Timeline for Why is there a unique hyperbolic simplex of largest area?
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13 events
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| Dec 30, 2011 at 19:29 | comment | added | Ian Agol | There is a proof of the maximality of regular simplices using symmetrization due to Peyerimhoff: ams.org/mathscinet-getitem?mr=1934304 One may give a purely synthetic argument for the maximality of ideal regular hyperbolic 3-simplices this way. I don't know if this counts as a "high-brow" proof though. | |
| Dec 30, 2011 at 16:03 | comment | added | Ian Agol | @Bruno: I think unknown (google) meant "$n$-tuples of distinct points" | |
| Dec 30, 2011 at 11:46 | comment | added | S. Carnahan♦ | @Bruno, I think you are using an inconvenient definition of "moduli space" when you say that the volume function is discontinuous. At least in your 3 point example, you should take a quotient by the group of conformal transformations of the boundary if you want to describe a space of configurations. Of course, you may have discontinuous behavior if you choose additional rigidifying structure in addition to a set of points, and if you choose a bad partial compactification. | |
| Dec 30, 2011 at 11:10 | comment | added | Bruno Martelli | Actually, this is more or less the only problem that can occur. If you restrict yourself to $(n+1)$-uples of points where at least 3 of them are distinct, then the volume function is continuous, see Luo arxiv.org/abs/math/0412208, Proposition 4.1 (or Ratcliffe). However, the configuration space is not compact once you removed these bad configurations, so it is not even a priori obvious that there exists a simplex of maximum volume. | |
| Dec 30, 2011 at 11:01 | comment | added | Bruno Martelli | You can construct analogous examples by taking any ideal simplex $\Delta$, and a hyperbolic isometry $f$ converging to one vertex $v$ of $\Delta$. The iterates $f(\Delta)$ converge to the degenerate simplex with point $v$, while of course their volume is constantly equal to the volume of $\Delta$. However, the volume is indeed continuous on the (non-compact) sub-space of non-degenerate simplexes, i.e. on $(n+1)$ points that are not contained in a hyperplane: the dihedral angles are well-defined here and depend continuously on the $(n+1)$ points. The volume is continuos on dihedral angles ... | |
| Dec 30, 2011 at 10:54 | comment | added | Bruno Martelli | For instance, take the three points $0, 1, x$ in $\partial H^2 = \mathbb R \cup \{\infty\}$. When $x$ is not $0$ or $1$ you get an ideal triangle of area $\pi$, when $x$ tends to $0$ or $1$ you get a degenerate triangle of area zero. | |
| Dec 30, 2011 at 10:35 | comment | added | Igor Rivin | @Bruno: why is it not continuous there? | |
| Dec 30, 2011 at 8:22 | comment | added | Bruno Martelli | (that is, those where the $n+1$ points are contained in a hyperplane) | |
| Dec 30, 2011 at 8:22 | comment | added | Bruno Martelli | Note that the volume is not a continuous function on the entire moduli space of the $n+1$ points: you need to remove the degenerate configurations. | |
| Dec 30, 2011 at 1:24 | comment | added | Ian Agol | Jungreis' paper may be relevant, although he indirectly is appealing to Haagerup-Munkholm. ams.org/mathscinet-getitem?mr=1452185 | |
| Dec 29, 2011 at 20:54 | history | edited | John Pardon | CC BY-SA 3.0 |
added 285 characters in body
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| Dec 29, 2011 at 19:23 | answer | added | Igor Rivin | timeline score: 19 | |
| Dec 29, 2011 at 16:57 | history | asked | John Pardon | CC BY-SA 3.0 |