1 Appendix to Notes 2, on Hyperbolic Geometry

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’sfirst 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 ? Comment on the “extendability” axiom. The wording in Euclid is a bit unclear. Recall that I stated it as follows: 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. A model of hyperbolic geometry: Consider an open disk, say of radius 1, in the earlier model of Euclidean geometry. Thus, let D = (α, ) 2 + 2 < 1 j n o Definition: A “point” is a pair (α, ) 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’sfirst 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. 1.1 Euclidean solid geometry Planes are mentioned in some of the 1st 20 definitions in chapter 1. They don’tappear 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.) 2 Euclid, Book 13 "Platonic Solids” What are they? Definitions: Although Platonic solids are three dimensional objects, we start with a definition in R2. 1. 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 f j ≤ ≤ ≤ ≤ g is a polygon (called the “unit square”). Its boundary is (x, y) (x, y) S and either x = 0, x = 1, j 2 o9r ( y = 0 or y = 1 ) but this set is not itself a polygon, at least not as a polygon is usually defined. Definitions: 1. polygon A polygon in R2 is a closed bounded subset of R2 with the fol- lowing 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 R3 is a subset of a plane which is a polygon in that plane. What “unusual” polygons are there with this definition? 2. A set in Rn, 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 Now we move to three dimensional objects. 4. A polyhedron is a connected closed bounded region in R3 whose boundary consists of a finite set of polygons. These polygons are called “faces” of the polyhedron. Polyhedra can also be convex or non-convex. convex polyhedron nonconvex polyhedron j Not a polyhedron 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’teven 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. Test of geometric intuition: Start with a cube. 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. Make the same cuts on a second side, oriented as shown. Do this again. Do this again. What is the result? 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. 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 diago- nal, 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" He is describing a tetrahedron, in which three equilateral triangles meet at each vertex: sum of angles at vertex: 3 60o = 180 < 360. Other possibilities: Four triangular faces meet at a vertex. This gives an octahedron sum of angles at vertex = 4 60 = 240 < 360. Five triangular faces meet at a vertex, giving an icosahedron (20 faces): sum of angles at vertex = 5 60 = 300 < 360 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.) How about square faces? The only possibility is a cube — three faces meet at a vertex. sum of angles at vertex= 3 90 = 270.. Four faces meeting at a point would give a sum of 360 There is one with pentagonal faces, three meeting at a vertex. This gives a dodecahedron (12 faces) sum of angles at vertex: 3 108 = 324 < 360. That’s all there can be (five), because any others would give too large an angle sum ( 360 ) ≥ Exactly what has been proved here? Theorem: There are no more than five platonic solids. Theorem: There are no more than five platonic solids. But how do we know there are any at all? The greatest idea of Greek mathematicians: Assertions such as the existence of any of the platonic solids need to be “proved”. 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!) 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. 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,y and let it have been cut at C such that AC is double CB.z 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.x 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? yBook 1, proposition 2 zprop 6.10 xproposition 46, book 1 ¶prop. 11.12 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.

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