Download Non-Euclidean Geometry

Non-Euclidean Geometry
§1.
Some History
Euclid (c. 300 B.C.) is considered the founder of Euclidean geometry, the classical
geometry which we all learned in high school and have been studying in this course so far.
He was building upon the work of previous Greek scholars such as Thales (640 - 546
B.C.), Plato (429 - 348 B.C.) and Pythagoras (c. 572 - 500 B.C.). Essentially,
he collected all the existing geometrical knowledge and based it upon a foundation of
what are today called ”axioms”. Because he derived every result in the Elements from
these “postulates” as he called them, he was really the founder of deductive thinking. His
influence was so great that over 2000 years later, his work is still studied in school.
Of course, there were flaws in his reasoning. Sometimes assumptions crept in which
were not derivable from his basic postulates. But what really puzzled scholars for hundreds
of years was his famous ”Fifth Postulate” about parallel lines. The other postulates are
very simple, in fact he called them ”self-evident truths”. But Postulate V is anything but
simple. It states:
Postulate V. If a straight line falling on two straight lines makes the interior angles on
the same side less than two right angles, then the two straight lines, if produced indefinitely,
meet on that side on which are the angles less than two right angles.
Because of the complicated nature of this postulate, or axiom, scholars were convinced
that it should be derivable from the other axioms. Clearly, Euclid knew he was assuming
something major; he certainly did not try to “prove” it from the previous axioms. For
centuries, it was felt that Postulate V was a “blemish” on geometry. Only in the past 150
years or so has it been proved that it was necessary to assume an axiom of this nature.
Why? Because we can assume the negative of such an axiom and obtain another geometry,
just as consistent as Euclidean geometry. This “non-Euclidean geometry” is “hyperbolic”
or “Lobatchewskian” geometry.
Playfair (1748 - 1819) introduced an axiom which is equivalent to Postulate V:
Playfair’s axiom. Given a line l and a point P not on l, there exists exactly one line m
through P which does not meet l, i.e. m is “parallel” to l.
P
m
l
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This is the form of the parallel postulate which we shall use instead of the more
complicated one of Euclid.
The first serious attempt to find an alternative to the fifth postulate was made by a
Jesuit priest Saccheri (1667 - 1773). He attempted a “proof by reductio ad absurdum”
or “proof by contradiction” as it is called today. Here is essentially what he attempted:
D
C
A
B
Start with 2 equal line segments AC and BD perpendicular to a line AB. Join CD. It is
easy to show that ! C = ! D. Saccheri then considered three cases:
(1) ! C = ! D = a right angle.
(2) ! C = ! D > a right angle.
(3) ! C = ! D < a right angle.
The first case is the Euclidean case. He wished to show the latter two cases were impossible.
With a great deal of effort, Saccheri was able to prove the following results about the
sum of the angles of a triangle in each case:
(1) implies that the sum of the angles equals two right angles, i.e. π.
(2) implies that the sum of the angles is greater than two right angles.
(3) implies that the sum of the angles is less than two right angles.
Saccheri showed that (2) is impossible because it contradicts one of Euclid’s previous
results (that the sum of two angles of a triangle is less than two right angles), but more
fundamentally, the problem is that possibility (2) implies that lines are finite in length, as
we have seen on the sphere.
However, Saccheri was unable to find a contradiction to (3). Finally, frustrated, he
stated that his results were “contrary to the nature of a straight line”. But while doing this,
he derived many results which were later shown to be valid in hyperbolic or Lobatchewskian
geometry.
Karl F. Gauss (1777 - 1855) was the first to realize that a geometry denying
Postulate V should be developed for its interest without expecting a contradiction. But
he was afraid of being ridiculed by the scientific world, so kept these ideas to himself until
others first published. Later we shall see some of his work.
We come now to the two real “fathers of non-Euclidean geometry”.
The Hungarian Bolyai, János (1802 - 1860) was the son of Bolyai, Farkas (1775 1856) who was also a mathematician and had been a fellow student of Gauss. Farkas had
spent a great deal of time and effort working with parallels and having become completely
frustrated with the question, warned his son not to waste his time on the matter. But
János persevered, and finally published his ideas as an appendix to his father’s textbook.
His idea was to study “Absolute Geometry” which is essentially the body of results which
can be derived from Euclid’s axioms without accepting or denying the famous Postulate
V. We shall see some of this work later.
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János’ work stopped when he heard about a Russian mathematician Nicholai Lobatschewsky (1793 - 1856). He derived many results by assuming the opposite of
Postulate V, so is really the founder of non-Euclidean or hyperbolic or Lotatschewskian
geometry, to use its many names. We shall also be seeing some of his results. It should be
pointed out that although Bolyai and Lobatschewski developed their work indenpendently,
Bolyai heard of Lobatschewski’s work and was discouraged from further pursuing his line
of research; on the other hand, there is no evidence at all that Lobatschewski, working
alone far away in Russia, ever heard of Bolyai or of his pioneering work.
One further point: in Euclidean geometry we can find a model of hyperbolic geometry
and in hyperbolic geometry we can find a model of Euclidean geometry. Thus each is as
consistent as the other (this is referred to as “relative consistency”).
§2.
Some Absolute Geometry
Here are a few properties of Absolute Geometry which are valid both in Euclidean
and in non-Euclidean geometry. They do not depend upon Postulate V.
Definition 1. Two rays AA" and BB " are said to be “parallel” if
(1) AA" and BB " do not intersect.
(2) Every ray drawn through A within the angle BAA" intersects the ray BB " .
A
A'
B
B'
To further explain this, consider all the rays through the point A, going toward the right.
Such a ray either intersects BB " or it does not intersect BB " . Consider a ray which does
not intersect BB " . As we rotate this ray clockwise, the “right parallel to the line BB "
through A” is the first ray which does not intersect BB " ; subsequent positions of the ray
do intersect BB " . Continue rotating this ray about A until it no longer intersects BB " .
The last ray which does not intersect BB " is the ”left parallel to BB " through A”.
O
M
A
N
B
Drop ON ⊥ AB. The angles N OL and BOM are equal by symmetry. The angle
N OL is called the “angle of parallelism” and depends upon the length c = ON. It is often
denoted by Π(c).
Now if Π(c) = π2 (or 90 degrees if you prefer), then this is the well-known parallel
postulate for Euclidean geometry. The left and right parallels form one straight line M L
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L
so that there is one and only one parallel through O to the line AB. But if Π(c) < π/2, we
have the non-Euclidean case; through O there are two “parallels” to AB, a “left parallel”
OM and a “right parallel” OL.
Forgetting for awhile this distinction, we can prove many properties of parallelism. In
particular, we can show that if a line l is parallel to line m, then m is also parallel to l. If
l and m are both parallel to line p, then l and m are parallel to each other. The details of
such a proof are omitted here. More interestingly, all the results known from high school
about congruent triangles still hold, because in their proof no mention is made of Postulate
V.
Once again, let AM and BM be parallel rays, and let " be an arbitrarily small angle,
i.e. an angle as small as you wish. Within the angle BAM (see the figure below), take a
ray from A making with AM an angle less than ". This ray must intersect BM in some
point C. On CM , take a point D1 with CD1 = CA, so that "CAD1 is an isosceles
triangle. Then we have:
!
AD1 B i.e.
!
AD1 C = ! CAD1 < ! CAM < ".
Repeat, forming another isosceles triangle AD1 D2 and again
!
.
AD2 B = ! D1 AD2 < ! D2 AM < "
A
x
o
M
x
B
C
o
D1
..
D2
Continuing this way, as points D1 , D2 , D3 , . . . tend to the end M (i.e. tend to “infinity”),
line ADn tends to the position AM and the angle ADn B tends to zero. This is essentially
Bolyai’s proof of:
Result 1. When two parallel lines are regarded as meeting at infinity, the angle of intersection must be considered as being equal to zero.
Definition 2. An “asymptotic triangle” consists of two parallel rays with a common “end”
and a common transversal.
The term “end” refers to a common direction for the two lines; i.e. in this case, the lines
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AM and BM are “right parallels”.
A
M
B
Asymptotic triangles behave much like ordinary triangles. For example,
Result 2. If two asymptotic triangles ABM, ABM " have AB = A" B " and
the ! B = ! B " .
A
!
A=
!
A" ,
B'
M
A'
B
M'
We shall consider asymptotic triangles more closely in the next section on Hyperbolic
Geometry.
One of the important differences between Euclidean and hyperbolic geometry is the
existence, in hyperbolic geometry, of lines which are neither parallel nor intersecting. They
are usually called “ultraparallel” lines. Such lines have a common perpendicular, as the
following result will show. In Euclidean geometry of course, only parallel lines have a
common perpendicular.
Result 3. IF two lines are neither intersecting nor parallel, they have a common perpendicular.
Notice the importance of the word IF. Since in Euclidean geometry two lines are either
intersecting or parallel, this result is really of no interest in Euclidean geometry, only in
hyperbolic geometry. Nevertheless, it is a theorem of absolute geometry. Let’s sketch a
proof of this result.
Proof: From A on the line AL, drop the perpendicular AB to the second line BM as
in the figure below. If AB is perpendicular to AL, then we are done: AB is a common
perpendicular line of AL and BM .
But if AB is not perpendicular to BM , suppose that L is on that side of AB for which
! BAL is acute. Since the two lines are neither intersecting nor parallel, there is a smaller
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angle BAM such that AM is parallel to BM . Let C and D be two points on BM in
the order B, C, D, M . By a well-known result of Euclid which does not depend upon
Postulate V, the internal angle at D of the triangle ACD is less than the external angle at
C. Hence, when BD increases from 0 to ∞, so that the ! DAL decreases from ! BAL to
! M AL, the ! ADB decreases from a right angle to zero. At the beginning of this process,
we have
! DAL < ! ADB
(since ! BAL is acute whereas ! ADB is a right angle). But at the end of this process, the
inequality is reversed (since ! M AL is positive.) Hence, there must be some intermediate
position for which
! DAL = ! ADB.
A
L
M
B
D
C
For such a point D (see the following diagram), we obtain two triangles OAE, ODF by
drawing EF perpendicular to BD through O, the midpoint of AD. Since these triangles
are congruent, EF is perpendicular not only to BD but also to AL.
A
E
L
x
O
F
x
D
B
M
Surprisingly, many of our Euclidean results on line reflections, point reflections, rotations, translations, and glides are valid in Absolute geometry, with few modifications.
The most important difference is that we must distinguish between the product of reflections in two parallel lines and the product of reflections in two ultraparallel lines. Even
more surprising, when we consider absolute geometry in three dimensions, the proof of the
existence of only five regular polyhedra is still valid.
§3.
Some Hyperbolic Geometry
When proving Result 3 (i.e. that two lines which are neither intersecting nor parallel
have a common perpendicular), we saw that if a transversal (say AD in the previous
diagram) meets two lines in such a way that the “alternate” angles OAE, ODF are equal,
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then the two lines are ultraparallel. This is in contrast to the Euclidean case where such
lines are parallel. This immediately tells us that in the right-angled asymptotic triangle
EF M of the diagram below, the external angle at E is greater than the internal angle at
F since this latter angle is < π2 whereas the angle at E is π2 .
F
M
E
Since the sum of the angles of the asymptotic triangle EF M is the sum of the internal
angles at E and F plus the “angle” at M , which is zero, we have
interior ! E + interior ! F + “angle” at M < interior ! E + external ! E + 0 = π.
So we see that the sum of the angles of this right-angled asymptotic triangle is less than
π.
The study of asymptotic triangles is quite interesting, but we shall content ourselves
here by showing two typical properties. See H.S.M. Coxeter “Introduction to Geometry”
pages 291 - 295 for further details.
Result 3. If two asymptotic triangles ABM, A" B " M " are such that
AB = A" B " , then the triangles are congruent.
Proof:
A
!
B =
!
B " and
B'
M
D
A'
D'
B
M'
We must show that ! B " A" M = ! BAM , so that the triangles are congruent.
For a proof by contradiction, suppose the angles are unequal and in fact ! A > ! A" . Then
let a ray AD be such that ! BAD = ! B " A" M " where D is the intersection of this ray and
BM . On B " M " , let D" be a point such that B " D" = BD. Then the triangles ABD, A" B " D"
are congruent, and so
!
B " A" D" = ! BAD = ! B " A" M "
which is the required contradiction, and so
congruent.
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!
BAM =
!
B " A" M " and the triangles are
Result 4. In an asymptotic triangle ABM , the exterior angle at A is greater than the
interior angle at B.
Proof: AM and BM are parallel lines with the common end M . Let AF be a line through
A such that ! F AG = ! M BA where G is a point on BN as shown in the diagram below.
We want to show that ! M AG > ! M BA.
G
F
A
Y
O
M
B
X
.
Let O be the midpoint of AB, drop the perpendicular OX to the line BM and let the line
OX intersect AF in the point Y . The triangles OBX, OAY are congruent because the
angles at O are equal, side OB = side OA, and the angles OBX, OAY are equal since
! OAY = ! F AG = ! M BA. Thus the angle OY A = ! OXB = π and so the line XY
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is also perpendicular to the line AF . Hence BM and AF are ultraparallel because they
have a common perpendicular line. Since AM is parallel to BM and AF is ultraparallel
to BM , this tells us that the ! M AG > ! F AG. Since this latter angle was constructed
equal to the ! M BA, we have external ! M AG > interior ! M BA as required.
A “doubly asymptotic triangle” ALM is formed by a line LM and a point A with
lines AL and AM parallel to LM , while a “trebly asymptotic triangle” KLM has its three
sides mutually parallel.
A
K
L
M
L
M
doubly asymptotic triangle
trebly asymptotic triangle
There are many other fascinating results in hyperbolic geometry which, while quite
elementary in nature, require considerable time to study in detail. In particular, we could
show that the area of any asymptotic triangle is finite, even a trebly asymptotic triangle;
that the sum of the angles in any triangle is less than π (this will be shown in the next
tutorial); and that the area of any triangle ABC is a constant multiple of its “angular
defect”, i.e.
Area of "ABC = µ(π − ! A − ! B − ! C);
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and, with considerbly more work, that the angle of parallelism is
Π(c) = 2 arctan e−c = arc cot(sinh c) = arccos(tanh c).
which was proved by Lobatschevski.
§4.
“Circles” in the Hyperbolic Plane
In the hyperbolic plane, two lines can be (i) intersecting or (ii) parallel or (iii) ultraparallel. Hence we define a “pencil of lines” rather differently from in the Euclidean
plan:
Definition 3. A pencil of lines consists of all lines which (i) intersect in a common point;
(ii) are parallel to a given line; or (iii) have a common perpendicular.
Let x, y be two lines of a pencil. Then if x and y intersect in a point P , the product
σy σx is a “rotation” ρP,α about the point P through some angle α. If x and y have a common perpendicular, the product σy σx is a “translation” τ along the common pependicular
of x and y through some distance. But if x and y are parallel, then we say the product
σy σx is a “parallel displacement”.
Let P be a pencil of lines through the point P . Then if a point X is reflected in all
the lines of the pencil P, we obtain a circle, just as in Euclidean geometry. A “circle” is
simply the locus of a point X when reflected in all the lines of an intersecting pencil. The
following diagram on the left shows only some of the images of a point X since only some
of the lines through P can be shown.
Similarly, let P be a pencil of parallel lines with a common “end” M . Then if a point
X is reflected in all the lines of the pencil P, we obtain a horocycle. A “horocycle” is
simply the locus of a point X when reflected in all the lines of a parallel pencil. Again,
only some of the images of a point X can be shown in the right hand diagram.
X
M
X
Pencil P of lines
intersecting in a point
Pencil P of parallel lines
with a common "end" M
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Continuing, we ask what would be the locus of a point which is at a fixed distance
from a given line t? In Euclidean geometry, this would be two straight lines parallel to the
given line t. But in hyperbolic geometry, this is an “equidistant curve”, or “hypercycle”,
having two branches, one on each side of its axis t. Either branch may be described as the
locus of a point X when reflected in all the lines of this pencil of perpendiculars to t. The
following diagram is an attempt to illustrate this, but it is very difficult to draw on the
Euclidean plane as the right angles do not appear to be right angles.
X
X'
Pencil P of lines with a common perpendicular
In Part II we shall study a model of the hyperbolic plane in which circles, horocycles
and hypercycles can be diagramed more convincingly.
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