Timeline for Lower central series vs torsion-free lower central series
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 1, 2021 at 1:31 | vote | accept | Irina | ||
| Feb 28, 2021 at 23:18 | answer | added | Andy Putman | timeline score: 5 | |
| Feb 27, 2021 at 22:53 | comment | added | YCor | At least it's quite coherent because at the Lie algebra level, the Malcev K-completion $G\otimes K$ is the same as $(G\otimes Q)\otimes_Q K$ for every field of characteristic zero $K$ (the best known case being $K=R$, since this yields a simply connected nilpotent Lie group). | |
| Feb 27, 2021 at 22:48 | comment | added | LSpice | Sorry, you're right; I don't know what I meant by $G \otimes_{\mathbb Z} \mathbb Q$, although it appears that @YCor charitably attached the best possible interpretation to it. | |
| Feb 27, 2021 at 22:37 | comment | added | YCor | The Malcev Q-completion can be defined purely group-theoretically (no reference to algebraic groups): I guess it's convenient to denote it $G\otimes Q$. I don't think in any case that $G$ is always meant to be the group of $Z$-points (which by the way would only be meaningful for a group defined over $Z$, not only over $Q$). | |
| Feb 27, 2021 at 20:50 | comment | added | Irina | @LSpice: $G$ is a nilpotent group, not an abelian group. I was assuming you were referring to the Malcev completion, which turns a finitely generated torsion-free nilpotent group into an algebraic group over $\mathbb{Q}$ in which the original group is the $\mathbb{Z}$-points. | |
| Feb 27, 2021 at 20:43 | comment | added | LSpice | $G$ is an Abelian group, hence a $\mathbb Z$-module, and $G \otimes_{\mathbb Z} \mathbb Q$ means to tensor as $\mathbb Z$-modules. It does wind up only seeing the torsion-free quotient of $G$. (For example, $\mathbb Z/n\mathbb Z \otimes_{\mathbb Z} \mathbb Q = \mathbb Z/n\mathbb Z \otimes_{\mathbb Z} n\mathbb Q = n\mathbb Z/n\mathbb Z \otimes_{\mathbb Z} \mathbb Q = 0$.) | |
| Feb 27, 2021 at 18:46 | comment | added | Irina | @LSpice: I only know what $G \otimes_{\mathbb{Z}} \mathbb{Q}$ means when $G$ is torsion-free, but I agree that once you kill off the irrelevant torsion subgroup your statement is equivalent to mine. | |
| Feb 27, 2021 at 18:44 | comment | added | Irina | @YCor: I don't see why that has to be hold. I agree that the crux of the problem is to prove that $\gamma_k(G)$ is finite-index in $\gamma_k^{tf}(G)$ for all $k$ (you could then use the additivity of Hirsch ranks in extensions), but once you get beyond small $k$ I don't see why the difference is precisely the torsion subgroup of that quotient. | |
| Feb 27, 2021 at 13:12 | comment | added | YCor | Yes. $G/\gamma_k^{\mathrm{tf}(G)}$ is the quotient of $G/\gamma_k(G)$ by its torsion subgroup, which is finite. So they have the same Hirsch rank. | |
| Feb 27, 2021 at 5:06 | comment | added | LSpice | $\newcommand\Z{\mathbb Z}\newcommand\Q{\mathbb Q}$Isn't $\gamma_k^{\text{tf}}(G) \otimes_\Z \Q/\gamma_{k + 1}^{\text{tf}}(G) \otimes_\Z \Q$ isomorphic to $\gamma_k(G \otimes_\Z \Q)/\gamma_{k + 1}(G \otimes_\Z \Q)$, so we're comparing the ranks of $\gamma_k(G \otimes_\Z \Q)/\gamma_{k + 1}(G \otimes_\Z \Q)$ and $\gamma_k(G)/\gamma_{k + 1}(G)$? | |
| Feb 27, 2021 at 5:05 | history | edited | LSpice | CC BY-SA 4.0 |
\text{}
|
| Feb 27, 2021 at 2:46 | history | asked | Irina | CC BY-SA 4.0 |