Timeline for A Dold-Thom style construction of a cohomology class from a sphere bundle
Current License: CC BY-SA 3.0
10 events
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| Feb 10, 2015 at 19:00 | comment | added | Eric Wofsey | @მამუკაჯიბლაძე: Sorry, you're right, "reduce" was the wrong word. I meant looking at the (not a priori discrete) space of pointed self-homotopy equivalences, which is a union of connected components in the space of all pointed self-maps. In the case of an Eilenberg-MacLane space, these components happen to be contractible, so you get the same thing (up to homotopy) as you would if you just took homotopy classes of equivalences. | |
| Feb 10, 2015 at 14:02 | comment | added | მამუკა ჯიბლაძე | @EricWofsey Sorry, I'm confused again - do you mean homotopy classes of self-homotopy equivalences? Then this would mean going in the "opposite direction" - not reducing the structure group but rather passing to the discrete quotient (of a larger group, but I think it can be also realized as quotient of homeomorphisms). Whereas reducing would mean passing to a subgroup or to a connective cover, no? For example, capturing principal bundles would mean reducing the structure group to the self-homeomorphisms given by the action of (a topological group model of) $K(\mathbb Z,n)$ on itself. | |
| Feb 10, 2015 at 9:01 | comment | added | Eric Wofsey | @მამუკაჯიბლაძე: Non-principal bundles will be classified by the classifying space of the group of self-homeomorphisms of $K(\mathbb{Z},n)$ in general. This might seem fairly unnatural (it depends on the specific $K(\mathbb{Z},n)$ space used); you could reduce the structure group to the group of self-homotopy equivalences to get something more canonical. But that group is just a discrete $\mathbb{Z}/2$, and the associated class in $H^1(-,\mathbb{Z}/2)$ is just the orientation class of your sphere bundle. | |
| Feb 10, 2015 at 8:34 | comment | added | მამუკა ჯიბლაძე | @EricWofsey AHA this seems to be the key! Hopefully I can figure out exactly what missing data makes the difference... And also, non-principal $K(\mathbb Z,n)$ bundles are still classified by something too, right? Does it immediately drop down to just $\mathbb Z/2$ coefficients?? | |
| Feb 8, 2015 at 19:22 | comment | added | Eric Wofsey | You don't (a priori) get something in $H^{n+1}$ because this construction gives a $K(\mathbb{Z},n)$-bundle, not a principal $K(\mathbb{Z},n)$-bundle. | |
| Feb 8, 2015 at 12:20 | comment | added | მამუკა ჯიბლაძე | Now that you say it - yes I agree it looks suspicious, even for $n=0$. But I don't understand where is the problem with that fibrewise construction. Maybe I must just honestly find out what happens with, say, double covering of the circle... | |
| Feb 8, 2015 at 12:17 | comment | added | Qiaochu Yuan | The group is connected if the original space is connected, but $S^n$ with a basepoint adjoined is not. Concerning the second part, this doesn't seem true for $n = 1$. | |
| Feb 8, 2015 at 11:57 | comment | added | მამუკა ჯიბლაძე | And concerning the second part - maybe what I obtain should not be called Euler class, but in any case I do obtain something in $H^{n+1}(\_,\mathbb Z)$, don't I? | |
| Feb 8, 2015 at 11:56 | comment | added | მამუკა ჯიბლაძე | Sorry I am confused by that - the basepoint becomes zero of the resulting topological group anyway, no? So the resulting group must still be connected I think. | |
| Feb 8, 2015 at 10:52 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |