Timeline for The homology of the universal covering space, why so difficult to compute
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 20, 2020 at 16:21 | comment | added | Phil Tosteson | @BenWieland I don't understand, representations of nilpotent groups are nilpotent, right? I think Tim is correct that his restriction is weaker. | |
| Mar 20, 2020 at 15:47 | comment | added | Phil Tosteson | @TimCampion I think that may be right. The precise condition I know is that $\pi_1 X$ acts unipotently on $H_*(\tilde X)$, but it seems reasonable that these may be equivalent. | |
| Mar 20, 2020 at 15:33 | comment | added | Ben Wieland | The fundamental group being nilpotent is a pretty minor restriction, a good special case. But that is not sufficient. This requires that the space $X$ be nilpotent, ie, that the action of the fundamental group on the homotopy groups be nilpotent, which is a severe restriction, particularly for this question. | |
| Mar 18, 2020 at 16:03 | comment | added | Tim Campion | @GSM At any rate, for a question of the form "Why is problem X hard?", an answer of the form "Problem X fits into a broad class of problems Y, all of which are hard." is a perfectly acceptable answer. It's an even better answer if it comes along with "the standard simplifying assumption on problems of type Y is Z, and if you assume Z, then X is not hard." If you want more information, you need to ask a more precise or perhaps a more general question. | |
| Mar 18, 2020 at 15:56 | comment | added | Tim Campion | @GSM So now two important additional bits of context have been raised which were absent from the initial question: (1) There is a solution to the problem if one puts restrictions on $\pi_1(X)$ and (2) These sort of restrictions seem to be necessary in a broad range of homotopical questions. In this light, I think your question would benefit from some editing and re-focusing. For instance, (and maybe this is too radical a change) it might be more to-the-point in view of (1) and (2) to ask something like "When should I expect a homotopical problem to require nilpotence-like restrictions?" | |
| Mar 18, 2020 at 15:10 | comment | added | user43326 | @GSM maybe you are right but this is already posted as answer and there are comments on it, so I would be reluctant to delete the answer. | |
| Mar 18, 2020 at 10:16 | comment | added | GSM | @user43326 imho you should post your answer as a comment. Please do understand my comment as a friendly one :-) | |
| Mar 18, 2020 at 10:08 | comment | added | GSM | @TimCampion you wrote: "So perhaps it's unreasonable to demand such a method in this case" I think it is exactly the motivation of the title of my question: "Why it is so difficult to compute" | |
| Mar 18, 2020 at 5:16 | comment | added | Tim Campion | Maybe this points at the issue though. There are precious few methods in algebraic topology which can deal with $\pi_1$ issues in a general, uniform way without assumptions like nilpotence. So perhaps it's unreasonable to demand such a method in this case. @PhilTosteson is that "$\pi_1(X)$ is nilpotent" or rather "$\pi_1(X)$ acts nilpotently on the higher homotopy"? | |
| Mar 17, 2020 at 21:55 | comment | added | GSM | As Phil Tosteson noticed, this spectral sequence is useful only in a very particular case. My question is about a general fundamental group... In general it is not useful as far as I understand, it was a motivation for my initial question. | |
| Mar 17, 2020 at 20:41 | comment | added | Phil Tosteson | This will only converge if the fundamental group is nilpotent. | |
| Mar 17, 2020 at 20:21 | history | answered | user43326 | CC BY-SA 4.0 |