[Math] Classificating the isometries of the Hyperbolic Plane

hyperbolic-geometryisometrylinear algebra

I just did the classification of the isometries of the Euclidean Plane using some results of linear algebra and euclidean geometry, guiding myself through this article:

http://repositorio.ul.pt/bitstream/10451/8852/1/ulfc104259_tm_Fernando_Bacalhau.pdf
( pages 25 to 38, sorry for it to be in portuguese, but I believe it is the standard proof using linear algebra).

The theorem says that every isometry of the euclidean plane is exactly one of the following:
(i) A translation by a vector.
(ii) A rotation.
(iii) A reflection through a line.
(iv) A glide reflection through a line and glided by a vector.

And then my teacher asked me to study and try to classify the isometries of the Hyperbolic plane in the same way, using the matrices of the Möbius transform.

I then proved that the isometries of the hyperbolic plane are Möbius Transforms, but I don't see how to apply this in the structure of the euclidean classification I did before. Will those results of linear algebra for the inner product still be valid in $\mathbb{H}^2$ ? How can I start this classification with Möbius transforms?

Any help would be appreciated.
Thanks.

Best Answer

Here's a way to approach this problem from the linear algebra point of view. And in this answer I'm going to side-step the issue of orientation reversing isometries, which are not represented by Möbius transformation, instead by anti-Möbius transformations. So, I will explain how to use linear algebra to classify orientation preserving (o.p.) isometries, equivalently to classify Möbius transformations.

The classification of o.p. isometries is invariant under conjugacy. To see some examples of this from a synthetic point of view (i.e. no linear algebra), consider an o.p. isometry $\phi$, another o.p. isometry $\psi$, and the conjugate o.p. isometry $\psi \phi \psi^{-1}$.

First, if $\phi$ has a fixed point $x$ then $\psi \phi \psi^{-1}$ also has a fixed point, namely $\psi(x)$. Proof: $$(\psi \phi \psi^{-1})(\psi(x)) = \psi \phi(x) = \psi(x) $$ Furthermore, if $\phi$ rotates by an angle $\theta$ around $x$ then $\psi$ also rotates by an angle $\theta$ around $\phi(x)$. Proof: take a ray $R$ based at $x$, and the angle from $R$ to $\phi(R)$ at $x$ equals $\theta$; so the ray from $\psi(R)$ to $\psi(\phi(R))$ at $\psi(x)$ equals $\theta$.

Second, and with similar proofs, if $\phi$ fixes some geodesic line $L$ then $\psi \phi \psi^{-1}$ also fixes a line, namely $\psi(L)$. Furthermore, if $\phi$ translates a distance $d$ along $L$ then $\psi \phi \psi^{-1}$ translates a distance $d$ along $\psi(L)$.

Third, a parabolic transformation is one which fixes a single point on the circle at infinity and fixes no finite point.

Here's where things get interesting. Now you have to convince yourself of two further issues.

  1. Every o.p. isometry either fixes a point and rotates about it by some angle, or fixes some geodesic line and translates along that line by some distance, or is a parabolic isometry, or is the identity.
  2. Any two o.p. isometries which are rotations by the same angle, or are translations by the same distance, are conjugate. A similar statement holds for parabolic isometries, which I'll hold off until the end.

It is possible to prove this synthetically, but Möbius transformations certainly help. Here's an outline.

As usual we represent Möbius transformations by elements of $PSL(2,\mathbb{R}) = SL(2,\mathbb{R}) / \pm I$. Here's some algebra facts which you can easily verify.

  • Given $M \in SL(2,\mathbb{R})$, $\text{trace}(M)$ is a conjugacy invariant in $SL(2,\mathbb{R})$. Also, its absolute trace $|\text{trace}(M)|$ is a conjugacy invariant in $PSL(2,\mathbb{R})$.
  • If $|\text{trace}(M)|$ is not equal to 2 then it is a complete conjugacy invariant, i.e. if $M_1,M_2$ have the same absolute trace not equal to 2 then they are conjugate in $PSL(2,\mathbb{R})$. The case of absolute trace equal to $2$ corresponds to parabolic transformations and is not quite as clean; I'll skip that until the end.

Now, blend the geometry and the algebra: for each value of the absolute trace that is not equal to $2$, construct an example, verify that it is one of the isometries in the above list, and verify that every distinct isometry in the above list corresponds to exactly one value of the absolute trace. Having done that, the classification is complete! (Outside of the issue that I have ignored the parabolic case where absolute trace equals 2).

For example, absolute trace equal to zero is represented by $\pmatrix{0 & 1 \\ -1 & 0} : z \mapsto -\frac{1}{z}$ which is a rotation about the fixed point $0+1i$ in the upper half plane, with rotation angle $\pi$. More generally, absolute trace in the interval $[0,2)$ is always a rotation about a fixed point, with angles varying over $(0,\pi]$ by a simple strictly monotonic continuous function $[0,2) \mapsto (0,\pi]$ (whose formula some people remember but I never can).

For another example, absolute trace equal to $\frac{5}{2}$ is represented by $\pmatrix{2 & 0 \\ 0 & 1/2}$ which is a translation along the line $\text{RealPart}(z)=0$ by a distance $\ln(4)$. More generally, absolute trace in the inverval $(2,\infty)$ is always a translation along a line, with translation distance varying over $(0,\infty)$ by a simple strictly monotonic continuous function $(2,\infty) \mapsto (0,\infty)$ (again, with an explicit formula).

A few last words about absolute trace equal to $2$. Each of these represents either the identity or an o.p. parabolic isometry. In the full group of generalized Möbius transformations there is just one conjugacy class of parabolic isometry. But in $\text{PSL}(2,\mathbb{R})$ there are two conjugacy classes, one rotating "positively" around the fixed point at infinity and the other rotating "negatively".

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