Timeline for Is Malcev completion an embedding?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 22 at 12:56 | comment | added | Qwert Otto | Wow, I didn't know that. Thank you so much! | |
| Apr 22 at 7:27 | comment | added | Adrien | You're right of course, but group-like elements in a Hopf algebra are linearly independant so that's fine. | |
| Apr 22 at 0:45 | comment | added | Qwert Otto | I'm sorry; I think that's not necessarily the case (at least in general). Even if $X\to V$ is an injection for a set $X$ and a vector space $V$, this may not induce an injective linear map. Are there other reason that I'm missing? | |
| Apr 21 at 20:29 | comment | added | Adrien | Sure, since the map $G\rightarrow \widehat{k[G]}$ is. | |
| Apr 21 at 15:32 | comment | added | Qwert Otto | Thank y'all for the answer and a counterexample. Under the residual torsion-free-nilpotency, is it also true that the map $k[G]\to\widehat{k[G]}$ is also injective? | |
| Apr 21 at 14:26 | comment | added | Uri Bader | For a specific example, fix a prime $p$ and an integer $n\geq 3$ and let $G$ be the corresponding congruence subgroup $\ker(\text{SL}_n(\mathbb{Z})\to \text{SL}_n(\mathbb{Z}/(p))$. Then $G$ is torsion free and, by conisdering deeper conruence subgroups, it is residually (finite-)nilpotent. However $G$ has no non-trivial torsion free nilpotent qoutient (e.g by proerty T), so $\hat{G}$ is trivial. | |
| Apr 21 at 11:15 | vote | accept | Qwert Otto | ||
| Apr 21 at 10:19 | history | answered | Adrien | CC BY-SA 4.0 |