Month: April 2017

Esher’s Limit Circle IV rendered on the complex upper half-plane

Posted on by arkajad

In the recent note “SL(2,R) generators and vector fields on the half-plane” we got acquainted with two particular simple symmetries of the non-Euclidean hyperbolic geometry of the complex upper half-plane \mathbb{H}. These two symmetries are horizontal translation and uniform dilation. The horizontal translational symmetry means that all geometric properties of objects are invariant when all parts of the object keep the same height (constant y), and only the x coordinate is translated by a fixed amount. The symmetry with respect dilations means that when we multiply all y coordinates by a certain positive number, then we should also multiply the x coordinates by the same number. We obtain this way not only an object that is similar to the original, but, moreover, it also has the same size!

In Hyperbolic angels and demonswe were playing with “equal sizes” of Esher’s angels and demons on the Poincare disk:

Escher’s Angels and Demons and hyperbolic geometry

Today we will use the Cayley transform and move the Esher’s art piece to the upper half-plane.

From “The Equation That Couldn’t Be Solved” by Mario Livio.

The exact name of the image we will be playing with is Circle Limit IV. We want to render this image on the upper half-plane using Cayley transform. The problem that we need to address is this: to render the whole circle, we would need the whole upper half-plane. But we can only show a finite part of the plane. Therefore we need to decide how to cut?

I created a little Mathematica program to help me with this decision:

Mathematica code used for deciding on the size of the complex half-plane version of Esher’s Limit Circle IV

I decided to use -3\leq x\leq 3 and 0<y\leq 4. The part of the disk that is not being used is then small enough not to worry about (on the right in the image, near 1 on the circle boundary ). That gives the ratio 3:2 of the image. I decided to create 1440×960 pixels image. For this I had to find on the net Esher’s Limit Circle with high enough resolution. I have found it here. I cropped it to remove border and ended up with 2272×2276 image. In principle it should be a square, but few pixels should not make a difference. So, here is my Mathematica code: importing the image, converting to table, converting pixels to coordinates, making Cayley transform to the disk, converting disk coordinates to pixels, reading the color, using it for coloring the pixel on the half-plane;

Mathematica code for converting Esher’s Limit Circle IV from the disk to the upper half-plane

And here is the end result:

Esher’s Limit Circle IV developed onto upper half-plane. Click on the image to open it in full resolution.

All these angels have the same size, when measured with the hyperbolic stick, that was possibly created by the hyperbolic devils, that are all of the same size as well.

SL(2,R) generators and vector fields on the half-plane

Posted on by arkajad

Whenever we have a matrix Lie group, we also have its infinitesimal generators. From generators we get group elements by exponentiation. Generators form a linear space. In fact they form algebra with respect to commutators. In our case, for the groups SU(1,1) and SL(2,R), the Lie algebras are three-dimensional. We can choose there three linearly independent generators that form a linear basis for the whole algebra. From exponentials of generators we can form one-parameter subgroups. When the group acts on some space, like SU(1,1) on the disk D, or SL(2,R) on the half-space \mathbb{H}, one parameter subgroups define orbits – streamlines of the corresponding vector fields. Here are the details.

Suppose \gamma(t) is a path in the group SL(2,R). Suppose at t=0 we have \gamma(0)=I. For all t we have \det \gamma(t)=1. Let \dot{\gamma}(0) be the derivative of \gamma(t) with respect to t at t=0. Then \mathrm{tr}(\dot{\gamma}(0))=0. This follows from the general identity valid for matrix functions t\mapsto A(t) with invertible A(t) (see e.g. here)

(1)   \begin{equation*}\frac{d}{dt}[\det\,A(t)]=\det\,A(t)\cdot \mathrm{tr }[A(t)^{-1}\,\frac{d}{dt}A(t)].\end{equation*}

When A(0)=I and \det\,A(t)\equiv 1, then \mathrm{tr }(\frac{d}{dt}A(t))|_{t=0}=0. We have already met this property when discussing the group SU(1,1) in Getting hyperbolic

The set of all tangent vectors \dot{\gamma}(0) at identity is called the Lie algebra of the group. In our case the Lie algebra of SL(2,R) consists of all real 2\times 2 matrices with trace zero. It is denoted sl(2,R). The elements of sl(2,R) are called “generators” of the group. If X is a generator, that is, in our case, if \mathrm{tr }(X)=0, then e^{tX} is a one-parameter subgroup of the group. That \det e^{tX}=1 follows from another useful identity that can be found in Wikipedia under the term The determinant of the matrix exponential:

(2)   \begin{equation*}\det e^X=e^{\mathrm{tr }X}.\end{equation*}

The Lie algebra of group is an “algebra” with respect to the commutator operation. In our case if X and Y are matrices in sl(2,R), then Z=[X,Y]=XY-YX is also in sl(2,R) because for any two matrices x,y we have that \mathrm{tr }(XY)=\mathrm{tr }(YX). Thus trace of a commutator is always zero.

In SU(1,1) straight lines on the disk we met these SU(1,1) generators (I am changing the sign of X_3 here)

(3)   \begin{eqnarray*} X_1&=&\begin{bmatrix}0&i\\-i&0\end{bmatrix},\\ X_2&=&\begin{bmatrix}0&1\\1&0\end{bmatrix},\\ X_3&=&\begin{bmatrix}-i&0\\0&i\end{bmatrix}. \end{eqnarray*}

In Getting real the Cayley transform was defined using the matrix \mathcal{C}

(4)   \begin{equation*}\mathcal{C}=\frac{1}{1+i}\begin{bmatrix}i&1\\-i&1\end{bmatrix}.\end{equation*}

(5)   \begin{equation*}\mathcal{C}^*=\mathcal{C}^{-1}=\frac{1}{1-i}\begin{bmatrix}-i&i\\1&1\end{bmatrix}.\end{equation*}

The transformation A\mapsto A'=\mathcal{C}^{-1}A\mathcal{C} maps the group SU(1,1) onto the group SL(2,R). The same transformation maps the Lie algebra su(1,1) onto the Lie algebra sl(2,R). In particular we obtain the following three generators in sl(2,R):

(6)   \begin{eqnarray*} X_1'&=&\mathcal{C}^{-1}X_1\mathcal{C}=\begin{bmatrix}0&1\\1&0\end{bmatrix},\\ X_2'&=&\mathcal{C}^{-1}X_2\mathcal{C}=\begin{bmatrix}-1&0\\0&1\end{bmatrix},\\ X_3'&=&\mathcal{C}^{-1}X_3\mathcal{C}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}. \end{eqnarray*}

We can take exponentials of these generators and construct one-parameter subgroups

(7)   \begin{eqnarray*} A_1'(t)&=&\exp tX_1'=\begin{bmatrix}\cosh t&\sinh t\\ \sinh t&\cosh t\end{bmatrix},\\ A_2'(t)&=&\exp tX_2'=\begin{bmatrix}\exp(-t)&0\\ 0&\exp(t)\end{bmatrix},\\ A_3'(t)&=&\exp tX_1'=\begin{bmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{bmatrix}. \end{eqnarray*}

From these we can define their orbits in the half-plane:

(8)   \begin{eqnarray*} z_1(t)&=& A_1'(t)\cdot z'=\frac{\cosh(t) z'+\sinh (t)}{\sinh(t) z'+\cosh(t)},\\ z_2(t)&=& A_2'(t)\cdot z'=\frac{e^t z'}{e^{-t}}=e^{2t}z',\\ z_3(t)&=& A_3'(t)\cdot z'=\frac{\cos(t) z'+\sin (t)}{-\sin(t) z'+\cos(t)}. \end{eqnarray*}

Differentiating with respect to t at t=0 we obtain vector fields. I will use the tilde symbol to denote these vector fields. I will be skipping primes from this place on.

(9)   \begin{eqnarray*} \tilde{X}_1(x,y)&=&(dz_1(t)/dt)|_{t=0}=(1-x^2+y^2,-2xy),\\ \tilde{X}_2(x,y)&=&(dz_2(t)/dt)|_{t=0}=(2x,2y),\\ \tilde{X}_3(x,y)&=&(dz_3(t)/dt)|_{t=0}=(1+x^2-y^2,2xy). \end{eqnarray*}

Here are streamlines of these three vector fields.

Streamlines of the vector field (1-x^2+y^2,-2xy) on the half-plane \mathbb{H}

Streamlines of the vector field (x,y) on the half-plane \mathbb{H}

Streamlines of the vector field (1+x^2-y^2,2xy) on the half-plane \mathbb{H}

While the second vector field produces very simple streamlines, the first and the third are kind of strange.
Looking at Eqs. (6) we see, that it may be wise to introduce the linear combinations as follows:

(10)   \begin{eqnarray*}Y_+&=&\frac{1}{2}(X_1'+X_3')=\begin{bmatrix}0&0\\1&0\end{bmatrix},\\ Y_-&=&\frac{1}{2}(X_1'-X_3')=\begin{bmatrix}0&1\\0&0\end{bmatrix}. \end{eqnarray*}

Now Y_+ and Y_- are nilpotents (that is their squares are zero), therefore taking exponentials is easy:

(11)   \begin{eqnarray*} e^{tY_+}&=&I+tY_+=\begin{bmatrix}1&0\\t&1\end{bmatrix},\\ e^{tY_-}&=&I+tY_-=\begin{bmatrix}1&t\\0&1\end{bmatrix}. \end{eqnarray*}

We are not going to worry about Y_-, but using Y_+ proves to be an excellent idea! The corresponding orbits on the half-plane are extremely simple

(12)   \begin{equation*}z_+(t)=z+t.\end{equation*}

The corresponding vector field is

(13)   \begin{equation*}\tilde{Y}_+(x,y)=(1,0).\end{equation*}

Here are the corresponding streamlines:

Streamlines of the vector field (1,0) on the half-plane \mathbb{H}

We will exploit \tilde{X}_2 and \tilde{Y}_+ in the future. Their interpretation is very simple. The first one describes uniform dilation. Expanding (or contracting) in horizontal and in vertical direction at the same rate. The second one is a uniform horizontal translation. The hyperbolic geometry of the plane needs to be invariant with respect to these two kinds of transformations. If that is so, then it is invariant with respect to the whole SL(2,R) group, because the third generator can be obtained from these two and their commutator!

This last statement was wrong. Calculating the commutator we get [X_2',Y_+]=2Y_+. Therefore Y_+ and X_2' do not generate the whole sl(2,R) algebra.

Deriving invariant hyperbolic Riemannian metric on the half-plane

Posted on by arkajad

The two-dimensional non-Euclidean hyperbolic geometry that we met is living on the unit disk in the complex plane. Strictly speaking it is living inside the disk, in the disk interior. The unit circle that forms the boundary of the disk is also important, but it is a different story. An interesting and even exciting story, but it does not belong to this particular series. Perhaps it will be enough to mention here that holomorphic functions on the disk attain their extrema on the boundary, thus the circle is a special case of so called Shilov boundary.

We used Cayley transform to move our scene from the disk to the complex half-plane – the set of complex numbers with positive imaginary part. The boundary circle (minus one point) is then mapped onto the real axis. Real axis is the boundary of the half-plane. With Cayley transform, which reads as

(1)   \begin{equation*}z'=\frac{z+1}{i(z-1)},\end{equation*}

one point on the boundary, namely z=1 is mapped into infinity. We will not worry about this issue, as we will be mainly interested in the geometry of the interior.

The inverse transform, from the half-plane to the disk is

(2)   \begin{equation*} z=\frac{z'-i}{z'+i}. \end{equation*}

These are the formulas written in terms of complex variables. But studying geometry we usually deal with real variables. If we write z=x+iy,\, z'=x'+iy', then Eq. (2) can be written as

(3)   \begin{eqnarray*} x&=&\frac{x'^2+y'^2-1}{x'^2+(1+y')^2},\\ y&=&\frac{-2x'}{x'^2+(1+y')^2}. \end{eqnarray*}

Certainly Eq. (2) looks less complicated than (3) – that is why often using complex notation is convenient. But not always.

In Following Einstein: deriving Riemannian metric on the Poincaré disk we have derived the formula for the line element ds of the invariant distance on the disk. In terms of real coordinates x,y on the disk the formula that we have derived reads:

(4)   \begin{equation*}ds^2=\frac{4(dx^2+dy^2)}{(1-x^2-y^2)^2}.\end{equation*}

How the formula for ds^2 will look in terms of the variables x',y' on the half-plane? Will it be more complicated or simpler?

It is prudent to use computer software to derive the results below!

“It is unworthy of excellent men to lose hours like slaves in the labor of calculation which could be relegated to anyone else if machines were used.”
— Gottfried Leibniz

Let us calculate. First we calculate the partial derivatives – and they look really awful:

(5)   \begin{eqnarray*} \frac{\partial x}{\partial x'}&=&\frac{4 x' (1+y')}{\left(x'^2+(1+y')^2\right)^2},\\ \frac{\partial x}{\partial y'}&=&\frac{2(1+y')^2-2x'^2}{\left(x'^2+(1+y')^2\right)^2},\\ \frac{\partial y}{\partial x'}&=&\frac{2(x'^2-(1+y')^2)}{\left(x'^2+(1+y')^2\right)^2},\\ \frac{\partial y}{\partial y'}&=&\frac{4 x' (1+y')}{\left(x'^2+(1+y')^2\right)^2}. \end{eqnarray*}

Then we compute dx^2+dy^2 and express it in terms of dx',dy':

    \begin{eqnarray*} dx^2+dy^2&=&(\frac{\partial x}{\partial x'}\,dx'+\frac{\partial x}{\partial y'}\,dy')^2+(\frac{\partial y}{\partial x'}\,dx'+\frac{\partial y}{\partial y'}\,dy')^2\\ &=&\frac{4 \left(dx'^2+dy'^2\right)}{\left(x'^2+(1+y')^2\right)^2}. \end{eqnarray*}

Certainly it is not a very simple formula.

Then we calculate the denominator of (4) and express it in terms of x',y':

(6)   \begin{equation*} (1-x^2-y^2)^2=\frac{16y'^2}{\left(x'^2+(1+y')^2\right)^2}. \end{equation*}

Also not very simple. But when we calculate the whole expression – a miracle happens:

(7)   \begin{equation*}ds^2=\frac{4(dx^2+dy^2)}{(1-x^2-y^2)^2}=\frac{dx'^2+dy'^2}{y'^2}. \end{equation*}

(8)   \begin{equation*}ds^2=\frac{dx'^2+dy'^2}{y'^2}. \end{equation*}

Beautiful!

Beautiful!

Why is it so that, after quite messy calculations, we arrive, at the end, at a very simple and beautiful end result? Probably the answer is that the end result must be invariant with respect to the very simple group, namely SL(2,R). So it must be simple. And it comes out even simpler than we would expect.

Next we need to exploit this simplicity …