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Hyperbolic geometry
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What is hyperbolic geometry. Hyperbolic geometry (or Lobachevsky geometry)
is the geometry that has the same axioms as Euclidean geometry, except for the fifth
postulate, the later being replaced with its negation. Recall that the fifth postulate
of Euclid is equivalent to the following statement: given a line l and a point A not
in l, there is at most one line passing though A and not intersecting l. We say “not
intersecting” instead of “parallel” because in hyperbolic geometry, being parallel
will mean more than just being disjoint.
The statement that there is always at least one line trough A not intersecting l is
a statement of absolute geometry (i.e. this statement does not depend on the fifth
postulate). Namely, if AB is the segment perpendicular to l, where B lies in l, then
the line passing through A and perpendicular to AB does not intersect l. Thus,
in hyperbolic geometry, there are at least two lines through A not intersecting the
line l. Then it follows that there are infinitely many such lines (take e.g. any line
between the two already considered).
Hyperbolic geometry appeared in early nineteenth century. Before that, most
mathematicians believed that the fifth postulate can be deduced from other axioms
of Euclidean geometry. Thus many unsuccessful attempts to prove the fifth postulate
appeared. However, it turned out that all these “proofs” contain mistakes, i.e. they
use implicitly some statement equivalent to the fifth postulate.
“Proof ” by Proclus. The first erroneous proof of the fifth postulate is probably
due to Proclus, a commentor of Euclid who lived in 5th century A.D. He argued
as follows. Suppose that there are two lines l1 and l2 passing through A and not
intersecting l. We can also assume that l1 is the line perpendicular to AB, where
AB is a segment perpendicular to l with point B lying on l. The distance between
l1 and l2 grows unboundedly as we go away from A. Sooner or later, this distance
will exceed the distance between the lines l1 and l. Therefore, at some moment, the
line l2 will intersect the line l. Q.E.D.
This is just a sketch of the argument, and every line here requires further explanation. However, the first claim, namely, that the distance between two intersecting
lines grows unboundedly as one moves away from the intersection point, does make
sense in absolute geometry (although it requires a more precise formulation) and
it is true. The problem is with the second part of the argument. One tends to
think that the distance between non-intersecting lines is constant. However, this
does not follow from the axioms of absolute geometry. It may well happen that
non-intersecting lines go away from each other and that the distance is not even
bounded. Actually, this is exactly what will happen in hyperbolic geometry: if two
lines have a common perpendicular, then this perpendicular is unique, and on both
sides of it, the distance between the lines grows unboundedly. So, although the
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distance between the lines l1 and l2 gets bigger and bigger, it will never exceed the
distance between l1 and l, which also gets bigger and bigger.
Equivalents of the fifth postulate. This unsuccessful attempt of Proclus can be
turned into a true theorem: the fifth postulate is equivalent to the statement that
the distance between two non-intersecting lines is bounded. There are many other
basic facts of Euclidean geometry that are actually equivalent to the fifth postulate.
One such fact is that the sum of all angles in any triangle is 180 degrees. We proved
this fact using the fifth postulate, but it is also possible to prove the fifth postulate
using this statement. The proof goes back to Legendre.
Consider a line l and a point A not belonging to l. Drop the perpendicular AB
from A to l, and, as before, denote by l1 the line passing through A and perpendicular
to AB. It is a theorem of absolute geometry that the line l1 does not intersect the
line l. Take a point C1 on the line l such that |BC1 | = |AB| (we will assume that
C1 is to the right of B). Then the triangle ABC1 is an isosceles triangle, hence
the angles ∠A and ∠C1 in this triangle are equal (this follows from SAS, which is
independent of the fifth postulate). We also know that the angle ∠B is 90◦ and that
the sum of all angles is 180◦ . It follows that both angles ∠A and ∠C1 in the triangle
ABC1 are 45◦ . The next step is to mark another point C2 to the right of C1 such
that |AC1 | = |C1 C2 |. Then the triangle AC1 C2 is isosceles. The sum of angles ∠A
and ∠C2 in this triangle is equal to 180◦ minus the angle C1 , which is 45◦ . Then
each of the angles ∠A and ∠C2 in the triangle AC1 C2 is 45◦ /2.
Continue this process. We will obtain a sequence of points C1 , C2 , . . . , Cn , . . . on
the line l such that |ACn | = |Cn Cn+1 |. The angle ∠A in the triangle ACn−1 Cn will
be 45◦ /2n−1 . We can compute the angle ∠BACn as follows:
∠BACn = ∠BAC1 + ∠C1 AC2 + ∠C2 AC3 + · · · + ∠Cn−1 ACn =
µ
¶
1 1
1
◦
= 45 1 + + + · · · + n−1 = 90◦ (1 − 1/2n ).
2 4
2
◦
We see that this angle tends to 90 and n goes to infinity. Thus we can obtain an
angle, that would be as close to the right angle as we wish.
Now assume that there is some other line l2 passing through the point A and
not intersecting the line l. Then we can choose a number n to be so big that
∠BACn = 90◦ (1 − 1/2n ) is bigger than the angle between AB and the line l2 . Then
l2 enters the triangle BACn through the vertex A. Since it enters the triangle, it
must also exit it by intersecting the side BCn (it cannot intersect two other sides
because otherwise we would have two lines intersecting in two different points). A
contradiction. Q.E.D.
Defect of a triangle. We just proved that if the sum of angles in ANY triangle
is 180◦ , then the fifth postulate holds. By working slightly harder, one can prove
a stronger statement: if the sum of angles in some triangle is 180◦ , then the fifth
postulate holds. Thus in hyperbolic geometry, the sum of angles in a triangle is
always strictly less than 180◦ . A further modification of Legendre’s proof will show
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that if the sum of angles in some triangle is at least 180◦ , then the fifth postulate
holds. In other words, the sum of angles in a triangle can never be greater than 180◦
(indeed, if it is at least 180 degrees, then we have the Euclidean geometry, hence it
is exactly 180).
Define the defect of a triangle ABC as follows:
δ(ABC) = 180◦ − ∠A − ∠B − ∠C.
Note that in Euclidean geometry, the defect of any triangle is zero. In a sense,
the defect measures the discrepancy between a hyperbolic triangle and a Euclidean
triangle. Probably the most important property of the defect is additivity: if a big
triangle is composed of several smaller triangles that do not overlap except at the
boundaries, then the defect of the big triangle is equal to the sum of defects of all
smaller triangles. Let us prove this statement in one particular case, where a triangle
ABC is subdivided into two triangles by a cevian BD, the smaller triangles being
ABD and DBC. Indeed, we have
δ(ABD) = 180◦ − ∠A − ∠ABD − ∠BDA,
δ(DBC) = 180◦ − ∠CDB − ∠DBC − ∠C.
Since ∠BDA + ∠CDB = 180◦ and ∠ABD + ∠DBC = ∠B, we have
δ(ABD) + δ(DBC) = 180◦ − ∠A − ∠B − ∠C = δ(ABC),
Q.E.D.
Legendre’s “proof ”. Another unsuccessful attempt to prove the fifth postulate
is due to Legendre and is based on the notion of defect. We know that the fifth
postulate is equivalent to the statement that in any triangle, the sum of all angles
is 180◦ . Thus it suffices to prove the latter. Consider any triangle ABC. We want
to show that the defect δ(ABC) is zero (then the sum of angles would be 180◦ ).
Let us do the following construction. On the side AC, Construct a triangle CAB1
that is congruent to ABC. This construction is possible due to basic properties of
isometries, it is independent of the fifth postulate. Through the point B1 , draw a
line A1 C1 intersecting both lines BA and BC, so that A1 belongs to the line BA
and C1 belongs to the line BC. We claim that the defect of the triangle A1 BC1 is
at least twice as much as the defect of the triangle ABC. Indeed, the big triangle
A1 BC1 is decomposed into four smaller triangles, among which two are copies of
ABC. So we just use the additivity of the defect and the fact that the defect is
always nonnegative. Keep doing this construction. We obtain a sequence of triangles
An BCn such that the defect of each next triangle is at least twice bigger than the
defect of the previous one. It follows that
δ(An BCn ) ≥ 2n δ(ABC).
Now assume that δ(ABC) is strictly positive. Then, sooner or later, the number
2n δ(ABC) will get larger than 180 degrees. This means that the defect of the triangle
An BCn would be larger than 180 degrees, which is clearly impossible because the
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sum of angles cannot be negative. The contradiction shows that actually we must
have δ(ABC) = 0, i.e. the sum of angles in ABC is 180◦ . But then this is true for
any triangle (we can always choose the angle ∠B to be acute). The fifth postulate
follows. Q.E.D.
Later Legendre found an error in his proof. The problem is with the claim that
one can always draw a line through the given point Bn−1 that intersects both lines
BA and BC. This innocent looking statement is actually equivalent to the fifth
postulate. In hyperbolic geometry, this statement would not be true.
Parallel lines and the angle of parallelism. Let us start discussing hyperbolic
geometry in a more positive sense. First we need to modify the notion of parallel
lines, since in hyperbolic geometry, there are much more non-intersecting lines than
in Euclidean geometry.
Consider a line l and a point A not in l. The point A belongs to infinitely many
lines not intersecting l. Drop the perpendicular AB to the line l. For each line
AC passing through A and not intersecting l, we will measure the angle ∠BAC
(we are thinking of the line l being horizontal and assuming that the point C is
always on the right of the line AB). It is possible that ∠BAC < 90◦ . There is the
following dichotomy: either there is a line AC not intersecting l with the smallest
possible angle ∠BAC (for a line not intersecting l), or there is a line AC intersecting
l with the biggest possible angle ∠BAC (for a line intersecting l). This dichotomy
is essentially a property of the set of real numbers. It follows from the following
statement (which is basically an axiom of the real calculus): Consider two subsets A
and B in the set of all real numbers such that the intersection of A and B is empty,
each element of A is strictly less than each element of B, and the union of A and
B consists of all real numbers. Then either A has a biggest element, or B has a
smallest element.
It is impossible to have a line AC intersecting l and having the biggest possible
angle ∠BAC. Assume the contrary. We can also assume that the point C belongs to
l. Then take any point C 0 on the line l that is farther away from B than C. The line
AC 0 would be another line intersecting l with ∠BAC 0 > ∠BAC. A contradiction.
We must conclude that there is a line AC not intersecting l with the smallest
possible angle ∠BAC. This line is called a line parallel to l. The corresponding angle
∠BAC is called the angle of parallelism. It only depends on the length x = |BA|,
and it is denoted by Π(x) (Π is the capital Greek letter pi). If we took C on the left
of the line AB, then we would get another line parallel to l. Thus there are exactly
two lines parallel to l: one is parallel to the right and one is parallel to the left.
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