Timeline for Is the map $\mathrm H^4(S_{24}) \to \mathrm H^4(M_{24})$ surjective?
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 9, 2018 at 17:03 | comment | added | Theo Johnson-Freyd | Results of Dutour Sikiric – Ellis and Gaberdiel – Persson – Ronellenfitsch – Volpato (and, self promotion, Johnson-Freyd – Treumann) in fact imply that the image of $H^4(S_{24}) \to H^4(M_{24})$ is precisely the $\mathbb Z/6$ subgroup. More sharply, $M_{24}$ lifts to the “spin alternating” group $2A_{24}$ (pullback of $S_{24} \to O(24) \leftarrow Spin(24)$), and for $n$ large $H^4(2A_n) = \mathbb Z/24$ generated by the “$p_1/2$”. Ibidim imply that $p_1/2$ generates $H^4(M_{24})$. | |
| Nov 30, 2016 at 1:35 | vote | accept | Theo Johnson-Freyd | ||
| Sep 26, 2016 at 22:22 | history | answered | Theo Johnson-Freyd | CC BY-SA 3.0 |