Conformally Euclidean geometry of the upper half-plane

In Deriving invariant hyperbolic Riemannian metric on the half-plane we have calculated the line element for the SL(2,R) invariant hyperbolic geometry on the upper half-plane:

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

Using this formula we can easily calculate the length of any segment of any curve. If (x(t),y(t)) is a parametric equation of the curve, then dx=\frac{dx}{dt}dt=\dot{x}(t)dt,\, dy=\frac{dy}{dt}\,dt=\dot{y}(t)\,dt, and so

(2)   \begin{equation*}ds=\sqrt{\frac{dx^2+dy^2}{y^2}}=\sqrt{\frac{\dot{x}^2+\dot{y}^2}{y^2}}dt,\end{equation*}

(3)   \begin{equation*}s(t_0,t_1)=\int_{t_0}^{t_1}\frac{\sqrt{\dot{x}(t)^2+\dot{y}(t)^2}}{y(t)}\,dt.\end{equation*}

In this last formula we have used the fact that in the upper half-plane we have y(t)>0.

From the studies of geometry of surfaces and its generalization developed mainly by Riemann we know that the formula for the line element allows us not only to find the length, but also the angles.

From Wikipedia:
Bernhard Riemann – see Modern Science Map

Riemannian geometry originated with the vision of Bernhard Riemann expressed in his inaugural lecture “Ueber die Hypothesen, welche der Geometrie zu Grunde liegen” (“On the Hypotheses on which Geometry is Based”). It is a very broad and abstract generalization of the differential geometry of surfaces in \mathbf{R}^3. Development of Riemannian geometry resulted in synthesis of diverse results concerning the geometry of surfaces and the behavior of geodesics on them, with techniques that can be applied to the study of differentiable manifolds of higher dimensions.

It allows us to calculate scalar products of tangent vectors. Let us rewrite Eq. (1) in the form:

    \[ds^2=\begin{bmatrix}dx&dy\end{bmatrix}\begin{bmatrix}g_{xx}&g_{xy}\\g_{yx}&g_{yy}\end{bmatrix}\begin{bmatrix}dx\\dy\end{bmatrix}= g_{xx}dx^2+2g_{xy}dx\,dy+g_{yy}dy^2,\]

using the symmetric (i.e. g_{xy}=g_{yx}) “Riemannian metric” matrix g

    \[g=\begin{bmatrix}g_{xx}&g_{xy}\\g_{yx}&g_{yy}\end{bmatrix}.\]

Thus, from Eq. (1), we have g_{xx}=1/y^2, g_{xy}=g_{yx}=0, and g_{yy}=1/y^2. Or

(4)   \begin{equation*}g=\frac{1}{y^2}\begin{bmatrix}1&0\\0&1\end{bmatrix}.\end{equation*}

In general, if \xi=(\xi_x,\xi_y) and \eta=(\eta_x,\eta_y) are two tangent vectors at z=x+iy, then their scalar product is given by a general formula

(5)   \begin{equation*}(\xi,\eta)_{z}=\begin{bmatrix}\xi_x& \xi_y\end{bmatrix}g\begin{bmatrix}\eta_x\\ \eta_y\end{bmatrix}=g_{xx}\xi_x\eta_x+g_{xy}(\xi_x\eta_y+\xi_y\eta_x)+g_{yy}\xi_y\eta_y,\end{equation*}

which in our case specializes to

(6)   \begin{equation*}(\xi,\eta)_{z}=\frac{\xi_x\eta_x+\xi_y\eta_y}{y^2}.\end{equation*}

The numerator is exactly the same as for the Euclidean scalar product. It has the consequence that the angles in the hyperbolic upper half-plane geometry are the same as in the Euclidean geometry. That it is so follow from the definitions. Angles in any Riemannian geometry are defined the same way as in the Euclidean geometry. If \xi, \eta are two tangent vectors at z, then the angle \phi between them is defined through the formula:

(7)   \begin{equation*}\cos\phi=\frac{(\xi,\eta)}{||\xi|| ||\eta||},\end{equation*}

where ||\xi||=\sqrt{(\xi,\xi)} and ||\eta||=\sqrt{(\eta,\eta)} are their norms. If we use the scalar product given by Eq. (6), then the denominator y^2 in (\xi,\eta) cancels out with the product of denominators is the norms. The end result is

(8)   \begin{equation*}\cos(\phi)=\frac{\xi_x\eta_x+\xi_y\eta_y}{\sqrt{\xi_x^2+\xi_y^2}\sqrt{\eta_x^2+\eta_y^2}}.\end{equation*}

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Therefore, when we look at the picture like the one obtained in the last post:

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

all hyperbolic geometry angles between various lines at any particular joining point are the same as the perceived ones by our eyes ones – contrary to the sizes, where SL(2,R) invariant geometry and Euclidean geometry we are used to are different.

Whenever we have two metrics, g and g' that differ a scalar factor g'=c\,g, we say that they are conformally related, or conformal to each other. In our case the metric g in Eq. (4) is proportional to the identity matrix, our geometry is conformally Euclidean.

In the next note we will start calculating “straight lines”, or “geodesics” of our geometry. Some of them are almost evident candidates: vertical lines, perpendicular to the real line. But what about the other ones?

Our geometry is a toy geometry, as simple as possible. The group S(1,1) (or SL(2,R) ) is too simple. But one step further there is SU(2,2) – which became the basis for the “chronometric cosmology” developed by the late mathematician Irving Ezra Segal.

From American Astronomical Society, Irving Ezra Segal (1918 – 1998)

… The number of astronomers convinced was small, with experimentalists providing a slightly more favorable reception than theoreticians. Occasionally, the controversy was quite theatrical. A younger colleague of Segal was assaulted on Massachusetts Avenue by an astrophysicist still fuming from a Segalian conversation, and the publication of all the correspondence to and from Cambridge would be delightful. Particularly vituperative referee reports were always posted on his office door, some years overflowing onto the walls. Tee-shirts emblazoned with “Save Energy—Stop the Expansion of the Universe” can still be seen in Harvard Square. Whatever the final judgment is on the chronometric theory, Segal’s contributions to mathematics and mathematical physics have already had profound influence. Segal will be remembered for his conviction that the proper role of mathematics is to illuminate physics and his constant, irascible charm. We will also remember the perfect host, pouring wine for many a dinner, discussing physics, mathematics, philosophy, the proper preparation of coffee, and life.

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

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

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.