Timeline for Conjugacy of torsion subgroups in Gl(n, Z) for small n [duplicate]
Current License: CC BY-SA 3.0
9 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Jul 8, 2013 at 9:02 | history | closed |
Ian Agol Theo Johnson-Freyd Andrey Rekalo Carlo Beenakker user10534 |
Duplicate of Number of conjugacy classes in GL(n,Z) | |
Jul 8, 2013 at 6:33 | comment | added | JHM | @Aakumadula: evidently,i see that if $A \in GL(n,Q)$ has order $k$, then the minimal polynomial $m_A$ of $A$ divides $\prod_{d|k} \Phi_d$. Over $\mathbb{C}$ the elementary divisors $d_1|\ldots|d_r=m_A$ then determine $A$ up to conjugacy in $GL(n,C)$. In your initial comment, were you suggesting something of how to determine the conjugacy type of $A$ in $GL(n, Q)$ relative to its elementary divisors? | |
Jul 8, 2013 at 4:08 | comment | added | Venkataramana | @Martel: No, for example, a product $\prod \Phi _d$ of cyclotomic polynomials of total degree $3$, has very few solutions. Surely these give conjugacy classes. The issue would be rational conjugacy vs integral conjugacy, but this is not insoluble. | |
Jul 8, 2013 at 3:15 | review | Close votes | |||
Jul 8, 2013 at 9:06 | |||||
Jul 8, 2013 at 3:01 | comment | added | JHM | @Aakumadala: I think you are describing torsion elements in,say, $GL(3, \mathbb{C})$ -- which is a different question than torsion in $GL(n,\mathbb{Z})$. E.g. one of these groups has elements of arbitrarily large finite order, while the other doesn't. | |
Jul 8, 2013 at 2:51 | comment | added | Venkataramana | conjugacy class of torsion elements should be easy: they look like companion matrices of products $f$ of cyclotomic polynomials, with no repeated roots. | |
Jul 8, 2013 at 1:46 | answer | added | Igor Rivin | timeline score: 2 | |
Jul 8, 2013 at 0:04 | review | First posts | |||
Jul 8, 2013 at 0:28 | |||||
Jul 7, 2013 at 23:47 | history | asked | Incognito | CC BY-SA 3.0 |