Hyperbolic Geometry

OPEN UNIVERSITY

Intro to Hyperbolic Geometry

Shirleen Stibbe www.shirleenstibbe.co.uk

My badge for this session

M203 Pure Mathematics Summerschool

Euclid's Parallel Postulate

Given any line l and a point P not on l, there is a unique line which passes

through P and does not meet l.

How to lose it

Define a non-Euclidean geometry, where either:

1) there are lots of lines through P

or

2) there are no lines through P

which don't meet l.

Hyperbolic geometry is the first kind. (The other is Elliptic.)

2

Poincarre model of Hyperbolic Geometry

Unit disc: D = {z : |z | < 1} = {(x, y): x2 + y2 < 1}

Unit circle: C = {z : |z | = 1} = {(x, y): x2 + y2 = 1}

Boundary point: a point on C [not in N-E geometry]

d-point: a point in D

d-line: part of a generalised circle which meets C at right angles, and lies in D

d-lines

boundary points

d-point

(x, y) or x + iy

C

D

3

Hyperbolic Parallelism

l1 l2

l3

Parallel lines: meet on C [not in D]

Ultra-parallel lines: do not meet at all

l2 & l3 are parallel

l1 & l2 are ultra-parallel

l1 & l3 are ultra-parallel

4

Some jaunty figures in D

(with some interesting properties)

Sum of angles < π

Sum of angles < 2π

Sum of angles = 0

Each side is parallel to the

other two

!

5

Given a d-line l and a d-point p not on l, there are:

a) exactly two d-lines through p which are parallel to l

b) infinitely many d-lines through p which are ultra-parallel to l.

Given a line l and a point p not on l, there is exactly one line through p which is parallel to l

p

p l1

l2

l p l

NB: l1 and l2 are both parallel to l but not to each other

E vs Non-E Parallel lines

E

Non-E

l

6

Parallel d-lines have none

E Parallel lines have lots

Non-E

?

Not possible

Angle sum of triangle would be π

Ultra-parallel d-lines have one

? Not possible

Angle sum of quadrilateral would be 2π

E vs Non-E Common Perpendiculars

7

Non-Euclidean reflections

Reflect in diameter l Reflect in d-line l

l

Properties:

1 the d-line l is left fixed by the non-E reflection in l. 2 d-lines are mapped to d-lines

3 magnitudes of angles are preserved, but orientation is reversed

4 reflections are indirect transformations

l

8

Direct Transformations

Reflection in l1 followed by reflection in l2

A

Lines meet at A ∈ D Rotation about A A is a fixed point

Lines are parallel Limit rotation No fixed points

1 Rotations

2 Translations

Lines are ultra-parallel

9

Direct azbbaz)z(M

++

= a, b ∈ C, |b| < |a|

Indirect azbbza)z(M

++

= a, b ∈ C, |b| < |a|

Matrix ⎟⎠

⎞⎜⎝

⎛=

abba

A

Why in this form?

Transformations must map the unit disc to itself

Therefore restricted to a subgroup of

the Mobius transformations that fix D.

Mobius transformations [General]

10

Direct zm1mzK)z(M

−−

= |K| =1, |m| < 1

Indirect

Matrix ⎟⎠

⎞⎜⎝

⎛

−

−=

1mm1

A

mz1mzK)z(M

−−

=

M(m) = 0 M(0) = -Km

|K| =1, |m| < 1

M(m) = 0 M(0) = -Km ̅

Origin lemma - very NB

Given any point A in D, there exists a non-E transformation sending A to 0.

Mobius transformations [Canonicall]

11

Non-Euclidean Distance

★★ Key idea: non-E transformations should preserve non-E distance!

Distance function: d(z, w) denotes the non-E distance between the points z and w in D.

If M is a non-E transformation, then:

d(z, w) = d(M(z), M(w))

distance between the points

distance between their images

Definition:

d(0, z) = tanh-1(|z|), z ∈ D

distance between z and the origin

Alternative form:

d(0, z) = ⎟⎟⎠

⎞⎜⎜⎝

⎛

−

+

z1z1

log21

e

12

13

Calculate d(v, w)

Example: Find d(i/2, -i/2)

1

2

3 Then d(i/2, -i/2) = d(0, -4i/5)

= tanh-1(4/5)

= 1.0986

z)2/i(12/iz)z(M

+−

=

5/i44/11i

)2/i(2/i12/i2/i)2/i(M −=

+−

=−+−−

=−

1 Find a transformation taking v to 0

v) m ,1K(zv1vz)z(M ==

−−

=

2 Calculate M(w)

3 Then d(v, w) = d(M(v), M(w))

= d(0, M(w))

= tanh-1(|M(w)|)

14

Going round in circles

Some useful notations for circles

In R2: K1 has equation: (x - a)2 + (y - b)2 = r12

Centre = (a, b), Radius = r1

In C: K2 = {z: |z - α| = r2} , z, α ∈ C, r2 ∈ R

Centre = α, Radius = r2

Note: if K1 = K2, then

α = a + ib, r1 = r2

In D: K3 = {z: d(z, β) = r3}, z, β ∈ C, r3 ∈ R

Centre = β, Non-E radius = r3

Note: r3 ≠ tanh-1(r2)

β ≠ α

15

Example: K is the non-E circle K = {z: d(z - 0.34i) = 0.61}

Find the Euclidean centre and radius of K.

Non-E: d(O, a) = 0.61 + 0.35 = 0.96 = tanh-1(|a|)

Euclidean: |a| = tanh(0.96) = 0.75 = 3/4

⇒ a = 3i/4 (imaginary, above the origin)

Non-E: d(O, b) = 0.61 - 0.35 = 0.26 = tanh-1(|b|)

Euclidean: |b| = tanh(0.26) = 0.25 = 1/4

⇒ b = -i/4 (imaginary, below the origin)

E-centre: c = 1/2(a + b) = 1/2(3i/4 – i/4) = i/4

E-radius: r = 1/2|a - b| = 1/2|3i/4 + i/4| = 1/2

E-circle: K = {z: |z - i/4| = 1/2

Draw a picture

Non-E centre: m = 0.34i, so |m| = 0.34

Non-E distance: d(O, m) = tanh-1(0.34) = 0.35 (< 0.61)

Non-E distances:

Origin to m: 0.35

Radius: 0.61

a, b on opposite sides of the origin

m

a

b

0.61

0.61 0.35

Other views 1: (Lobachevsky)

Unit Disk viewed as a Pseudosphere

16

Other views 2:

Project the unit disc onto a dome world

17

L

tanh-1(L)

Non-Euclidean length: d(0, z) = tanh-1(|z|)

1,1212log2

12

12log2)(

21

2

2

212

<⎟⎟⎠

⎞⎜⎜⎝

⎛

−+

+−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−−

+−−−=

t

t

tttf

e

e

Formula for dome outline:

Other views 2:

Lengths: Unit disc vs projection onto dome world

18

M.C. Escher's Circle Limit III, 1959

19

Other views 3: Escher