Hyperbolic Geometry

Intro to Hyperbolic Geometry Shirleen Stibbe

www.shirleenstibbe.co.uk

OPEN UNIVERSITY

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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

C (x, y) or x + iy

d-point

D

boundary points

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 < Ď&#x20AC;

Sum of angles < 2Ď&#x20AC;

Each side is parallel to the other two

!

Sum of angles = 0 5

E vs Non-E Parallel lines E

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

Non-E 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.

l2 l1

NB:

p

l

p

l

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

6

E vs Non-E Common Perpendiculars Parallel lines have lots

E

Non-E

Parallel d-lines have none

Not possible ?

Angle sum of triangle would be π

Ultra-parallel d-lines have one

Not possible

?

Angle sum of quadrilateral would be 2π 7

Non-Euclidean reflections

l

l

Reflect in d-line

l

Reflect in diameter

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

8

Direct Transformations Reflection in l1 followed by reflection in l2 1

Rotations

A

Lines meet at A â&#x2C6;&#x2C6; D Rotation about A A is a fixed point

Lines are parallel Limit rotation No fixed points 2

Translations

Lines are ultra-parallel

9

Mobius transformations [General]

Direct

M( z) =

az + b bz + a

a, b ∈ C, |b| < |a|

Indirect

az + b M( z) = bz + a

a, b ∈ C, |b| < |a|

Matrix

⎛ a b ⎞ A = ⎜ ⎟ ⎝ b 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. 10

Mobius transformations [Canonicall]

Direct

M( z) = K M(m) = 0

Indirect

M( z) = K

|K| =1, |m| < 1

M(0) = -Km

z −m 1 − mz

M(m) ̅ =0 Matrix

z −m 1 − mz

|K| =1, |m| < 1

M(0) = -Km

⎛ 1 − m ⎞ A = ⎜ ⎟ ⎝ − m 1 ⎠

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

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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: ⎛ 1 + z ⎞ d(0, z) = 1 loge ⎜ ⎟ ⎜ ⎟ 2 ⎝ 1 − z ⎠ 12

Calculate d(v, w) 1

Find a transformation taking v to 0 z−v M( z) = (K = 1, m = v) 1− v z

2

Calculate M(w)

3

Then d(v, w)

= d(M(v), M(w)) = d(0, M(w)) = tanh-1(|M(w)|)

Example: Find d(i/2, -i/2) 1 M( z) =

z − i/ 2 1 + (i / 2)z

2 M ( −i / 2) = − i / 2 − i / 2 = − i = − 4i / 5 1 + i / 2 ( −i / 2) 1 + 1/ 4 3 Then d(i/2, -i/2)

= d(0, -4i/5) = tanh-1(4/5) = 1.0986

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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 14

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 centre: m = 0.34i, so |m| = 0.34 Non-E distance: d(O, m) = tanh-1(0.34) = 0.35 (< 0.61) a

Draw a picture

Non-E distances:

0.61

Origin to m: 0.35

m

0.35

0.61

b

Radius: 0.61 a, b on opposite sides of the origin

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 15

Other views 1: (Lobachevsky) Unit Disk viewed as a Pseudosphere

16

Other views 2: Project the unit disc onto a dome world

17

Other views 2: Lengths: Unit disc vs projection onto dome world Non-Euclidean length: d(0, z) = tanh-1(|z|) L

tanh-1(L)

Formula for dome outline:

⎛ f (t ) = 2 − t − 1 2 log e ⎜ ⎜ ⎝ ⎛ 1 − 2 + 2 log e ⎜⎜ ⎝ 2

2 − t 2 + 1 ⎞⎟ 2 − t 2 − 1 ⎟⎠ 2 + 1 ⎞ ⎟⎟, 2 − 1 ⎠

t <1 18

Other views 3: Escher

M.C. Escher's Circle Limit III, 1959 19