Download MAT 360 Lecture 10

MAT 360
Lecture 10 Hyperbolic
Geometry
Homework next week (Last hw!)
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Chapter 4: Problem 10.
Chapter 5: Problem 8
Chapter 6: Problems 2, 3, 5, 14
Extra credit (deadline at the end of the
course).
• Chapter 4: Problems 8 and 9
• Chapter 5: Problem 1
• Chapter 6: Problem 15.
To son Janos:
For God’s sake, please give it [work on
hyperbolic geometry] up. Fear it no less
than sensual passion, because it, too,
may take up all your time and deprive
you of your health, peace of mind and
happiness in life.
Wolfgang Bolyai
Hyperbolic axiom
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There exist a line l and a point P not in l
such that at least two parallels to l pass
through P.
Lemma:
If the hyperbolic axiom holds (and all the
axioms of neutral geometry hold too)
then rectangles do not exist.
Exercise:
 Find another formulation of this lemma.
 Prove it
Proof of lemma
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If hyp axiom holds then Hilbert’s parallel
postulate fails because it is the negation
of hyp. Axiom.
Existence of rectangles implies Hilbert’s
parallel postulate
Therefore, rectangles do not exist.
Formulation: Assume neutral
geometry.
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If rectangles exist then hyperbolic axiom does
not hold.
If Euclid’ v does not hold then rect. do not exist.
(any statement equivalent to Euclic V holds )
then rect. do not exist.
If hyp axiom holds then if three angles of a
quadrilateral are right then the fourth angle is
not right
Recall:
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In neutral geometry, if two distinct lines l
and m are perpendicular to a third line,
then l and m are parallel. (Consequence
of Alternate Interior Angles Theorem)
Universal Hyperbolic Theorem
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In hyperbolic geometry, for every line l and
every point P not in l there are at least two
distinct parallels to l passing through p.
Corollary: In hyperbolic geometry, for every line l
and every point P not in l there are infinitely
many parallels to l passing through p.
Theorem
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If hyperbolic axiom holds then all
triangles have angle sum strictly smaller
than 180.
Can you prove this theorem?
Recall
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Definition: Two triangles are similar if
their vertices can be put in one-to-one
correspondence so that the
corresponding angles are congruent.
Similar triangles
Recall Wallis attempt to “fix” the
“problem” of Euclid’s V:
Add postulate: “Given any triangle ΔABC,
and a segment DE there exists a
triangle ΔDEF similar to ΔABC”
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Why the words fix and problem are
surrounded by quotes?
Theorem
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In hyperbolic geometry, if two triangles
are similar then they are congruent.
In other words, AAA is a valid criterion
for congruence of triangles.
Corollary
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In hyperbolic geometry, there exists an
absolute unit of length.
Recall
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Quadrilateral □ABCD is a Saccheri
quadrilateral if
• Angles <A and <B are right angles
• Sides DA and BC are congruent
The side CD is called the summit.
Lemma: In a Saccheri quadrilateral □ABCD,
angles <C and <D are congruent
Definition
Let l’ be a line.
 Let A and B be points not in l’
 Let A’ and B’ be points on l’ such that the
lines AA’ and BB’ are perpendicular to l’
We say that A and B are equidistant from l’
if the segments AA’ and BB’ are
congruent.
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Question
Question: If l and m are parallel lines, and
A and B are points in l, are A and B
equidistant from m?
Theorem
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In hyperbolic geometry if l and l’ are
distinct parallel lines, then any set of
points equidistant from l has at most two
points on it.
Lemma
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The segment joining the midpoints of the
base and summit of a Saccheri
quadrilateral is perpendicular to both the
base and the summit and this segment is
shorter than the sides
Theorem
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In hyperbolic geometry, if l and l’ are
parallel lines for which there exists a pair
of points A and B on l equidistant from l’
then l and l’ have a common
perpendicular that is also the shortest
segment between l and l’.
Theorem
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In hyperbolic geometry if lines l and l’
have a common perpendicular segment
MM’ then they are parallel and MM’ is
unique. Moreover, if A and B are points
on l such that M is the midpoint of AB
then A and B are equidistant from l’.
Hyperbolic Geometry Exercises
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Show that for each line l there exist a
line l’ as in the hypothesis of the previous
theorem. Is it there only one?
Let m and l be two lines. Can they have
two distinct common perpendicular
lines?
Let m and n be parallel lines. What can
we say about them?
Where in the proof are we using the
hyperbolic axiom?
Theorem
Let l be a line and let P be a point not on
l. Let Q be the foot of the perpendicular
from P to l.
Then there are two unique rays PX and
PX’ on opposite sides of PQ that do not
meet l and such that a ray emanating
from P intersects l if and only if it is
between PX and PX’.
Moreover, <XPQ is congruent to <X’PQ.
Crossbar theorem
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If the ray AD is between rays AC and AB
then AD intersects segment BC
Dedekind’s Axiom
Suppose that the set of all points on a line is
the disjoint union of S and T,
SUT
where S and T are of two non-empty subsets
of l such that no point of either subsets is
between two points of the other. Then there
exists a unique point O on l such that one of
the subsets is equal to a ray of l with vertex
O and the other subset is equal to the
complement.
Definition
Let l be a line and let P be a point not in
l.
The rays PX and PX’ as in the statement
of the previous theorem are called
limiting parallel rays.
The angles <XPQ and X’PQ are called
angles of parallelism.
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Question
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Given a line l, are the “angles of
parallelism” associated to this line,
congruent to each other?
Theorem
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Given lines l and m parallel, if m does
not contain a limiting parallel ray to l then
there exist a common perpendicular to l
and m.
Definition
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Let l and m be parallel lines.
If m contains a limiting parallel ray (to l)
then we say that l and m are asymptotic
parallel.
Otherwise we say that l and m are
divergently parallel.
Janos Bolyai
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I can’t say nothing except this: that out
of nothing I have created a strange new
universe.