Last Lecture …

Hyperbolic tilings

In the last lecture we had a look at tilings of the hyperbolic plane. We want to tile the plane with regular $p$-gons, such that at each vertex $q$ polygons meet. Then we do a little calculation and see that for $\frac1p+\frac1q < \frac12$ there exist such tilings of the hyperbolic plane.

These tilings can be generated by a software written by Martin von Gagern from TU Munich. It is written in Java and can be downloaded from his website. I created an Escher like picture with it 🙂

HeavenHell Continue reading

End of term get-together

Hello everybody,

Tomorrow, Thursday the 14th, we will meet at 7 pm at “Cafe A” to have a good time together and welcome the end of the semester. Cafe A is the students cafe at the ground floor of the architecture building.

Everybody is welcome and I’m looking forward to seeing you tomorrow,

Lecture 26-28

Sorry, but I didn’t find the time to blog the lectures so far but I scanned my notes. Sorry for the bad quality but I will try to improve it. But for the moment these are the available scans:

Lecture_26   Lecture_27  Lecture_28

Comment … I just printed one of the files and it looks terrible but somehow scanning pencil written notes is not as easy as I expected …

Lecture26 Lecture27 Lecture28

Now it’s much better – I hope 🙂


The certificates for those who have the required amount of points in their homework are available at our secretary’s office MA 873 starting on Tue 12.02 (office hours 9:30-11:30). With the certificate you can register for the oral exam at the Prüfungsamt and sign up for a date at the secretary’s office.

Available dates so far are 21.02. and 26.02.2013. Additional dates will available before shortly before the start of the next semester.

Lecture 25

Theorem (Hyperbolic Pythagoras).
Let $\triangle ABC$ be a hyperbolic triangle with angles $\alpha$, $\beta$ and $\gamma=\frac\pi2$.
Then the lengths of the edges $a$, $b$ and $c$ satisfy
\[\cosh c=\cosh a\cdot\cosh b\,.\]


We calculate the length in the upper half-plane model after normalizing the triangle to the following:


We use the formula derived for the hyperbolic distance in the upper half-plane using complex variables:

\cosh d(z_1,z_2)-1
=\frac{\lvert z_1-z_2\rvert^2}{2\operatorname{Im}z_1\operatorname{Im}z_2}\,,

So for the length of the edges of the triangle we obtain:
\cosh a&=1+\frac{\lvert B-C\rvert^2}{2\operatorname{Im}B\operatorname{Im}C}\\
\cosh c&=1+\frac{\lvert\mathrm e^{\mathrm i\phi}-k\mathrm i\rvert^2}{2k\sin\phi}\\
&=1+\frac{\left(\mathrm e^{\mathrm i\phi}-k\mathrm i\right)\left(\mathrm e^{-\mathrm i\phi}+k\mathrm i\right)}{2k\sin\phi}\\
\cosh b&=\frac{1^2+1}{2\cdot1\cdot\sin\phi}\\

This calculation implies the required equality.

What is the relation of hyperbolic and Euclidean triangles?

Looking at “small” hyperbolic triangles (i.e. triangles with small edge lengths and area) hyperbolic triangles behave similar to Euclidean triangles.
If the area is very small, then the sum of its interior angles is almost $\pi$.
The hyperbolic Pathagoras’ Theorem becomes the Euclidean Pythagoras’ Theorem for small edge lengths:
The Taylor series for $\cosh$ is $\cosh x=\sum\frac{f^{(k)}(0)}{k!}x^k=1+\frac12x^2+\dotsb$.
Thus $\cosh c=\cosh a\cdot\cosh b$ becomes
which gets $c^2=a^2+b^2$ for small $a$, $b$ and $c$.

Hyperbolic motions in $H_+^2$

Consider the following maps:
\[g\colon\hat\C\to\hat\C\,,\ z\mapsto\frac{az+b}{cz+d}\,,\]
where $ad-bc>0$ and $a,b,c,d\in\R$.
Here, $g$ maps $\infty$ to $\frac ac$ and $-\frac dc$ to $\infty$.
\[\tilde g\colon\hat\C\to\hat\C\,,\ z\mapsto\frac{-a\bar z+b}{-c\bar z+d}\,.\]

    • The maps $g$ and $\tilde g$ are bijective (the inverse is just the transformation corresponding to $\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}$).
    • These maps preserve the upper half-plane:
      We have
      &=\frac1{2\mathrm i}\left(\frac{az+b}{cz+d}-\frac{a\bar z+b}{c\bar z+d}\right)\\
      &=\frac1{2\mathrm i\lvert cz+d\rvert^2}\left(ac\lvert z\rvert^2+adz-ac\lvert z\rvert^2-ad\bar z+bc\bar z+bd-bcz-bd\right)\\
      &=\frac{ad-bc}{2\mathrm i\lvert cz+d\rvert^2}(z-\bar z)\\
      &=\frac{ad-bc}{\lvert cz+d\rvert^2}\operatorname{Im}z>0\,.
      Similarly for $\tilde g$.
    • These maps are isometries:
      \frac{\lvert g(z_1)-g(z_2)\rvert^2}{2\operatorname{Im}g(z_1)\operatorname{Im}g(z_2)}
      &=\frac{\lvert cz_1+d\rvert^2\lvert cz_2+d\rvert^2}{(ad-bc)^2}\frac1{\operatorname{Im}z_1\operatorname{Im}z_2}\frac{(ad-bc)^2\lvert z_2-z_1\rvert^2}{2\lvert cz_1+d\rvert^2\lvert cz_2+d\rvert^2}\\
      &=\frac{\lvert z_1-z_2\rvert^2}{2\operatorname{Im} z_1\operatorname{Im}z_2}\,.
      Again, a similar calculation for $\tilde g$ yields that $\tilde g$ is also an isometry.
    • Hyperbolic lines are mapped to hyperbolic lines.

An arbitrary hyperbolic isometry of $H_+^2$ is of the form
\[z\mapsto\frac{az+b}{cz+d}\quad\text{or}\quad z\mapsto\frac{-a\bar z+b}{-c\bar z+d}\,,\]
where, again, $ad-bc>0$ and $a,b,c,d\in\R$.

Let $\phi\colon H_+^2\to H_+^2$ be an isometry. We can construct a map $g$ which is one of the following, such that $g\circ\phi$ maps the imaginary axis onto itself and $g\circ\phi(\mathrm i)=\mathrm i$ and $g\circ\phi((0,\mathrm i))=(0,\mathrm i)$:

      • $z\mapsto\frac{z-u}v$ with $\phi(\mathrm i)=u+\mathrm i v$,
      • $z\mapsto\frac v{\bar z-u}$,
      • $z\mapsto\alpha\frac{z-x}{z-y}$ where $\phi(0)=x$, $\phi(\infty)=y$.


Then $g\circ\phi$ is an isometry, since $g$ and $\phi$ are. Further $g\circ\phi$ is the identity on the imaginary axis, as it is isometric and maps the imaginary axis onto itself preserving the intervals $(0,\mathrm i)$ and $(\mathrm i,\infty)$.
Now consider $x+\mathrm i y$ with $g\circ\phi(x+\mathrm i y)=u+\mathrm i v$.
We have $d(k\mathrm i,x+\mathrm i y)=d(g\circ\phi(k\mathrm i),u+\mathrm i v)=d(k\mathrm i,u+\mathrm i v)$ for all $k\in\R^+$.
\[\frac{\lvert k\mathrm i-(x+\mathrm i y)\rvert^2}{2ky}=\frac{\lvert k\mathrm i-(u+\mathrm i v)\rvert^2}{2kv}\]
and hence
for all $k$.
This implies $y=v$ and $u=\pm x$.
So $g\circ\phi$ is either the identity or it the reflection with respect to the imaginary axis, $z\mapsto-\bar z$.
As $\phi=g^{-1}\circ g\circ\phi$, $\phi$ is of the given type.