Timeline for Special groups, special resolutions and group cohomology
Current License: CC BY-SA 4.0
38 events
| when toggle format | what | by | license | comment | |
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| S Apr 8 at 21:01 | history | bounty ended | CommunityBot | ||
| S Apr 8 at 21:01 | history | notice removed | CommunityBot | ||
| S Mar 31 at 19:19 | history | bounty started | GSM | ||
| S Mar 31 at 19:19 | history | notice added | GSM | Canonical answer required | |
| Dec 15, 2022 at 18:56 | comment | added | Peter Kropholler | @JeremyRickard Let $\Delta$ denote the set of elements of $G$ that have finitely many conjugates: this is a characteristic subgroup and is (locally finite)-by-abelian. Since our group $G$ in the question has finite cohomological dimension over $\mathbb Z$, it follows $\Delta$ is torsion-free abelian of finite rank. So some subgroup $H$ of finite index in $G$ centralises $\Delta$. Now your original line of reasoning works for $H$ so I think you successfully show $G$ must be virtually abelian. That is an interesting reduction. | |
| Dec 15, 2022 at 13:46 | comment | added | Jeremy Rickard | @PeterKropholler When I thought that I could identify $\mathbb{Z}$ as $\mathbb{Z}[G]\otimes_{\mathbb{Z}[Z(G)]}\operatorname{coker}(f)$, that gave a contradiction since $\mathbb{Z}[G]$ is a free $\mathbb{Z}[Z(G)]$-module of rank strictly greater than $1$. But in fact we only have $\mathbb{Z}\cong\mathbb{Z}[G]\otimes_{Z(\mathbb{Z}[G])}\operatorname{coker}(f)$, and I couldn't see how to proceed, since $\mathbb{Z}[G]$ may not be a free $Z(\mathbb{Z}[G]$-module. | |
| Dec 15, 2022 at 10:57 | comment | added | Peter Kropholler | @JeremyRickard I'm sorry you deleted your answer because it looked plausible save for one point I noticed that the centre of the group ring can be a little larger than the group ring of the centre. Does this really cause a lot of difficulty or was there some other problem with your plan? | |
| S Oct 11, 2022 at 12:06 | history | bounty ended | CommunityBot | ||
| S Oct 11, 2022 at 12:06 | history | notice removed | CommunityBot | ||
| Oct 3, 2022 at 13:47 | comment | added | Jeremy Rickard | @GSM I'm not sure. It's possible that my proof could be fixed, but I haven't managed yet. | |
| Oct 3, 2022 at 11:45 | comment | added | GSM | @JeremyRickard but do you still think that there is no such non-abelian group verifying condition 1 and 2? | |
| S Oct 3, 2022 at 10:51 | history | bounty started | GSM | ||
| S Oct 3, 2022 at 10:51 | history | notice added | GSM | Canonical answer required | |
| S Oct 3, 2022 at 4:55 | history | bounty ended | GSM | ||
| S Oct 3, 2022 at 4:55 | history | notice removed | GSM | ||
| Oct 3, 2022 at 4:55 | vote | accept | GSM | ||
| Oct 3, 2022 at 6:19 | |||||
| Oct 2, 2022 at 21:24 | comment | added | Jeremy Rickard | @GSM I just realised that my answer doesn’t work. If you unaccept it then I’ll delete it. | |
| Oct 2, 2022 at 15:50 | vote | accept | GSM | ||
| Oct 3, 2022 at 4:55 | |||||
| Oct 2, 2022 at 14:25 | comment | added | Fernando Muro | @GSM this doesn’t quite answer my question. | |
| Oct 2, 2022 at 13:46 | comment | added | GSM | @FernandoMuro the resulotion I am looking for has to verify 1 and 2 in the same time. | |
| Oct 2, 2022 at 12:37 | comment | added | Fernando Muro | @GSM in 2 you don’t want the resolution to be free or projective, do you? | |
| Oct 2, 2022 at 8:48 | comment | added | GSM | @FernandoMuro there is a big class satisfying condition 1. Essentially my question was about non-abelian groups satisfying 1 and 2. | |
| Oct 2, 2022 at 0:07 | comment | added | Fernando Muro | What big class of groups satisfies the two conditions? | |
| S Oct 1, 2022 at 15:44 | history | bounty started | GSM | ||
| S Oct 1, 2022 at 15:44 | history | notice added | GSM | Canonical answer required | |
| Sep 30, 2022 at 8:51 | history | edited | GSM | CC BY-SA 4.0 |
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| Sep 29, 2022 at 19:12 | comment | added | GSM | @BenjaminSteinberg sure, that will be an easy exercise. But I want a finite free resolution of ZG-modules that turns out to be a finite resolution of ZG-bimodules. | |
| Sep 29, 2022 at 18:43 | comment | added | Benjamin Steinberg | It seems to be more natural to look at free bimodule resolutions of Z rather than resolutions by ZG. 1. is equivalent to 2. if you wanted free bimodule resolutions | |
| Sep 29, 2022 at 18:01 | history | edited | GSM | CC BY-SA 4.0 |
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| Sep 29, 2022 at 16:31 | history | edited | GSM | CC BY-SA 4.0 |
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| Sep 29, 2022 at 16:24 | comment | added | GSM | @Z.M let say that free module in this question is given by a finite direct sum of Z[G]. The multiplication on the right and on the left are the obvious ones. | |
| Sep 29, 2022 at 16:21 | comment | added | Z. M | How do you view a free left $R:=\mathbb Z[G]$-module as a $R$-bimodule? Note that the "obvious way" is not well-defined since it depends on how you see a module to be free. For example, let $M$ be the free left $R$-module $R$ of rank 1. For a unit $u\in R$, the right multiplication $u\colon R\to M$ is another way to see $M$ as a free $R$-module, while the "obvious" right multiplications do not coincide (if $u$ is not in the center). | |
| Sep 29, 2022 at 16:06 | history | edited | GSM | CC BY-SA 4.0 |
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| Sep 29, 2022 at 16:04 | comment | added | GSM | @abx oups, I will edit my question, to avoid commutativity. Thanks for your remark. | |
| Sep 29, 2022 at 16:02 | comment | added | abx | No. Take $G=\mathbb{Z}$. Then $\mathbb{Z}[G]$ is the ring $R=\mathbb{Z}[t,t^{-1}]$ of Laurent polynomials, and $\mathbb{Z}$ admits the resolution $0\rightarrow R\xrightarrow{\ \times (t-1)\ }R\rightarrow \mathbb{Z}\rightarrow 0$. | |
| Sep 29, 2022 at 15:58 | history | edited | GSM | CC BY-SA 4.0 |
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| Sep 29, 2022 at 15:42 | history | edited | YCor | CC BY-SA 4.0 |
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| Sep 29, 2022 at 15:25 | history | asked | GSM | CC BY-SA 4.0 |