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SashaKolpakov
  • Member for 11 years, 3 months
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Different Metrics for Baire Space and their induced Topologies
edited the misspell $f_n(m)=0$, whenever $m\neq n$
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Torus minimizer of Willmore energy
a little tex editing (now all formulas ok)
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Integral orthogonal group for indefinite ternary quadratic form
@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.
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Integral orthogonal group for indefinite ternary quadratic form
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 $\geq 15$.
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Integral orthogonal group for indefinite ternary quadratic form
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.
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Integral orthogonal group for indefinite ternary quadratic form
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.
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Integral orthogonal group for indefinite ternary quadratic form
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.
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Integral orthogonal group for indefinite ternary quadratic form
Well, my little wild speculation is really too wild there: the groups is non-reflective! Again.
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Integral orthogonal group for indefinite ternary quadratic form
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 ...)
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Integral orthogonal group for indefinite ternary quadratic form
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?
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References on complete intersections in Grassmanian
I added the "reference-request" tag
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