# Find the fixed geodesic of an orientation-preserving isometry of the $3D$ hyperboloid model

Let $\mathcal{I}^3\subset\mathbb{R}^4$ be the standard hyperboloid model for hyperbolic $3$-space and consider the usual $\mathrm{SO}(3,1)$ action of $\mathrm{PSL}(2,\mathbb{C})$ on $\mathcal{I}^3$. Given a matrix $\gamma=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\in\mathrm{PSL}(2,\mathbb{C})$, there is a unique geodesic $g\in\mathcal{I}^3$ preserved by the action of $\gamma$. Particularly, $\gamma$ translates along and rotates around $g$ (and these quantities can be written in terms of the trace $a+d$).

I want to know how to determine the axis $g$ in terms of the entries of $\gamma$. One way to go about this would be to use the fact that $g$ is the intersection $P\cap\mathcal{I}^3$ where $P$ is a uniquely determiend Euclidean plane passing through the origin in $\mathbb{R}^4$. Perhaps we can write down two vectors in terms of $a,b,c,d$ that span this plane, using the fact that the extended action of $\gamma$ to $\mathbb{R}^4$ preserves the plane.

By comparison, in the upper half-space model $\mathcal{H}^3$, the fixed godesic will be the Euclidean half-circle or half-line orthogonal to $\partial\mathcal{H}^3$ that connects the points $\dfrac{a-d\pm\sqrt{(a-d)^2+4bc}}{2c}$, where we interpret these as $0$ and $\infty$ when $c=0$. But in the hyperboloid model, points on the boundary are not so accessible. So how do we describe the fixed geodesic there?

• Thank you to @BenoîtKloeckner for commenting the answer to my preliminary question about this, which I've deleted after seeing the answer was very obvious. – j0equ1nn Dec 7 '17 at 16:26

The fixed points in the upper half space model map to vectors on the lightcone, which span the plane that intersects with the $\mathcal{I}^3$ at $g$. The details to compute this are here.
• That works fine for points but is awkward to do on an entire geodesic in the upper half-plane. I'm looking for a way of characterizing the plane in $\mathbb{R}^4$ that intersects with $\mathcal{I}^3$ to give $g$. Since this depends only on the matrix entries, there should be some function $\mathrm{PSL}(2,\mathbb{C})\rightarrow\mathbb{R}^4\times\mathbb{R}^4$ that outputs this pair of vectors. – j0equ1nn Dec 7 '17 at 20:39
• That makes sense, but let's say I give you a specific matrix. Can we say exactly what the pair of light like vectors are? I have not found any literature on a conformal map from $\overline{\mathcal{H}^3}$ to $\mathcal{I}^3\cup\big(\big\{p\in\mathbb{R}^{3,1}\mid\mathrm{n}(p)=0\big\}/\sim\big)$. – j0equ1nn Dec 7 '17 at 21:59