Introduction to Hyperbolic Geometry by rebeccaGerritY

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									Chapter 17

Introduction to Hyperbolic Geometry

In this chapter, we begin our study of hyperbolic geometry. In addition to the nine postulates of
neutral geometry, throughout the rest of the book we assume the following postulate.

   Postulate 10H (The Hyperbolic Parallel Postulate). For every line                  and every
   point P that does not lie on , there are at least two different lines containing P and parallel
   to .

    The term hyperbolic geometry refers to the axiomatic system consisting of the undefined
terms and postulates of neutral geometry together with the hyperbolic parallel postulate. Of
course, every theorem in neutral geometry is also a theorem in hyperbolic geometry, including the
Saccheri–Legendre theorem and the theorem on additivity of defects from the preceding chapter.
    In addition, the work we did in the preceding chapter quickly yields several other theorems in
hyperbolic geometry: Because the Euclidean parallel postulate is false in hyperbolic geometry, each
of the postulates that we showed to be equivalent to the EPP is also false. In particular, we have
the following important theorem.
Theorem 17.1 (Hyperbolic Angle-Sum Theorem). In hyperbolic geometry, every convex
polygon has positive defect.
Proof. Because the EPP is false, Theorem 16.11 shows that the weak angle-sum postulate is also
false; therefore, there does not exist a triangle with zero defect. By the Saccheri–Legendre theorem,
therefore, every triangle has positive defect. The result for convex polygons with more than three
vertices is then proved by induction on the number of vertices, just as in the proof of Corollary
16.3.

Corollary 17.2. In hyperbolic geometry, there does not exist a rectangle.
Proof. This follows either from the preceding theorem, because a rectangle would have zero defect,
or from Theorem 16.10, which shows that the existence of a rectangle is equivalent to the EPP.

    One of the oddest features of hyperbolic geometry follows from the fact that Wallis’s postulate is
false: We cannot construct similar triangles with arbitrary scale factors. In fact, the next theorem
shows that we cannot construct similar triangles at all in hyperbolic geometry (except for congruent
triangles, which are similar for silly reasons).

                                                 205
206                         CHAPTER 17. INTRODUCTION TO HYPERBOLIC GEOMETRY

Theorem 17.3 (AAA Congruence Theorem). In hyperbolic geometry, if there is a correspon-
dence between the vertices of two triangles such that all three pairs of corresponding angles are
congruent, then the triangles are congruent.

Proof. Suppose ABC and DEF are triangles such that ∠A ∼ ∠D, ∠B ∼ ∠E, and ∠C ∼ ∠F .
                                                                 =        =               =
If any side of ABC is congruent to the corresponding side of DEF , then the two triangles are
congruent by ASA. So suppose for the sake of contradiction that all three pairs of corresponding
sides are noncongruent: AB = DE, AC = DF , and BC = EF . Since there are three inequalities
and each one must be either “greater than” or “less than,” at least two of them must go in the
same direction; without loss of generality, let us assume that AB < DE and AC < DF .

                                                               D

                                                       P           Q

                                       A


                              B            C   E                       F

                         Fig. 17.1: Proof of the AAA congruence theorem.


    Choose points P ∈ Int DE and Q ∈ Int DF such that DP ∼ AB and DQ ∼ AC (Fig.
                                                                   =                =
17.1). It follows from SAS that DP Q ∼ ABC, and therefore ∠DP Q ∼ ∠B ∼ ∠E and
                                           =                                  =      =
∠DQP ∼ ∠C ∼ ∠F . By the polygon cutting theorem, P QF E is a convex quadrilateral (in fact,
        =      =
a trapezoid by the corresponding angles theorem). Since ∠EP Q is supplementary to ∠DP Q and
therefore also to ∠E, and ∠F QP is supplementary to ∠DQP and thus to ∠F , it follows that
P QF E has angle sum equal to 360◦ . This contradicts the hyperbolic angle-sum theorem.

    This theorem is remarkable because it implies that it is impossible to construct a scale model
or a scaled image of anything—in a hyperbolic world, a photograph would necessarily distort the
shape of its subject unless it were actual size.
    Another way of interpreting the AAA congruence theorem is that it guarantees the existence of
an “absolute” standard of length, independent of any arbitrary conventions. In Euclidean geometry,
the distances given to us by the distance postulate are in a certain sense arbitrary: We could
multiply all distances by a positive constant c, and all areas by c2 , and the theorems of Euclidean
geometry would still be true with this new distance scale. (This just corresponds to changing the
units of our measuring scale, as from feet to meters.) Thus if two remotely separated civilizations in
a Euclidean universe wanted to agree on a unit of length, they would have to rely on some physical
phenomenon having the same properties in both locations, such as the wavelength of light emitted
from a certain kind of atom.
    However, in hyperbolic geometry things are very different. If our hypothetical civilizations lived
in a hyperbolic world, they could simply agree that the universal unit of length is, say, the side
length of a 45◦ -45◦ -45◦ triangle. Because the AAA theorem guarantees that all such triangles are
congruent, there would be no ambiguity about what this length is. (Of course, there is also no
SACCHERI AND LAMBERT QUADRILATERALS                                                                207

ambiguity about what a 45◦ angle is either, since it can be described as either of the angles created
by the bisector of a right angle.)


Saccheri and Lambert Quadrilaterals
After triangles, rectangles are probably the second most important figures in Euclidean geometry.
Since rectangles are nonexistent in hyperbolic geometry, it is natural to look for other figures that
might play analogous roles.
    There are two obvious ways that one might attempt to construct a rectangle without benefit of
the Euclidean parallel postulate, each of which leads to a useful class of quadrilaterals in hyperbolic
geometry. One way is to start with a segment AB, construct congruent perpendicular segments
                                  ←→
AD and BC on the same side of AB (Fig. 17.2), and join C and D to form a quadrilateral ABCD.
The other is to start with two segments AB and BC meeting perpendicularly at B, construct
perpendiculars to them at A and C, and let D be the point where the two perpendiculars meet
(Fig. 17.3).

              D                     C                             D
                                                                                   C



                  A             B                                     A           B

      Fig. 17.2: A Saccheri quadrilateral.               Fig. 17.3: A Lambert quadrilateral.


    In Euclidean geometry, both constructions would yield rectangles, as you can check; but in
hyperbolic geometry, the remaining angles cannot be right angles. Thus we make the following
definitions. A Saccheri quadrilateral is a quadrilateral with a pair of congruent opposite sides
that are both perpendicular to a third side, and a Lambert quadrilateral is a quadrilateral with
three right angles.
    Saccheri quadrilaterals are named after Giovanni Saccheri (whom we met in Chapter 1 and
heard from again in the preceding chapter), and Lambert quadrilaterals after Swiss mathematician
Johann Heinrich Lambert (1728–1777). Although the nomenclature is well established in Western
tradition, it is not historically accurate: Both types of quadrilaterals were studied extensively many
centuries earlier by Islamic mathematicians such as Omar Khayyam.
    When we draw figures in hyperbolic geometry, it is generally impossible to make all the line
segments appear straight while making all distances and angles appear correct, simply because the
plane in which we are drawing our pictures is a Euclidean one (at least to a very close approxima-
tion). We will generally attempt to make the angles look correct, and to make the lines look straight
whenever possible. But sometimes lines have to be drawn as curves in order to make the angles
appear correct, and it is usually impossible to make all distances appear correct. Another useful
                                                       e
way to visualize hyperbolic figures is in the Poincar´ disk model, in which all angles are correct but
most lines look like arcs of circles. Fig. 17.4 illustrates Saccheri and Lambert quadrilaterals in the
        e
Poincar´ model. In these figures, the hyperbolic “lines” containing the sides are shown as dashed
208                           CHAPTER 17. INTRODUCTION TO HYPERBOLIC GEOMETRY

lines or curves, to make it clear that each side is part of a diameter or a circle intersecting the unit
circle perpendicularly.




                                                                              e
                 Fig. 17.4: Saccheri and Lambert quadrilaterals in the Poincar´ model.


   In this section, we will study some of the basic properties of Saccheri and Lambert quadrilaterals.
We begin with some terminology. For a Saccheri quadrilateral, we make the following definitions:

   •    The   two congruent opposite sides are called the legs.
   •    The   side perpendicular to the legs is called the base.
   •    The   side opposite the base is called the summit.
   •    The   two angles adjacent to the summit are called the summit angles.
   •    The   segment joining the midpoints of the summit and the base is called the midsegment.

In a Lambert quadrilateral:

   • The angle that is not a right angle is called the fourth angle.
   • The vertex of the fourth angle is called the fourth vertex.

      Here are a few elementary properties, whose proofs we leave for you to carry out.

Theorem 17.4 (Properties of Saccheri Quadrilaterals). Every Saccheri quadrilateral has the
following properties:

 (a)    Its diagonals are congruent.
 (b)    Its summit angles are congruent and acute.
 (c)    Its midsegment is perpendicular to both the base and the summit.
 (d )   It is a parallelogram.
 (e)    It is a convex quadrilateral.

Proof. Problem 17.1.

      Analogously, for Lambert quadrilaterals, we have the following properties.

Theorem 17.5 (Properties of Lambert Quadrilaterals). Every Lambert quadrilateral has the
following properties:

 (a) Its fourth angle is acute.
 (b) It is a parallelogram.
 (c) It is a convex quadrilateral.
APPLICATIONS OF SACCHERI QUADRILATERALS TO PARALLELISM                                            209

Proof. Problem 17.2.

   A somewhat less elementary property of Lambert quadrilaterals is given in the next theorem.
Theorem 17.6. In a Lambert quadrilateral, either side between two right angles is strictly shorter
than its opposite side.
Proof. Suppose ABCD is a Lambert quadrilateral with fourth vertex D. We need to show that
AB < CD and BC < AD. Since the argument is the same in both cases, we will only prove
the second inequality. By trichotomy, there are three possibilities: BC = AD, BC > AD, or
BC < AD.
   If BC were equal to AD, then ABCD would be a Saccheri quadrilateral with AB as its base;
but then the summit angle ∠C would be acute by Theorem 17.4, which is a contradiction.

                                       D
                                                              C

                                                              E


                                           A              B

                                 Fig. 17.5: Proof of Theorem 17.6.


   Suppose BC > AD. Let E be the point on BC such that BE = AD. Then ABED is a
Saccheri quadrilateral with base AB, and it follows from Theorem 17.4 that its summit angle
∠BED is acute. However, ∠BED is an exterior angle for DEC, so it must be greater than the
remote interior angle ∠C, which is a right angle. This is a contradiction, so the only remaining
possibility is that BC < AD.

    There are also corresponding inequalities for Saccheri quadrilaterals, which can be proved by
using the fact that the midsegment of a Saccheri quadrilateral divides it into two Lambert quadri-
laterals.
Theorem 17.7. In a Saccheri quadrilateral, the base is strictly shorter than the summit, and the
midsegment is strictly shorter than either leg.
Proof. Problem 17.3.


Applications of Saccheri Quadrilaterals to Parallelism
Saccheri quadrilaterals turn out to be essential tools for studying parallel lines in hyperbolic geom-
etry. In this section, we discuss several ways in which hyperbolic parallels are dramatically different
from their Euclidean counterparts.
    One essential feature of parallel lines in Euclidean geometry is that they are equidistant. In
fact, many people think of equidistance as a defining characteristic of parallel lines. In hyperbolic
geometry, things are very different, as we will see. First, we note that Theorem 12.4 (which is
210                         CHAPTER 17. INTRODUCTION TO HYPERBOLIC GEOMETRY

valid in neutral geometry, despite the fact that it was erroneously delayed until Chapter 12) shows
that the existence of two equidistant points on the same side is enough to guarantee that lines are
parallel. However, as the next theorem shows, in hyperbolic geometry, parallel lines are far from
equidistant.

Theorem 17.8. Suppose       and m are two different lines. No three distinct points on   are equidis-
tant from m.


                                        P         Q    R




                                                                 m
                                        F         G    H

                    Fig. 17.6: Two lines cannot have three equidistant points.



Proof. Let and m be distinct lines, and suppose P , Q, and R are three distinct points on such
that d(P, m) = d(Q, m) = d(R, m). Without loss of generality, we may assume that P ∗ Q ∗ R.
Since and m are not the same line, these distances cannot be zero, so none of the points P, Q, R
lies on m. At least two of them have to be on the same side of m, so Theorem 12.4 implies that
    m. Therefore all three points are on the same side of m (because otherwise and m would have
to intersect).
    Let F , G, and H be the feet of the perpendiculars to m from P , Q, and R, respectively (Fig.
17.6). By the same argument as in the proof of Theorem 12.4, P QGF , QRHG, and P RHF are
all Saccheri quadrilaterals with bases F G, GH, and F H, respectively. It follows that the two
summit angles of each quadrilateral are congruent: ∠F P Q ∼ ∠P QG, ∠GQR ∼ ∠QRH, and
                                                               =                   =
∠F P Q = ∼ ∠QRH. By transitivity, therefore, ∠P QG ∼ ∠GQR. Since these two angles form a
                                                         =
linear pair, they are both right angles. But they are also summit angles of Saccheri quadrilaterals,
so they are acute. This is a contradiction.

    Another familiar feature of parallel lines in Euclidean geometry is that they always have in-
finitely many common perpendiculars, because any line perpendicular to one of two parallel lines is
also perpendicular to the second (Theorem 12.8). In hyperbolic geometry, common perpendiculars
are much harder to come by.

Theorem 17.9 (Uniqueness of Common Perpendiculars). If and m are parallel lines that
admit a common perpendicular, then the common perpendicular is unique.

Proof. Problem 17.4.

    In fact, as we will see in the next chapter, there exist parallel lines that admit no common
perpendiculars. The next theorem gives two useful conditions that are sufficient to guarantee that
a common perpendicular exists.
APPLICATIONS OF SACCHERI QUADRILATERALS TO PARALLELISM                                             211

Theorem 17.10. Suppose and m are parallel lines. If either of the following conditions is
satisfied, then and m have a common perpendicular.

 (a) There are two distinct points on that are equidistant from m.
 (b) There is a transversal for and m that makes congruent alternate interior angles.


                                                                                        t
              P       M       Q                                     X    F      P

                                                                        M

                                        m
               F      N       G                                     Q       G   Y           m



       Fig. 17.7: Two equidistant points.           Fig. 17.8: Congruent alternate interior angles.



Proof. Suppose first that (a) holds, and let P and Q be distinct points on such that d(P, m) =
d(Q, m) (Fig. 17.7). Because and m are parallel, P and Q lie on the same side of m. Let F and
G be the feet of the perpendiculars to m from P and Q, respectively. Then P QGF is a trapezoid
by the trapezoid lemma, and the assumption that d(P, m) = d(Q, m) implies that it is a Saccheri
                                                             −
                                                            ←→
quadrilateral. If M N is the midsegment of P QGF , then M N is perpendicular to both and m by
Theorem 17.4(c).
    Now suppose (b) holds, and let t be a transversal meeting at P and m at Q, and making a
pair of congruent alternate interior angles (Fig. 17.8). If t is perpendicular to either line, then the
assumption of congruent alternate interior angles guarantees that it is perpendicular to both, and
we are done; so henceforth, we may assume that t is not perpendicular to either or m. There are
two pairs of alternate interior angles: Because each angle in one pair is supplementary to one of the
angles in the other pair, and supplements of congruent angles are congruent, we may conclude that
both pairs of alternate interior angles are congruent. One of those pairs consists of acute angles;
choose points X ∈ and Y ∈ m such that the congruent acute angles are ∠XP Q and ∠P QY .
    Let M be the midpoint of P Q, let F be the foot of the perpendicular from M to , and G the
foot of the perpendicular from M to m. Because t is not perpendicular to or m, neither F nor
                                                                             −
                                                                            −→
G lies on t. Right triangles cannot have obtuse angles, so F must lie on P X (and not its opposite
                           −
                          −→
ray), and G must lie on QY . Therefore F P M and GQM are congruent by AAS.
    The fact that ∠XP Q and ∠P QY are alternate interior angles means that F and G lie on
                            −
                           −→        −
                                    −→
opposite sides of t. Since M P and M Q are opposite rays, the partial converse to the vertical angles
                       −
                      −→         −
                                −→
theorem shows that M F and M G are opposite rays. In particular, this means that F , M , and G
                              ←→
are collinear, and therefore F G is a common perpendicular for and m.

    In the next two chapters, we will delve much more deeply into the theory of parallels in hyper-
bolic geometry.
212                         CHAPTER 17. INTRODUCTION TO HYPERBOLIC GEOMETRY

Problems
17.1. Prove Theorem 17.4 (properties of Saccheri quadrilaterals).
17.2. Prove Theorem 17.5 (properties of Lambert quadrilaterals).
17.3. Prove Theorem 17.7 (side inequalities in Saccheri quadrilaterals).
17.4. Prove Theorem 17.9 (uniqueness of common perpendiculars).

								
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