Horosphere of hyperbolic space in Minkowski space

Question

In their paper about decomposition of cusped hyperbolic manifold Epstein and Penner state that given a point $x\in L$ (where with $L$ they denote the positive component of the light-cone) that the subset of Minkowski space given by: $O_x=\{y\in I^n\;|\;-1=\langle x,y\rangle\}$ is a horosphere centered in the point at infinity identified with the ray of $x$. My question is basically this one (Light cone as the space of horospheres of $\mathbb{H}^n$) but I don't fully understand the argument given in the answer. What is not immediately obvious to me is what follows: when he acts transitively by choosing the plane span($e_1,e_2$) he is just checking the intersection of one line and the horosphere, why should it be trivial that it generalizes immediately? I guess that what he is saying is that by working with transformation in $O(n,1)$ and preserving the ray $x$ he preserves the point at infinity, the thing is that I don't see why it should preserve length on such a line (from this it would follow that it also preserves the horosphere and I think I could finish from this). Basically, I can work out the simplified example, but I am not sure if I understand why I can reduce myself to such a case. Thanks in advance, and please let me know if something is not clear in my question.

Answer 1

I guess that what he is saying is that by working with transformation in $O(n,1)$ and preserving the ray $x$ he preserves the point at infinity, the thing is that I don't see why it should preserve length on such a line.

Distance in the hyperbolid model is defined in terms of the inner product of the Minkowski space, while Lorentz transformations are defined as those that leave that inner product unchanged. So the preservation of length is a direct consequence of the choice of transformation group.

when he acts transitively by choosing the plane $\operatorname{span}(e_1,e_2)$ he is just checking the intersection of one line and the horosphere, why should it be trivial that it generalizes immediately?

I see the answer from Malkoun talking the plane $\operatorname{span}(e_1,e_{n+1})$, not $\operatorname{span}(e_1,e_2)$. Typo?

I would think about this as a sequence of individual transformations. Lorentz transformations if you focus on the model, hyperbolic isometries for a more model-agnostic terminology. Given any horosphere with any ideal point as its center, and any geodesic converging on that center, you can apply these steps to get to the described situation.

1. Perform a rotation which fixes the $e_{n+1}$ axis, and which puts the center into the plane $\operatorname{span}(e_1,e_{n+1})$ with a positive first coordinate. Since the time-like dimension is fixed, this is both a hyperbolic rotation and a Euclidean rotation. For $n=2$ that rotation is uniquely defined, but for higher dimensions there is some variations there about how you orient things. The choice doesn't really matter.

2. Perform a limit rotation which fixes the center of the horosphere and moves the line in question into the plane $\operatorname{span}(e_1,e_{n+1})$. A limit rotation will preserve the horosphere, and can be defined to move any single point on the horosphere to any other. For $n=2$ this would again be uniquely defined, with more leeway for higher dimensions.

3. Perform a hyperbolic translation along the geodesic that corresponds to the plane $\operatorname{span}(e_1,e_{n+1})$. This will fix your line, but will change the position of the horosphere in your model. In the embedding space this results in a parallel translation of the hyperplane defining the horosphere. (Don't get me wrong: the Lorentz transformation for this translation is not a parallel translation of the whole hyperboloid. It's just that when you compare horosphere before and after the translation that you find them to lie in parallel hyperplanes.) This is where the choice of $x=e_1+e_{n+1}$ instead of one of its multiple comes into play: by choosing that specific vector, the answer fixes one specific hyperplane, thus effectively using the result of this translation.

Each of these transformations was an operation which preserves hyperbolic lengths and angles. Taken together transformed the fully generic initial configuration into the final position that the answer uses to demonstrate orthogonality. So the proof is without loss of generality: the property shown in the special case must also hold in the general case because the transformations between these cases preserve that property of orthogonality.

If I find the time, I might provide some illustrations of the transformation steps using the Poincaré disk model, since I don't have tools at hand to quickly get a 3d illustration for the hyperboloid case. But if someone else wants to illustrate this, please be my guest.