MAT 360

Lecture 10 Hyperbolic Geometry

What is the negation of Hilbert’s Axiom?

There exists a line l and a point P not on l such that

there are at least two parallels to l through p.

Hyperbolic axiom 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.

Proof of lemma If hyp axiom holds then Hilbert’s parallel

postulate does not hold because it is the negation of hyp. Axiom.

Existence of rectangles implies Hilbert’s parallel postulate

Therefore, rectangles do not exist.

Recall: 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 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 If hyperbolic axiom holds then all

triangles have angle sum strictly smaller than 180.

Can you prove this theorem?

Recall 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”

Why the words fix and problem are surrounded by quotes?

Theorem In hyperbolic geometry, if two triangles

are similar then they are congruent. In other words, AAA is a valid criterion for

congruence of triangles.

All triangles have angles 22.5 degrees, 60 degrees, and 90

degrees.

Therefore,all

triangles are

congruent

The distancewe “see”

in this picture is not the

one given by

Theorem 4.3

Congruent Hexagons

Corollary In hyperbolic geometry, there exists an

absolute unit of length.

Definition Quadrilateral □ABCD is a Saccheri

quadrilateral if • Angles <A and <B are right angles• Sides DA and BC are congruentThe side CD is called the summit.

Lemma In a Saccheri quadrilateral □ABCD, angles <C

and <D are congruent

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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.

QuestionQuestion: If l and m are parallel lines, and

A and B are points in l, are A and B equidistant from m?

Theorem 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 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 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 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 Show that for each line l there exist a line

l’ as in the hypothesis of the previous theorem. Is it the 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?

TheoremLet 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.

Where in the proof are we using the hyperbolic axiom?

Crossbar theorem (Recall) If the ray AD is between rays AC and AB

then AD intersects segment BC

Dedekind’s AxiomSuppose that the set of all points on a line is

the disjoint union of S and T, S U Twhere 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.

Question Given a line l, are the “angles of

parallelism” associated to this line, congruent to each other?

Theorem 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 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 I can’t say nothing except this: that out

of nothing I have created a strange new universe.