Deriving invariant hyperbolic Riemannian metric on the half-plane

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 …

The disk and the hyperbolic model

The unit disk in the complex plane, together with geometry defined by invariants of fractional linear SU(1,1) action, known as the Poincaré disk, that is the arena of hyperbolic geometry. But why “hyperbolic”? It is time for us to learn, and to use. In principle the answer is given in Wikipedia, under the subject “Poincaré disk model”. There we find the following picture

Poincare disk and hyperboloid

with formulas:

Transformation between hyperboloid and disk

We want to derive these formulas ourselves. Let us first introduce our private notation. The hyperboloid will live in a three-dimensional space with coordinates X,Y,T. This is the space-time of Special Relativity Theory, but in a baby version, with Z coordinate suppressed.

The light cone in our space-time has the equation T^2-X^2-Y^2=0. Of course we assume the constant speed of light c=1. Inside the future light cone (the part with Tgeq 0) there is the hyperboloid defined by T=\sqrt{1+X^2+Y^2}. The coordinates (X,Y,T) of events on this hyperboloid satisfy the equation

(1)   \begin{equation*}T^2-X^2-Y^2=1.\end{equation*}

As can be seen from the picture above, every straight line passing through the point with coordinates X_0=Y_0=0, T_0=-1 and a point with coordinates (X,Y,T) on the hyperboloid, intersects the unit disk at the plane T=0 at a point with coordinates (x,y). We want to find the relation between (X,Y,T) and (x,y).

Given any two points, P=(X_0,Y_0,T_0), Q=(X_1,Y_1,T_1), the points (X,Y,T) on the line joining P and Q have coordinates (X(u),Y(u),T(u)) parametrized by a real parameter u as follows:

(2)   \begin{equation*}(X(u),Y(u),T(u))=(1-u)(X_0,Y_0,T_0)+u(X_1,Y_1,T_1).\end{equation*}

For u=0 we are at P, for u=1 we are at Q, and for other values of u we are somewhere on the joining line. Our P has coordinates (0,0,-1), our Q has coordinates (X,Y,T) on the hyperboloid, and we are seeking the middle point with coordinate (x,y,0). So we need to solve equations

(3)   \begin{eqnarray*} x&=&(1-u)0+uX=uX,\\ y&=&(1-u)0+uY=uY,\\ 0&=&(1-u)(-1)+uT. \end{eqnarray*}

From the last equation we find immediately that u=1/(1+T), and the first two equations give us

(4)   \begin{eqnarray*} x&=&\frac{X}{1+T},\\ y&=&\frac{Y}{1+T}. \end{eqnarray*}

We need to find the inverse transformation. First we notice that

    \[x^2+y^2=\frac{X^2+Y^2}{(1+T)^2}=\frac{T^2-1}{(1+T)^2}=\frac{T-1}{T+1}.\]

Therefore 1-(x^2+y^2)=\frac{2}{T+1} and so

    \[T+1=\frac{2}{1-x^2-y^2},\]

    \[T=\frac{1+x^2+y^2}{1-x^2-y^2}.\]

Using Eqs. (4) we now finally get

(5)   \begin{eqnarray*} X&=&\frac{2x}{1-x^2-y^2},\\ Y&=&\frac{2y}{1-x^2-y^2},\\ T&=&\frac{1+x^2+y^2}{1-x^2-y^2}. \end{eqnarray*}

Thus we have derived the formulas used in Wikipedia. Wikipedia mentions also that the straight lines on the disk, that we were discussing in a couple of recent posts, are projections of sections of the hyperboloid by planes. We will not need this in the future. But we will use the derived formulas for obtaining the relation between SU(1,1) matrices and special Lorentz transformations of space-time events coordinates. This is the job for the devil of the algebra!

SU(1,1) parametrization

In the last post, while getting hyperbolic, we have met the Lie group SU(1,1) – the group of 2\times 2 complex matrices A of determinant one, \det A=1, and such that

(1)   \begin{equation*} AGA^*=G,\end{equation*}

where

(2)   \begin{equation*}G=\begin{bmatrix}1&0\\0&-1\end{bmatrix},\end{equation*}

and star A^* is Hermitian conjugated matrix to A.

It was our first meeting here with SU(1,1), and there was no enough time to decide whether we want to make friends with this group or not. Today we will look at SU(1,1) closer, in particular at her shape. I think we will like her … She has, as we will see below, the shape of the inside of a solid torus, donut-like shape:

How do we get this donut? One method is to visit a backery, another method is by doing some little algebra. We will choose the second method. In fact, we are going to get not only the shape of the group itself, but also the shape of the disk that the group is acting on! It will be the cross-section of the torus:

Let us take a matrix A from SU(1,1). It is a complex 2\times 2 matrix, so let us write it as

(3)   \begin{equation*}A=\begin{bmatrix}\lambda&\mu\\ \nu&\rho\end{bmatrix}.\end{equation*}

From \det A=1 we get

(4)   \begin{equation*}\lambda\rho-\mu\nu=1.\end{equation*}

On the other hand we have

(5)   \begin{equation*}AGA^*=\begin{bmatrix}\lambda&\mu\\ \nu&\rho\end{bmatrix}\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}\bar{\lambda}&\bar{\nu}\\ \bar{\mu}&\bar{\rho}\end{bmatrix}=\begin{bmatrix}\lambda\bar{\lambda}-\mu\bar{\mu}&\lambda\bar{\nu}-\mu\bar{\rho}\\\nu\bar{\lambda}-\rho\bar{\mu}&\nu\bar{\nu}-\rho\bar{\rho}\end{bmatrix},\end{equation*}

and so from the determining Eq. (1) we must have

(i)   \begin{align*} |\lambda|^2-|\mu|^2&=1,\\ |\rho|^2-|\nu|^2&=1,\tag{ii}\\ \lambda\bar{\nu}-\mu\bar{\rho}&=0,\tag{iii}\\ \nu\bar{\lambda}-\rho\bar{\mu}&=0.\tag{iv} \end{align*}

Eq. (iv) is just a complex conjugation of (iii), so it does not bring us any new information. The whole information is contained in the first three equations.

From the first two equations we deduce that, since |\lambda|^2=1+|\mu|^2\geq 1 and |\rho|^2=1+|\nu|^2\geq 1, therefore \lambda and \rho are non-zero. From Eq. (iii), dividing by non-zero \bar{\rho}, we get

(6)   \begin{equation*} \mu=\frac{\lambda\bar{\nu}}{\bar{\rho}},\end{equation*}

which we substitute into Eq. (4)

    \[\lambda\rho-\frac{\lambda|\nu|^2}{\bar{\rho}}=1,\]

or, multiplying both sides with \bar{\rho},

    \[\lambda|\rho|^2-\lambda|\nu|^2=\bar{\rho}.\]

Then, using Eq. (ii), we get \lambda=\bar{\rho}, thus

(7)   \begin{equation*}\rho=\bar{\lambda}.\end{equation*}

Therefore, from Eq. (6), we get \mu=\bar{\nu}, or

(8)   \begin{equation*}\nu=\bar{\mu}.\end{equation*}

So, we obtain that every matrix A of SU(1,1) is of the form

(9)   \begin{equation*}A=\begin{bmatrix}\lambda&\mu\\ \bar{\mu}&\bar{\lambda}\end{bmatrix}\end{equation*}

where

(10)   \begin{equation*}|\lambda|^2-|\mu|^2=1.\end{equation*}

Conversely, every matrix of this form is in SU(1,1).

Now, we will determine the shape of our beauty.

Since |\lambda|\geq 1, we can write

(11)   \begin{equation*} \lambda=|\lambda|e^{i\theta},\end{equation*}

where \theta is the uniquely defined phase, 0\leq\theta<2\pi.

Let

(12)   \begin{equation*} z=\frac{\bar{\mu}}{\lambda}.\end{equation*}

Then, using Eq. (10)

(13)   \begin{equation*} |z|^2=\frac{|\mu|^2}{|\lambda|^2}=\frac{|\lambda|^2-1}{|\lambda|^2}=1-\frac{1}{|\lambda|^2}<1.\end{equation*}

Thus z is in the interior of the unit disk in the complex plane.

Conversely, given \theta and z, with |z|<1 we can determine uniquely \lambda and \mu. From Eq. (13) we set

(14)   \begin{equation*}|\lambda|=\frac{1}{\sqrt{1-|z|^2}},\end{equation*}

then
Finally, using Eq. (12), we set

(15)   \begin{equation*}\mu=\bar{\lambda} \bar{z}.\end{equation*}

Summarizing, the shape of SU(1,1) is that of the Cartesian product of the circle parametrized by \theta, 0\leq\theta<2\pi, and of the interior of the unit disk, parametrized by complex numbers z with |z|<1. That is the interior of the solid torus, that is the donut without its skin.

Pretty and tasty. Not necessarily healthy, though….