Lecture 17

Hyperbolic Geometry

Lorentz Spaces

The vector space $\mathbb{R}^{p+q}$ with the bilinear form
$$ \left<x,y\right>_{p,q} = \sum_{i=1}^{p} x_i y_i – \sum_{i=p+1}^{p+q} x_i y_i $$
is the Lorentz-space $\mathbb{R}^{p,q}$. For our purposes the space $\mathbb{R}^{n,1}$ with scalar product (non-degenerate symmetric bilinear form)
$$ \left<x,y\right>_{n,1} = \sum_{i=1}^{n} x_i y_i – x_{n+1} y_{n+1} $$
is the most important case.

Reminder The orthogonal transformations are denoted by $$ O(p,q) = \{f \in GL(p+q,\mathbb{R}) \;|\; \left<f(v),f(w)\right>_{p,q} = \left<v,w\right> \;\forall v,w \in \mathbb{R}^{p+q}\}. $$
Example $\mathbb{R}^{2,1} \leadsto \mathbb{R}^3$ with $\left<x,y\right>_{2,1} = x_1y_1 + x_2y_2 – x_3y_3$

•  $\left<x,x\right>_{2,1} = x_1^2 + x_2^2 – x_3^2 = -1$

time-like vectors $\left<x,x\right>_{2,1} < 0$

•  $\left<x,x\right>_{2,1} = 0$

light-like vectors $\left<x,x\right>_{2,1} = 0$

•  $\left<x,x\right>_{2,1} = 1$

space-like vectors $\left<x,x\right>_{2,1} > 0$

Hyperbolic Spaces

Definition The $n$-dimensional hyperbolic space is
$$ H^n = \{ x\in \mathbb{R}^{n,1} \;|\; \left<x,x\right>_{n,1} = \sum_{i=1}^{n} x_i^2 – x_{n+1}^2 = -1,\; x_{n+1} > 0 \}. $$
The length of a curve $\gamma : [a,b] \rightarrow H^n$ is defined by
$$ \mathrm{length}(\gamma) = \int_a^b \sqrt{\left<\gamma'(t),\gamma'(t)\right>_{n,1}}\:dt.$$

Why is $\left<\gamma'(t),\gamma'(t)\right>_{n,1} > 0$?

The tangent space at a point $p \in H^n$ is the orthogonal complement $p^\perp$ of $p$, $\gamma : [a,b] \rightarrow H^n$ with $\gamma(t_0) = p$.
\begin{align*}\gamma \in H^n &\Longrightarrow \left<\gamma(t),\gamma(t)\right>_{n,1} = -1 \\&\Longrightarrow 2\left<\gamma(t),\gamma'(t)\right>_{n,1} = 0.\end{align*}
The scalar product of $p$ is $\left<p,p\right>_{n,1} = -1$. Construct a basis $\{p,b_1,\ldots,b_n\}$ with $\left<p,b_i\right>_{n,1} = 0$. Then the matrix of $\left<\cdot,\cdot\right>_{n,1}$ has the form
$$\left( \begin{array}{rc} -1 & 0 \\ 0 & \left( \begin{array}{cc} & \\ & \end{array} \right) \end{array} \right) \Longrightarrow \mathrm{signature}(\left<\cdot,\cdot\right>_{n,1}|_{p^\perp})\text{ is }(n,0).$$
$\leadsto$ On the tangent spaces of the hyperbolic space embedded in $\mathbb{R}^{n,1}$ we obtain a positive-definite scalar product by restricting $\left<\cdot,\cdot\right>_{n,1}$.

$\longrightarrow$ Riemannian manifolds.

We can measure angles between curves:

$$ \cos\alpha = \frac{\left<\gamma’_1(0), \gamma’_2(0)\right>_{n,1}} {\sqrt{\left<\gamma’_1(0), \gamma’_1(0)\right>_{n,1} \left<\gamma’_2(0), \gamma’_2(0)\right>_{n,1}}} $$
for $\gamma_1,\gamma_2$ in hyperbolic space.

The Hyperbolic Line

The hyperbolic line $H^1 \subset \mathbb{R}^{1,1}$ is given by $\{x\in \mathbb{R}^{1,1} \;|\; x_1^2 – x_2^2 = -1,\; x_2>0\}$. We want to measure the length of the curve $\gamma$ in $H^1$ connecting two points $p,q$.

A parametrization of $H^1 \subset \mathbb{R}^{1,1}$ is given by $\gamma : \mathbb{R} \longrightarrow H^1,\quad t \longmapsto {\sinh t \choose \cosh t}$. Then

$$\left<\gamma(t), \gamma(t)\right>_{1,1} = \sinh^2t – \cosh^2t = -1.$$
Let $p = \gamma(s_1)$ and $q = \gamma(s_2)$ where $s_1 < s_2$. Then, since $\gamma'(t) = {\cosh t \choose \sinh t}$,
\mathrm d(p,q) &= \mathrm{length}(\gamma|_{[s_1,s_2]}) = \int_{s_1}^{s_2} \sqrt{\left<\gamma'(t),\gamma'(t)\right>_{1,1}}\:dt = \int_{s_1}^{s_2} \underbrace{\sqrt{\cosh^2t – \sinh^2t}}_{=1}\:dt \\ &= s_2 – s_1.
$\leadsto$ arc-length parametrization of $H^1$.

For $p$ and $q$ we have
\left<p,q\right>_{1,1} &= \left<\gamma(s_1), \gamma(s_2)\right>_{1,1} \\
&= \sinh s_1\sinh s_2 – \cosh s_1\cosh s_2 \\
&= -\cosh(s_2 – s_1) \\ \\
\Longrightarrow \cosh(\mathrm d(p,q)) &= – \left<p,q\right>_{1,1} \\
\mathrm d(p,q) &= \mathrm{arcosh}(-\left<p,q\right>_{1,1}).

Hyperbolic lines in $H^n$

Definition A hyperbolic line in $H^n$ is the non-empty intersection of $H^n$ with a 2-dim. subspace $U$ of $\mathbb{R}^{n,1}$.

Proposition The restriction of $\left<\cdot,\cdot\right>_{n,1}$ to $U$ has signature $(+-)$.

Proof Since $U \cap H^n\ne\emptyset$ there exists $u\in U$ with $\left<u,u\right> = -1$. Let $\{u,v\}$ be a basis of $U$. Define $$w = v + \left<v,u\right>u = v – \frac{\left<v,u\right>}{\left<u,u\right>}u.$$
\left<u,w\right> &= \left<u,v\right> + \left<v,u\right>\underbrace{\left<u,u\right>}_{-1} = 0 \\ \\
\Longrightarrow& w\in u^\perp \\
\Longrightarrow& \left<w,w\right> > 0
since the signature of $\left<\cdot,\cdot\right>|_{u^\perp}$ is $(++\ldots+)$.


So we can identify $H^n \cap U \subset \mathbb{R}^{n,1}$ naturally with $H^1 \subset \mathbb{R}^{1,1}$.

Lecture 16


Definition. Let $q$ be a quadratic form on $\mathbb{R}^n$ or $\mathbb{C}^n$. Then the corresponding quadric is $Q = \{[v] \in \RP^n \mid q(v) =0\}$

A conic is a quadric in $\RP^2$.

How many indefinite non-degenerate quadratic forms exists in $\mathbb{R}^{n+1}$ up to change a basis?

Classification by the signature: (+++..+), (+++..+ -), …, (+ – …-), (- – – ..-), should be indefinite $\Rightarrow$ n indefinite quadrtic forms (+++..+ -), … , (+ – …-).

The number of non-empty non-degenerate quadrics in $\mathbb{R}P^n$ is $\lceil \frac{n}{2} \rceil$


$n=2$: In $\RP^2$ there exist only one indefinite non-degenerate quadric/conic up to projective transformations. It has signature ( + + – ) resp. ( – – +).

$n=3$: In $\RP^3$ there exist $\lceil \tfrac32 \rceil = 2$ different quadrics up to projective transformations:

  1.  (+ + + – ), ( – – – +), $Q_1 = \{[v] \in \mathbb{R}P^3 \mid v_1^2 + v_2^2 + v_3^2 – v_4^2 = 0 \}$
  2.  ( + + – – ), ( – – – +),  $Q_2 = \{[v] \in \mathbb{R}P^3 \mid v_1^2 + v_2^2  –  v_3^2  –  v_4^2 = 0 \}$


1. is a unit sphere (or ellipse, paraboloid or 2-sheeted hyperboloid) depending on the choice of affine coordinate. To obtain a sphere as affine image of the quadric we can choose the affine coordinates  $u_1 = \frac{v_1}{v_4}$, $u_2 = \frac{v_2}{v_4}$, and $u_3 = \frac{v_3}{v_4}$.

2. is one-sheeted hyperboloid or hyperbolic paraboloid.

Degenerate Quadratic forms/quadrics.

Let b be a degenerate bilinear form. Then the space $\{ u \in U \mid b(u, v) = 0 \forall v \in V \} = \ker(b) $ is a subspace of $V$. Consider a subspace $U_1 \subset V$ such that $V = \ker(b) \oplus U_1$, then $b\mid_{U_1}$ defines a non-degenerate quadric $Q_1$ in $P(U_1)$. The quadric $Q$ defined by $b$ is the union of lines through points in the non-degenerate quadric $Q_1$ defined by $b|_{U_1}$ and points in $ P(\ker(b))$ if $Q_1 \neq \emptyset$ (see Exercise 8.1).

Example of degenerate conic.

Consider the following bilinear form in $\mathbb{R}P^2$:

b (\begin{pmatrix} x_1\\  x_2\\x_3\end{pmatrix} , \begin{pmatrix} x_1\\ x_2\\ x_3\end{pmatrix}) = {x_1}^2 – {x_2}^2.

Then $\ker (b) = span \{e_3\}$ and $\mathbb{R}^3 = \ker(b) \oplus \underbrace {span\{e_1, e_2\} }_{U_1}$.  Projectively, $P(\ker(b))$ is a point and the quadric defined by $b|_{U_1}$ in $P(U_1)$ consists of two points. So the degenerate/singular conic definined by $b$ in $\RP^2$ consists of two crossing lines.


Proposition. A non-degenerate non-empty quadric $Q \subset \RP^n$ determines the corresponding bilinear form up to a non-zero scalar multiple.

Proof. Let $b$  and $\tilde b$ be two linear forms defining $Q$. Consider an orthonormal basis w.r.t. $b$, i.e. $\{ e_1 .. e_p, f_1 .. f_q \} $, such that:

  1. $b(e_i, e_i) = 1$ for all $i = 1, \ldots, p$;
  2. $b(f_j, f_j) = -1$ for all $j = 1, \ldots, q$; and
  3. $b(e_i, e_k) = b(f_j, f_l) = b(e_i, f_j) = 0$ for all
    $i, k = 1, \ldots, p$ with $i \neq k$ and
    $j, l = 1, \ldots, q$ with $j \neq l$.

Consider the vectors $e_i \pm f_j$. Then

& b( e_i \pm f_j, e_i \pm f_j) = \underbrace{b(e_i, e_i)}_{=1} \pm \underbrace{2b(e_i, f_j)}_{=0} + \underbrace{b( f_j, f_j)}_{-1} = 0 \\
\Rightarrow& 0 = \tilde b( e_i \pm f_j, e_i \pm f_j) =  \tilde b(e_i, e_i) \pm \tilde 2b(e_i, f_j) + \tilde b ( f_j, f_j)\\
\Leftrightarrow& \pm 2 \tilde b (e_i, f_j) = – \tilde b(e_i, e_i) – \tilde b(f_j, f_j) \\
\Rightarrow& \tilde b (e_i, f_j) = 0 \\
\Rightarrow& \tilde b(e_i, e_i) = – \tilde b (f_j, f_j).

So setting $\tilde b (e_1, e_1) = \lambda \neq 0$ implies that $\tilde b(e_i, e_i) = \lambda$ for all $i = 1, \ldots, p$ and $\tilde b(f_j, f_j) = – \lambda$ for all $j = 1, \ldots, q$.


Orthogonal Transformations

Definition. Let $b$ be a non-degenerate symmetric bilinear form on $\mathbb{R}^{n+1}$. Then $F: \mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+1} $ is orthogonal (wrt. $b$) if $ b(F(v), F(w)) = b(v,w) \forall v,w \in \mathbb{R}^{n+1}$.

The group of orthogonal transformations for a bilinear form of signature $(p, q)$ with $p+q = n+1$ is denoted by $ O(p,q)$. If $q=0$ we obtain the “usual” group of orthogonal transformations $O(n+1) = O(n+1, 0)$

If $Q \subset \RP^n$ is a non-degenerate quadric defined by $b$ and  $F: \mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+1} $ is an orthogonal transformation, then the map $f: \mathbb{R}P^n \rightarrow \mathbb{R}P^n$ with $[x] \mapsto f([x]) = [F(x)]$ maps the quadric onto itself: $f(Q) = Q$.

Proposition. If the signature of a non-degenerate quadric $Q$ is $(p,q)$ with $p \neq q$ and $p \neq 0 \neq q$, then any projective transformation with $ f(Q) = Q$ is induced by an orthogonal transformation $F$ w.r.t. $b$ that defines $Q$.

Proof. Let $b$ be the bilinear form defining $Q$ and $F$ the linear transformation defining the projective transformation $f$. Then $\tilde b(v,w):= b(F(v), F(w))$ is another bilinear form defining $Q$.

According to the previous proposition there exists $\lambda \neq 0$ such that

\[b=\lambda \tilde b.\]

If $\lambda > 0 $ then $\frac{1}{\sqrt \lambda} F$ is an orthogonal transformation, since $ b( \frac{1}{\sqrt \lambda} F(v), \frac{1}{\sqrt \lambda} F(w)) = \frac{1}{\lambda} b (F(v), F(w)) = \frac{1}{\lambda}\tilde b (v,w) = b(v,w)$.

So we need to show that $\lambda > 0$.

If $\{ e_1, \ldots, e_{n+1}\}$ is an orthogonal basis of $\mathbb{R}^{n+1}$ w.r.t. $b$ then $\{F(e_1), \ldots, F(e_{n+1}\}$ is also orthogonal with

  • If $b(e_i, e_i) = 1$ then $\tilde b(e_i, e_i) = \lambda$
  • $b(e_i, e_i) = -1 \Rightarrow \tilde b(e_i, e_j) = – \lambda$.

But the signature is invariant w.r.t. to change of basis (i.e. w.r.t. F). So since $ p \neq q$ we have that the $\lambda $ is positive. $\square$

If $p = q$ (neutral signature), for example $p=q=2$ for a quadric  in $\RP^3$, then there exists a projective transformation preserving the quadric not induced by an orthogonal transformation. Let $b(x,x) = {x_1}^2 + {x_2}^2 – {x_3}^2 – {x_4}^2$. Then the map:

f : \RP^3 \rightarrow \RP^3 ,

yields $b(F(v), F(w)) = – b (x,x) = {x_3}^2 + {x_4}^2 – {x_1}^2 – {x_2}^2$. So $f$ preserves the quadric but it is not induced by an orthogonal transformation $F$.




Lecture 6

Cross-ratio (Doppelverhältnis)

We are looking for invariants with respect to projective transformations.

Definition: Let $ P_{i} =[v_i]= \ssqvector{x_{i} \\ y_{i}}$, $i=1,\ldots,4$, be four distinct points on a projective line $ \RP^{1} $. Then the cross-ratio of these points is

&= \frac{\det(v_1v_2)}{\det(v_2v_3)}\frac{\det(v_3v_4)}{\det(v_4v_1)}\\
&= \dfrac{x_{1}y_{2}-x_{2}y_{1}}{x_{2}y_{3}-x_{3}y_{2}} \dfrac{x_{3}y_{4}-x_{4}y_{3}}{x_{4}y_{1}-x_{1}y_{4}}.\\

If $y_{i} \neq 0$, we may introduce affine coordinates $ u_{i} = \frac{x_{i}}{y_{i}}$. This yields

cr(P_{1},P_{2},P_{3},P_{4}) =
&= \dfrac{y_{1}y_{2}( \frac{x_{1}}{y_{1}}-\frac{x_{2}}{y_{2}} ) }{y_{2}y_{3}( \frac{x_{2}}{y_{2}}-\frac{x_{3}}{y_{3}} )} \dfrac{y_{3}y_{4}( \frac{x_{3}}{y_{3}}-\frac{x_{4}}{y_{4}} ) }{y_{4}y_{1}( \frac{x_{4}}{y_{4}}-\frac{x_{1}}{y_{1}} )} \\
&= \dfrac{u_{1}-u_{2}}{u_{2}-u_{3}} \dfrac{u_{3}-u_{4}}{u_{4}-u_{1}}\,.

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Lecture 5

Projective transformations

Let $V$, $W$ be two vectorspaces over the same field and of the same dimension and $F\colon V \rightarrow W$ a linear isomorphism. In particular $ker(F) = \{0\}$, so F maps 1-dimensional subspaces to 1-dimensional subspaces.

Hence $F$ induces a map from $P(V)$ to $P(W)$.

Definition: A projective transformation $f$ from $P(V)$ to $P(W)$ is a map defined by a linear isomorphism $F\colon V \rightarrow W$ such that

f([v]) = [F(v)] \quad \forall [v] \in P(V)\,.
\end{equation*} Continue reading

Lecture 4

Definition: Two triangles $\triangle_1 = \triangle(A_1, B_1, C_1)$ and $\triangle_2 = \triangle(A_2, B_2, C_2)$ are in perspective w.r.t. a point $S$ if

S = (A_1 A_2) \cap (B_1 B_2) \cap (C_1 C_2)

The triangles are in perspective w.r.t. a line $\ell$ if

A’ = (B_1 C_1) \cap (B_2 C_2)\\
B’ = (A_1 C_1) \cap (A_2 C_2)\\
C’ = (A_1 B_1) \cap (A_2 B_2)\\

lie on $\ell$.

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Lecture 3

Definition. Let $P(V)$ be a projective space of dimension $n$. Then $n+2$ points in $P(V)$ are said to be in general position if no $n+1$ of them are contained in a $(n-1)$-dimensional projective subspace. In terms of linear algebra this implies that no $n+1$ representative vectors are linearly dependend, i.e. every $n+1$ are linearly independent.

Examples. If $n = 1$ we have to consider $n+2 = 3$ points on a projective line. These three points are in general position as long as they are disjoint.
Points in general position on a projective line

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