Timeline for Integral orthogonal group for indefinite ternary quadratic form
Current License: CC BY-SA 3.0
23 events
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| Sep 11, 2013 at 10:37 | history | edited | few_reps | CC BY-SA 3.0 |
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| Sep 8, 2013 at 21:28 | comment | added | few_reps | The five generators are the reflections in the planes supporting the faces of the convex hull of the following points : A=(1,1,1,1) , B=(0,1,1,1), C=(0,1,1,2), D=(O,O,1,1), E=(1,1,1,3); note that A,B,C,E are on a common plane, and D is a cusp. The reflection in the face BCD is the S1 of Fuchs, the reflection in the face CED is his S4. The 3 other reflections are permutations of the basis vectors. | |
| Sep 8, 2013 at 19:38 | comment | added | Will Jagy | @few_reps, the four generators mentioned are each a jump of one variable. What is the fifth generator? Note that they stop at "root" solutions, quoted as Theorem 1.3 in the Fuchs Bulletin piece. | |
| Sep 8, 2013 at 19:31 | comment | added | few_reps | @Will : Finally, the group of autochronous automorphisms of the quadratic form $w^2+x^2+y^2+z^2=2wx+2wy+2wz+2xy+2xz+2yz$ is a reflection group (on 5 generators), but in Sasha's reference, It is written that the apollonian group is a subgroup of the latter, not the full of it as I had understood. | |
| Sep 7, 2013 at 18:49 | comment | added | Will Jagy | few_reps, you should be able to comment anywhere now. Did the email with Sasha work out? | |
| Sep 5, 2013 at 22:42 | comment | added | few_reps | Well, I'm gonna send you an email (I made a typo : there are only 5 generators). | |
| Sep 5, 2013 at 22:31 | comment | added | SashaKolpakov | @few_reps: This is very interesting. Please let me know how it looks like. In Fuchs' paper, I found four generators for the Apollonian group and a geometric description of the corresponding arrangement of reflection planes, tangent to each other at infinity. | |
| Sep 5, 2013 at 22:20 | comment | added | SashaKolpakov | Actually (silly me again!), in his recent paper, John McLeod says that Vinberg's algorithm (as MacLeod applies it in arxiv.org/abs/1007.2299) produces an infinite-volume polytope in dimension $14$ and does not halt in dimensions $\geq 15$. | |
| Sep 5, 2013 at 22:02 | comment | added | few_reps | @Sasha : I still don't have enough reps to comment your answer. I seem to find that the Apollonian group is a reflection group (with 6 generators). | |
| Sep 5, 2013 at 21:55 | comment | added | SashaKolpakov | All right, that's silly me: what Vinberg says in his original paper is that if finding $m$ normals and the respective half-spaces $H^{-}_i$, $i=1,...,m$, we have that the volume of $P=\cap^m_{i=1} H^{-}_i$ is finite, then the algorithm halts at the $m$-th step. | |
| Sep 5, 2013 at 21:34 | comment | added | SashaKolpakov | I'm afraid here my ignorance comes into play more than yours: what is usually done by people using Vinberg's algorithm is that they search for a representation $O(q) = \Gamma\rtimes H$, with $H$ finite and $\Gamma$ generated by reflections. Then either they find a finite-volume polytope generating $\Gamma$ by reflections in its sides or $\Gamma$ is not finitely generated by reflections (so the algorithm does not stop, finding more and more normals for new "facets"). I will look into some papers for what you ask about and get back to you. | |
| Sep 5, 2013 at 21:14 | comment | added | few_reps | @Sasha : excuse my ignorance, but I am pretty sure to have heard once that it may happen that the algorithm terminates, but the polyhedron obtained has infinite volume ... and hence cannot be of finite index in the automorphism group of the quadratic form ... Would you have a reference ? | |
| Sep 5, 2013 at 19:30 | vote | accept | Will Jagy | ||
| Sep 5, 2013 at 19:30 | comment | added | Will Jagy | In that case, I was never going to get near this myself, and did the right thing asking for help. | |
| Sep 5, 2013 at 19:26 | comment | added | few_reps | It cannot : its rational homology would be trivial, and its not the case. | |
| Sep 5, 2013 at 19:15 | comment | added | Will Jagy | Thanks. So, this group is not generated by reflections? | |
| Sep 5, 2013 at 15:51 | comment | added | SashaKolpakov | If this group is non-reflective then Vinberg's algorithm should not stop (indeed, otherwise the group would be reflective though). The fundamental polygon has 10 reflective sides and 4 sides identified in pairs by the infinite-order isometries, in'nit? Sorry if I got it wrong. Actually, I would be happy with any description, not necessarily a picture (however, a picture could be very beautiful ...) | |
| Sep 5, 2013 at 14:58 | comment | added | few_reps | The method uses Voronoi cells ... Note that this group is non-reflective ... and I believe (but might be wrong) that Vinberg algorithm should not finish. If I had an easy way to plot hyperbolic triangles, I could draw the fundamental polygon, but I just tried with Sage and it did not work ... | |
| Sep 5, 2013 at 12:45 | comment | added | SashaKolpakov | What kind of methods do you use? I just wonder if one has a kind of Vinberg's algorithm, producing nice polytopes/polyhedra. Can you visualise the fundamental domain (Poincare's polygon) for this action in the upper half-plane? What are the angles of the fundamental polygon? | |
| Sep 5, 2013 at 12:12 | history | edited | few_reps | CC BY-SA 3.0 |
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| Sep 5, 2013 at 9:47 | history | edited | few_reps | CC BY-SA 3.0 |
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| Sep 5, 2013 at 8:43 | review | First posts | |||
| Sep 5, 2013 at 8:49 | |||||
| Sep 5, 2013 at 8:24 | history | answered | few_reps | CC BY-SA 3.0 |