Download 1 Geomtery and the Axiomatic Method

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Geomtery and the Axiomatic Method
1.1
Early Origins of Geometry,
1.2
Thales and Pythagoras (1 lecture, inc course intro)
Questions
1. Why taking class?
2. How many teachers-in-training?
3. What is geometry? (meaning? Geo-metros. . . )
4. What is an axiom?
Overview of Geometric History
Ancient Egypt, Babylonia, China: basic rules for easy practical shapes. Length, volume, etc. No
equations. Also surveying: tax collection. India: religious symbology, designs of altars, etc.
Ancient Greece Began process of abstraction: ‘math for the sake of math.’ Famous people: Thales of
Miletus (c.600BC), Pythagoras of Samos (c.580BC) (both on west coast/islands off present day
Turkey). Both travelled widely and absorbed much from other cultures (e.g. Egypt, etc.) Began
deductive reasoning, if not quite from first principles.
Euclid of Alexandria (c.300BC) collected much work, especially of geometry. Elements became most
read book in history (except Bible). Standard textbook in all disiplines until early 1900s. Standardized basic axioms and proved everything else from them.
Modern Development Perspective in Renaissance, to projective geometry and other non-Euclidean
geometries. More modern idea (Klein’s Erlangen Program 1872) that geometry is the study +
classification of symmetry (i.e. group theory).
Thales’ Theorems
1. A circle is bisected by a diameter.
2. The base angles of an isoceles triangle are equal.
3. The pairs of angles formed by two intersecting lines are equal.
4. Two triangles are congruent if they have two angles and the included side equal.
5. An angle inscribed in a semicircle is a right angle.
His ‘proofs’ were not what we’d understand by the term. Part deductive and part inductive
reasoning based on examples. Discuss and investigate these.
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1.4
The Rise of the Axiomatic Method (1/2 lecture)
Deductive System Contains four things:
1. Undefined terms: accepted without any further definition/explanation. E.g., length, width,
line, point, etc.
2. Postulates/Axioms: logical statements accepted as true without proof. E.g., ∃ set with no elements.1
3. Defined terms: concepts defined in terms of 1 and 2. E.g., Area, associativity.
4. Theorems: logical statements deduced from 1–3.
Example. Chess.
1. Pieces (as black/white objects) and the board
2. Rules for how each piece moves
3. Concepts such as ‘check’, ‘en-passant’
4. “Black can win in 5 moves”
Example. Semigroup
1. Elements of a set G and a binary operation
2. (A1) Closure: ∀ a, b ∈ G, ab ∈ G.
(A2) Associativity: ∀ a, b, c ∈ G, a(bc) = ( ab)c.
3. Square a2 = aa, commutativity, etc.
4. E.g. a3 is well-defined. (R, +) is a semigroup.
1.5
Properties of Axiomatic Systems (1/2 lecture)
Definition 1.1. A model of an axiomatic system is an interpretation of the undefined terms such that
the axioms are all true.
Examples.
1. (Z, +) is a model of a semigroup.
2. Triangle and {1, 2, 3} under compositions are models of S3 .
Idea Any theorem in an axiomatic system is true in any model
Consistency No two statements (theorem/axioms) contradict each other.
Independence No axiom is a theorem of the others.
1 Euclid:
Axiom globally true statment, Postulates about the stuff with which we’re studying.
2
Completeness One cannot add any new axiom (in terms of the undefined terms) to the system so
that the resulting system is consistent and independent.
Example. Semigroup axioms consistent + independent but not complete. E.g.,
A3 (identity): ∃e ∈ M such that ae = ea = a
is a new axiom. The three are consistent and independent. Call structure a monoid.
Argument over whether adding inverse is really a new axiom (not in terms only of undefined terms).
If make ( G, e, ·) the monoid, then inverse is a new axiom.
Even group not complete. Could add commutativity as an axiom.
In general telling that a system is complete is close to impossible! Essentially every possible
logical statement formed by the system would have to be decidable (T or F) in terms of the system.
Example. A1 There are exactly 18 students in the class
A2 Every student earns a letter grade A–F
A3 All students have the same name
A4 Colin and Arturo both get the same grade
Gödel
Theorem 1.2. Given a consistent system containing the natural numbers, the consistency of the system cannot
be proved within the system.
Theorem 1.3. Any such system contains undecideables: statements which are neither provable nor disprovable within the system.
In ZFC, the Continuum hypothesis is such a statement.
1.6
Euclid’s Axioms (in handout) (1 lecture)
Undefined Terms E.g., point, line.
Axioms/Postulates A1 a = b and b = c =⇒ a = c (= is transitive)
A2 a = b and c = d =⇒ a + c = b + d
A3 a = b and c = d =⇒ a − c = b − d
A4 Things that coincide are equal (in magnitude)
A5 The whole is greater than the part
P1 Can create a line by joining two points
P2 Can extend a finite line continuously
P3 Can create a circle given a point and a radius
P4 All right angles are equal
P5 If a straight line crosses two others and the angles on one side sum to less than two right
angles then the two lines meet on that side
Consistency Never demonstrated, or shown not to be
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Completeness No. Euclid makes many hidden assumptions (e.g., continuity, existence of points)
Basic Theorems à la Euclid Built on Postulates with very pictorial proofs: e.g.
• Any line segment can be bisected (Thm I.10)
• If two straight lines cut one another, opposite angles are equal (Thm I.15)
• An angle in a semicircle is right-angle (Thm III.31)
You can find an English translation of Euclid’s proofs at
http://math.furman.edu/ jpoole/euclidselements/euclid.htm
Parallel Lines, their construction and uniqueness
Definition. Two lines are parallel if they do not meet
Theorem (I. 16). If one side of a triangle is protruded, then the
exterior angle is larger than either of the opposite interior angles.
In the language of the picture (although Euclid never quantified
angles), we have δ > α and δ > β.
Proof. Take the midpoint M of AC and draw the bisector.
Extend it the same distance beyond M to E. Connect CE.
The opposite angles at M are equal and so the we have two
congruent triangles: AMB ∼
= CME. It follows that the angle
indicated is also α. Clearly this is less than δ.
Bisect the other edge to see that β < δ.
α
δ
β
A
E
M
α
B
C
Theorem 16 essentially constructs a parallel line to AB through a given point C not on the line
AB. It remains to prove that this line really is parallel.
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Theorem (I. 27). Suppose that a line falls on two other lines in
such a way that the indicated angles are equal. Then the two lines
are parallel.
Proof. Suppose the lines were not parallel. Then they must
meet on one side. WLOG suppose they meet on the right
side at point C. But Theorem I. 16 says that the angle α at B,
being external to the triangle ABC must be greater than the
angle α at A. This is a contradiction.
α
α
A
α
α
B
C
It follows that the line CE constructed in the proof of Theorem I. 16 really is parallel to AB. The
following theorem is an immediate corollary of the last.
α
Theorem (I. 28). Suppose that a line falling on two other lines
makes the same angles. Then the two lines are parallel.
α
Up to here only the first four of Euclid’s postulates are used. Thus in any model for which the first
four postulates are true, it is possible to create parallel lines just by measuring angles. The parallel
postulate (P5) is what tells us that, in Euclidean Geometry, this is the only way to create parallel
lines. In other geometries, e.g., Hyperbolic Geometry, Euclid’s first four postulates are satisfied,
when parallels can be constructed, however with the fifth postulate, these parallels are not unique.
Indeed in Hyperbolic Geometry, there are infinitely many parallel lines through a given point.
Theorem (I. 29). Suppose that a line falls on two parallel lines. Then the alternate angles are equal.
This is exactly the converse to Theorem I. 27.
Proof. We must prove that α = β. Suppose not and WLOG
that α > β. But then β + γ < α + γ and so β + γ is less than
a straight edge. By the parallel postulate, the parallel lines
`1 , `2 meet on the left side of the picture. A contradiction,
since parallel lines do not meet.
γ
α
β
Angles in a triangle add to 180◦ Finally, Euclid is in a position to prove the most famous result
about triangles: that its interior angles sum to a straight edge. Euclid words this slightly differently.
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`1
`2
A
Theorem (I. 32). If one side of a triangle is protruded, the exterior
angle is equal to the two interior and opposite angles
B
Proof.
1. Construct CE parallel to BA
D
C
A
E
2. Then ∠ ABC = ∠ECD and
∠ ACE = ∠ BAC
3. ∴ ∠ ACD = ∠ ACE + ∠ECD
= ∠ BAC + ∠ ABC
B
C
D
Playfair’s Postulate Euclid’s Parallel postulate is a little hard to work with, given that it is stated in
the negative. A more modern interpretation is given by Playfair’s postulate:
Given a line and any point not on that line, there exists exactly one parallel line through said point.
Theorem. Playfair’s postulate is equivalent to the Parallel postulate (in the presence of Euclid’s other axioms).
Proof. Suppose that Euclid’s axioms/postulates all hold and that we are given a line and a point not
on that line. Theorem I. 27 constructs a parallel line throught the point, while Theorem I. 29 says that
there is only one such. This is Playfair’s postulate.
Conversely, assume that Euclid’s postulates P1–P4 are true and that P5 is false. We prove the contrapositive of the required statement. The parallel postulate may be written
∀ lines `, m,
` ∩ m = ∅ =⇒ angles made by a crossing line sum to a straight edge.
The negation of this is, the assumption that the parallel postulate is false, is as follows:
∃ lines `, m, such that ` ∩ m = ∅ and angles made by a crossing line do not sum to a straight edge.
Assume that we have two such parallel lines `, m as in the
picture. The falseness of the parallel postulate assures us
that α + γ is less than a straight edge. Now, since the first 28
theorems of Book I of the Elements are true, we may choose
any point A on m and use Theorem 28 to construct a parallel
line `ˆ to ` through A.
α A
γ
α
The lines `ˆ and m are necessarily different because α + γ is less than a straight edge. We have therefore constructed two parallel lines to ` through A, whence Playfair’s postulate is false.
We have therefore shown that Playfair’s postulate and the Parallel postulate are equivalent.
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m
`ˆ
`
Pythagoras’ Theorem The crowning achivement, and penultimate Theorem, of Book I is Pythagoras’ Theorem (Thm I. 47). Indeed Book I was probably structured in order to efficiently prove this
result. Here is a modernized version of Euclid’s proof (Euclid did not use the word ‘Area’, instead
suggesting, for instance, that ‘the parallelogram BOLD is double the triangle ABD’). Perhaps the reason that Euclid gave so much importance to this theorem comes from the fact that the Pythagoreans’
‘proof’ was incorrect due to their lack of understanding about incommensurable lengths. Euclid
probably wanted to correct this and to put such an important theorem on a sound logicial footing.
H
K
G
A
1. Construct squares on each of the sides of the
right-angled 4 ABC.
2. Since the side lengths are the same, we see that
4 ABD is congruent to 4 FBC
3. Area of 4 formula =⇒ 4 ABD =
F
B
O
C
L
BOLD
4. Similarly 4 FBC = 12 ABFG
5. ∴ ABFG =
BOLD
6. Similarly ACKH =
D
1
2
OCEL
7. Now sum together the rectangles to obtain the
square BCED
E
Book II: Geometric algebra We tend to think of Pythagoras’ Theorem as a statement in Geometric
Algebra: i.e., c2 = a2 + b2 , but this is not how Euclid saw it. To Euclid, Geometric Algebra was about
solving equations to find unknowns. Of course the concept of algebra hadn’t been invented yet, so
the very problems had to be stated geometrically. Such concerns comprise the bulk of Book II of the
Elements, and are mostly attributable to the Pythagoreans.
Here is an example.
Theorem (II. 11). A straight line can be divided so that the rectangle contained by the whole and one of the
segments is equal to the square on the remaining segment.
In our modern language we would say: given AB = a, find H on AB, so that AH = x, with
x2 = a( a − x ). Here is Euclid’s proof.
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G
F
1. Construct square ABDC on AB
H
A
B
2. Bisect AC; call midpoint E
3. Connect EB
4. Extend AC
5. Lay off EF = EB on AC extended
E
6. Construct square FGH A
C
D
In our modern language,
with our understanding of Pythagoras’, we see that Euclid has con√
5−1
structed H so that x = 2 a. We can easily check that this solves the equation x2 = a( a − x ). Note
that this is a construction of the golden ratio:
√
√
5−1
5+1
AH : HB = 1 :
=
: 1.
2
2
For many centuries after Euclid, the meaning of ‘solve’ was exactly this: construct a solution.
Euclid’s Axioms
Undefined Terms E.g., point, line.
Axioms/Postulates A1 a = b and b = c =⇒ a = c (= is transitive)
A2 a = b and c = d =⇒ a + c = b + d
A3 a = b and c = d =⇒ a − c = b − d
A4 Things that coincide are equal (in magnitude)
A5 The whole is greater than the part
P1 Can create a line by joining two points
P2 Can extend a finite line continuously
P3 Can create a circle given a point and a radius
P4 All right angles are equal
P5 If a straight line crosses two others and the angles on one side sum to less than two right
angles then the two lines meet on that side
Consistency Never demonstrated, or shown not to be
Completeness No. Euclid makes many hidden assumptions (e.g., continuity, existence of points)
8
Basic Theorems à la Euclid Built on Postulates with very pictorial proofs: e.g.
• Any line segment can be bisected (Thm I.10)
• If two straight lines cut one another, opposite angles are equal (Thm I.15)
• An angle in a semicircle is right-angle (Thm III.31)
You can find an English translation of Euclid’s proofs at
http://math.furman.edu/ jpoole/euclidselements/euclid.htm
9
Parallel Lines, their construction and uniqueness
Definition. Two lines are parallel if they do not meet
Theorem (I. 16). If one side of a triangle is protruded, then the
exterior angle is larger than either of the opposite interior angles.
In the language of the picture (although Euclid never quantified
angles), we have δ > α and δ > β.
Proof. Take the midpoint M of AC and draw the bisector.
Extend it the same distance beyond M to E. Connect CE.
The opposite angles at M are equal and so the we have two
congruent triangles: AMB ∼
= CME. It follows that the angle
indicated is also α. Clearly this is less than δ.
Bisect the other edge to see that β < δ.
α
δ
β
A
E
M
α
B
C
Theorem 16 essentially constructs a parallel line to AB through a given point C not on the line
AB. It remains to prove that this line really is parallel.
Theorem (I. 27). Suppose that a line falls on two other lines in
such a way that the indicated angles are equal. Then the two lines
are parallel.
Proof. Suppose the lines were not parallel. Then they must
meet on one side. WLOG suppose they meet on the right
side at point C. But Theorem I. 16 says that the angle α at B,
being external to the triangle ABC must be greater than the
angle α at A. This is a contradiction.
α
α
A
α
α
B
C
It follows that the line CE constructed in the proof of Theorem I. 16 really is parallel to AB. The
following theorem is an immediate corollary of the last.
α
Theorem (I. 28). Suppose that a line falling on two other lines
makes the same angles. Then the two lines are parallel.
α
Up to here only the first four of Euclid’s postulates are used. Thus in any model for which the first
four postulates are true, it is possible to create parallel lines just by measuring angles. The parallel
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postulate (P5) is what tells us that, in Euclidean Geometry, this is the only way to create parallel
lines. In other geometries, e.g., Hyperbolic Geometry, Euclid’s first four postulates are satisfied,
when parallels can be constructed, however with the fifth postulate, these parallels are not unique.
Indeed in Hyperbolic Geometry, there are infinitely many parallel lines through a given point.
Theorem (I. 29). Suppose that a line falls on two parallel lines. Then the alternate angles are equal.
This is exactly the converse to Theorem I. 27.
Proof. We must prove that α = β. Suppose not and WLOG
that α > β. But then β + γ < α + γ and so β + γ is less than
a straight edge. By the parallel postulate, the parallel lines
`1 , `2 meet on the left side of the picture. A contradiction,
since parallel lines do not meet.
γ
`1
α
β
`2
Angles in a triangle add to 180◦ Finally, Euclid is in a position to prove the most famous result
about triangles: that its interior angles sum to a straight edge. Euclid words this slightly differently.
A
Theorem (I. 32). If one side of a triangle is protruded, the exterior
angle is equal to the two interior and opposite angles
B
Proof.
1. Construct CE parallel to BA
D
C
A
E
2. Then ∠ ABC = ∠ECD and
∠ ACE = ∠ BAC
3. ∴ ∠ ACD = ∠ ACE + ∠ECD
= ∠ BAC + ∠ ABC
B
C
Playfair’s Postulate Euclid’s Parallel postulate is a little hard to work with, given that it is stated in
the negative. A more modern interpretation is given by Playfair’s postulate:
Given a line and any point not on that line, there exists exactly one parallel line through said point.
Theorem. Playfair’s postulate is equivalent to the Parallel postulate (in the presence of Euclid’s other axioms).
Proof. Suppose that Euclid’s axioms/postulates all hold and that we are given a line and a point not
on that line. Theorem I. 27 constructs a parallel line throught the point, while Theorem I. 29 says that
there is only one such. This is Playfair’s postulate.
Conversely, assume that Euclid’s postulates P1–P4 are true and that P5 is false. We prove the contrapositive of the required statement. The parallel postulate may be written
∀ lines `, m,
` ∩ m = ∅ =⇒ angles made by a crossing line sum to a straight edge.
11
D
The negation of this is, the assumption that the parallel postulate is false, is as follows:
∃ lines `, m, such that ` ∩ m = ∅ and angles made by a crossing line do not sum to a straight edge.
Assume that we have two such parallel lines `, m as in the
picture. The falseness of the parallel postulate assures us
that α + γ is less than a straight edge. Now, since the first 28
theorems of Book I of the Elements are true, we may choose
any point A on m and use Theorem 28 to construct a parallel
line `ˆ to ` through A.
m
`ˆ
α A
γ
α
`
The lines `ˆ and m are necessarily different because α + γ is less than a straight edge. We have therefore constructed two parallel lines to ` through A, whence Playfair’s postulate is false.
We have therefore shown that Playfair’s postulate and the Parallel postulate are equivalent.
Pythagoras’ Theorem The crowning achivement, and penultimate Theorem, of Book I is Pythagoras’ Theorem (Thm I. 47). Indeed Book I was probably structured in order to efficiently prove this
result. Here is a modernized version of Euclid’s proof (Euclid did not use the word ‘Area’, instead
suggesting, for instance, that ‘the parallelogram BOLD is double the triangle ABD’). Perhaps the reason that Euclid gave so much importance to this theorem comes from the fact that the Pythagoreans’
‘proof’ was incorrect due to their lack of understanding about incommensurable lengths. Euclid
probably wanted to correct this and to put such an important theorem on a sound logicial footing.
H
K
G
A
1. Construct squares on each of the sides of the
right-angled 4 ABC.
2. Since the side lengths are the same, we see that
4 ABD is congruent to 4 FBC
3. Area of 4 formula =⇒ 4 ABD =
F
B
O
C
L
BOLD
4. Similarly 4 FBC = 12 ABFG
5. ∴ ABFG =
BOLD
6. Similarly ACKH =
D
1
2
OCEL
7. Now sum together the rectangles to obtain the
square BCED
E
12
Book II: Geometric algebra We tend to think of Pythagoras’ Theorem as a statement in Geometric
Algebra: i.e., c2 = a2 + b2 , but this is not how Euclid saw it. To Euclid, Geometric Algebra was about
solving equations to find unknowns. Of course the concept of algebra hadn’t been invented yet, so
the very problems had to be stated geometrically. Such concerns comprise the bulk of Book II of the
Elements, and are mostly attributable to the Pythagoreans.
Here is an example.
Theorem (II. 11). A straight line can be divided so that the rectangle contained by the whole and one of the
segments is equal to the square on the remaining segment.
In our modern language we would say: given AB = a, find H on AB, so that AH = x, with
x2 = a( a − x ). Here is Euclid’s proof.
F
G
1. Construct square ABDC on AB
A
H
B
2. Bisect AC; call midpoint E
3. Connect EB
4. Extend AC
5. Lay off EF = EB on AC extended
E
6. Construct square FGH A
C
D
In our modern language,
with our understanding of Pythagoras’, we see that Euclid has con√
5−1
structed H so that x = 2 a. We can easily check that this solves the equation x2 = a( a − x ). Note
that this is a construction of the golden ratio:
√
√
5−1
5+1
AH : HB = 1 :
=
: 1.
2
2
For many centuries after Euclid, the meaning of ‘solve’ was exactly this: construct a solution.
2
2.1
Euclidean Geometry
Angles, Lines and Parallels (1 lecture)
Basic Notions (undefined terms) Point, Line
Basic Definitions Segment AB (all points between A, B)
~ extension of the segment AB beyond B
Ray AB:
~ AC.
~ Denote ∠ BAC
Angle with vertex A: two rays AB,
Vertical angles: angles with a common vertex whose sides form two lines (opposite)
Supplementary angles: if sum to a straight edge (share a side in common and other sides point in
opposite directions)
13
Congruence To Euclid, meant same magnitude/measure, as in can move one object on top of the
other and have them match up. So angles, lengths, triangles. Difficult idea without group theory! At
elementary level think of degree measures being the same.
Degree measure denoted m∠ ABC.
Length of segement denotes AB.
Right angle has congruent supplementary angle.
Intersecting lines are perpendicular if one of the angles made is a right angle.
Parallel lines do not intersect.
Bisector of a segment is a point/ray splitting magnitude in half (of segment/angle)
Example. Define a midpoint. Define a triangle given three points.
Playfair’s Postulate Given a line and a point not on the line, it is possible to construct one and only
one line through the given point parallel to the line.
Theorem 2.1. Playfair’s postulate (with A1–4, P1–4) is logically equavalent to Euclid’s axioms.
Recall Euclid 5: If have line falling on two lines with angles on one side summing to less than
straight angle, then lines meet on that side.
Simpler case: suppose angles on one side of the line are α, β. Then
α + β 6= 180 =⇒ ` ∩ `ˆ 6= ∅.
Contrapositive: ` ∩ `ˆ = ∅ =⇒ α + β = 180. I.e. parallel =⇒ α + β = 180.
Proof. For both parts, let A 6∈ ` be given. May construct a parallel line through A by previous
theorems: draw line through A and crossing ` and translate angle to A.
1. Assume that Euclid’s axioms hold. Euclid says angles on each side sum to straight edge. If had
another parallel line then angles the same, so impossible. Thus Playfair true.
2. Conversely, by Playfair there are no other parallel lines. Already have one, which satisfies
interior angles result, so Euclid true.
NOTE: Spherical geometry (line = geodesic) satisfies only E2, E4. Can’t make circles of arbitrary
radius and can’t define a line by two points (for antipodal points get two choices). Thus cannot
construct a parallel line to start with. Spherical geometry contains no parallel lines.
Hyperbolic geometry satisfies E1–E4 so can create an initial parallel line, but can also create infinitely
many others.
14
2.2
Congruent Triangles and Pasch’s Axiom (1 lecture)
Back to the start. . .
Definition. Two triangles are congruent if one can be laid on top of the other so that edges and angles match.
We write
4 ABC ∼
= 4 XYZ
Means that AB = XY, etc., and the angles similarly.
Strictly, if the two triangles differ by an isometry: translations/rotation/reflection: Euclid did
not have such notions. Let the set of all triangles in the plane be T . The set of isometries E(2) =
R2 o O(2) acts on T (R2 is the normal subgroup of E(2)) Say that
T1 , T2 are congruent iff ∃ g ∈ E(2) : g ◦ T1 = T2 .
Basic triangle theorems: three bits of info being equal are enough to see that two triangles are
congruent.
Theorem (I. 4). Side-Angle-Side. If these are equal in two triangles then they are congruent. Trace round the
triangle: if two sides have the same length and the angle between the same magnitude, then. . .
Proof. Given two such triangles ABC, XYZ with AB = XY, BC = YZ, ∠ ABC = ∠XYZ, lay AB on
XY at Y. Then A is on X since the lengths are equal. Since the angles are equal, BC lies on YZ. SInce
the lengths are equal, C lies on Z. Thus AC = XZ.
Theorem (I. 26). Angle-Side-Angle.
Theorem (I. 26). Angle-Angle-Side.
Theorem (I. 8). Side-Side-Side.
Anything else? Of only three things to choose, only possibilities are SSA and AAA. Both false.
Think about why.
Definition. An isosceles triangle has two sides congruent (’legs’). Base angles are the two equal angles.
Theorem (I. 5). Base angles equal
Already seen: our proof uses a bisector, then SAS on the right angled triangles.
Problem 1 Euclid doesn’t define bisectors till proposition 9. . . His proof different, more complicated.
Very easy to get into circular aguments.
Problem 2 How do we know that the bisector of the angle intersects the other line? Similar problem
to Euclid’s first theorem on building a triangle on a line. How do we know that circles intersec?
Leads to need for Pasch’s Axiom (1882)
Let A, B, C be non-collinear points and ` a line not passing through any of them. If ` passed through
the side AB then it must pass through a point on AC or on BC, but not on both.
PICTURE!!! Essentially two choices for `. With this axiom, the bisecting line must intersect.
Hilbert included this, and the SAS theorem as axioms in his own system (to avoid groups. . . ). Big
challenge: sometimes a proof is not valid in a different system even though the result is!
15
2.4
Measurement and Area in Euclidean Geometry (1/2 lecture)
Euclid (indeed everyone in antiquity) had no notion of area or even length: instead about relative
measurement. I.e., rod b is twice the length of rod a. We hear ‘congruent’ and think that the numerical
measures of the quantity are equal. This is an extra imposition of structure and one reason it’s been
so hard for mathematicians to properly define number!
Modern math defines measure axiomatically to get around the problem and because it’s so useful.
Modern uses of words like square literally come from Euclid constructing a physical square.
Defining area without measure/number
Definition. A parallelogram consists of four points no three of which are collinear and so that the sides do not
cross and opposite sides are parallel.
Theorem. Opposite sides are congruent (lots of AAS after splitting parallelogram in two)
Definition. A rectangle is a parallelogram with all angles being right angles. Its area is its base times height
(pick two adjacent sides and call one the base one the height).
Still required in ancient times a fundamental notion of length to get a number.
Exercise Defn indep of which is base and which height (draw two rectangles rotated. . . )
Definition. Two figures have the same area if they are equivalent in that they can be cut into a finite number
of polygonal pieces which are congruent. (could be done with triangles)
Theorem. Call one side of a parallelogram the base. Drop a perpendicular to the opposite side. Then area is
base times length of perpendicular.
Proof: draw a rectangle with said base/height and add/subtract triangles to make conguruent.
Theorem (I. 41). (Area of triangle is half base times height) If a parallelogram and triangle have the same base
and height then the parallogram is twice the triangle.
Can see the problem of why people tried to square the circle.
Cevians + triangle centers
After Giovanni Ceva (1647–1734): a cevian is an line joining a vertex to the opposite side of the triangle
Theorem (Ceva’s Theorem). Given a triangle 4 ABC and cevians AX, BY, CZ, then
BX CY AZ
= 1 ⇐⇒ the cevians meet at a common point P.
XC YA ZB
16
Proof. (⇐) Pairs of triangles with same altitudes, so areas
proportional to their bases: i.e.,
A
a( ABX )
BX
a( PBX )
=
=
a( AXC )
XC
a( PXC )
a( ABX ) − a( PBX )
a( ABP)
BX
=⇒
=
=
a( AXC ) − a( PXC )
a( APC )
XC
Z
P
Alternately, draw parallel lines and think about it: 4 ABP
and 4 APC have the same base and altitudes proportional
to BX, XC. . .
Now multiply corresponding results together to get 1.
(⇒) This time define P as the intersection of BY and CZ,
then ray AP meets BC at X 0 . Want to prove that X = X 0 .
Know that
BX 0 CY AZ
=1=
X 0 C YA ZB
BX − X 0 X
=⇒ 0
=
X X + XC
2.5
Y
BX CY AZ
BX 0
BX
=⇒ 0 =
XC YA ZB
XC
XC
BX
(now solve for X 0 X = 0)
XC
B
X
A
C
Y
Z
P
B
X0
C
X
Similar Triangles (1/2 lecture)
Definition. Triangles are similar if corresponding pairs of sides are in the same ratio. Write 4 ABC ∼
4 XYZ
Note: side lengths proportional but angles the same. Thus AAA theorem holds (later).
Kleinian viewpoint: set of triangles under action of conformal group R2 × R+ × O(2) (isometries
and dilations) has equivalence classes the sets of similar triangles.
Theorem (AAA). Suppose a line intersects two sides of a triangle. Then the smaller triangle is similar iff the
line is parallel to the third side of the triangle.
Needs lemmas. . .
A
F
Lemma 2.2. Draw ` parallel to BC and construct F so that EF ⊥
BD
CE
AB
AC
AB. Then
=
, whence
=
AD
AE
AD
AE
D
E
B
`
C
BD
a( BED )
a(CED )
CE
Just observe that
=
=
=
(by mirroring). Add 1 to each side to finish.
AD
a( AED )
a( AED )
AE
Lemma 2.3. The converse is also true.
Use Playfair to construct unique parallel through D, then formular from lemma holds, etc. . .
Now can prove the AAA theorem: Take two triangles with the same angles: if different sizes then
bung one inside the other. Angles same implies parallel bases so sides in proportion. Now repeat
17
with each of the other angles.
Can use to define trig functions:
Definition 2.4. Given an angle α = ∠ ABC, construct D on AC so that 4 ADB is right-angled at D.
Then
sin α :=
AD
,
BD
cos α :=
BD
.
AB
Theorem 2.5. sin and cos are well-defined.
Proof. RTP: if α = β then sin α = sin β, etc. If α = β in two right-angled triangles, then triangles are
similar (AAA), so corresponding sides have the same ratio lengths.
2.6
Circle Geometry (1 lecture)
A
chord
Definition. A chord is a line joining two points on a circle. A
diameter is a chord passing through the center. An arc is the segment of the circular edge between the chord points (major or minor
by length).
∠ AOB: central angle
∠ APB: inscribed angle
Recall Thales: a diameter divides a circle into congruent
pieces.
minor arc
diameter
B
O
C
major arc
P
Theorem. Given three non-collinear points, ∃ unique circle through them.
Proof. Idea: AB, AC chords, their perpendicular bisectors meet at the center.
Draw parallelogram: proves distance from central point the same.
Uniqueness: the center must be on the perpendicular bisectors because these are the loci of the points
equidistant from each pair of points.
Theorem. Angle at the center of a circle is twice the inscribed angle.
Proof. Isosceles triangles
Corollary 2.6. If two triangles in a circle have the same arc then they are congruent.
Angles in a semi-circle are right angles.
Theorem. Draw a quadrilateral in a circle: its opposite angles are supplementary.
(Just draw the center and compare angles at center to outside: angles at center make up whole
revolution)
18
Definition. A line is tangent to a circle if it intersects it only once.
Theorem. ` is tangent to a circle iff it is perpendicular to the radius at the point of tangency.
Proof. First suppose line ` through T is perpendicular to the radius OT.
WTP: all other points on ` are outside circle.
Since angles in 4 add to a straight edge, Exterior angles theorem (Euclid) says
∀ P ∈ ` \ T,
∠OPT < ∠OTP =⇒ OP > OT.
P
T
O
Thus P outside circle, whence line intersects only at T: line is tangent.
Conversely, suppose ` is tangent at T but not perpendicular to OP.
Construct perpendicular to ` through O and call intersection P.
Again apply exterior angle theorem: since ∠OPT is a right angle,
∠OTP < ∠OPT =⇒ OT > OP,
P
T
O
whence P is inside the circle. The line must therefore exit the circle
again, so is not a tangent.
Theorem. Given a point outside a circle, there are exactly two lines on this point which are tangent to the
circle.
Proof. Could try to apply Intermediate value theorem! but too much sledgehammer. . . Draw circle
centered at midpoint of OP and compute angles: the two intersection points are the tangency points.
3
3.1
Analytic Geometry
The Cartesian Co-ordinate System (1/2 lecture)
Descartes (1596–1650) “La Geometrie” and Fermat (1601–1655). Introduced axes, variables x, y, z,
constants a, b, c. Decoupled geometry from algebra: now x2 didn’t have to mean a physical square.
Basic Notions Assume everything we know about continuity, etc., on the real line. Perpendicular
axes. Origin. Co-ordinates of a point. Distance function (by Pythagoras’) gives x2 + y2 = r2 for a
circle.
Important Axioms later. Decartes/Fermat’s view was that analytic geometry was built on top of
Euclid. An alternative way of describing things. But still allowed to assume anything from Euclid.
When doing naive analytic geometry, use whichever method is easier.
Example: Let A = ( x, y) and Bp
= (v, w). Then C := ( x + v, y + w) defines a parallelogram OACB.
Proof: calculate distances: BC = x2 + y2 = OA, etc. Now use side side side with diagonal to see
that have two congruent triangles (Euclid). Thus everything parallel.
19
3.2
Vector Geometry (1/2 lecture)
Vectors ‘Invented’ by William Rowan Hamilton (1805–1865) Irish mathematician (invented quaternions, Hamiltonian reformulation of classical mechanics).
Also attributable to Oliver Heaviside (1850 onwards) and Gibbs
Hamilton was really about quaternions which work well algebraically, better than do vectors themselves. By c.1900 everyone was using vectors.
Problem: multiplication in 3 dimensions didn’t work, unlike complex numbers as a model for 2D
(later). Adding a 4th dimension sorted everything out.
Definition. A vector has length and direction. All parallel directed segments with same length/direction are
the same vector (or represent the same vector2 ).
Standard representation: place tail at origin and give co-ords of head. Various notations to make
easier. Right now: everything 2D.
Algebraic definition of addition/scalar mult. See that addition has a nice picture.
Theorem. Adding fourth point to A, B like in previous section to give parallelogram is very easy to check!
Theorem. The medians of a triangle meet at a point 1/3 of the way along each median.
Proof. Let the triangle be spanned by u, v, with tails at O. Define the point M along the median
bisecting the opposite side by
~ = 2 · 1 ( u + v ) = 1 ( u + v ).
OM
3 2
3
Now observe that any point on the median on an adjacent side has position vector
1
1
1
1
v+α u− v
or
u+β v− u .
2
2
2
2
Observe that α = β =
3.4
1
3
gives OM.
Angles in Co-ordinate Geometry (1/2 lecture)
Relies on defining sin θ, cos θ (as before but extend to all angles). Indeed some of the earliest measures
of angle were actually measure of trig function (usually cotangent bizarrely). Then get
x = r cos θ,
y = r sin θ
Define radian measure of an angle as that where the radius equals the arc. Thus a natural unitless
unit!
Theorem (Cosine Rule). (Take Pthag as read) c2 = a2 + b2 − 2ab cos C
2 So
each vector is an equivalence class of directed line segments.
20
Proof. Let the base be b and dropa perpendicular from the top of a down to b. Call height h and split
b = x + y, where x = a cos θ. Then, using Pthag twice:
c2 = h2 + y2 = a2 − x2 + y2 = a2 + (y + x )(y − x ) = a2 + b( x + y − 2x )
= a2 + b2 − 2bx = a2 + b2 − 2bx cos C
Situation works for all possible arrangements of triangles: just take care of sign of x, y. See pics:
a
h
h
c
a
c
a
c
C
C
x<0
y>0
x>0
b
y<0
b
y>0
b
Theorem (Sine Rule).
C
sin A
a
=
sin B
b
=
sin C
c
=
1
d
h
x>0
where d is diameter of circumscribed circle.
Proof. 1d makes this easy! Draw circulscribed circle and make a triangle with a diameter and one side
a. This is right angled. Either:
Angle A since shares same arc, or
Angle π − A since opposite angles in circumscribed quadrilateral are supplementary.
Then sin(π − A) = sin A = da .
Definition 3.1. Dot product: u · v = u1 v1 + u2 v2
Theorem 3.2. u · v = |u| |v| cos θ
Proof. Just use cosine rule on u, v, v − u triangle.
α
Can now easily obtain cos(α − β) = · · · from dot product on the vectors ( cos
sin α ),
cos β
sin β
, etc. Now
even/oddness of sine cos, and sin α = cos(90 − α) covers the sine equivalents. Homework.
3.5
The Complex Plane (1/2 lecture)
History Euler (1707-1783) extended R to place where equation x2 = −1 had a solution. Symbol i.
Allows a fuller description of 2D geometry encapsulated algebraically: i.e. rotations and reflections
are algebra.
Basic Ideas Argand diagram, Addition, (real) scalar multiplication, modulus (length), multiplication, conjugate (reflection).
Polar Form z = x + iy = |z| (cos θ + i sin θ ) = |z| eiθ (definition for ez essentially follows power
series). Call θ = arg(z).
Now multiplication is geometric: multiply modulus and add arguments (mod 2π).
21
Complex Functions Real and imaginary parts, etc. So everything works as aexpected.
Theorem (Fund Thm of Algebra). Let p(z) be a non-constant polynomial. Then there is a complex number
a with p( a) = 0.
Long division, etc., shows that can completely factorize over C.
The point at infinity and the extened complex plane Suppose that a sequence (zn ) of complex
numbers has modulus increasing without bound:
∀ M > 0, ∃ N ∈ N such that |zn | > M.
Certainly limn→∞ |zn | = ∞ in the real sequence sense. In complex geometry, agree to create anew
point ∞ and define ∞ := limn→∞ zn for any such sequence. The resulting set is the extended complex
plane or Riemann Sphere
S 2 = C ∪ { ∞ } = C = C∞ .
Example. f : S2 → S2 : z 7→ z−1 2 is bijective.
Certainly f |C\{2} : C \ {2} → C \ {0} is bijective, with inverse f −1 (z) = 2 + 1z . Now observe that in
S2 , we have f (2) = ∞ and f (∞) = 0. Similarly f −1 (∞) = 2 and f −1 (0) = ∞.
Stereographic Projection Can visualize S2 as a sphere in R3 , and C as the equatorial plane. In
this identification, ∞ = (0, 0, 1) is the north pole. Draw picture: If ∞ = (0, 0, 1) and point in C is
z = |z| eiθ , then quickly see that all points on the line joining them are


 
0
|z| cos θ
0 + t  |z| sin θ  .
1
−1
Simply need this having length 1: whence t = 0,
( X, Y, Z ) =
| z |2 − 1
,
,
| z |2 + 1 | z |2 + 1 | z |2 + 1
2x
2y
2
.
| z |2 +1
Thus intersection point has co-ords
!
Very important tool for transferring spherical geometry to planar.
Theorem. Circles in S2 map to circles or lines in C (line being a circle through ∞)
2x
2y
x 2 + y2 − 1
Proof. We have the correspondence ( X, Y, Z ) =
,
,
. Every circle
x 2 + y2 + 1 x 2 + y2 + 1 x 2 + y2 + 1
d
in S2 is the intersection of a plane aX + bY + cZ = d with the sphere. Since √ 2 | |2 2 is the distance
a +b +c
of the plane from the origin, we have a non-empty intersection iff d2 < a2 + b2 + c2 .
Now assume that z = x + iy is projected onto said plane in the sphere. Then
2ax + 2by + c( x2 + y2 − 1) = d( x2 + y2 + 1) ⇐⇒ (d − c)( x2 + y2 ) − 2ax − 2by + (c + d) = 0.
There are three cases:
22
1. (c = d = 0)
In this case we have ax + by = 0 which covers all lines through the origin.
2. (c = d 6= 0)
In this case we normalize the equation of the plane so that both c = d = 1. Then
ax + by = 1,
which covers all straight lines not passing through the origin.
3. (c 6= d) In this case we divide the equation of the plane through by d − c so that we may assume
that we have d − c = 1. Then
x2 + y2 − 2ax − 2by + 2c + 1 = 0 ⇐⇒ ( x − a)2 + (y − b)2 = a2 + b2 − 2c − 1.
Note that
d2 < a2 + b2 + c2 ⇐⇒ (c + 1)2 < a2 + b2 + c2 ⇐⇒ 1 + 2c < a2 + b2
⇐⇒ a2 + b2 − 2c − 1 > 0,
whence we certainly have the equation of a circle in C. We have complete freedom over a
and b and can therefore choose any center for this circle. The freedom of choice of c allows
us to select any desired radius. Conversely, given center ( a, b) and radius r, we may choose
c = 21 ( a2 + b2 − r2 − 1) and d = 21 ( a2 + b2 − r2 + 1), then
aX + bY + cZ = d.
Example. ( x − 7)2 + (y + 2)2 = 49 corresponds to a = −7, b = 2, c = 21 (49 − a2 − b2 + 1) = − 23 ,
d = c − 1 = − 52 . We therefore have the intersection of the sphere and the plane
3
5
−7X + 2Y − Z = − ⇐⇒ 14X − 4Y + 3Z = 5.
2
2
Note that a2 + b2 + c2 > d2 , as required.
Angles and Area Consider r( x, y) =
1
x 2 + y2 +1
2x
2y
x 2 + y2 −1
as a parameterization of the sphere. The
tangent vectors at r( x, y) are

 2


2( x2 + y2 + 1) − 4x2
y − x2 + 1
1
2

 −2xy 
=
−4xy
rx = 2
( x + y2 + 1)2
( x 2 + y2 + 1)2
2
2
2
2
2x ( x + y + 1) − 2x ( x + y − 1)
2x


−2xy
2
 x 2 − y2 + 1
ry = 2
( x + y2 + 1)2
2y
Take dot products:
4
4
2
2
2
2 2
2
(
y
−
x
+
1
)
+
4x
y
+
4x
= 2
( x + y2 + 1)2
( x 2 + y2 + 1)4
2
,
r x · ry = 0.
|r x | = ry = 2
x + y2 + 1
rx · rx =
23
Consider two small changes in ( x, y): ( x, y) + ( dx1 , dy1 ) and ( x, y) + ( dx2 , dy2 ). The resulting tangent vectors on the surface of S2 are then
du = r x dx1 + ry dy1 ,
dv = r x dx2 + ry dy2 .
The angle between these satisfies:
du · dv
cos θ =
=q
| du| | dv|
=q
2
|r x |2 dx1 dx2 + ry dy1 dy2
q
2
2
|r x |2 dx21 + ry dy21 |r x |2 dx22 + ry dy22
dx1 dx2 + dy1 dy2
q
,
dx21 + dy21 dx22 + dy22
thus the angles are equal! Stereographic projection preserves the angles between intersecting curves.
More generally, the metric on S2 is
( x2
4
( dx2 + dy2 )
+ y2 + 1)2
which is conformal to the flat metric on C. Lengths are preserved near the unit circle, but are reduced
hugely for large x, y.
The area element is


−4xy2 − 2x ( x2 − y2 + 1)
4
2
2
2

 dx dy
−2y(y − x + 1) − 4x y
dS = r x × ry dx dy = 2
2
4
( x + y + 1) (y2 − x2 + 1)( x2 − y2 + 1) − 4x2 y2 

−2x ( x2 + y2 + 1)
4
2
2
dx dy


−
2y
(
y
+
x
+
1
)
= 2
( x + y2 + 1)4 2
2
2
2
(1 − x − y )(1 + x + y ) 

−2x
q
4
4
2
2 2


−2y
= 2
dx dy = ( x2 + y2 + 1)3 (1 + x + y ) dx dy
( x + y2 + 1)3 2
2
1−x −y
4
= 2
dx dy
( x + y2 + 1)2
One can check that this makes sense: the lower half of the sphere corresponds to x2 + y2 ≤ 1. Its area
(using polar co-ordinates) is then
ZZ
x 2 + y2 ≤1
dS =
Z 2π Z 1
0
0
1
4
−2 r dr dθ = 2π · 2
= 2π.
(r 2 + 1)2
r + 1 0
Similarly, the upper half of the sphere has area
ZZ
x 2 + y2 ≥1
dS =
Z 2π Z ∞
0
1
→∞
4
−2 r
dr
dθ
=
2π
·
= 2π.
(r 2 + 1)2
r 2 + 1 1
The sphere has area 4π as expected.
24
Rational points Suppose that x, y ∈ Q. Then clearly X, Y, Z ∈ Q. Conversely, if X, Y, Z ∈ Q, then
Z = 1−
x2
2
∈ Q =⇒ x2 + y2 + 1 ∈ Q.
+ y2 + 1
But then x, y ∈ Q. Thus sterographic projection maps rational points in the plane to rational points on
the sphere. Indeed all rational points on the sphere can be found this way. Restricting this to points on
the line y = 0, the stereographic projection is a bijection S1 → R ∪ ∞, the one-point compactification
of R. We see that the rational number x = mn ∈ R corresponds to the rational point
( X, Z ) =
2x
x2 − 1
,
x2 + 1 x2 + 1
=
m2 − n2
2m
,
m2 + n2 m2 + n2
.
This exactly gives us the standard parameterization of the primitive Pythagorean triples:
a2 + b2 = c2 =⇒ a = 2m, b = m2 − n2 , c = m2 + n2
25
26
3.6
Birkhoff’s Axiomatic System for Analytic Geometry (1 lecture)
Note: George David Birkhoff published these in 1932, he is my great-great-grand-advisor!
Assume the usual properties of real numbers (continuity, etc.)
Undefined terms Point, line
Let the set of points be S
Distance: ∃d : S × S → R+
Angle: ∃m∠ : S × S × S → [0, 2π )
Postulates/Axioms
Ruler Points on any line ` are in 1–1 correspondence with all real numbers x ∈ [0, d( A, B)] in such a
way that if x A , x B correspond to A, B, then | x A − x B | = d( A, B).
Euclidean Given points A 6= B, ∃ a unique line containing them.
Protactor The rays emanating from a point O can be put into a 1–1 correspondence with the set
[0, 2π ) so that if a, b correspond to rays OA, OB, then m∠ AOB = a − b mod 2π. This correspondence is continuous in A, B.
Note that this says that m∠ AOB = −m∠ BOA mod 2π.
SAS similarity Suppose that two triangles share an equal angles, and that the sides adjacent to said
angle are in the same ratio. Then the remaining angles are equal, and the final sides are in the
same ratio.
Definitions
Betweenness B between A and C if d( A, B) + d( B, C ) = d( A, C )
Segment AB is the points A, B and all those between
Basic shapes Triangles, circles, etc.
Theorems All the standard results from Euclidean Geometry
Analytic Geometry is a model of these axioms
Define “point” as an ordered pair ( x, y) of real numbers.
Distance: d( A, B) = usual square root formula.
Line: all points satisfying a linear equation ax + by + c = 0
Angle: Define vectors as ordered pairs then ‘angle’ via
cos θ =
v·w
|v| |w|
Can define cosine by power series, so no geometric meaning required.
Lemma 3.3. Angle is well-defined w.r.t. Birkhoff’s axioms
27
Proof. Given a point, identify a ray with a direction vector w = ( yx ) of length 1. Two choices:
x
x
√
√
w=
,
ŵ =
.
− 1 − x2
1 − x2
Fix a particular direction w = 10 and calculate:
cos θ = p
x
x 2 + y2
= x ∈ [−1, 1]
Since cos−1 is continuous, obtain a unique 0 ≤ θ ≤ π. Extend to π < θ < 2π by using ŵ.
Thus have bijective correspondence between angles
and [0, 2π ).
Now check the rest of the postulate: if w1 = yx11 and w2 = yx22 , then check that
cos θ1 =
Define
sin θ1 =
x1
,
| w1 |
cos θ2 =
x2
.
| w2 |
y1
,
| w1 |
sin θ2 =
y2
.
| w2 |
Show algebraically (using just dot product and above formulae) that the angle between these vectors
is θ = θ1 − θ2 , since
cos(θ1 − θ2 ) = cos θ1 cos θ2 + sin θ1 sin θ2 .
Birkhoff’s Euclidean Axiom is also satisfied. If ( x1 , y1 ) and ( x2 , y2 ) lie on the line ax + by + c = 0
then
a( x1 − x2 ) + b(y1 − y2 ) = 0,
whence we may choose a = y1 − y2 , b = x2 − x1 . Using either point we see that c = x1 y2 − x2 y1 ,
whence the desired line has equation
(y1 − y2 ) x + ( x2 − x1 )y + x1 y2 − x2 y1 = 0.
Ruler postulate essentially trivial. Fix points P, Q on a line ` then all points on the line (betweenness) are
P+
t
( Q − P ).
d( P, Q)
Then, given A ∈ ` define x A = t. Palpably | x A − x B | = d( A, B).
Finally the cosine rule holds (only depends on dot product on a triangle) whence SAS also holds:
c2 = a2 + b2 − 2ab cos C
says that if one scales a, b by the same factor, then c must scale by that factor. Repeated applications
of the rule prove that the corresponding angles are the same.
Weaknesses/benefits of Birkhoff’s axioms: needs real numbers! Includes continuity, etc. Easier
to compute with. Less easy to see pictures.
28
4
Constructions
4.1
Euclidean Constructions (1 lecture)
4.3
Constructibility (1 lecture)
Pre-midterm goal is the following:
Theorem 4.1. The constructible numbers form a field C such that Q ⊆ C ⊆ C.
(Put together: only practicalities pre-midterm, fields stuff after midterm)
Definition 4.2. Let S = { P0 , . . . , Pn } be a finite set of points in a plane Π. Then,
(a) A line ` is constructible from S if it passes through two distinct points of S.
(b) A circle is constructible from S if it has centre in S and radius = distance between two points in S.
Definition 4.3.
1. Let P0 , P1 , be two points in a plane Π. Then a point P ∈ Π is constructible
from P0 , P1 if there is a finite sequence { P0 , P1 , . . . , Pn } with Pn = P and s.t., for any set S j =
{ P0 , P1 , . . . , Pj }, the point Pj+1 is one of the following three cases,
(a) The intersection of two lines constuctible from S j .
(b) in the intersection of two circles constructible from S j .
(c) in the intersection of a line and a circle constructible from S j .
2. A angle is consrtuctible if it is the angle between two constructible lines.
3. ( x, y) ∈ R2 is constructible if it is constructible from (0, 0) and (1, 0).
4. z ∈ C is constructible if (<z, =z) is constructible.
Proposition 4.4. Let ` be a line and P, Q be two points in `. Then the perpendicular lines to ` through P, Q
and the mid-point of P, Q are all constructible.
Proof.
.
..........................
.
............ .... ....................................
...................
......................
.
.
.
.
.
.
.. . ..
...... .... ..........
....
....
....
....
....
...
....
....
....
.. ...
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P2
P
Q
~ is constructible.
~ = m PQ
Proposition 4.5. `, P, Q as before. Then for any m ∈ Z, the point R s.t. PR
29
Proof. Do m times , a circle of radius | PQ| moving m to the right each time.
...................................................... ............................................................... ...............................................................
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P
Q
P2
P3
P4
Corollary 4.6. Any integer is constructible.
Theorem 4.7. Let ` be a line and P a point not in `. Then the line through P perpendicular to ` and the line
through P parallel to ` are both constructible.
Proof. By previous proposition, we can draw a circle about P of sufficient radius so that it intersects `.
Call the two intersection points P1 , P2 . Can now erect a perpendicular bisector to P1 , P2 by proposition
4.4. This is the first of the required lines. We can appeal to the same proposition to erect a further
perpendicular to this new line at P giving us the second required line.
~ for q ∈ Q is constructible
~ = q PQ
Theorem 4.8. ` a line, P, Q points in `. Then any point R ∈ ` s.t. PR
from P, Q.
Proof. We can construct a perpendicular to ` through P, call
it m. Let S 6= P be constructible on m. Now construct n, the
perpendicular to m through S. Now define T s.t. |ST | =
~ = pST.
~
| PQ|. Now, ∀ p ∈ Z+ , we can construct X s.t. SX
Now define the points Y and Z s.t.,
~ ) ∩ m,
{Y } = ( XQ
Then | PZ | =
1
p
~ ) ∩ `.
{ Z } = (YT
p
and can therefore construct q for integer q.
q
~
~ = q PQ.
∴ ∀ p ∈ Q, we have constructed R s.t. PR
p
.
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...
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S
n
T
X
m
P
Z
Q
`
Y
Corollary 4.9. Any rational number is constructible.
Theorem 4.10. C is a field.
Proof. C ⊆ C by default. We are reuqired to show that C is closed under the field operations of C.
For addition: Suppose a, b ∈ C . Now construct a circle of radius |b| centered at a, and draw the
parallel to 0b through a. The intersections are a ± b.
For multiplication: similar picture to division theorem (save for midterm!). Draw line from 1 to q + i,
intersect y-axis, then draw through p to pq + i.
For division: already done.
Thus C is a subfield of C.
30
Post Midterm
Theorem 4.11. Suppose the field F is constructible. Then all the points constructible in one step from F are
solutions of either linear or quadratic equations with coefficients in F.
Proof. There are three cases: One can build a new point only by the intersection of two lines joining
points in F, two circles with center and radius in F, or a line and a circle joining similarly. All such
have equations of the form
az + bz + c = 0,
(z − a)(z − a) = r2
where a, b, c, r ∈ F (indeed r ∈ R). It can be checked the the intersections in each case involve solving
a linear or quadratic equation for z. By the fundamental theorem of algebra such an equation always
has a root.
Corollary 4.12. If F is a constructible field and P is a point constructible in one step from F, then
FP := { a + bP : a, b ∈ F}
is constructible, and has characteristic 1 or 2 over F.
If F is a constructible field built in a finite number of such steps from Q, then [F : Q] = 2n for some n ∈ N.
If z ∈ C is constructible, then it has minimal polynomial over Q of degree 2n for some n ∈ N.
Theorem 4.13. Let l, m be lines intersecting at P. Then the line through P bisecting an angle between l, m is
constructible.
Proof. Draw a circle of constructible radius about P. Call the intersections with l, m P1 , P2 . Can bisect
P1 , P2 with a line through P by previous work.
Corollary 4.14. If cos α is constructible, then ± cos α2 , ± sin α2 are also constructible.
√
Proposition 4.15. If ρ ∈ R+ is consrtructible, then ρ is also.
Proof. If ρ is constructible, then
−1 <
2ρ
n2
2ρ
n2
2ρ
n2
− 1 is constructible for all n ∈ Z. Choose large enough n so that
− 1. Then,
− 1 < 1. Put cos α =
r
√
ρ
1 + cos α
α
=
,
cos =
2
2
n
√
is constructible and hence so is ρ.
Remark It can be shown that if z2 is constructible for z ∈ C, then z is constructible. Similarly, if
a, b (6= 0) are constructible then so are a − b, ab−1 . It follows that the set of all constructible numbers
forms a field F s.t. Q ⊂ F ⊂ C.
Definition 4.16. Suppose K is an extension field of Q. If there is a finite sequence of field extensions
F0 ⊂ F1 ⊂ · · · ⊂ Fr s.t. F0 = Q, K ⊆ Fr and [ Fj : Fj−1 ] = 2, ∀ j, then K is called a constructible extension
of Q.
Theorem 4.17. Let x, y ∈ R. The following are equivalent.
31
(a) x, y are both constructible.
(b) x + iy ∈ C is constructible.
(c) Q( x, y) is a constructible extension of Q.
(d) Q( x + iy) is a constructible extension of Q.
Proof. “(a) ⇐⇒ (b)” To construct ( x, 0), (y, 0), draw perpendiculars from ( x, y) to the coordinate axes.
Then draw a circle about (0, 0) of radius |y| which will intersect the x-axis at (y, 0). The argument
can easily be reversed.
“(b)⇒(c)” Let ( x, y) be constructible and P0 , P1 , . . . , Pn be as in definition 4.3 with Pn = ( x, y). Let
the subfields K0 , K1 , . . . , Kn of R be defined by K0 = Q and Ki = Ki−1 ( xi , yi ) for each i. There are
three cases to consider depending on which instance of the definition we are in for each i. Direct
calculation shows that [Ki : Ki−1 ] = 1 in the first case (two straight lines) and 2 in the other two cases.
Now ( x, y) = ( xn , yn ) and so Q( x, y) ⊆ Kn . Therefore Q( x, y) is constructible.
“(c)⇒(a)” Suppose Q = F0 ⊆ F1 ⊆ · · · ⊆ Fr with K ⊆ Fr , [ Fj : Fj−1 ] ∈ {1, 2}. If α ∈ Fj \ Fj−1 (arbitrary)
then the minimal polynomial of α over Fj−1 has degree 2. Thus ∃b, c ∈ Fj−1 s.t. α2 + 2bα + c = 0. α
can thus be written in terms of b, c ∈ Fj−1 by field operations and the square root. Hence if b, c are
constructible then so is α. Apply recursively.
“(c)⇒(d)” Suppose the extension field structure of before. Since i2 = −1 we have that [ Fr (i ) : Fr ] is
either 1 or 2. Thus Q( x + iy) ⊆ Q( x, y, i ) ⊆ Fr (i ) and so Q( x + iy) is a constructible extension of Q.
“(d)⇒(b)” The argument is very similar to that of “(c)⇒(a)”.
Corollary 4.18. If z ∈ C is constructible, then [Q(z) : Q] = 2r for some r ∈ Z+ .
Proof. Clear from theorem and definition.
2r = [ Fr : Q(z)][Q(z) : Q] = 2 p 2q
by prime factorisation.
Theorem 4.19. It is impossible to duplicate a cube (draw a cube of double the volume of a given cube).
Proof.
Start with a cube of length 1. √
If we could duplicate the cube, we
would require the length
√
√
3
3
3
2 to be constructible. Consider √
Q( 2). The minimal polynomial of 2 is x3 − 2 (irreducible
√ by
Eisenstein with p = 2). Hence [Q( 3 2) : Q] = 3 which is not an integral power of 2. Therefore 3 2 is
not constructible.
Theorem 4.20. It is impossible to trisect an angle (in general).
Proof. Conside the angle π3 . Trisecting being possible is equivalent to cos π9 being constructible.
Using the triple angle formula,
1
π
π
π
√ = cos = 4 cos3 − 3 cos .
3
9
9
2
We therefore wish to solve p(α) = 8α3 − 6α − 1 = 0 for α constructible. Claim that p is irreducible
over Q. Change variable to γ = α1 . Then γ3 + 6γ2 − 8 = 0 ⇒ γ is a divisor of 8. All divisors of 8 fail
to generate 0 so p(α) is irreducible and is thus the minimal polynomial of α over Q.
∴ [ Q ( α ) : Q ] = 3 6 = 2r .
32
Remark If we assume the theorem that π is transcendental then it can easily be shown that ‘squaring the circle’ is impossible (would require π1 to be constructible).
5
Transformational Geometry
5.1
Euclidean Isometries (1 lecture)
Klein’s Idea: Geometry of a set is the study of invariants (properties preserved under a group of
transformations of the set)
In Euclidean Geometry, these transformations are ‘isometries’.
Definition 5.1. A function f : R2 → R2 is a Euclidean Isometry if it preserves lengths: i.e.,
∀ P, Q ∈ R2 , we have | f ( P) f ( Q)| = | PQ|
−→
Considering this in terms of vectors (pairs of points): for all vectors if u = AB, then f (u) :=
−−−−−−→
f ( A ) f ( B ).
Book approach: very affine (no origin) Prove that isometries are 1–1, onto, preserves lines, then angles. Hard phrasing as very abstract. Eventually does fixed points: no fixed points gives translations,
one fixed is rotation, two is reflection, three (or more) is the identity.
Instead: Do with vectors. What does an isometry do to a triangle? If A, B, C are the corners of
a triangle, it follows that f ( A), f ( B), f (C ) is a congruent triangle. What does this say for vectors?
−→
−
→
−→
If u = AB and v = BC, then the third side of the triangle is u + v = AC. Clearly f (u + v) =
f (u) + f (v). Otherwise said: f is a linear transformation on vectors.3
Now compute:
1
| f (v) + f (w)|2 − | f (v)|2 − | f (w)|2
2
1
= | f (v + w)|2 − |v|2 − |w|2
2
1
= | v + w |2 − | v |2 − | w |2
2
= v·w
f (v) · f (w) =
Therefore dot products are preserved. Hence:
Theorem 5.2. Isometries preserve angles (and parallel lines, etc.)
With a little 3A, can think of isometries as linear transformations of the vector space R2 which,
upon choosing the usual basis, become orthogonal matrices. Check calculations to see that

2
2

a + c = 1
a b
and isometry ⇐⇒
ab + cd = 0
c d

 2
b + d2 = 1
−−−−−−→ −−−−−−→
−→ −→
−→
−→
care: f (OA + OB) = f (OA) + f (OB) = f (O) f ( A) + f (O) f ( B). Can only think of this as additive on points if
f (O) = O is a fixed point.
3 Take
33
Put together: get b = ±c and usual cos/sin breakdown of rotations/reflections.
Abstractly: E(2) = T (2) o O(2) (semidirect product with translations normal subgroup). Indeed,
if ( A, c) : v 7→ Av + c, then ( A, c)−1 = ( A−1 , − A−1 c), and ( A, c) ◦ ( B, d) = ( AB, Ad + c), whence
( A, c) ◦ ( I, d) ◦ ( A, c)−1 = ( A, Ad + c) ◦ ( A−1 , − A−1 c) = ( AA−1 , − AA−1 c + Ad + c)
= ( I, Ad)
is a translation. Thus translations form a normal subgroup.
7
Non-Euclidean Geometry
7.1
Background and History (1/2 lecture)
Recall Euclid’s Postulates:
1. Can construct a line between points
2. Can extend a line indefinitely
3. Can describe a circle given a center and distance
4. All right angles congruent
5. Parallel postulate (equivalently Playfair’s postulate: given line and a point not on the line, there
exists a unique parallel line through the point)
Describing a geometry as non-Euclidean typically means ignoring the the parallel postulate.
For much of history many believed that the parallel postulate was a theorem of the others. Just
couldn’t find a proof of this.
Early approaches: Giovanni Saccheri (1667–1773) + Johann Lambert (1728–1777) tried assuming
the opposite of parallel postulate to obtain a contradiction to any theorem based only on the first four
postulates (1st half Elements I). To no avail. One approach: assume a quadrilateral with base-adjacent
sides congruent and base right angles. Parallel postulate required to prove that the quadrilateral is
a rectangle. They tried assuming that the angle(s) at the top were not right angles. Contradictions
with obtuse angles, but could not find any assuming acute angles. Both went a long way, but gave
up, objecting to the unnatural ideas they were having!
Bolyai–Lobashevsk Postulate In the early 1800s James Boylai, Carl Friedrich Gauss and Nikolai
Lobashevsky (independently) defined an axiomatic geometry based on the first four of Euclid’s postulate plus an alternative to the parallel postulate:
Given a line and a point not on the line, there exist more than one parallel line through the point
The axiomatic system based on these was slowly believed to be constistent and this is now known
as hyperbolic geometry. Felix Klein indeed called this the hyperbolic postulate. Consistency proved in
late 1800s by Beltrami, Klein and Poincaré. All these created models of hyperbolic geometry inside
Euclidean geometry by defining point, line, etc weirdly.
34
7.2
Models of Hyperbolic Geometry (1 lecture)
The Poincaré Model (actually Beltrami)
Definition 7.1. The Poincaré disk is the set of points inside the unit circle {( x, y) : x2 + y2 < 1}.
The undefined terms in the Poincaré model are as follows:
Points Points in the Poincaré disk
Lines Any straight line or circular arc meeting the unit circle at right angles (call these hyperbolic
lines)
Theorem 7.2. This is a model of hyperbolic geometry
Proof. Check the postulates.
1. Can certainly join any two points in the Poincaré disk by a hyperbolic line. If the two points
lie on a diameter, then this diameter is the required line. Otherwise. . . note that all circles intersecting the unit circle orthogonally have the form x2 + y2 + ax + by + 1 = 0 with a2 + b2 > 4.
2
2
Indeed this has square radius r2 = a +4 b − 1. Given two points on such a circle, this becomes a
pair of linear equations for a, b. . .
2. We need to define distance: this is tricky and we do it below.
3. Any finite hyperbolic line does not intersect the circle and thus may be extended.
4. Automatically true since angle has the same meaning as in Euclidean geomtery
5. Draw some! Finding all is a challenge.
Hyperbolic Distance
Definition 7.3. Given two points P, Q in the Poincaré disk, there is a unique hyperbolic line on which
they lie. Suppose that this arc intersects the unit circle at R, S. Then the hyperbolic distance between
P and Q is defined to be
| PS| | QR| d( P, Q) = ln
| PR| | QS| Note that this is independent of the labelling of P, Q and R, S. Observe also that limP→S d( P, Q) = ∞
so that all points in the disk are infinitely far from the circle!
Lemma 7.4. The distance function centered at P is increasing along any hyperbolic line.
Proof. Let | Q| R = x and | QS| = y, along the same arc. Then y decreases as x increases. If both are
positive, then
d
d
1 1 dy
d( P, Q) =
ln x − ln y = − ·
>0
dx
dx
x y dx
35
IVT says circles of any radius can be drawn. Very hard to draw!
Lemma 7.5 (Make HW). A circle intersects the unit circle orthogonally iff x2 + y2 + ax + by + 1 = 0 with
2
2
a2 + b2 > 4. Indeed this has square radius r2 = a +4 b − 1
√
Proof. Draw picture: if radius r then orthogonality says that center must be a distance 1 + r2 from
origin. Thus equation is
p
p
( x − 1 + r2 cos θ )2 + (y − 1 + r2 sin θ )2 = r2
Now multiply out to see that
p
p
x2 + y2 − 2 1 + r2 cos θx − 2 1 + r2 sin θy + 1 = 0
√
√
Thus a = −2 1 + r2 cos θ and b = −2 1 + r2 sin θ, whence
a2 + b2 = 4(1 + r2 ) > 4 ⇐⇒ r > 0.
Conversely, if a2 + b2 > 4, then choose θ via tan θ =
b
a
and r2 =
a2 + b2
4
− 1.
√
Problem Find the hyperbolic distance between (1/2, ± 5/12).
Solution We need to find the hyperbolic line on which these points lie. Thus:
√
1
10
5
1
=⇒ b = 0, a = −
+
+ a ± 5/12b + 1 = 0
4 12 2
3
The center is therefore at ( 53 , 0).
The intersection with the unit circle is when
10
3
4
− x + 2 = 0 =⇒ x =
=⇒ y = ±
3
5
5
PR 2 Compute Distance: Since | PR| = | QS| and | PS| = | QR|, we only need d( P, Q) = ln | |2 . Now
| PS|
!2
r
√
5
4
3 1 2
4
+
−
−
= (4 − 15),
| PR| =
5 2
12 5
15
!2
r
√
5
3 1 2
4
4
2
−
+
+
= (4 + 15)
| PS| =
5 2
12 5
15
2
Then
√
√
15
√ = ln(31 + 8 15) ≈ 4.12687
d( P, Q) = ln
4 − 15
p
√
Other sides: lengths | PR| = 2/3, etc, gives ln(5 + 2 (6)) ≈ 2.292
Hyperbolic triangle with these edges: 2.292 : 2.292 : 4.12687, Euclidean triangle with these edges:
0.5 : 0.5 : 0.707. Relative ratios: 1 : 1 : 1.8 hyper and 1 : 1 : 1.414. Euclidean.
Little work: calculate angles in triangle to be 90 : 24.03 : 24.03. Angles in triangle sum to ≈
138.07◦ !
Can be shown that hyperbolic area is equal to (radian defect), so that an ideal triangle with corners
on the circle has zero angles and area π. Consequently, area of Poincare disk is infinite.
4+
36
The Klein model
Similar approach, except that lines are taken to be straight line segments within the unit circle and
the distance function is
1 | PS| | QR| dK ( P, Q) = ln
2
| PR| | QS| where R, S are the points where the chord meets the circle. Problem is that angles aren’t preserved,
so need a different concept of angle.
Perpendicularity comes from following idea: Take a line (chord) and find the tangents to the
unit circle where it meets. Any chord which passes through the intersection of these tangents is
orthogonal to the original line. Measuring other angles is difficult!
General Theorem: hyperbolic geometry is negatively curved. Impossible to find a mapping f :
model → R2 which preserves both concepts of straight line and angle. Poincare’s model preserves
angles, Klein’s straight lines, but cannot do both. Similar problem in map-making.
7.3
Basic Results of Hyperbolic Geometry (2 lectures)
All first 28 theorems in Euclid are true (neutral geometry)! So can use all sorts of stuff like SAS,
etc. Can also use anything from metric geometry: we have a well-defined distance function: any
function f : H → H such that ∀ A, B, d( f ( A), f ( B)) = d( A, B) preserves theorems. Harder question
to identify isomtries explicitly (essentially SO(1, 1)).
Theorem 7.6 (Fundamental Thm of Parallels in Hyperbolic Geometry). Given a hyperbolic line ` and a
point P not on `, there are exactly two parallel lines m, n through P with the following properties:
1. Every line through P lying within the angle made by one of the parallels m, n and the perpendicular from
P to ` must intersect `, while all others are parallel to `.
2. m, n make acute angles with the perpendicular from P to `.
Proof. Drop a perpendicular from P to ` at Q. Since angles are continuous, IVT says that there exists
an angle ∠QPC so that the line through PC separates those angles intersecting ` from those with
don’t. Now reflect this angle across PQ. Reflections preserve parallels and angles (they are isometries!). Gives the other line PD. Thus 1 is proved.
For 2, suppose for contradiction that ∠QPC = 90◦ . Then so is ∠QPD. But then DPC lie on a
hyperbolic line parallel to `. Hyperbolic postulate says ∃ at least one more line through P parallel to
`. This must lie inside one or other of the angles and so intersects `. Contradiction.
If ∠QPC > 90◦ then the extension of PC through P lies inside the angle ∠QPD and thus intersects `.
Contradiction.
Note important ability to use reflections abstractly without computation!
Definition 7.7. The two parallels are called limiting parallels or asymptotic parallels.
Other parallels are ultraparallel.
37
Example Find the limiting parallels to x2 + y2 + 2x + 3y + 1 = 0 through (0, 0).
12
5
, y = 0, − 13
.
Find intersection with boundary: x2 + y2 = 1 =⇒ 2x + 3y = −2. Get x = −1, 13
2
2
All arcs through (0, 0) are diameters (0 + 0 + a0 + b0 + 1 6= 0), whence the limiting parallels are:
y = 0 and y = − 12
5 x.
Repeat for through ( 15 , 0). Obtain x2 + y2 + ax + by + 1 = 0 includes this point. Whence
2
1
1
26
+ a + 1 = 0 =⇒ a = −
=⇒ b = 0.
5
5
5
Example Find the angles in an equilateral triangle centered at the origin and of Euclidean radius r.
Suppose that one corner is on the x-axis. The hyperbolic arcs are all (Euclidean) congruent, so just
think about one of them. The center of one of the circls will be on the line making angle π/3 with the
positive x-axis. Such circlewith radius R has equation
p
√ p
x 2 + y2 − 1 + R2 x − 3 1 + R2 y + 1 = 0
√
If this passes through (r, 0), then r2 − 1 + R2 r + 1 = 0, whence
2
R =
1
r+
r
2
−1
Calculate
dy
r + r −1 − 2x
−(1 − r2 )
√
=
=√
dx
2y − 3(r + r −1 )
3(r 2 + 1)
Thus the angle internal to the triangle is
θ = 2 tan−1 √
1 − r2
3(r 2 + 1)
Sanity check: limr→0+ θ =
π
3
and limr→1− θ = 0.
More strictly: if triangle has hyperbolic radius d, then r =
θ = 2 tan−1
e d −1
e d +1
= tanh d2 . Can then be seen that
sech(d/2)
√
3
Theorem 7.8. If f : D → D is an isometry of the Poincare disk onto itself, then
f (z) = β
z−α
,
αz − 1
|α| < 1, | β| = 1
Consider moving a point w to the origin: this accomplished by taking α = w: any β will do.
1+|z
Theorem 7.9. The distance from O to z is d(0, z) = ln 1−|z||
38
Thus if we are a Euclidean distance r from the origin, then the hyperbolic distance δ is related to
it via
1+r
eδ − 1
δ = ln
,
r= δ
1−r
e +1
Some calculus: we want to estimate the arc-length of a curve z(t) from z(t) = w to z(t + ∆t). Take
an isomtery moving w to the origin:
g(z) = −
z−w
z−w
=
wz − 1
1 − wz
Thus w = z(t) mapped to O and z(t + ∆t) mapped to
z(t + ∆t) − w
z(t + ∆t) − z(t)
=
1 − wz(t + ∆t)
1 − z(t)z(t + ∆t)
0
z (t)∆t
z0 (t)∆t
=
≈
1 − z(t)(z(t) + ∆tz0 (t))
1 − z(t)(z(t) + ∆tz0 (t))
!
0 ( t ) ∆t
z
(
t
)
z
1
+
≈ z0 (t)∆t
1 − z(t)z(t) (1 − z(t)z(t))2
P=
≈
z0 (t)
1 − |z(t)|
2
([ a − bx ]−1 ≈ a−1 + ba−2 x + · · · )
∆t
as ∆t → 0.
We may assume that the path z(t) is locally a hyperbolic line, whence the distance OP is the same
as the distance z(t) → z(t + ∆t) in the limit. Since ln(1 + x ) ≈ x we see that
1 + | P|
2 |z0 (t)|
d(O, P) ≈ ln
∆t
≈ 2 | P| ≈
1 − | P|
1 − |z(t)|2
Theorem 7.10. If z(t) for t ∈ [t0 , t1 ] is a parameterized hyperbolic curve in the Poincare disk, then its arclength is
Z t1
t0
2 |z0 (t)|
1 − |z(t)|
2
dt
Definition 7.11. A geodesic is a path of shortest length between two points.
Theorem 7.12. The geodesics in hyperbolic geometry are the hyperbolic lines (i.e. these have shortest hyperbolic distance between points).
Proof. Given points A, B, move A to zero and B to P via an isometry which leaves P on the x-axis.
The hyperbolic arc from A to B is then transformed to the straight line segment from O to P. It is
enough therefore for us to show that the shortest hyperbolic path from O to P = ( a, 0) is the x-axis.
Parameterize a curve z(t) = x (t) + iy(t) from O to P and compute the arc-length:
Z t1 p 02
Z t1 √ 02
Z t1
2 x + y 02
2 x
2 |z0 (t)|
dt =
dt ≥
dt
2
2
2
2
t0 1 − x − y
t0 1 − x
t0 1 − | z ( t )|
since we are increasing the numerator and decreasing the denominator. But this RHS is exactly the
hyperbolic distance as ( x (t), 0) parameterizes the straight line from O to P.
39
Area Suppose we’re at a point P( x, y) and we increase the x, y co-ordinates by a small amount
dx, dy respectively. Since angle is preserved, this will produce (approximately) a rectangle with
side-lengths
2 dx
,
1 − x 2 − y2
2 dy
1 − x 2 − y2
It follows that the area element is the product of these and so the area of a region D in the Poincare
disk is
4 dx dy
=
(1 − x 2 − y2 )2
ZZ
D
Theorem 7.13. If δ = ln
the area is
ZZ
D
ZZ
1+r
1−r
D
,
4r dr dθ
(1 − r 2 )2
r=
e δ −1
e δ +1
are the hyperbolic and euclidean distances respectively, then
sinh(δ) dδ dθ
Circles Suppose that a circle has hyperbolic radius δ. Then we may perform an isometry to move
its center to the origin. It would then be parameterized by
cos θ
z(t) = r
sin θ
where r =
Z 2π
0
e δ −1
.
e δ +1
Its circumference is then
2r
4πr
1 3
1 5
dθ =
= 2π sinh(δ) = 2π δ + δ + δ + · · · > 2πδ
1 − r2
1 − r2
3!
4!
Its area would be
Z 2π Z δ
0
0
2
2
sinh(δ) dδ dθ = 2π (cosh(δ) − 1) = π δ + δ4 + δ6 + · · ·
4!
6!
2
> πδ2
Many results in hyperbolic geometry can be obtained by replacing lengths with hyperbolic versions: e.g. consider the hyperbolic distances in a right-triangle of hyperbolic side lengths α, β, γ
where γ is the hypotenuse. By isometries we can place the corners at (0, 0), ( a, 0), (0, b), where
a=
eα − 1
eα + 1
b=
eβ − 1
eβ + 1
What relationship do we have? Calculating first that
cosh(d(z, w)) = cosh ln
|1 − zw| + |z − w|
|1 − zw| − |z − w|
=
|1 − zw|2 + |z − w|2
|1 − zw|2 − |z − w|2
we quickly see that for our triangle, we have
1 + a2 b2 + a2 + b2
1 + a2 1 + b2
|1 − iab|2 + | a − ib|2
cosh γ =
=
·
= cosh α cosh β
=
1 + a2 b2 − a2 − b2
1 − a2 1 − b2
|1 − iab|2 − | a − ib|2
40
This is the hyperbolic Pythagorean theorem. Thought about with power series:
1 4
1 2
1 4
1 2
1 4
1 2
1+ β + β +···
1+ γ + γ +··· = 1+ α + α +···
2
4!
2
4!
2
4!
which, by some cancelling, yields
γ2 +
1 4
1
γ + · · · = α2 + β2 + (12α2 β2 + α4 + β4 ) + · · ·
12
12
which is Pythagoras’ to 2nd order.
9
9.1
Fractal Geometry
In search of a natural geometry (1/2 lecture)
Idea Reality is not built of spheres, planes, solids, etc. These just approximations. So geometry
seems dry. In practice our notions of concepts like length/area change as we change the scale of
what we’re looking at.
Meaning of ‘close to’ in mathematics is difficult!
Length of a coastline Famous paper “How Long Is the Coast of Britain? Statistical Self-Similarity
and Fractional Dimension” by Benoit Mandelbrot (1967) argued that this question doesn’t have a
good answer. Strict response from the Ordnance Survey is ‘It depends’. CIA (arrogantly!) states 7723
miles, though no evidence as to why.
Fractal Foundation Pics
As length R of ruler goes to zero, perimeter (number of rulers needed) goes to infinity. Plotting
log(1/R) against log N seems to give a straight line!
log N ≈ D log(1/R) = − D log R = log R− D =⇒ N ≈ R− D
This number D is the fractal dimension of the coastline. For the UK this is approximately 1.25.
Questions How do we know such a D exists? We don’t: it’s empirical!
What happens for a smooth curve? If we did this for a circle: p
x2 + y2 = 1: approx arc-length by taking rulers of length R = 2(1 − cos θ ), where θ angle at the
center of the circle. If have N rulers, then
2π
2π
2
θ=
=⇒ R = 4 1 − cos
N
N
1
Now as R → 0, clearly N → ∞. But how? Well cos 2π
N ≈ 1− 2 ·
2
R ≈2
2π
N
2
=
4π 2
N2
41
2π 2
,
N
whence
Thus
2 log R = log 4π 2 − 2 log N =⇒
log N
log 4π 2
≈ 1+
→1
log 1/R
log 1/R
So the fractal dimension of a circle (indeed any smooth curve) is 1.
Will see why this idea of fractal dimension useful later.
Self-similarity (1/2 lecture)
9.2
As a concept, ‘fractal’ is hard to define. Refers to a collection of objects with one or more common
properties. Mostly that structure is apparent at huge magnification.4
Picture of fern: subfronds look similar to the whole. Many fractals share this sort of self-similarity.
The Cantor Set Defined as C0 = [0, 1], now delete middle third of interval. Rinse and repeat.
Resulting set is C . It is self-similar in that it is bijective with half of itself. Facts: length (Cn ) =
(2/3)n → 0 but contains infinitely many points (indeed uncountably many - in bijection with [0, 1]!).
Bijective to a line should mean dimension 1, no length should mean dimension 0? How to fix!
Sierpinski triangle/gasket Similar idea. Start with triangle/square and delete middle chunk(s)
repeatedly. Area of result is zero. But clearly contains infinitely many lines.
Fractal Dimension (1/2 lecture)
9.3
Consider building a new figure from an old by taking N copies of the original scaled by a magnification factor r.
Point
Have no size so just get N points
Line Original line is made up of N copies of line scaled by r =
1
N
√
Square Original square made of N (=perfect square) copies of square scaled by r = 1/ N
Cube
N (=perfect cube) copies scaled by r =
Observe that N =
1 D
r
1
√
3
N
where D is the dimension.
Definition 9.1. A figure is self-similar if it is built from a number N of copies of itself scaled by a
magnification factor r < 1.
log N
log N
The fractal dimension of the figure is D = log(1/r) = − log r .
4 In
calculus, if you zoom in to any differentiable point you get a straight line.
42
Examples: Cantor set has D =
log 2
log 3
≈ 0.6309
log 3
log 2 ≈ 1.58496
log 8
Sierpinski gasket D = log 3 ≈ 1.8928
log 20
Menger Sponge D = log 3 ≈ 2.7268
Sierpinski triangle D =
Fractal trees: whole made of three parts each a copy of the original scaled by 0.4 = 2/5. Thus
log 3
D = log(5/2) ≈ 1.199
Second picture D =
log 7
log(1/0.3)
≈ 1.616
43
44
45
9.4
The Koch snowflake
Discuss Koch snowflake. Area, length, number of edges, etc.
9.5
Contraction Mappings and the Space of Fractals (1 lecture)
Thusfar: only dealt with fractals which are self-similar where the whole is made up of N pieces each
scaled by the same factor.
Example The Cantor set is invariant under the mappings
x
.
3
S1 ( x ) =
S2 ( x ) =
x 2
+
3 3
in that both S1 , S2 are 1–1, and C = S1 (C) + S2 (C).
Definition 9.2. A contraction mapping is a function S on a subset of Rn such that ∃c ∈ [0, 1) with
|S( x ) − S(y)| ≤ c | x − y|
Idea is that fractals may be generated by contraction mappings applied to an initial shape.
Cantor set At each stage Cn+1 := S1 (Cn ) ∪ S2 (Cn ). Can think of Cantor set as being the limit under
this contraction mapping.
E.g.:
1
4
∈ C since S1 (S2 ( 14 )) =
1
4
Indeed can apply this contraction mapping to any interval: Cantor set is in this sense an attractor.
Theorem 9.3. Let S1 , . . . , Sn be contraction mappings in Rn with ratios c1 , . . . , cn . Define a transformation
S on H (the set of compact subsets of Rn ) by
S( D ) =
n
[
Si ( D )
i =1
Then, S is a contraction mapping on H, with contraction ratio c = max{ci }, and S has a unique fixed set
F ∈ H given by
F = lim Sk ( E)
k→∞
for any non-empty E ∈ H.
Koch curve Define four contraction mappings, each with factor c = 1/3.
!
√
√
x 1
3
1
3
1
S1 ( x, y) =
,0 ,
S2 ( x, y) =
x−
y+ ,
x+ y
3
2
2
3 2
2
√
√
√ !
1
3
1
3
1
3
x 2
S3 ( x, y) =
x−
y+ ,
x− y+
,
S4 ( x, y) =
+ ,0
2
2
2 2
2
6
3 3
I.e. S2 is rotate 60◦ and shift right, S3 is rotate −60◦ and shift right/up.
Can start with any interval (indeed any curve!) and will get the Koch curve in the limit!
46
47
Sierpinski carpet
48
49
50
Fractals and contraction maps examples
View the animations using acrobat reader or they probably won’t work!
The Koch curve
Define four contraction mappings, each with scale factor c = 1/3.
!
√
√
x 3
3
1
1
1
S1 ( x, y) =
,0 ,
S2 ( x, y) =
x−
y+ ,
x+ y
3
2
2
3 2
2
√
√
√ !
3
1
3
1
3
x 2
1
x−
y+ ,
x− y+
,
S4 ( x, y) =
+ ,0
S3 ( x, y) =
2
2
2 2
2
6
3 3
Thus S1 just scales by 1/3,
S2 scales by 1/3 then rotates 60◦ counter-clock-wise and shifts right,
S3 scales by 1/3 then rotates 60◦ clockwise and shifts right and up,
S4 sclaes by 1/3 and shifts right.
Now iterate the contraction map S := S1 ∪ S2 ∪ S3 ∪ S4 starting with any interval (indeed any
compact subset of the plane) and we will get the Koch curve in the limit!
At each step we apply each of the four maps S1 , S2 , S3 , S4 to the entire picture and union the results
together.
Starting with the line segment [0, 1]
Starting with a parabola
51
Starting with a circle!
For those of you curious how these were made, here’s the code in Asymptote: It was then embedded into pdf using the animate package for LATEX
import graph;
import animate;
size(220);
transform
transform
transform
transform
S=scale(1/3);
T=shift((1/3,0))*rotate(60)*S;
U=shift((1/2,sqrt(3)/6))*rotate(-60)*S;
V=shift((2/3,0))*S;
//line
path[] P={(0,0)--(1,0)};
//parabola
//real f(real x){return x*(1-x);}
//path[] P={graph(f,0,1,20,operator --)};
52
//circle
//path[] P=shift((0.5,0))*scale(0.5)*unitcircle;
int N=6;
animation A;
save();
draw(P[0]);
A.add();
restore();
for(int i=1; i<=N; ++i){
save();
P.push(S*P[i-1]--T*P[i-1]--U*P[i-1]--V*P[i-1]);
draw(P[i],linewidth(0.2));
A.add();
restore();
}
label(A.pdf("controls",multipage=false),fontsize(5));
53
The Sierpinski Carpet
This is produced similarly: at each stage we take the entire picture, scale it by a factor of 1/3 and
make eight copies. This is most obvious in the first animation, where the eight squares in the second
step comprise the entire original square minus the middle ninth.
54
The same animation starting with a circle
And starting from a triangle
55
Different Scale Factors
Finally a fractal with two different scale factors. We make four copies of the original line: two which
are scaled by 1/2 and two which are are scaled by 1/3 and rotated. Pause at the first iteration to see
this clearly.
56
57
9.6
Fractal Dimension
log N
Idea: D = log(1/r) where r is the similarity ratio: I.e. take N copies of the whole object, each with
lengths multiplied by r < 1 to create the next step.
How to make sense of this when there are multiple (perhaps infinite!) copies of different scales?
Definition 9.4. Let A be compact and define the closed e-ball centered at x ∈ A by
B( x, e) = {y ∈ A : d( x, y) ≤ e}
Say that
M
S
n =1
B( xn , e) is an e-covering of A if A ⊆
M
S
n =1
B( xn , e): that is,
∀ x ∈ A, ∃n : x ∈ B( xn , e)
A.
Draw pic! Idea: given A compact and e > 0, there exists a minimal number M of e-balls covering
Definition 9.5. Let A be compact. The minimal e-covering number for A is
(
)
N ( A, e) = min
M:A⊆
M
[
B( xn , e)
n =1
Example: square side length 1, e = 21 seems like N ( A, 12 ) = 4.
Note: N well-defined since compact means that A can be covered by a finite number of open
balls (∃ finite subcover of every covering), now close these. Thus N is the minimum of a non-empty
set of positive integers.
As e decreases, expect N ( A, e) to increase. Should be same sort of relationship as for dimension.
E.g., consider a line length 1. For given 0 < e will need:
e
N ([0, 1], e)
( 12 , 1]
1
1 1
2
(4, 2]
( 61 , 14 ]
3
1 1
(8, 6]
4
1 1
( 10
, 8]
5
1
1
( 2m , 2m−1 ]
m
log N
Observe that log(1/e) → 1 is the dimension of the line. Similar trick happens in other dimensions
though hard to compute precisely.
Definition 9.6. Given a compact set in Rn , its fractal dimension is the limit
DF = lim
e →0
log N ( A, e)
log(1/e)
Theorem 9.7 (Box-counting). Let A be compact and cover Rn by square boxes of side length
be the number of boxes intersecting A. Then
DF = lim
n→∞
log Nn ( A)
log 2n
58
1
2n .
Let Nn ( A)
Idea: number of boxes covering A can be trapped between two sequences of e-balls both converging to the same thing.
Theorem 9.8 (Iterated Function System). Let {Sn }nM=1 be an IFS and A its attractor, where each Sn has
scale factor cn ∈ (0, 1). Suppose that at each stage of the construction, portions of the fractal meet only at
boundary points. Then the fractal dimension is the uniquie number D such that
M
∑ cnD = 1
n =1
log M
− log M
log r = log(1/r ) as
scaled by 13 , we’d need to
Ex: if all scale-factors cn = r are the same, then Mr D = 1 =⇒ D =
For a fractal say made of 2 copies scaled by
D
D
1
1
2
+2
= 1 =⇒ D ≈ 1.6055
2
3
Fern
S1 : c 1
S2 : c 2
S3 : c 3
1
2
and two copies
Create a fern fractal by the three contraction maps
= 43 , rotate by 5◦ clockwise and shift up 1/4,
= 41 , rotate by 60◦ counter-clockwise and shift up 1/4,
= 41 , rotate by 60◦ clockwise and shift up 1/4,
Dimension is solution to ( 34 ) D + 2
1 D
4
= 1, i.e. D = 1.3267.
59
before.
solve