3
$\begingroup$

A matrix $X=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathrm{PSL}_2(\mathbb{C})$ acts isometrically on the upper half-space model $\mathbb{H}^3$ via isometric extension of the Mobius transformation on $\widehat{\mathbb{C}}=\partial\mathbb{H}^3$, which is defined by $X(z)=\dfrac{az+b}{cz+d}$. $X$ is called loxodromic if $tr(X)=a+d\in\mathbb{C}\smallsetminus[-1,1]$. The action of such an isometry is by translation along and rotation about a unique invariant axis in $\mathbb{H}^3$. This action can be described completely by a specification of

  1. a pair of fixed points on $\widehat{\mathbb{C}}$ (the axis being the geodesic between these),
  2. the translation length, and
  3. the angle of rotation.

For 1: The fixed points can be directly solved for in terms of the entries, and in fact this is related to the classification of the type of isometry by trace, using the characteristic polynomial. As far as I know the expression in $a,b,c,d$ however is not linked to a convenient matrix property. Is it?

For 2: The translation length can be computed from the trace. In fact it's $2\mathrm{cosh}^{-1}(\frac{tr(X)}{2})$ where if necessary we use the complex logarithm.

For 3: What about the rotation amount? If we only know the trace, can we determine this? If not, is there some other nice minimal matrix property that gives this information? My feeling is that if $tr(X)=x+iy$ where $x,y\in\mathbb{R}$ then we should have something like a translation length determined by $x$ and a rotation angle determined by $y$.

If this is well-known, forgive me, I am not finding it in my references. If this is the case please feel free to migrate to stackexchange.

$\endgroup$
  • $\begingroup$ I haven't checked any references. But I think that one just has to compute the derivative of $X$ at one of the fixed points on the boundary. The angle is its phase. If you want a right handed screw-motion, then you should take the angle at the attractive fixpoint. $\endgroup$ – Sebastian Goette Dec 24 '15 at 14:01
2
$\begingroup$

Complex translation length $\lambda$ is given by $tr X = 2\cosh \lambda,$ where $\Re \lambda > 0,$ and is the translation length, whilst the imaginary part is the rotation angle.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I found a reference for this: Maclachlan-Reid (2003), p.372. $\endgroup$ – j0equ1nn Jan 1 '16 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.