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Let $\Gamma$ be a Kleinian group and let $\mathbb{H}^3$ be the upper half-space model for hyperbolic 3-space. Then $\mathbb{H}^3/\Gamma$ is an orientable hyperbolic 3-orbifold (with the group action defined in the usual way). A common goal in working with such things is to find canonical decompositions for classes of these orbifolds, for instance so that we may estimate their volume.

If $D\subset\mathbb{H}^3$ is discrete, infinite, and $\Gamma$-invariant, then $D$ can be used as the set of vertices in a polyhedral decomposition of $\mathbb{H}^3$ which passes to a polyhedral decomposition of $\mathbb{H}^3/\Gamma$. In particular, if $\Gamma$ is non-elementary then we can achieve such a $D$ by taking the orbit under $\Gamma$ of some point in $\mathbb{H}^3$, and in that case we even inherit an algorithm for how to draw the edges and faces of the decomposition.

If it's a canonical decomposition we are after, why not just do this using the point $(0,0,1)$? Not only is it the canonical choice but it's not hard to compute points in its orbit. I guess my question is, since this seems like an easy solution, what is stopping people from doing it this way? Is there some reason why this is not useful? Are there obstructions I'm overlooking?

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  • $\begingroup$ In what sense is $(0, 0, 1)$ the canonical choice? What happens if it has non-trivial stabilizer? $\endgroup$ – Igor Rivin Apr 24 '15 at 12:57
  • $\begingroup$ I see it as canonical because it's the only point in the space (not on the boundary) with one one and zeros otherwise. If it has non-trivial stabilizer, the orbit is still discrete, infinite, and $\Gamma$ invariant, provided $\Gamma$ is non-elementary. If that causes problems with how to assign faces/edges, well it so happens that only elliptical elements fix that point so that would not be an issue for any manifold. $\endgroup$ – j0equ1nn Apr 25 '15 at 21:00
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    $\begingroup$ It is canonical, if you're after a classification of discrete subgroups of $PSL_2(\mathbb{C})$. But Kleinian group theorists usually study such groups up to conjugacy, in which case this does not make a canonical choice for a conjugacy class. One may single out finitely many orbits of points (if it's cofinite volume) by taking points of maximal injectivity radius, for example. $\endgroup$ – Ian Agol Apr 25 '15 at 22:09
  • $\begingroup$ Thanks Ian, that's true because what I said starts with a fixed matrix representation for the group. I'm curious now about how the decomposition would change as one varies representatives of the conjugacy class but sticks with this same point to take the orbit of. Could that give a canonical matrix representation to use? $\endgroup$ – j0equ1nn Apr 26 '15 at 16:59
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No. (See Ian Agol's comment.)

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