Timeline for $\operatorname{Spaces}/BG$ $\sim$ $\operatorname{Spaces}^G$ $\sim$ $??(\Omega G)$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 23 at 3:02 | vote | accept | მამუკა ჯიბლაძე | ||
| Mar 22 at 17:18 | answer | added | Dmitri Pavlov | timeline score: 6 | |
| Mar 22 at 17:06 | comment | added | მამუკა ჯიბლაძე | Great! That's my first suggestion which I could not make precise! Can you make an answer out of this? | |
| Mar 22 at 17:05 | comment | added | Dmitri Pavlov | @მამუკაჯიბლაძე: Yes, for example you can take the category of spaces S equipped with a homomorphism of ∞-groups A→Ω(Aut(S)). | |
| Mar 22 at 17:02 | comment | added | მამუკა ჯიბლაძე | @DmitriPavlov Yes I agree. Let $A$ be braided. Is it possible to construct some $\mathscr X(A)$ out of $A$ itself, without passing to $BA$ first? | |
| Mar 22 at 16:02 | comment | added | Dmitri Pavlov | If A is braided (and not merely homotopy commutative), we can simply construct the delooping BA as an ∞-group and then take the category of spaces with an action of BA. Since BΩG≃G, this recovers the category of G-spaces. If A is merely homotopy commutative, I have an impression there is not enough data left to reconstruct G-spaces, since G can be reconstructed from the ∞-category of G-spaces, but G cannot be reconstructed from the ∞-group ΩG, unless we equip ΩG with a braiding. (Indeed, the ∞-group ΩG has exactly the data needed to reconstruct G as a space, not ∞-group.) | |
| Mar 22 at 9:05 | comment | added | მამუკა ჯიბლაძე | @GregoryArone Interesting suggestion. For any space $X$ one might ask about either $E\to X$ giving rise to an internal group in the homotopy category of $\operatorname{Spaces}/X$ (but then one must suitably interpret pullbacks over $X$ - presumably one must go for homotopy pullbacks), or just principal bundles over $X$, wrt arbitrary topological groups? | |
| Mar 22 at 8:37 | comment | added | Gregory Arone | If $A$ is a topological group, you could still ask what is the category of topological groups with an action of $A$ is equivalent to. Perhaps you can show that it is equivalent to the category of group bundles over $BA$. But $B^2\!A$ does not exist, so there is no guess to make about that. I guess you are hoping that if $A$ has enough commutativity for $B^2\!A$ to exist, then you can show an equivalence to the category of space bundles over $B^2\!A$. | |
| Mar 22 at 7:07 | history | asked | მამუკა ჯიბლაძე | CC BY-SA 4.0 |