hyperbolic geometry
DESCRIPTION
Slides from a lecture by Shirleen Stibbe given at an Open University pure mathematics summerschool.TRANSCRIPT
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OPEN UNIVERSITY
Intro to Hyperbolic Geometry
Shirleen Stibbe www.shirleenstibbe.co.uk
My badge for this session
M203 Pure Mathematics Summerschool
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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.)
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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
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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
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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
!
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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
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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
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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
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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
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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]
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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]
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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:
d(0, z) = ⎟⎟⎠
⎞⎜⎜⎝
⎛
−
+
z1z1
log21
e
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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)|)
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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
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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
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Other views 1: (Lobachevsky)
Unit Disk viewed as a Pseudosphere
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Other views 2:
Project the unit disc onto a dome world
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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
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M.C. Escher's Circle Limit III, 1959
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Other views 3: Escher