Timeline for Torsion in GL_n(Z)
Current License: CC BY-SA 2.5
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 23, 2014 at 20:23 | vote | accept | Andy Putman | ||
| Oct 6, 2010 at 22:00 | answer | added | vytas | timeline score: 2 | |
| Oct 6, 2010 at 20:41 | answer | added | Ian Agol | timeline score: 3 | |
| Oct 6, 2010 at 16:14 | answer | added | Alex B. | timeline score: 11 | |
| Oct 6, 2010 at 6:05 | comment | added | Torsten Ekedahl | The answer to 2 is definitely no. This is true already for elements of order a prime $\ell$: Conjugacy classes of matrices of order $\ell$ without eigenvalue $¡$ and with $n=\ell-1$ correspond to the elements of the class group of $\ell$'th roots of unity. Such matrices become conjugate even over the $p$-adic integers. By conjugating one of them with a suitable integer matrix their mod $p$-reductions become equal (using the surjectivity of $\mathrm{SL}_n(\mathbb Z)\to\mathrm{SL}_n(\mathbb Z/p)$. | |
| Oct 6, 2010 at 5:49 | comment | added | Alex B. | In your situation, the proof of the above result will construct all elements of GL_n(Z) of order k, whenever k is cube-free. If it's not cube-free, anything you can say would be interesting. I can put this up as an answer, if you would like me to. | |
| Oct 6, 2010 at 5:47 | comment | added | Alex B. | The striking result you are looking for reads, in full generality, as follows: a finite group has finitely many isomorphism classes of indecomposable integral representations if and only if for any prime $p$, all its Sylow $p$-subgroups are cyclic of order at most $p^2$. This result is effective in the sense that when there are finitely many isomorphism classes, the proof constructs them for you. But as far as I know, there is nothing remotely resembling a classification or even any hope of obtaining one, if this condition is violated. | |
| Oct 6, 2010 at 5:38 | comment | added | Victor Protsak | Andy, see mathoverflow.net/questions/27175/representation-theory-over-z, which in particular contains a reference to the theorem in Curtis and Reiner. | |
| Oct 6, 2010 at 5:29 | comment | added | Andy Putman | Another addendum to the above comments on the literature. The "classification" alluded to above is known for elements of prime order (it appears somewhere in Curtis and Reiner, but my copy is at my office so I can't tell you where). I also seem to recall seeing a paper working out elements of prime squared order, but I've been told that this is as far as is known in general. | |
| Oct 6, 2010 at 5:08 | history | edited | Andy Putman | CC BY-SA 2.5 |
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| Oct 6, 2010 at 5:07 | comment | added | Andy Putman | @Scott : Whoops! In my defense, my original draft had SL instead of GL... | |
| Oct 6, 2010 at 5:07 | comment | added | Andy Putman | @Pete : By the way, if you could completely answer my questions then you could classify all finite subgroups of GL_n(Z). That's why I expect that there are interesting partial answers (known to someone, but again not me). | |
| Oct 6, 2010 at 5:07 | comment | added | S. Carnahan♦ | I think Tom Church was making a subtle correction to your claim that $\phi_p$ is surjective. | |
| Oct 6, 2010 at 5:03 | comment | added | Andy Putman | And I've done some searching of the literature, but I didn't find answers to the above. Of course, I suspect that they are known to someone, but I'm just a simple topologist. | |
| Oct 6, 2010 at 5:02 | comment | added | Andy Putman | By "classify" I mean give a reasonable way to list (up to conjugacy) all the order $k$ elements for some fixed $k$. You can grind it out for small $k$ (but even $k=2$ is somewhat nontrivial), and you have remarkable results like Feit's theorem (Theorem 23 in your manuscript), but I get the impression that it's a jungle in general. | |
| Oct 6, 2010 at 4:59 | comment | added | Pete L. Clark | Added: of course every finite group $G$ occurs as a subgroup of $\operatorname{GL}_{|G|}{\mathbb{Z}}$, so a sufficiently exacting classification implies a classification of all finite groups (so, yes, "hopeless"). But my point is that the questions you ask are much, much weaker than a complete classification, so you should probably check out the existing literature on the subject (if you haven't already). | |
| Oct 6, 2010 at 4:56 | comment | added | Pete L. Clark | I'm not sure what you mean by "hopeless" and "classify" in your first sentence. Are you aware that there are a lot of results describing finite subgroups of $\operatorname{GL}_n(\mathbb{Z})$? For some of them -- with and without proof -- see e.g. Section 3 of math.uga.edu/~pete/8410Chapter9.pdf. (I don't know offhand if they address your specific questions.) | |
| Oct 6, 2010 at 4:55 | comment | added | Andy Putman | @Tom : Hence the word "interesting" <grin>. | |
| Oct 6, 2010 at 4:54 | comment | added | Tom Church | A trivial observation (which I am positive you already know) is that any $y\in \text{GL}_n(\mathbb{Z}/p)$ which comes from $x\in \text{GL}_n(\mathbb{Z})$ has $\det y=\pm 1\in (\mathbb{Z}/p)^\times$. | |
| Oct 6, 2010 at 4:49 | history | edited | Andy Putman | CC BY-SA 2.5 |
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| Oct 6, 2010 at 4:39 | history | asked | Andy Putman | CC BY-SA 2.5 |