org/wiki/Lusternik-Schnirelmann_category for a definition. | |
| Jan 10, 2016 at 4:00 | comment | added | Theo Johnson-Freyd | @AndréHenriques Well, I agree the interesting theory is "$KO [x] / (x^2 - 1)$ for $x$ of degree 4 mod 8", and not "$KO[x]/x^2$", and I'm happy to believe that the latter is the one I suggested, although I don't know enough K-theory to eyeball why. | |
| Jan 9, 2016 at 17:59 | comment | added | André Henriques | Theo: the construction you're suggesting is indeed an $E_\infty$ structure on $KO+KSp$, but it is one for which the map $KSp \wedge KSp\to KO$ is zero. We want one such that the map $KSp \wedge KSp\to KO$ induces an equivalence $KSp \wedge_{KO} KSp\to KO$. | |
| Jan 8, 2016 at 19:07 | comment | added | Theo Johnson-Freyd | Isn't $KO + KSp$ something like $[S^4,KO]$ (not base-point preserving)? That would be $E_\infty$ because $KO$ is. | |
| Jan 8, 2016 at 15:46 | comment | added | Qiaochu Yuan | @André: well, all I can show on my own is that $KO + KSp$-cohomology has products, coming from the product in usual $KO$-cohomology (and using the fact that $KSp$ is $KO$ shifted by $4$). I don't know how to upgrade this to an $E_{\infty}$ structure either. | |
| Jan 8, 2016 at 11:30 | comment | added | André Henriques | Qiaochu: are you saying that $KO + KSp$ is $E_\infty$? I certainly don't know how to prove that that is the case. | |
| Jan 8, 2016 at 2:19 | comment | added | Qiaochu Yuan | The obvious candidate for the "K-theory" you want is $KO + KSp$, which indeed has products and is $4$-periodic. | |
| Jan 7, 2016 at 23:08 | history | answered | Theo Johnson-Freyd | CC BY-SA 3.0 |