Looking at Non-Euclidean Geometry and the Poincare Disk

Thanks to Bernhard Riemann and Nikolai Lobachevsky, a rich alternate geometry was developed beyond the limits of Euclidean geometry.  It included using generalizations of the respective  triangles for which the sum of angles could be greater than 180 degrees (the Riemannian case) or less than 180 degrees.  The basic comparison of the two geometries is depicted in Fig. 1.This non-Euclidean geometry allowed itself to be paired to the advanced math of tensor calculus. (Which then was applied by Einstein to formulate his theory of general relativity).

As can be ascertained by inspection - looking carefully at the meridian circles and latitude parallels in the Riemannian sphere in Fig.2,-   there are no two parallel lines in the Euclidean sense, since any two geodesics (curves of shortest path) must intersect. Thus, the sum of the angles of the triangle formed by 3 geodesics will always total > 180 degrees.  Looking at the spherical geometry, it’s also easy to see why positively curved Riemannian geometry was the first to be developed – because it was based on the already familiar geometry of the sphere. (From which we use spherical trigonometry to obtain distances and angles, e.g. using a law of sines of the form: sin a/ sin(A) = sin b/ sin (B) where common letters refer to sides and capital letters to angles.)




















The proper terms for the respective spaces were: elliptic (for Riemannian or +-curved), and hyperbolic (for Lobachevskian or (-)-curved.

Interestingly,  all three geometries – Euclidean, elliptic and hyperbolic are actually related to each other via a model known as the Poincare disk (Fig. 3).

What Poincare achieved in his disk model is depicting a hyperbolic plane within a Euclidean plane. The difference is that straight lines in Poincare’s disk once they are the same as geodesics (great circles) are warped. Thus, all straight lines on the disk appear to be curved lines – as shown – unless a particular line traverses the midpoint of the disk, in which case the apparent bending or warp is reduced to zero. So it appears as a straight line.

Another unique difference of the Poincare disk model from an ordinary circle (which some will be tempted to assume or see) is that the boundary of his model is infinitely far away or we would say “at infinity”.  Another peculiar aspects concern how right angles are made and where they are made. For example, the geodesic shown in the Poincare disk model makes right angles at either boundary. Similarly, any two geodesics that intersect (imagine another straight line 90 degrees different in orientation from that shown also passing through the disk center) form EUCLIDEAN angles at the point of their intersection.

Thus, the disk model is extremely valuable in that it permits discussion of non-Euclidean geometry in Euclidean terms. Let’s carry this forward looking at the curve PQ on the disk, and also the intermediary points A, B along it. Let’s assume we’d like to find the length x = AB, a segment of the curve PQ. Then one can show:

x = ln [(AQ/AP)/ (BQ/ BP)]

which uses a property common to natural logarithms called the ‘cross ratio’.  Note the above definition contains embedded within the definition for points (A,B) , line (AB= x), length x, as well as “angle” – defined in terms of the Euclidean angle subtended at the point of intersection by tangents to the lines viewed as circular arcs in the putative Euclidean plane.

 

Earlier, I showed the form the law of sines takes for spherical (elliptic) geometry. In the case of hyperbolic geometry it takes a different one from that (understandable given that the sum of angles in a hyperbolic triangle < 180 degrees!).  First, let’s recall the law of sines for the Euclidean case: 

 

a/ b = sin(A)/ sin(B).

 

Now for the hyperbolic case we have:

 

sin (A)/ sin(B) = {(sinh(a/k)/ sinh(b/k)}

where A, B, a, b again have the standard definitions as per angles and sides, and k denotes a positive constant that is confused with curvature!)  To be more specific, the value of k depends on the choice of the units of measurement. The value of k also expresses a definite length, all other factors being equal.

 For example, consider a number of concentric horocycles (such as QABP in Fig. 3) at a distance x from each other. Let their corresponding arcs have lengths: a1, a2, a3…..an. The ratio of each one to the next is then exp (x/k) where a1, a2, a3 etc. form a geometrical progression with the quotient:

 

exp (- x/ k) = 1  / (exp(x/k))

In other words, if the first arc is 1, the 2^nd is 0.9, then the third is (0.9)² = 0.81 and the fourth is (0.9)³  =  0.729 and so on.

 For those who may not know, “sinh” defines the hyperbolic sine. From calculus texts this is defined based on the exponential function, exp(x) as:

 

sinh(x) = ½ (exp(x) – exp(-x))

 with coordinates taken on the “unit hyperbola”: x² – y² = 1

 Similarly, the hyperbolic cosine (cosh(x)) is defined:

 cosh(x) = ½ (exp(x) + exp(-x))

 

In fact, when formulations are made one can obtain identities which parallel those of the standard trig functions. For example, in the latter we have:  cos²(x) + sin²(x) = 1, and in the hyperbolic sense one can prove:
 

 cosh²(x) - sinh²(x) = 1

 

 The hyperbolic cosine is useful to know when venturing into hyperbolic geometry because it is used to express a relation associating the 3 sides of a right triangle in hyperbolic space:

 

cosh(c/k) = cosh (a/k) cosh(b/k)

Many more detailed investigations of non-Euclidean geometry are possible, and also seeing how they link up with Euclidean geometry in a more general formalism. The basis for doing this entails taking a more comprehensive analytic (e.g. 'manifold') approach,  wherein we associate lines and points with a geometry. A simplified form for this may be expressed using the matrix formulation in 4(a). 

Where the left side denotes a line [u1, u2, u3]. If one finds that the point specified by (x1, x2, x3)  is the same as for another point specified (y1,y2, y3) such that the matrix relation in 4(b) is true then we can say that the correlation constitutes a polarity. It is possible that a suitable choice of polarity can be found for any combination of points for a line [u1, u2, u3] such that: u1 = x1, u2 = x2 and u3 = cx3.  In this case, if c = +1, the geometry is elliptic (e.g. Riemannian), and if c =-1 it is hyperbolic.   This gets us into the area of projective geometry.

 

 

 

 

 

 

Problems for Math Mavens:

1. Consider a triangle in hyperbolic space as computed by an interstellar ship traversing it. The dimensions are: a = 1 pc, b = 2 pc and c is unknown where pc denotes parsec (1 pc = 3.26 light years).  Using this and the equation associating the 3 sides of a right triangle in hyperbolic space, with k = 1, find the dimension of c. Thence, find the ratios of the sines of the angles, e.g. (sin A/ sin B)

2. Consider the Lobachevskian surface shown in Fig. 1.  Assume  it displays concentric horocycles such that exp (- x/ k) =      1 / (exp(x/k)) and when the ratios for x/k are taken the progression is of the form: 1, 0.5, 0.25, 0.125 etc.  Use this to deduce the curvature k, if k = -1/ k².  Is the non-Euclidean surface shown above a Riemannian or Lobachevskian space? Give reasons.

A Mathematical Diversion: Magic Discs

Geometry, as we've seen in previous blogs, comes in many different forms. Two that I already explored have been plane geometry (to do with lines and planes in the context of linear algebra), and non -Euclidean geometry, in the context of Einstein's General Theory of Relativity. In this blog I take a look at projective geometry- a whole vast sub-discipline of math- in the context of "magic discs", which are simple representations of it.

Magic discs are useful because they keep consideration to a finite number of points. One then explores projective "n-spaces" - denote them as P^n(F_q) over some field F. When this is done, one finds that P^n(F_q) has exactly: 1 + q + q^2 + q^3 + .......q^n = (q^(n+1) -1)/(q -1) different points. Magic discs enter because they enable some very elegant constructions. When I first taught these to advanced 2nd formers, during a Peace Corps math teaching stint- I encouraged them to visualize the magic disc by making cardboard cutouts. The cutouts were done for different diameters, which were then numbered with evenly spaced marks around the circumference. Having done this, the students used pins to attach them to firm backboards, and the circles could then be rotated.

The method for enumerating a given cardboard disc was always the same: i.e. mark 1 + q + q^2 equally spaced points around the circumference matching marks on the cardboard circles to the backboard. Then label them in an anti-clockwise direction by the numbers: 0, 1, 2 ...q(1 + q) . Remember at all times that 'q' is a power of a prime. Say, for example, that q = 2, then one will use a clock face that is marked off starting from '0' (on the immediate right of the circle). The total number of equally spaced points to be marked off is computed as:

1 + 2 + 2^2 = 1 + 2 + 4 = 7

Since the first one is always marked at the '0' point, then the others will be: 1, 2, 3, 4, 5 and 6. The next job is to partition this circular field into (1 + q) points so that for q it will be 3 points. One finds that, apart from the 0 point, the other positions will always be such that for any selection of two marked points there is one position of the disc that "works", that is the selected distances end up in points that are coincident with two special points on the disc. The spacings for q = 2 will then be obtained from: 1 + 2 + 4, or more simply 1,2, 4. In other words, starting at the zero point, mark one space over to reach the number 1 on the background, then mark 2 more to reach the number 3 on the background, then mark 4 more to reach the number 2 + 4 = 0 where we began. In many ways, this procedure is similar to what we saw in the earlier blog (last year) to do with groups and "clock face" arithmetic. See, e.g.

http://brane-space.blogspot.com/2010/04/looking-at-groups.html


In Fig. 1, is shown the magic disc resolution for the case of q = 3. And this leads to q(1 + q) marked off numbers in toto, or 3(1 + 3) = 3 x 4 = 12. And we confirm that the numbers go from 0 - 12 on the clock face. (Or in terms of the physical model, the numbers appearing on the backboard). The number of special points spanning the circle is similarly: 1 + q = 1 + 3 = 4 in all. The trick is then to identify them. The partition that works is by successive spacings of: 1, 2, 6 and 4 in succession, i.e. 1 added to 0, then 2 added to 1 (3), then 6 added to 3 (9) and finally 4 added to that ...bringing us back to 0.

Lastly, in Fig. 2 we have a much larger disc for the case of q = 5. Here, the total numbers marked off will be: q (1 + q) = 5 (1 + 5) = 30, in all. The number of special points to partition the circle will be 1 + q = 1 + 5 = 6, which will yield six partitioned spaces. These will be obtained from: 1, 2, 7, 4, 12 and 5. In other words, 1 added to 0, then 2 added to that (3), then 7 added to that (10), then 4 added to that (14), then 12 added to that (26) and finally 7 added to that ....which takes us back to 0.