The next dates for Oral Exams in Geometry I are Friday 31.05 and 05.07.2013. You can register for the exam in the secretary’s office (Mathias Kall, MA 873).

# Author Archives: Thilo Rörig

# Orals exams on 03.05.2013

The next date for Oral Exams in Geometry I is Friday 03.05.2013. You can register for the exam

in the secretary’s office (Mathias Kall, MA 873) starting from Monday 15.04.2013.

# 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 🙂

# 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 …

Now it’s much better – I hope 🙂

# Exams

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 21 added

Sorry for the delay, but I didn’t find the time earlier to add Lecture 21.

# Lecture 21

**Stereographic projection**

**Definition. **Denote the unit sphere in $\mathbb{R}^{n+1}$ by $\mathbb{S}^{n}=\left\{{x \in \mathbb{R}^{n+1}|(x,x)=1}\right\}$ and its north pole by $\mathbf{N}=e_{n+1}$.

The *stereographic projection* is a map $\sigma\colon \mathbb{S}^{n}\to\mathbb{R}^{n} \cup \left\{{\infty}\right\}$ from $\mathbf{N}$ to the plane

$E=\left\{{x \in \mathbb{R}^{n+1}|x_{n+1}=0}\right\} \cong \mathbb{R}^{n}$ through the equator:

\[

\sigma(X)=

\left\{

\begin{array}{ll}

\infty & \textrm{, } X = \mathbf{N}\\

l_{NX}\cap E & \textrm{, } X \neq \mathbf{N}\\

\end{array}

\right.

\]

Analytically $\sigma$ is given by

\[

\sigma(X)= \sigma\ \left(\begin{pmatrix}

x_1\\\vdots\\x_{n+1}

\end{pmatrix} \right)

= \frac{1}{1-x_{n+1}}\begin{pmatrix}

x_1\\\vdots\\x_{n+1}

\end{pmatrix}

\]

since

\[N+\lambda(X-N)=e_{n+1}+\lambda \begin{pmatrix}

x_1\\\vdots\\x_{n+1}

\end{pmatrix}

= \begin{pmatrix}

y_1\\\vdots\\y_{n}\\0

\end{pmatrix}

\Rightarrow \lambda=-\frac{1}{x_{n+1}-1}=\frac{1}{1-x_{n+1}}, X\neq N.

\]

**Sphere inversion**

**Definition. **The *sphere inversion* $i_{M,r}\colon \mathbb{R}^{n} \cup \left\{{\infty}\right\} \to \mathbb{R}^{n} \cup \left\{{\infty}\right\}$ in the sphere

$\mathbf{S}_{M,r}=\left\{{x \in \mathbb{R}^{n}|\left\|x-M\right\|=r}\right\}$ is given by the following conditions:

- $X’$ lies on the ray $\overrightarrow{MX}$,
- $\left\|M-X\right\|\left\|M-X’\right\|=r^2$,
- $M\longleftrightarrow\infty$.

**Properties of sphere inversions**

- $i_{M,r}$ is an involution,
- $\mathbf{S}_{M,r}$ is fixed,
- affine subspaces containing M are mapped onto themselves.

**Lemma.**

**Proof.**

Since

\[

\left\|M-A\right\|\left\|M-A’\right\|=r^2 = \left\|M-B\right\|\left\|M-B’\right\| \iff

\frac{\left\|M-A\right\|}{\left\|M-B\right\|}=\frac{\left\|M-B’\right\|}{\left\|M-A’\right\|}

\]

we get that $\Delta AMB$ and $\Delta B’MA’$ are similar. In particular:

\[

\measuredangle MA’B’=\measuredangle ABM.

\measuredangle MAB=\measuredangle A’B’M.

\]

$\Box$

**Corollary.** Circles through M are mapped to lines. In particular, the line is parallel to the tangent to the circle at M.

**Remark. **All lines are thought to contain the point $\infty$ and are considered as circles of infinte radius.

**Corollary. **The stereographic projection is the restriction of the sphere inversion $i_{M,r}$ with $M=e_{n+1}$ and $r=\sqrt{2}$.

**Proof.**

- The equator of $\mathbb{S}^{n}$ is fixed.
- $\mathbb{S}^{n}$ is mapped to a plane (Corollary) through the equator (1).
- Both inversion and stereographic projection preserve the rays.
- The intersection points of a ray with $\mathbb{S}^{n}$ and $\mathbb{R}^{n}$ are unique, hence $i_{M,r}\Big|_{\mathbb{S}^{n}}=\sigma$.
- $N=M\longrightarrow\infty$ by $i_{M,r}$.

$\Box$

**Theorem.** An inversion in a sphere:

- maps spheres/hyperplanes to spheres/hyperplanes,
- is a conformal map, i.e. it preserves angles.

**Proof.**

(i) Consider the following sketch:

Hence the angles $\alpha$ and $\beta$ satisfy

\[

(\pi-\alpha)+\beta+\frac{\pi}{2}=\pi \Rightarrow \alpha-\beta=\frac{\pi}{2}.

\]

(ii) Conformality follows from the following picture

$\Box$

**Remark. **In general, the center of a sphere is *not* mapped to the center of its image by an inversion.

**Corollary.**

- The stereographic projection maps spheres through the north pole to hyperplanes and spheres not through the north pole to spheres.
- All spheres/hyperplanes are images of spheres on $\mathbb{S}^{n}$ under $\sigma$ (because $i_{0,1}$ is an involution).
- $\sigma$ is conformal.

**Formula for the sphere inversion.**

Consider the second figure showing the general idea of a sphere inversion. Since $X’$ lies on the ray $XM$ we have $X’=M+\lambda(X-M), \lambda > 0$ and from the second requirement for the sphere inversion we obtain

\begin{align*}

&&\underbrace{\left\|X’-M\right\|}_{X’-M=\lambda(X-M)} \left\|X-M\right\|&=r^2\\

\Rightarrow&& \lambda \left\|X-M\right\| \left\|X-M\right\|&=r^2\\

\Rightarrow&& \lambda = \frac{r^2}{\left\|X-M\right\|^2}.

\end{align*}

Hence the sphere inversion is given by

\[

i_{M,r}=M+\frac{r^2}{\left\|X-M\right\|^2}(X-M).

\]

# Lecture 18

Theorem.The shortest piecewise continously differentiable curve in $H^n$ connecting two points $p$ and $q$ is the hyperbolic line segment between them. It’s length is $$d(p,q) = arcosh(-\langle p,q\rangle).$$

# Lecture 15

### Pole-Polar Relationship

- b is non-degenerate symmetic bilinear form (e.g.scalar product on $\mathbb{R}^n$)
- orthogonal complement of a vector subspace $U\subseteq V$

\[U^{\perp}= \left\lbrace v\in V\mid b\left(u,v\right) = 0,\forall u\in U\right\rbrace \] - $dimU^{\perp} + dim U = dim V$, but $U^{\perp} \oplus U \neq V$.
- If $P\left( U\right)$ is a projective subspace of $P\left(V\right)$,then $P\left(U^{\perp}\right)$ yields another subspace of $P\left(V\right)$.
- in $\RP^2 = P(\mathbb{R}^3)$ we obtain:

Point $\left[ P\right] \leftrightarrow$ line $P\left( \left\lbrace p\right\rbrace ^{\perp}\right)$

Line $P\left( U\right)\leftrightarrow$ point $P\left(U^{\perp}\right)$

# Lecture 14

### Polarity: Pole – Polar Relationship

Let $b: V \times V \rightarrow \mathbb{F}$ be a symmetric non-degenerate bilinear form.

**Definition.** Let $U \leq V$ be a vector subspace. Then

\[

U^{\perp} = \{ v \in V \: | \: b(u,v) = 0, \: \forall u \in U \}.

\]

If $U = \{u_1,…,u_k\}$ then

\[

U^{\perp} = ( \text{span} \; U )^{\perp} = \{ v \in V \: | \: b(u_i,v) = 0, \: \forall i = 1,…,k \}.

\]

We call $U^{\perp}$ the *orthogonal complement* of $U$.

**Remark.** dim $U^{\perp} =$ dim $U^0 = n – k$, if dim $U = k$ and dim $V = n$.