2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – j0equ1nn
    Commented Dec 7, 2017 at 16:26

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$
    – j0equ1nn
    Commented Dec 7, 2017 at 20:39
  • $\begingroup$ @j0equ1nn The two endpoints you mention in the OP give two (light-like) vectors. Their span is your geodesic. $\endgroup$
    – Igor Rivin
    Commented Dec 7, 2017 at 21:34
  • $\begingroup$ 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 $\mathcal{I}^3\cup\big(\big\{p\in\mathbb{R}^{3,1}\mid\mathrm{n}(p)=0\big\}/\sim\big)$. $\endgroup$
    – j0equ1nn
    Commented Dec 7, 2017 at 21:59
  • $\begingroup$ It's in the link in my answer. $\endgroup$
    – Igor Rivin
    Commented Dec 7, 2017 at 22:27
  • $\begingroup$ Ah okay, I see what you mean. Thanks. $\endgroup$
    – j0equ1nn
    Commented Dec 7, 2017 at 23:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .