Riemannian metric in dimensions is given by symmetric matrix The components of this matrix are, in general, functions of coordinates of the points of the space under consideration. Knowing the metric, we can calculate scalar product of any two tangent vectors at the same point. Knowing the scalar product of any two vectors we can calculate the metric tensor. It goes as follows. Suppose we have coordinates . Sometimes the coordinates have their names, like but then it is often convenient to number them, for instance , and to refer to them through their indices. When we have coordinates, we have vectors tangent to the coordinate lines. They are usually denoted as or even simply . This is a standard notation used in differential geometry texts. So, for instance, is vector tangent to the coordinate line , that is the line when is varying while are kept constant. More precisely: is a vector field, since we can draw the coordinate of through every point. Therefore when I write I mean the vector field or I mean one particular vector at one particular point that should be evident from the context. When it needs to be specified, we can write – which means that we are specifying particular point
If we have scalar product defined at every point of our space, then the metric tensor at a point is given by the matrix
Vectors tangent to the coordinate lines form a basis in the tangent space at every point. When we say that a given has components , that means that – where we use Einstein convention implying summation over the dummy index. Thus we may write:
Let us now find the vector fields tangent to the coordinate lines:
These are tangent vectors at . According to the prescription described in Riemannian metric on SL(2,R) we need to shift them to the identity, that is we need to calculate , and then take trace of their products:
I did these calculations using computer. Here are the results:
The metric is not diagonal. That means the coordinate lines are not all perpendicular to each other. Moreover, the third line is “light-like”. We have In Minkowski space-time it would mean that it is a trajectory of an object moving with the speed of light. At the group identity we have At this point we have
We are going to discuss Riemannian metric on SL(2,R), which will be more difficult than Riemannian metric on the Poincare disk or on the upper half-plane , because SL(2,R) is three-dimensional. In fact it will be a pseudo-Riemannian metric, because SL(2,R) with its natural metric looks more like space-time (one minus and two pluses, or one plus and two minuses), while (and ) looks like space alone (all pluses).
In order to introduce the metric, to calculate geodesics and Curvature, we need to introduce coordinates. Well, in fact we don’t have to, but we will. It was the French mathematician Élie Cartan Cartan who invented the technique of doing geometry without coordinates, using the so called “moving frame” instead. That is sometimes very useful. But for now we will follow the old-fashioned straightforward way, we will use coordinates.
The group SL(2,R) is isomorphic to SU(1,1) via Cayley transformation, and we have already parametrized SU(1,1) in SU(1,1) parametrization and SU(1,1) decomposition , so in principle we could use what we already have. But we can also try something new, and then see how it relates to the old. Here I am going to use parametrization suggested by Keith Conrad in his notes on SL(2,R). Here is the beginning of his paper “DECOMPOSING ”
We are introducing three one-parameter subgroups of SL(2,R) called
Then, as Conrad is showing, every SL(2,R) matrix decomposes uniquely into the product . But we will change the order. Instead of writing , we will write There is a tradition in group theory to use the KAN order. If you search Google for “KAN decomposition”, you will find a lot of information. If you look for “NAK decomposition” – you will find much less, mainly written by eccentrics.
We prefer NAK, because it better corresponds to the polar decomposition of SU(1,1) that we have already discussed.
So, we write as , that is, after multiplying the three matrices, as
Now are coordinates of the element of the group. Of course we would like to know how they relate to coordinates that we have introduced in the group SU(1,1). For instance in Right circular orbits in SU(1,1) we were representing SU(1,1) as a torus. How our new parameters relate to the parameters of the torus?
To answer this question we need to Cayley transform back to SU(1,1) and then express and of the torus in terms of of the NAK decomposition. I did it, using any algebraic software it is a simple task. The end result is that is the same (that is why I have chosen NAK instead of KAN, while is an algebraic function of . With we have:
Thus after Cayley transform to SU(1,1) become coordinates on the unit disk – the cross-section of the torus. Here are their lines transformed to the disk:
“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.