Timeline for The homology of the universal covering space, why so difficult to compute
Current License: CC BY-SA 4.0
29 events
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| Mar 27, 2020 at 21:57 | comment | added | Manuel Rivera | What @RyanBudney is saying is similar to what I proposed below. However, the cool observation from our work is that this process is a purely algebraic one, namely it can be obtained by applying a series of functors to the ( $E_2$ ) coalgebra of singular chains without knowing that it came from a space! Moreover, the construction can be understood as part of an abstract homotopy theory (coalgebras under Koszul weak equivalences). | |
| Mar 26, 2020 at 22:12 | comment | added | GSM | @RyanBudney In some sense I wanted to use the information in the chain complex $C_{\ast}(X)$ and deduce a chain model $C_{\ast}(\tilde{X})$ without allowing a purely topological manipulation as lifting cells. My idea was that the homology of the covering space is encoded in $C_{\ast}(X)$ + extra structure such as "cocommutative" structure of the chain. I know my justification is maybe a little bit artificial :) | |
| Mar 26, 2020 at 22:06 | comment | added | Ryan Budney | The universal cover of a CW-complex is a CW-complex, you obtain the CW-structure by taking all lifts of all the cells of the original space. Once you have a CW-complex, you can form the cellular chain complex, just like how one computes the homology of any CW-complex, as presented in introductory algebraic topology textbooks. What about this does not answer your question? | |
| Mar 26, 2020 at 22:03 | comment | added | GSM | @RyanBudney Dear Ryan, I guess I just did not really understand your comment, It is probably my poor knowledge! I really would appreciate to understand more your comment! Could you please write it down your spectral sequence ? | |
| Mar 26, 2020 at 21:48 | comment | added | Ryan Budney | @GSM: Could you explain what you do not like about my "collapsed spectral sequence" argument, i.e. just lifting the CW complex to the universal cover, and "computing" with that? | |
| Mar 26, 2020 at 17:40 | history | edited | GSM | CC BY-SA 4.0 |
added 272 characters in body
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| Mar 26, 2020 at 9:14 | comment | added | HJRW | @GSM The problem is as follows: "Is there an algorithm that takes as input a finite 2-complex $X$, guaranteed to be aspherical, and decides whether or not $\pi_1X$ is trivial?". The problem is related to the Andrews--Curtis conjecture, and thence to the smooth 4-dimensional Poincare conjecture. | |
| S Mar 26, 2020 at 9:02 | history | bounty ended | GSM | ||
| S Mar 26, 2020 at 9:02 | history | notice removed | GSM | ||
| Mar 26, 2020 at 9:01 | vote | accept | GSM | ||
| Mar 26, 2020 at 3:08 | answer | added | Manuel Rivera | timeline score: 6 | |
| Mar 25, 2020 at 22:55 | comment | added | GSM | @HJRW I am curious about your statement " the triviality problem for aspherical presentations is a famous open problem" could you give more details about this open problem, please ? thank you. | |
| Mar 25, 2020 at 22:47 | comment | added | HJRW | @RyanBudney -- that's a great way of putting it. Of course, by "knowing" the fundamental group, people often mean they know a presentation for it. But the undecidability of the triviality problem shows that, even if we know a presentation, we don't know the universal cover, in the sense that we don't know if it's proper or not. Of course, the triviality problem for aspherical presentations is a famous open problem, so it's hard to answer the question this way, given the stipulation that $\pi_2$ is also known. | |
| S Mar 25, 2020 at 15:08 | history | suggested | sound wave | CC BY-SA 4.0 |
"now" instead of "know", however I removed it since "know" already appears before the enumerate list, moreover in this way there is symmetry wrt to first item of the list
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| Mar 25, 2020 at 14:36 | review | Suggested edits | |||
| S Mar 25, 2020 at 15:08 | |||||
| Mar 21, 2020 at 16:05 | comment | added | GSM | @RyanBudney I would like to illustrate the words "know" and "compute" by an example. Lets say we have the inverse problem, we have a group G acting very nicely on a connected space $X$. Suppose I "know" every group homology of $X$ and the induced action of $G$ on the homology of $X$, then theoretically (spectral sequence) allows us to "compute" the homology of the orbit space $X/G$. | |
| Mar 20, 2020 at 20:45 | comment | added | Ryan Budney | Your question is somewhat problematic. What do you mean by "computing"? What does it mean to "know" $\pi_1(X)$? If you "know" the fundamental group, presumably you "know" the cellular chain complex for the universal cover -- that is a collapsed spectral sequence that computes the homology, i.e. the best kind. So what is your objection to that? | |
| Mar 20, 2020 at 14:59 | answer | added | Phil Tosteson | timeline score: 5 | |
| Mar 20, 2020 at 14:45 | answer | added | user153879 | timeline score: 7 | |
| S Mar 20, 2020 at 12:19 | history | bounty started | GSM | ||
| S Mar 20, 2020 at 12:19 | history | notice added | GSM | Canonical answer required | |
| Mar 18, 2020 at 10:22 | comment | added | HJRW | Absent knowlede of $\pi_2(X)$, the problem of computing $H_2(\widetilde{X})\cong\pi_2(X)$ is known to be undecidable. Perhaps there's a similar undecidibility result for higher-dimensional homology groups, even with knowledge of the higher homotopy groups? | |
| Mar 18, 2020 at 9:28 | history | edited | GSM | CC BY-SA 4.0 |
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| Mar 17, 2020 at 21:01 | comment | added | Connor Malin | I suppose it doesn’t actually do anything besides tell you you can compute the homology of the universal cover via local coefficients | |
| Mar 17, 2020 at 20:21 | answer | added | user43326 | timeline score: 5 | |
| Mar 17, 2020 at 19:19 | history | edited | GSM | CC BY-SA 4.0 |
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| Mar 17, 2020 at 19:11 | comment | added | GSM | @ConnorMalin I'm not sure this spectral sequence would be helpful, unless i'm wrong... | |
| Mar 17, 2020 at 19:04 | comment | added | Connor Malin | There is a spectral sequence (the Serre spectral sequence) which involves homology of the base space with local coefficients, but as far as I know the easiest way to compute homology with local coefficients is via homology of its covers. | |
| Mar 17, 2020 at 18:52 | history | asked | GSM | CC BY-SA 4.0 |