Lecture 10 Hyperbolic Geometry MAT 360 Lecture 10 Hyperbolic Geometry

Homework next week (Last hw!) 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 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 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. 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: 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.

Corollary In hyperbolic geometry, there exists an absolute unit of length.

Recall 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.

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 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 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 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, S U T 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.

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.