Timeline for Finitely generated matrix groups whose eigenvalues are all algebraic
Current License: CC BY-SA 4.0
8 events
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| Apr 2, 2019 at 16:25 | comment | added | Geoff Robinson | That is true, but it does not a priori get you into a representation over a cyclotomic field. It does get you into some number field. The content of Schur's theorem is that finitely generated periodic linear groups over complex numbers are in fact finite. | |
| Apr 2, 2019 at 15:03 | comment | added | YCor | If I'm not wrong, the fact that every finite subgroup is conjugate into the algebraic closure of $\mathbf{Q}$ is immediate from basic theory (which basically works over an arbitrary algebraically closed field of characteristic zero, and in particular by counting, every irreducible is defined over the algebraics). | |
| Apr 2, 2019 at 14:59 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Typo
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| Apr 2, 2019 at 14:46 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Corrected text
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| Apr 2, 2019 at 14:42 | comment | added | Geoff Robinson | @YCor: You are right, I was careless. I will re-edit or delete. Schur's theorem is of course correct, but the eigenvalues being roots of unity does not give periodicity, as you say. And I did not say what I meant in the first part either. | |
| Apr 2, 2019 at 14:26 | comment | added | YCor | and also, that all elements have only eigenvalues of finite order doesn't imply being finite: just take the cyclic subgroup generated by a nontrivial unipotent element. | |
| Apr 2, 2019 at 14:22 | comment | added | YCor | $k$ cyclotomic (including $k=\mathbf{Q}$) doesn't mean that eigenvalues have finite order... | |
| Apr 2, 2019 at 13:38 | history | answered | Geoff Robinson | CC BY-SA 4.0 |