Timeline for Finite subgroup of $Gl(n,\mathbb Z)$ and congruences
Current License: CC BY-SA 3.0
25 events
| when toggle format | what | by | license | comment | |
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| Sep 14, 2011 at 7:41 | vote | accept | Wox | ||
| Sep 10, 2011 at 1:04 | answer | added | Ralph | timeline score: 1 | |
| Sep 9, 2011 at 12:47 | comment | added | Wox | $G/T$ with elements $(t_q,q)T$ is isomorph with a finite subgroup $Q$ of $GL(n,\mathbb Z)$ because $q$ must be an element of the automorphism group of the $\mathbb Z$-lattice in Euclidean vector space which is isomorphic to $T$. We know that there are only finite many conjugacy classes of finite subgroups in $GL(n,\mathbb Z)$. If the possible translations for each Q are rational between 0 and 1 with maximal denominator $|Q|$ for a specific origin, then we find finite many space groups up to origin shift and conjugacy of their quotient group. | |
| Sep 7, 2011 at 23:28 | comment | added | Ralph | Can you give some motivation, why the existence of a $t_q$ with rational componets of max. denominator $m$ shows that there are, up to isomorphism, only finitely many groups $G$ ? | |
| Sep 7, 2011 at 15:56 | answer | added | Wox | timeline score: 0 | |
| Sep 5, 2011 at 14:22 | history | edited | Wox | CC BY-SA 3.0 |
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| Sep 5, 2011 at 13:11 | history | edited | Wox | CC BY-SA 3.0 |
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| Sep 5, 2011 at 13:05 | history | edited | Wox | CC BY-SA 3.0 |
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| Sep 5, 2011 at 12:59 | history | edited | Wox | CC BY-SA 3.0 |
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| Sep 5, 2011 at 12:45 | history | edited | Wox | CC BY-SA 3.0 |
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| Sep 5, 2011 at 9:30 | history | edited | Wox | CC BY-SA 3.0 |
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| Sep 5, 2011 at 8:12 | comment | added | Wox | @Ralph: you're right, fixed it. | |
| Sep 5, 2011 at 8:09 | history | edited | Wox | CC BY-SA 3.0 |
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| Sep 4, 2011 at 16:16 | history | edited | Wox | CC BY-SA 3.0 |
Indeed: mod Z^n on the second row
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| Sep 2, 2011 at 23:07 | history | edited | Ralph |
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| Sep 2, 2011 at 22:43 | answer | added | Ralph | timeline score: 5 | |
| Sep 2, 2011 at 18:33 | comment | added | darij grinberg | Ah, right. Another of my $0$-$1$ mixups. | |
| Sep 2, 2011 at 18:22 | comment | added | Geoff Robinson | Isn't the point that your matrix has the form $mE$ for some idempotent matrix $E$, and is also integral? | |
| Sep 2, 2011 at 16:41 | comment | added | Ralph | No, it isn't the inverse, since $(1+ q + \dots + q^{m-1})(1-q)=0$. But this shows that the columns of $1-q$ are in the "kernel" of $1+q+\dots +q^{m-1}$. | |
| Sep 2, 2011 at 16:05 | comment | added | darij grinberg | I don't get it. Isn't $q+q^2+q^3+...+q^m=1+q+q^2+...+q^{m-1}$ the inverse of $1-q$ (where $1$ means the identity matrix), and thus invertible over $\mathbb Q/\mathbb Z$ as well? | |
| Sep 2, 2011 at 15:55 | history | asked | Wox | CC BY-SA 3.0 |