Timeline for Groups of matrices in which all elements have all eigenvalues equal in modulus
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 13, 2016 at 14:33 | answer | added | YCor | timeline score: 1 | |
| Jul 13, 2016 at 11:45 | vote | accept | Ian Morris | ||
| Jul 13, 2016 at 4:21 | answer | added | Venkataramana | timeline score: 4 | |
| Jun 9, 2016 at 9:14 | comment | added | Uri Bader | @GeoffRobinson I wrote the proof as promised (even though it is not as simple as I wanted it to be). | |
| Jun 7, 2016 at 16:18 | answer | added | Misha | timeline score: 6 | |
| Jun 7, 2016 at 14:27 | comment | added | Uri Bader | @GeoffRobinson, the proof I see involves taking Zariski closure and getting a group satisfying $\text{tr}(g)=\text{tr}(g^{-1})$. This means in particular that every element has 1 as an ev. I can see that either our group $G$ is compact, or is conjugated into $\text{SO}(2,1)$ which in turn reduces the problem to dim=2. I am not happy with it because I suppose I (or someone else) could come with a simpler proof after some thinking. If it doesn't happen I will write the argument above. | |
| Jun 7, 2016 at 13:23 | comment | added | Geoff Robinson | @UriBader : If you see how to fix it, you should! It would be useful. | |
| Jun 7, 2016 at 11:46 | comment | added | Uri Bader | @friedrichknop I deleted an earlier answer gave, as it was not complete. I see how to fix it, but I don't want it to be hanged there like that. Thank you for your comment. | |
| Jun 7, 2016 at 11:02 | comment | added | Neil Strickland | By the other comments, we can assume $\det=1$ and that all eigenvalues are on the unit circle. This can be formulated as a condition on the coefficients of the characteristic polynomial, which depend continuously on the matrix, so it will be harmless to replace $G$ by $\overline{G}$, so we can assume that $G$ is closed. Now $G$ is subconjugate to $SO(3)$ iff it is compact (because we can use integration wrt Haar measure to find an invariant inner product). So the real problem is to prove compactness. | |
| Jun 7, 2016 at 10:48 | answer | added | Uri Bader | timeline score: 8 | |
| Jun 7, 2016 at 10:03 | history | edited | Ian Morris | CC BY-SA 3.0 |
added 43 characters in body
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| Jun 6, 2016 at 19:45 | comment | added | Ian Morris | @Anton: I speculate that the group case might be easier than the semigroup case partly just because it is a proper subcase, but also because in the group case one has access to tools like Haar measure and maximal compact subgroups. | |
| Jun 6, 2016 at 19:29 | comment | added | Ian Morris | @Anton: firstly it's unpublished, and a published result (or one which I can verify myself) would reassure me of its correctness. Secondly, in general and where possible I would prefer to use an argument which I understand. Thirdly, I hope in the process to learn some methods with which to understand related problems. | |
| Jun 6, 2016 at 19:06 | comment | added | Geoff Robinson | You can also assume that every element of $G$ has $1$ as an eigenvalue: using Denis Serre's comment if an element $g$ of $G$ has three real eigenvalues, none of which is $1$, all must be $-1$, and the elements of $G$ of determinant $1$ form a normal subgroup of $G$ of index $2$ which is a direct factor of $G$. If $g$ has two non-real eigenvalues, then the other is $\pm 1$ and if it is not $1$, then you can assume $-I$ to $G$ without affecting the hypotheses, so $G$ is again a direct product of its determinant $1$ matrices and $<-I >$. | |
| Jun 6, 2016 at 17:44 | comment | added | user1688 | If it follows from the paper you cite, why don't you just cite that paper? And why do you expect the group case to be easier? | |
| Jun 6, 2016 at 16:56 | comment | added | Denis Serre | Because $\det$ is a morphism, as well as $|\det|^{-1/3}$, you may replace $G$ by $H$, which consists of the matrices $|\det A|^{-1/3}A$. This reduces the question to the case where the spectral radius of $A$ equals $1$ for every $A\in G$. | |
| Jun 6, 2016 at 14:24 | history | asked | Ian Morris | CC BY-SA 3.0 |