Timeline for Structure of second homotopy group of a compact CW complex
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 10 at 15:58 | comment | added | Dave Benson | A paper of Ould Houcine proves that there is a finitely presented group whose centre is $\mathbb{Q}$. I wonder whether this or related constructions produce a finite CW complex with $\pi_2$ isomorphic to $\mathbb{Q}$. | |
| Dec 28, 2023 at 9:31 | comment | added | Maxime Ramzi | @R.vanDobbendeBruyn : oh yeah, that makes sense, thanks ! | |
| Dec 27, 2023 at 23:41 | comment | added | R. van Dobben de Bruyn | @SFeesh for now it is referring to the construction of this answer. If I knew how to rule out $\mathbf Q$ in general, I would probably post it as a separate answer. But even in the case described here, my argument has some gaps in it. | |
| Dec 27, 2023 at 23:38 | comment | added | R. van Dobben de Bruyn | @MaximeRamzi the difficulty in the non-commutative setting is that one-sided ideals need not be two-sided, so getting $\mathbf Q$ as a cyclic module does not mean it becomes an algebra. (To show it is cyclic if it is finitely generated as $\mathbf Z[\pi_1(X)]$-module (say by $x_1,\ldots,x_k \in \mathbf Q$) take a generator of the subgroup of $\mathbf Q$ generated by $x_1,\ldots,x_k$.) | |
| Dec 27, 2023 at 22:42 | comment | added | SFSH | @R. van Dobben de Bruyn what do you mean by "this way" and "in this construction"? Are you referring to the setting of a compact CW complex, or something else? | |
| Dec 27, 2023 at 19:40 | comment | added | Maxime Ramzi | @R.vanDobbendeBruyn : $\mathbf Q$ is also not finitely generated as an associative $\mathbf Z$-algebra, simply because it is a filtered colimit of the $\mathbf Z[1/p, p \in F]$. I'm confused as to how you deduce that $\pi_2(Y)$ has to be a cyclic module if it is $\mathbf Q$ ? | |
| Dec 27, 2023 at 17:46 | comment | added | R. van Dobben de Bruyn | I think it is unlikely that you could get $\pi_2(Y)\cong\mathbf Q$ this way, for ring theoretic reasons. In this construction, $\pi_2(Y)$ is finitely generated over $\mathbf Z[\pi_1(Y)]$, which in turn is a finitely presented $\mathbf Z$-algebra. This is impossible at least when $\pi_1(Y)$ is abelian (if $\pi_2(Y)\cong\mathbf Q$, then it is a cyclic $\mathbf Z[\pi_1(Y)]$-module, hence isomorphic to $\mathbf Z[\pi_1(Y)]/I$ for an ideal $I$, so $\mathbf Q$ would be a finitely generated $\mathbf Z$-algebra). I think this is also impossible if $\pi_1(Y)$ is non-abelian, but I don't have a proof. | |
| Dec 27, 2023 at 15:44 | vote | accept | SFSH | ||
| Dec 27, 2023 at 15:49 | |||||
| Dec 27, 2023 at 13:06 | history | answered | Oscar Randal-Williams | CC BY-SA 4.0 |