Timeline for Pre-images of unipotent elements in $\operatorname{SL}_{n}(A)$
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 27, 2013 at 13:17 | vote | accept | Olivier | ||
| Nov 27, 2013 at 13:16 | comment | added | Olivier | @JimHumphreys So I found the set-up and Tim's answer confusing Hopefully, everywhere in my question, it is specifically indicated that I care about non trivial (that is non identity) unipotents; most especially, in my main question this is ensured by the fact I want the sought for unipotent element to reduce to a non-identity element modulo the maximal ideal. | |
| Nov 26, 2013 at 21:38 | comment | added | YCor | @Jim: I don't understand what you mean: it is clear that (over an algebraically closed field) the conditions $\sigma-1$ nilpotent and ``all eigenvalues are 1" are equivalent. And it is also clear that in a finite matrix group over a field of characteristic zero, there no non-identity unipotent elements. | |
| Nov 26, 2013 at 15:07 | comment | added | Jim Humphreys | @Olivier: Usually "unipotent" just means that all eigenvalues are 1, so there are typically a lot of such elements in a finite matrix group including the identity element. So I found the set-up and Tim's answer confusing. | |
| Nov 26, 2013 at 10:50 | comment | added | Olivier | @JimHumphreys I meant an element $\sigma$ such that $\sigma-1$ is nilpotent, but I now realize this is not quite what I want if $A$ is not reduced, or perhaps even a domain (I wouldn't say the diagonal matrix $(1+\alpha)$ with $\alpha$ nilpotent is unipotent). | |
| Nov 26, 2013 at 10:48 | history | edited | Olivier | CC BY-SA 3.0 |
Clarified the question
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| Nov 24, 2013 at 23:18 | comment | added | Jim Humphreys | @Olivier: Can you say more explicitly what you mean by "unipotent" element in a matrix group? | |
| Nov 23, 2013 at 11:22 | answer | added | Tim Dokchitser | timeline score: 5 | |
| Nov 22, 2013 at 14:21 | comment | added | YCor | btw In Theorem I, and in the questions, you probably assume that subgroups are closed. Also I don't think completeness is relevant to the question and Boston's result. | |
| Nov 22, 2013 at 14:19 | answer | added | YCor | timeline score: 3 | |
| Nov 22, 2013 at 9:23 | comment | added | YCor | A simple example for the second question is $A=\mathbf{Z}_p[[t]]$ (with $\mathbf{Z}_p$ the ring of $p$-adic integers), $G=\mathrm{SL}_n(\mathbf{Z}_p)$. In general, I don't think you expect better than a result assuming surjectivity of the reduction modulo $pA+\mathfrak{m}^2$. | |
| Nov 22, 2013 at 9:21 | comment | added | Olivier | @TimDokchitser Dear Tim, thanks a lot for this great counterexample. Any idea about non-finite subgroups? Also, will you consider posting this as an answer? After all, it does completely answer the most general version of question I. | |
| Nov 22, 2013 at 9:09 | comment | added | Tim Dokchitser | It seems there are counterexamples to the first question, at least for $p=5$. The group $\text{SL}_2({\mathbb F}_5)$ has a 2-dimensional symplectic representation with character in ${\mathbb Q}(\sqrt 5)$ and Schur index 2, so it looks like it is realizable over $A={\mathbb Z}_5(\zeta_5)$. So $\text{SL}_2({\mathbb F}_5)$ injects in $\text{SL}_2(A)$ and reduces, I suppose, onto $\text{SL}_2(A/{\mathfrak m})$. But it is just a finite group, so it has no unipotent elements. | |
| Nov 22, 2013 at 8:40 | history | edited | Olivier | CC BY-SA 3.0 |
Provided reference
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| Nov 22, 2013 at 8:06 | comment | added | Olivier | @YvesCornulier Cher Yves, bien sûr. J'ai aussi pris la liberté de t'envoyer l'article par mail. | |
| Nov 21, 2013 at 21:50 | comment | added | YCor | could you provide a link or reference to Boston's result? | |
| Nov 21, 2013 at 20:48 | history | edited | Olivier | CC BY-SA 3.0 |
Corrected mistakes.
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| Nov 20, 2013 at 14:11 | history | edited | Olivier | CC BY-SA 3.0 |
Provided an equivalent framing of the question and corrected the tags accordingly.
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| Nov 20, 2013 at 14:00 | history | edited | Olivier | CC BY-SA 3.0 |
Slight clarification.
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| Nov 19, 2013 at 23:10 | history | asked | Olivier | CC BY-SA 3.0 |