1 Appendix to notes 2, on Hyperbolic geometry:

Transcription

1 1230, notes 3 1 Appendix to notes 2, on Hyperbolic geometry: The axioms of hyperbolic geometry are axioms 1-4 of Euclid, plus an alternative to axiom 5: Axiom 5-h: Given a line l and a point p not on l, there are at least two distinct lines which contain p and do not intersect l. Theorem: Assuming Euclid s first four axioms, and axiom 5-h, and given l and p not on l, there are infinitely many lines containing p and not intersecting l. Proof?

2 Comment on the extendability axiom. The wording in Euclid is a bit unclear. follows: Recall that I stated it as A line can be extended forever. This is basically what Euclid wrote. Hilbert was more precise, and adopted an axiom also used by Archimedes: Axiom: If AB and CD are any segments, then there exists a number n such that n copies of CD constructed contiguously from A along the ray AB willl pass beyond the point B.

3 A model of hyperbolic geometry: Consider an open disk, say of radius 1, in the earlier model of Euclidean geometry. Thus, let D = { (α, β) α 2 + β 2 < 1 } Definition: A point is a pair (α, β) in this set. Definition: A line is any diameter of this disk, or the set of points on a circle which intersects D and which meets the boundary of D in two places and at right angles to the boundary of D at both of these places. Notice that a diameter can be considered part of such a circle of infinite radius. However, there a lot to be said before we can claim that this model satisfies Euclid s first four axioms. In particular, work is required to define what congruent line segments are. It is necessary to define some sort of distance between points which is different from Euclidean distance. With this definition, it turns out that a line segment near the boundary can look very short to us, and yet according to the definition of distance, be congruent (same length) as a segment near the middle of the disk which looks a lot longer to us. It is in this sense that a line can be extended indefinitely (to infinity ). This will be discussed in more detail if time permits later.

4 1.1 Euclidean solid geometry Planes are mentioned in some of the 1st 20 definitions in chapter 1. They don t appear again until chapter 11, where there are new definitions (e.g. "solid angle", definition 11), but no new postulates. (Hilbert included several axioms about planes.)

5 2 Euclid, Book 13 "Platonic Solids What are they?

6 Definitions: Although Platonic solids are three dimensional objects, we start with a definition in R polygon. (class exercise) As an example of a polygon we have, for example, triangles, squares, etc. Please note that I am talking here about the boundary and everything inside. For example, the set S = {(x, y) 0 x 1 and 0 y 1} is a polygon (called the unit square ). Its boundary is { (x, y) (x, y) S and either x = 0, x = 1, o9r y = 0 or y = 1 } but this set is not itself a polygon, at least not as a polygon is usually defined.

7 Definitions: 1. polygon A polygon in R 2 is a closed bounded subset of R 2 with the following properties: (a) It has a non-empty interior. (b) Its boundary consists of a finite set of line segments. (c) Each of these segments intersects exactly two of the others, one at each of its endpoints. There are no other intersections between these segments. (d) The boundary is connected. A polygon in R 3 is a subset of a plane which is a polygon in that plane. What unusual polygons are there with this definition?

8 2. A set in R n, is convex if each intersection of this set with a (straight) line is either empty, a single point, or a segment of that line. 3. A polygon is regular if it is convex, each edge is congruent to each other edge and all the angles between adjacent edges are equal. convex, regular non-convex irregular

9 Now we move to three dimensional objects. 4. A polyhedron is a connected closed bounded region in R3 whose boundary consists of a nite set of polygons. These polygons are called faces of the polyhedron. Polyhedra can also be convex or non-convex. convex polyhedron j nonconvex polyhedron

10 Not a polyhedron

11 But, all definitions are arbitrary, and authors often differ, according to the setting in which they are working. The definition I gave probably allows for some weird looking polyhedra. Some authors don t even attempt to define polyhedron, but simply define convex polyhedron, which would just add the conditon that the set is convex to what I have above, but would very much simplify what is allowed.

12 Test of geometric intuition: Start with a cube.

13 Looking down from the top, cut vertical slices (parallel to the vertical edges) along the four slanted lines, thus cutting off the four vertical edges.

14 Make the same cuts on a second side, oriented as shown.

15 Do this again.

16 Do this again. What is the result?

17 Definition: A Platonic solid, or regular polyhedron, is a convex polyhedron such that all faces are congruent to each other, as are all solid angles, and each face is a regular polygon.

18

19 Plato: The first will be the simplest and smallest construction,and its element is that triangle which has its hypotenuse twice the lesser side. When two such triangles are joined at the diagonal, and this is repeated three times, and the triangles rest their diagonals and shorter sides on the same point as a centre, a single equilateral triangle is formed out of six triangles; and four equilateral triangles, if put together, make out of every three plane angles one solid angle, being that which is nearest to the most obtuse of plane angles; and out of the combination of these four angles arises the first solid form which distributes into equal and similar parts the whole circle in which it is inscribed. See Elements, chapter 11 for defintion of "solid angle"

20 He is describing a tetrahedron, in which three equilateral triangles meet at each vertex: sum of angles at vertex: 3 60 o = 180 < 360. Other possibilities:

21 Four triangular faces meet at a vertex. This gives an octahedron sum of angles at vertex = 4 60 = 240 < 360.

22 Five triangular faces meet at a vertex, giving an icosahedron (20 faces): sum of angles at vertex = 5 60 = 300 < 360

23 Can six triangular faces meet at a vertex? Angle sum = 6 60 = 360, so the surface would be flat and we don t get a three dimensional solid. (See Euclid, Book 11, Propositions 20,21.)

24 How about square faces? faces meet at a vertex. The only possibility is a cube three sum of angles at vertex= 3 90 = Four faces meeting at a point would give a sum of 360

25 There is one with pentagonal faces, three meeting at a vertex. This gives a dodecahedron (12 faces) sum of angles at vertex: = 324 < 360.

26 That s all there can be (five), because any others would give too large an angle sum ( 360 )

27

28 Exactly what has been proved here?

29 Theorem: There are no more than five platonic solids.

30 Theorem: There are no more than five platonic solids. But how do we know there are any at all?

31 The greatest idea of Greek mathematicians: Assertions such as the existence of any of the platonic solids need to be proved.

32 The greatest idea of Greek mathematicians: Assertions such as the existence of any of the platonic solids need to be proved. For example, do the Platonic solids exist without the parallel postulate? (This may be the universe in which we live!)

33 The greatest idea of Greek mathematics: Assertions such as the existence of any of the platonic solids need to be proved. Earlier example: The "Pythagorean theorem". The result was discovered many times, in many places, but as far as we know, without proof before Pythagoras.

34 Construction of a cube. Proposition 15, Book 13. To construct a cube, and enclose it in a sphere. Let the diameter AB of the given sphere be laid on, and let it have been cut at C such that AC is double CB. And Let CD have been drawn from C at right angles to AB. And let the semi-circle ADB have been drawn on AB. And let DB have been joined. And let the square EFGH, having its side equal to DB, be laid out. And let EK, FL, GM, and HN have been drawn from points E, F, G, and H, respctively, at right angles to the plane of square EFGH. And let EK, FL, GM, and HN, equal to one of EF, FG, GH, and HE, have been cut off from EK, FL, GM, and HN, respectively. And let KL, LM, MN, and NK have been joined. Thus a cube contained by six equal squares has been constructed. Question: Did this require the parallel postulate? Book 1, proposition 2 prop 6.10 proposition 46, book 1 prop

35 Definition: Let P be a convex polyhedron. Then the dual of P is the polyhedron with vertices at the centers of the faces of P, and edges connecting the centers of any two faces with a common edge. Proposition: If P has v vertices, f faces, and e edges, then the dual of P has f vertices, e edges, and v faces. Construction of an Octahedron: Book 11, prop 14 or: The polyhedron whose vertices are the centers of the faces of a cube is an octahedron. (The "dual" of the cube".) Also, the polyhedron whose vertices are the centers of the faces of an octahedron is a cube.) Construction of a tetrahedron: Proposition 13, Book 13. (pg. 519 in Fitzpatrick edition, 37 lines) or: Prove that an appropriate set of 4 vertices of a cube form a tetrahedron. Construction of a Dodecahedron: Proposition 17. (pg. 530, 88 lines) Construction of an icosahedron: Prop 16, (pg lines)

36 Woodworker s proof that a dodecahedron exists In the slicing method we leave, at the end, one line segment on each of the original faces of the cube. We assume that these are of equal length and symmetrically placed.

37 Consider three of the cut planes as shown: Here we are assuming that each of the cutting planes goes from a center line of one face to a point a distance d 2 from the center line of another face, with orientation as shown. We need to calculate d in terms of the length s of a side of the cube. In fact, d will the length of an edge of our dodecahedron. We calculate the intersection of the three planes (yellow dot). This is a straight forward problem in linear algebra. After some further algebra it turns out that in order for that point to be the distance d from the three other points we must have ( sd s 2 ) 2 + s 4 + (s d) 4 = d 2 (4s 2d) 2 (1)

38 If we can find a positive solution to this equation which is less than s then we will have proven that a dodecahedron exists. In fact, some algebra leads to four solutions, d = s 2 d = 1 5 s (< 0) 2 d = d = 3 5 s, 2 ( golden mean, s, but we want d < s s (> 1) only feasible value, = 2 golden mean s )

39 d = 3 5 s 2 Here is one of the pentagons (yellow dots) : The golden mean can be constructed by methods of Euclid (straight edge and compass; book 2, proposition 11), and so in essence this

40 method can be said to be Euclidean. But algebra makes it easier to prove this.

41 The Icosahedron is the dual of the dodecahedron, and vice-versa.

42 Better method for constructing a dodecahedron: Start with the icosahedron, using a construction due to Luca Pacioli, (See Stillwell, page 21-22) First define the "golden ratio" as usual: 1 x = x 1 x x = = Construct a rectangle with this ratio of the longer side to the shorter. (Book 2, proposition 11, plus book 6, prop 30.) 2. Construct three perpendicular rectangles of this size as shown (Book 11, prop. 11)

43 Prove that ABC is an equilateral triangle. 3 Beyond the Platonic Solids 4 The Archimedian solids Faces are all regular polyhedra, but they are not all the same polyhedron. Archimedes proved that there are exactly 13 of these.

44 4.1 Euler s formula e = v + f 2. Applies to any simply connected polyhedron. Really a theorem in "topology"! Plausibility argument (??): Consider a polyhedron with only four vertices. Then this is a tetrahedron (possibly irregular). Show that if one vertex is added, then another edge is added also, so v + f e is unchanged. Add an edge. This adds a face, so again v + f e is unchanged. Check its value for a tetrahedron Proofs: There are many. For example, see page 471 of Stillwell. (There are many formulas due to Euler. The best known are this one and e iπ = 1.) This can be used to prove (again) that there are only 5 regular polyhedra.