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?



Does this count as a canonical decomposition for non-elementary hyperbolic 3-orbifolds?
No. (See Ian Agol's comment.)


