Activity 1-18: Hyperbolic Geometry

1. www.carom-maths.co.uk Activity 1-18: Hyperbolic Geometry

2. Who was Euclid? He was Greek, living at around 300BC. The statue of him on the left (something of a guess, perhaps) is in the Oxford University Museum of Natural History. We don’t know much about the man himself, but he left behind The Elements, one of the most important mathematical set of books of all time.

3. In these 13 books, he gathered together the maths of his day, and systematised it for the first time. In doing so, he introduced the idea of deductive proof, employing the axiomatic method. The idea here was to choosea list of axioms, that one would take as being self-evidently true, and then to deduce from these more complicated theorems.

4. Two important things: Thelist of axioms should be as small as possible while still being large enough to prove every theorem (the list should be COMPLETE). The list should NOT allow contradictory theorems to be proved from it (the list should be CONSISTENT.) Euclid carefully built his edifice of theorems upon the foundations given by his axioms. Theorem 47 was Pythagoras’s.

5. So what did Euclid choose as his axioms? He started with five common notions. 1. a = b and c = b means a = c. 2. a = b and c = d means a + c = b + d. 3. a = b and c = d means a - c = b - d. 4. Things which coincide with one another are equal to one another. 5. The whole is greater than the part. These seem to be hard to argue with so far...

6. He then chose five axioms. • Every two points lie on exactly one line. 2. Any line segment with given endpoints may be continued in either direction. 3. It is possible to construct a circle with any point as its centre and with a radius of any length. 4. All right angles are equal. 5. Given a line L and a point not on L, there is one and only one line which contains the point and which is parallel to L.

8. It is this fifth axiom, the Parallel Postulate, that caused a lot of trouble. The question was, could this be derived from the other axioms and common notions as a theorem? If so, it could be removed from the list of axioms, which would then be a smaller yet still (hopefully) complete set. Euclid himself seems to be unsure on this question. He certainly seems to go out of his way to avoid using the Parallel Postulate in his opening theorems.

9. Many, many mathematicians attempted to prove the Parallel Postulate to be either a necessary axiom, or a theorem, but all failed. John Wallis, English, (1616 –1703) Wallis showed that the Parallel Postulate was equivalent to saying that the angles in a triangle add to two right angles.

10. So the question became, • can you replace the Parallel Postulate with: • the angles in a triangle add to more than two right angles? • the angles in a triangle add to less than two right angles? Giovanni Saccheri, Italian, (1667 – 1733) Saccheri proved that the first of these led to a contradiction. The second of these, however, proved much harder to deal with. Saccheri searched for a contradiction, but never found one.

11. But then in the 19th century, Gauss, LobachevskyandBolyaifound a solution. Carl Friedrich Gauss German, (1777-1855) Nikolai Lobachevsky, Russian, (1792-1856) Janos Bolyai, Hungarian, (1802-1860)

12. They replaced the Parallel Postulate with another axiom, without removing the idea that the axioms should be consistent. The revised fifth axiom looked like this: Given a line and a point not on that line, there are infinitely many lines parallel to the given line through the point. All the other axioms were the same. This new geometry was shown to be consistent, and so another geometry could stand alongside Euclid’s as a possible way of modelling the universe.

13. This proved Euclid to have been remarkably far-sighted in including his fifth axiom in his list. The work above shows that it cannot be derived from the others, and is thus essential. So what does Non-Euclidean Geometry look like? Well, there are a number of non-Euclidean geometries. We will look at here hyperbolic geometry, as investigated by Gauss, LobachevskyandBolyai, which can be modelled simply by the following geometry-in-a-circle.

14. I use here the Non-Euclid software available from Joel Castellanos’ssite at the University of New Mexico. Download this now by clicking below. NonEuclid Interactive Constructions in Hyperbolic Geometry http://www.cs.unm.edu/~joel/NonEuclid/NonEuclid.html Task: try out the various constructions and commands here. What do we mean by point? By line? Can you make a triangle? What can you say about the angles in this triangle? Can you make a square, and what do we mean by square here? What do circles look like?

15. The picture shows two (straight) ‘lines’ in this geometry; these are arcs of circles that meet the light blue circle edge at right angles. A straight line in any geometry is the shortest distance between two points (the geodesic). We have an unusual idea of distance in operation here that means these circular arcs are indeed the shortest distance between the given points. We are free to interpret the notions of ‘point’ and ‘line’ in Euclid’s axioms as we wish, as long as we are consistent.

16. If we regard our black circle as the unit circle, centre O, in the Argand diagram, the distance between the points z1 and z2 is If z2 = 1 and z1 = -1, then d(z1, z2) = tanh-1(1) = . The distance function (or metric) here is such that towards the edge of the circle, distances get bigger and bigger. The distance across the entire circle is infinite.

17. The area of each triangle here, for example, is the same. What are the angles here? 90o, 45o and 30o, which add to 165o. The angle-sum of a triangle in hyperbolic geometry is less than 180o.

18. A miraculous fact: circles in this geometry, (even with our strange idea of distance), look exactly like circles in Euclidean geometry. Their centres, however, don’t: the less central that the hyperbolic centre of the circle is, the more it diverges from the Euclidean centre.

19. Can we draw a square here? Well, we can see clearly from cutting it into two triangles that a ‘square’ cannot have four 90o angles. It is more a case of drawing ‘a regular quadrilateral’, one with all sides equal, and all angles equal. Draw a line AB. Draw rays from A and B at 80o to AB.

20. Add a circle centre A, radius AB, and a circle centre B, radius AB, and mark the intersection points with the rays, E and F. Add a circle, centre F, radius FB = AB, and drag A so that this circle passes through E. ABFE is then a regular quadrilateral.

21. How can we prove all the angles are equal to 80o here? We can measure them, and they seem to be, but is that enough? It would be nice to know which if any of Euclid’stheorems we can use In this new geometry. Let’s check out Euclid’sfive axioms. (The common notions are certainly true here.)

22. Every two points lie on exactly one line. Pick two points A and B. Add a third C. Draw the Euclidean circle through A, B and C. Now vary C – how many circles ABC are perpendicular to the black circle? Exactly one. 2. Any line segment with given endpoints may be continued in either direction. Clearly true. 3. It is possible to construct a circle with any point as its centre and with a radius of any length. Given that distances become arbitrarily large towards the edge of the black circle, this is true. 4. All right angles are equal. Clearly true.

23. 5. Given a line L and a point not on L, there is one and only one line which contains the point and which is parallel to L. Ah! Two lines are parallel if they don’t meet when extended to infinity. We have a line BC, and a point A, and there are lots of lines through A parallel to BC.

24. So we have a viable geometry that differs from Euclid’s only in its final axiom. So given that Euclid’s first four axioms are true here, any of Euclid’stheorems that uses these, and not the fifth axiom, will hold. (Remember, for his early results, Euclid did not use the Parallel Postulate.) This includes the theorem that the two base angles of an isosceles triangle are equal (Proposition 5).

25. We can also use Proposition 4, which tells us that if two triangles have two pairs of sides that are equal, and the included angles are equal too, then the triangles are congruent. Thus the quadrilateral ABFE is regular.

26. With thanks to:To Joel Castellanos and his team for their superb Non-Euclid applet and site.The Open University, for helpful historical material. Carom is written by Jonny Griffiths, hello@jonny-griffiths.net