Timeline for Relation between $\mathbb{Z}\pi_1 (X)$-module $\pi_n (X)$ and $\mathbb{Z}$-module $H_n (X)$ or $\mathbb{Z}\pi_1 (X)$-module $H_n (\tilde{X})$
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| May 18, 2023 at 15:03 | vote | accept | M.Ramana | ||
| May 10, 2023 at 15:58 | history | edited | M.Ramana | CC BY-SA 4.0 |
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| May 10, 2023 at 15:39 | comment | added | M.Ramana | @FernandoMuro Sorry, you are right. I'll try to improve my question better. | |
| May 10, 2023 at 15:30 | comment | added | Fernando Muro | @M.Ramana yes, you are, for obvious reasons. BTW your edited question does not improve too much because most examples you’ve been given are simply connected. | |
| May 10, 2023 at 15:09 | history | edited | M.Ramana | CC BY-SA 4.0 |
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| May 10, 2023 at 14:00 | history | edited | M.Ramana | CC BY-SA 4.0 |
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| May 10, 2023 at 13:57 | comment | added | M.Ramana | Dear @FernandoMuro What about homotopy groups $\pi_i (X)$ for $(2\leq i\leq n)$ when $X=S^n$. I think they are always free $\mathbb{Z}\pi_1 (X)$-module. Am I correct? | |
| May 10, 2023 at 8:45 | comment | added | Fernando Muro | @IgorKhavkine homotopy types can be captured by algebraic data, that's correct, in several ways actually, but homotopy groups don't suffice in any single way, not even when regarded as modules over the fundamental group. Postnikov systems are more complex. | |
| May 10, 2023 at 8:42 | comment | added | Igor Khavkine | @FernandoMuro You're probably right that I gave an inaccurate summary of what a Postnikov system is supposed to be. I wouldn't be able to tell you the precise definition either. ^^) But in my understanding, in the references that I linked to, Postnikov did claim to have captured the complete homotopy type with algebraic data. If I'm wrong about that too, then I would happily be corrected. | |
| May 10, 2023 at 4:20 | comment | added | Fernando Muro | @IgorKhavkine that’s not the Pistnikov system. Also, there’s no way to recover a homotopy type from the homotopy groups equipped with the action of the fundamental groups. If that were true, simply connected spaces would be determined by their homotopy groups. | |
| S May 9, 2023 at 14:53 | history | suggested | J. W. Tanner | CC BY-SA 4.0 |
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| May 9, 2023 at 13:28 | review | Suggested edits | |||
| S May 9, 2023 at 14:53 | |||||
| May 9, 2023 at 10:38 | comment | added | Igor Khavkine | The collection of all the homotopy groups together with some data about the action of $\pi_1$ on them has been called a Postnikov system, from which one can in principle reconstruct the full homotopy type, and the homology groups in particular. However, it seems that reconstruction result has not proven to be practically useful. | |
| May 9, 2023 at 8:03 | comment | added | Fernando Muro | @M.Ramana you're not, $S^2$ is simply connected. | |
| May 9, 2023 at 5:26 | answer | added | Chris Gerig | timeline score: 4 | |
| May 9, 2023 at 5:24 | comment | added | M.Ramana | @FernandoMuro Yes, they are. but not as $\mathbb{Z}\pi_1(S^2)$-module. Am I right? | |
| May 9, 2023 at 5:23 | comment | added | M.Ramana | @ChrisGerig I don't know this sequence. Could you explain me a little bit more about it please? | |
| May 9, 2023 at 5:20 | comment | added | Fernando Muro | The homology groups of $S^2$ are all free while it has infinitely many non free homotopy groups. | |
| May 9, 2023 at 5:16 | comment | added | M.Ramana | @TheThinWhistler For example, you can take $X$ to be a finite wedge of $S^2$. Then $H_2 (X)$ is a free abelian group and $\pi_2 (X)$ is a free $\mathbb{Z}\pi_1 (X)$-module. Hence I think that there is a relation between them at least when they are free. | |
| May 9, 2023 at 5:09 | comment | added | The Thin Whistler | What makes you believe that such a relation could exist? | |
| May 9, 2023 at 4:43 | history | asked | M.Ramana | CC BY-SA 4.0 |