Timeline for Spectral sequence for H-space bundles
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 28, 2012 at 17:49 | history | edited | David White | CC BY-SA 3.0 |
Typos + added a tag
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| Feb 10, 2012 at 20:15 | answer | added | Mark Grant | timeline score: 4 | |
| Dec 2, 2011 at 19:45 | comment | added | Vitali Kapovitch | @Johannes sorry, my answer was quite wrong and I deleted it. | |
| Dec 2, 2011 at 12:40 | comment | added | Johannes Ebert | Just having a fibre that admits an H-space structure does not yield anything. Maybe you assume that there is a multiplication $E \times_B E \to E$ which is fibre-preserving (and hence turns every fibre into an H-space)? | |
| Dec 2, 2011 at 11:21 | comment | added | fred137 | I am sorry for asking a question that is not very precise. But I am still interested if there is a theory with some extra conditions imposed on the compatibility of the H-space structure and bundle structure. | |
| Dec 2, 2011 at 11:01 | comment | added | Oscar Randal-Williams | @Mark: Actually, that's a bad example, as every $S^1$-bundle admits the structure of a principal $U(1)$-bundle. But the point stands with e.g. $S^3 = SU(2)$. | |
| Dec 2, 2011 at 8:48 | comment | added | Mark Grant | In order for the question to admit a positive answer, I think you would need the $H$-space structure on $F$ to be compatible with the fibre bundle structure in some way. Otherwise, you could take, say, any bundle with fibre $S^1$ and ask whether the complex multiplication can help you compute (co)homology of the total space, which seems absurd. | |
| Dec 2, 2011 at 8:31 | history | asked | fred137 | CC BY-SA 3.0 |