Timeline for Which homotopy 2-types are H-spaces?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 23, 2022 at 11:24 | vote | accept | Blazej | ||
| Jan 9, 2022 at 2:50 | comment | added | Fernando Muro | Chris, but not always with a symmetric braiding. | |
| Jan 9, 2022 at 2:37 | answer | added | Fernando Muro | timeline score: 5 | |
| Jan 6, 2022 at 22:21 | comment | added | Chris Schommer-Pries | @QiaochuYuan I think that the map to $H^3(B \pi_1, \pi_2)$ is always zero. The $H^4(B^2\pi_1, \pi_2)$ classes correspond to certain braided monoidal categories, and I am pretty sure these can be realized with trivial associators - so the map should be zero. | |
| Jan 6, 2022 at 19:30 | answer | added | Tyler Lawson | timeline score: 13 | |
| Aug 9, 2021 at 16:19 | comment | added | Tim Campion | @Qiaochu Yuan If $X$ has a delooping $BX$, then $X$ admits a grouplike $A_\infty$ structure, which is much stronger than being an $H$-space, right? | |
| Oct 7, 2020 at 19:37 | comment | added | Qiaochu Yuan | A slightly stronger and more natural condition (I don't know if it's equivalent to being an $H$-group in this case) is that $X$ has a delooping $BX$; this means not only that $\pi_1$ is abelian and acts trivially on $\pi_2$ but that the Postnikov invariant must arise from a Postnikov invariant in $H^4(B^2 \pi_1, \pi_2)$. This is exactly the space of quadratic maps $\pi_1 \to \pi_2$, although I don't know how to describe the map to $H^3(B \pi_1, \pi_2)$ in these terms. | |
| Oct 7, 2020 at 13:19 | review | First posts | |||
| Oct 7, 2020 at 13:23 | |||||
| Oct 7, 2020 at 13:09 | history | asked | Blazej | CC BY-SA 4.0 |