Timeline for Subgroups of $GL_n(\mathbb Z)$ with finite coinvariants
Current License: CC BY-SA 3.0
9 events
when toggle format | what | by | license | comment | |
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Jul 30, 2012 at 18:30 | history | edited | Wilberd van der Kallen | CC BY-SA 3.0 |
improved phrasing
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Jul 27, 2012 at 17:14 | vote | accept | Igor Belegradek | ||
Jul 27, 2012 at 17:14 | vote | accept | Igor Belegradek | ||
Jul 27, 2012 at 17:14 | |||||
Jul 27, 2012 at 17:13 | history | edited | Igor Belegradek | CC BY-SA 3.0 |
details added
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Jul 27, 2012 at 16:43 | comment | added | user6976 | @Igor: $e_{i,j}(m)=e_{i,j}(1)^m$. | |
Jul 27, 2012 at 15:22 | comment | added | Igor Belegradek | Oops. So how does one prove the claim in the first sentence of your answer? | |
Jul 27, 2012 at 15:19 | comment | added | Wilberd van der Kallen | That will not work. They do not lie in $GL_n(\mathbb Z)$. | |
Jul 27, 2012 at 14:57 | comment | added | Igor Belegradek | Thank you! In fact, it is enough to consider the diagonal matrices $E_i(m)$ which have $m$ at $(i,i)$ entry and $1$ on all other diagonal entries. I am still in the process of proving that the group generated by $E_1(m), \dots, E_n(m)$ sits in any given finite index subgroup for large $m$ but it does sound believable. | |
Jul 27, 2012 at 14:19 | history | answered | Wilberd van der Kallen | CC BY-SA 3.0 |