I was wondering if someone could explain the interpretation of the following results. In hyperbolic geometry, we say that lengths are invariant under the action of Mob($\mathbb{H}$) if given any piecewise-differentiable curve $f:[a,b]\rightarrow \mathbb{H}$ and an element $\gamma \in $ Mob($\mathbb{H})$

$length_\rho (f) = length_\rho (\gamma \circ f)$,

namely the arclength of a curve $f$ in $\mathbb{H}$ with respect to some metric $\rho(z)$ is the same even after $\gamma$ has been applied to it. So I am looking for some conditions on this metric $\rho$, and since I know that $length_\rho (f) = \int_f \rho(z) |dz|$, after doing some manipulations I arrive at the condition

$\rho(z) - \rho\Big(\gamma(z)\Big) |\gamma'(z)| = 0$.

Now I know that the group of all Möbius transformations that preserve $\mathbb{H}$ is generated by $g(z) = az+b$, $a,b \in \mathbb{R}$, $a>0$, $h(z) = -\frac{1}{z}$ and $B(z) = -\bar{z}$.

Now we look at the case when $\gamma$ is a translation by $b$ units, and we arrive at the fact that

$\rho(z) = \rho(z+b)$, namely the metric $\rho(z)$ in $\mathbb{H}$ is invariant under any translation by a real number, and so dependso only on $Im(z)$. Doing next with $\gamma = az$, we get the condition combined with the previous one that

$\rho(z) = \frac{c}{Im(z)}$, where $c$ is some constant. We can show that this form of $\rho(z)$ is consistent with $h(z) = -\frac{1}{z}$ and $B(z) = -\bar{z}$, but we'll assume that for now.

So here's the question. Does this result mean that if lengths in the hyperbolic plane are invariant under Möbius transformations, then they are precisely the ones that have the element of arc length as $\frac{c}{Im(z)} |dz|$?

But then I can arrive at this metric for the hyperbolic plane without invoking Möbius transformations, namely looking at the pseudosphere and see how if I try to map this pseudo sphere into something flat. A natural coordinate system to choose for the pseudo sphere would be coordinates $(\theta, \rho)$, namely latitude and longitude. So say for a given latitude the radius of my pseudo sphere is $R$ then the length it subtends on the surface by an angle $d\rho$ is $R d\rho$. But if we map it to something flat then the length is just $d\rho$, so the lengths must have been "shrunk" by a factor of $R$. Further investigation reveals that I can arrive at the same hyperbolic metric without invoking Möbius transforms.

What's the Connection?


  • $\begingroup$ So $\rho$ is your metric, but then what is $\rho(z)$? And what model of the hyperbolic plane are you using? $\endgroup$ – user641 Apr 14 '11 at 4:28
  • $\begingroup$ @SteveD, sorry should not have said that, but $\pho(z) dz$ is some element of arc length, and $\rho(z)$ is the amount of " scaling". I am using the upper half plane model. $\endgroup$ – user38268 Apr 14 '11 at 4:56
  • $\begingroup$ Sorry I meant $\rho (z)$. $\endgroup$ – user38268 Apr 14 '11 at 5:08
  • $\begingroup$ When you speak about "the pseudosphere", what do you have in mind? Do you mean a certain spindle-like rotational surface in three-space or an abstract manifold? $\endgroup$ – Christian Blatter Apr 14 '11 at 8:51
  • $\begingroup$ @ChristianBlatter No not an abstract manifold, i'm not that advanced. It's a surface of revolution lying in $\mathbb{R}^3$ obtained by rotating a tractix about the $z-axis$. $\endgroup$ – user38268 Apr 14 '11 at 10:39

Let us begin with the pseudosphere: Classical differential geometry deals with smooth surfaces in $3$-space where lengths are lengths of curves lying on the surface. The theory develops various notions of curvature, and in due course the question arises whether there is a surface of revolution having constant Gaussian curvature $\kappa=-1$. Indeed there is: You have to rotate a tractrix about the $z$-axis. The resulting spindle-like surface $S$ is called a pseudosphere but looks nowhere like an ordinary sphere. In particular it has a "rim" along the equator (one of the principal curvatures is $\infty$ there), and this implies that $S$ is incomplete: There are geodesics on $S$ that cannot prolonged forever but brutally end on the rim. One more thing: $S$ has the topology of a cylinder extending to infinity at one end; therefore there are loops on $S$ which cannot be contracted to a point.

Now the hyperbolic plane $H$: It is the simplest domain on ${\mathbb C}$ one can think of. Furthermore in the realm of ${\mathbb C}$ one has the notion of conformal map. So the next question is: What are the conformal self-maps of $H$? By means of Schwarz' Lemma one finds the group $G$ of Moebius transformations you have listed. This group is richer than the groups of euclidean self-maps of $H$. Is there something (like angles, areas and the like) that is preserved by all maps $f\in G$? It turns out that one such invariant is the so-called hyperbolic line element $ds:={|dz|\over y}$, where $z=x+i y$, $y>0$. Now $H$ provided with this line element (or conformal metric $\rho(z)={1\over y}$) is what one nowadays calls a Riemannian manifold on which we can do differential geometry as we did before on surfaces in $3$-space. In particular one can compute the Gaussian curvature of $H$. Since the Moebius group is transitive on $H$ the local geometry around each point is the same for all points, therefore the Gaussian curvature is constant. The computation gives $\kappa\equiv-1$. But this $H$ has a much richer structure than the pseudosphere $S$; e.g., it is complete and allows of all sorts of regular tessellations (whereas the ordinary sphere admits only the five Platonic solids).

Since $S$ and $H$ both have constant Gaussian curvature $-1$ there has to be a connection between them. As you have indicated in your question you have constructed a (local) isometry $\psi:\ S\to H_1$ between $S$ and a part $H_1$ of $H$ such that to a rotation $\phi\mapsto \phi+ \delta$ of $S$ corresponds the translation $z\mapsto z+\delta$ of $H_1$. You should look at the inverse $\psi^{-1}:\ H_1\to S$ of this map. It is defined on all of $H_1:=\{z\in H\ |\ y\geq1\}$ and makes $H_1$ the so-called universal cover of $S$, insofar as any two points $z$ and $z+2k\pi$, $\ k\in{\mathbb Z}$, of $H_1$ map onto the same point on $S$. What one has gained is this: The covering surface $H_1$ is simply connected whereas $S$ is not.


Your Answer

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