Timeline for Integral orthogonal group for indefinite ternary quadratic form
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 5, 2013 at 21:37 | comment | added | Will Jagy | Sasha, thank you. The general reason I am interested in isometry groups for indefinite integral forms is a very long story that culminated in this: mathoverflow.net/questions/69444/… The information in my self-answer is known to a very, very few, such as Richard Borcherds. | |
| Sep 5, 2013 at 21:07 | comment | added | SashaKolpakov | I'm sorry, I messed up everything: I think that here "reflective" means that the maximal subgroup of $O(q)$ generated by reflections is finite-index in $O(q)$. Indeed, the group $O(q)$, with $q$ as in your question, is generated by 10 reflections and 2 isometries of infinite order, and is not reflective (because of the homology argument above), according to few_reps. The Apollonian group is reflective but has infinite co-volume. | |
| Sep 5, 2013 at 19:24 | comment | added | Will Jagy | In case of interest, the Apollonian group shares the main phenomenon in the Markov equation $x^2 + y^2 + z^2 = 3 x y z,$ in that the sum of the squares is equal to a symmetric polynomial with each term "squarefree," so movements are accomplished by "Vieta Jumping" ; Apollonian is $$ w^2 + x^2 + y^2 + z^2 = 2 w x + 2 w y + 2 w z + 2 x y + 2 x z + 2 y z $$ | |
| Sep 5, 2013 at 19:13 | comment | added | Will Jagy | Thanks for trying to explain. What does it mean for a group to be "non-reflective?" I'm confused; as you say, the Apollonian group is evidently generated by four reflections...Also, it appears the group in my problem above cannot even be generated by reflections? | |
| Sep 5, 2013 at 16:22 | comment | added | SashaKolpakov | Well, my little wild speculation is really too wild there: the groups is non-reflective! Again. | |
| Sep 5, 2013 at 7:12 | history | answered | SashaKolpakov | CC BY-SA 3.0 |