This is a continuation from Dedekind tessellation or circles all the way down. I am studying the very interesting paper by J. Kocik, aka Jurek, “A note on the Dedekind tessellation”. It is a pity that the paper is unpublished. I am not sure how much of its content I am allowed to reveal. I will take a risk and reveal just one goodie from this paper.
Here is the goodie:
And I will explain now how I produced this picture.
As I have explained it in Dedekind tessellation or circles all the way down, we have a nice tessellation of the upper half-plane. The upper half-plane is tessellated by circles which, in hyperbolic geometry are “straight lines”. Using Cayley transform we can move these circles to the unit disk. Kocik, in his paper, gives the algorithm for constructing the centers and the radii of these circles. I used Mathematica to implement this algorithm as follows:
I did not follow exactly Kocik’s algorithm. Moreover I have rotated the data from his paper by 90 degrees, so that it agrees with the Cayley transform I was using before. I stored the list of positions and radii of the circles that I was using in a file: c200.m.
The red circle is the unit circle. Strictly speaking the Dedekind tessellation should be restricted to the inside of the unit circle. But It is amazing to plot the circles also in the outside region. After all the unit circle is just a boundary between two universes. Why should we restrict ourselves to only one side? The universe we are living in is, perhaps, also a boundary between different multidimensional universes of hyperdimensional physics.
“Turtles all the way down” is an expression of the infinite regress problem in cosmology posed by the “unmoved mover” paradox. The metaphor in the anecdote represents a popular notion of the model that Earth is actually flat and is supported on the back of a World Turtle, which itself is propped up by a column of turtles. Questioning what the final turtle might be standing on, the anecdote humorously concludes that it is “turtles all the way down”
After a lecture on cosmology and the structure of the solar system, William James was accosted by a little old lady.
“Your theory that the sun is the centre of the solar system, and the earth is a ball which rotates around it has a very convincing ring to it, Mr. James, but it’s wrong. I’ve got a better theory,” said the little old lady.
“And what is that, madam?” Inquired James politely.
“That we live on a crust of earth which is on the back of a giant turtle,”
Not wishing to demolish this absurd little theory by bringing to bear the masses of scientific evidence he had at his command, James decided to gently dissuade his opponent by making her see some of the inadequacies of her position.
“If your theory is correct, madam,” he asked, “what does this turtle stand on?”
“You’re a very clever man, Mr. James, and that’s a very good question,” replied the little old lady, “but I have an answer to it. And it is this: The first turtle stands on the back of a second, far larger, turtle, who stands directly under him.”
“But what does this second turtle stand on?” persisted James patiently.
To this the little old lady crowed triumphantly. “It’s no use, Mr. James – it’s turtles all the way down.”
—J. R. Ross, Constraints on Variables in Syntax 1967
The Dedekind tessellation is a mathematical version of the flat Earth and turtles all the way down. Instead of the flat Earth we have the Poincaré disk, and instead of turtles we have circles – all the way down.
I have mentioned this subject at the end of Geodesics on the upper half-plane – parametrization, two days ago. In the meantime prof. Kocik was kind enough to send me his unpublished paper “A note on the Dedekind tessellation“. Looking inside I have found that I have made a mistake in my Mathematica program that was supposed to create my own version of the tessellation. After fixing my mistake (I have forgotten one term in my formula for the centers of the circles) I was able to create my own poor man version of the Dedekind tessellation (Kocik gives a simple method of computing the centers and the radii of all circles in the Dedekind tessellation!), and here I am going to explain, in details, how I did it.
Instead of the disk it is better to use the upper half-plane, where the group SL(2,R) acts by fractional linear transformations. Let me recall the action. With
the action on the upper half-plane, the set of complex numbers with is given by
Remark: In many places a different convention is being used, with defined as There is a simple relation between the two actions: we can obtain one action from another by composing it with the similarity transformation
The group SL(2,R) has a discrete subgroup SL(2,Z), often called the modular group, the group of matrices with entries that are integers. For instance the following two matrices:
Dedekind tessellation of the upper half-plane is an analogue of the tessellation of the usual Euclidean plane into squares of unit side length. All the plane can be nicely covered by translations of just one square, namely by translations by an integer amount in horizontal and vertical direction. In the case of the upper half-plane instead of a square we have a triangle, so called fundamental domain. In the picture below, taken from Condrad’s paper, we see the (grayed) standard fundamental domain consisting of complex numbers with and
Sure it does not look like a triangle, but it is a triangle, with the top vertex at infinity! The bottom side is a part of the unit circle – therefore a geodesic in the hyperbolic geometry we are dealing with.
By acting on this one triangle with matrices from the group SL(2,Z) we can nicely triangulate the whole upper half-plane – that is called the Dedekind tessellation. In Wikipedia article on Modular Group, in the section on Tessellation of the hyperbolic plane, it is accompanied with the following image:
Looking at this image it occurred to me that instead of acting on the fundamental domain I can as well act on the unit half-circle and on the two vertical lines. In order to do this I need a formula how elements of SL(2,Z) transform the unit circle and the two vertical lines. And I am interested in the resulting circles, I do not care about vertical lines as they do not add much to the tessellation. Using mathematical software (I use Mathematica most of the time) the task was not too difficult.
Suppose we have a circle of radius with center at on the real axis. We apply the transformation to this circle using matrix in SL(2,R). The result is, as it can be verified without much difficulty using any computer algebra software, the circle with
We are interested in transformations of one particular circle with and Then the formulas simplify:
Of course we are interested only in the cases where the denominator is non zero.
Now suppose we have a vertical line at . It is transformed into a circle with
We are interested in two cases and only when the denominator is non zero.
Using generators above, taking their inverses and products, in different orders, I generated 35416 SL(2,Z) matrices. They can be downloaded as the file sl2.zip. The file has the format:
Separate matrices look as .
I will describe now, step by step, how i have created Fig. 1 above, using Mathematica. It is certainly not an optimal way, but my description may help to understand how to create Dedekind tessellation using different methods.
From the two-dimensional disk we are moving to three-dimensional space-time. We will meet Einstein-Poincare-Minkowski special relativity, though in a baby version, with x and y, but without z in space. It is not too bad, because the famous Lorentz transformations, with length contraction and time dilation happen already in two-dimensional space-time, with x and t alone. We will discover Lorentz transformations today. First in disguise, but then we will unmask them.
We have the group SU(1,1) acting on the disk with fractional linear transformations. With and in SU(1,1)
the fractional linear action is
By the way, we know from previous notes that is in SU(1,1) if and only if
Having the new point on the disk, with coordinates we can use Eq. (1) to calculate the new space-time point coordinates This is what we will do now. We will see that even if depends on in a nonlinear way, the space-time coordinates transform linearly. We will calculate the transformation matrix and express it in terms of and We will also check that this is a matrix in the group SO(1,2).
The program above involves algebraic calculations. Doing them by hand is not a good idea. Let me recall a quote from Gottfried Leibniz, who, according to Wikipedia
He became one of the most prolific inventors in the field of mechanical calculators. While working on adding automatic multiplication and division to Pascal’s calculator, he was the first to describe a pinwheel calculator in 1685 and invented the Leibniz wheel, used in the arithmometer, the first mass-produced mechanical calculator. He also refined the binary number system, which is the foundation of virtually all digital computers.
I used Mathematica as my machine. The same calculations can be certainly done with Maple, or with free software like Reduce or Maxima. For those interested, the code that I used, and the results can be reviewed as a separate HTML document: From SU(1,1) to Lorentz.
Here I will provide only the results. It is important to notice that while the matrix has complex entries, the matrix is real. The entries of depend on real and imaginary parts of and
Here is the calculated result for :
In From SU(1,1) to Lorentz it is first verified that the matrix is of determinant 1. Then it is verified that it preserves the Minkowski space-time metric. With defined as
Since the transformation preserves the time direction. Thus is an element of the proper Lorentz group .
Remark: Of course we could have chosen with on the diagonal. We would have the group SO(2,1), and we would write the hyperboloid as It is a question of convention.
In SU(1,1) straight lines on the disk we considered three one-parameter subgroups of SU(1,1):
We can now use Eq. (7) in order to see which space-time transformations they implement. Again I calculated it obeying Leibniz and using a machine (see From SU(1,1) to Lorentz).
Here are the results of the machine work:
The third family is a simple Euclidean rotation in the plane. That is why I denoted the parameter with the letter In order to “decode” the first two one-parameter subgroups it is convenient to introduce new variable and set The group property is then lost, but the matrices become evidently those of special Lorentz transformations, transforming and , leaving unchanged, and transforming and leaving unchanged (though with a different sign of ). Taking into account the identities
In the following posts we will use the relativistic Minkowski space distance on the hyperboloid for finding the distance formula on the Poincare disk.