Geodesics on the upper half-plane – Part 1 Killing vectors

According to Wikipedia

In differential geometry, a geodesic is a generalization of the notion of a “straight line” to “curved spaces”.

The first line in the online Encyclopedia of mathematics is similar

The notion of a geodesic line (also: geodesic) is a geometric concept which is a generalization of the concept of a straight line (or a segment of a straight line) in Euclidean geometry to spaces of a more general type.

Wolfram’s MathWorld is somewhat more original:

A geodesic is a locally length-minimizing curve. Equivalently, it is a path that a particle which is not accelerating would follow. In the plane, the geodesics are straight lines. On the sphere, the geodesics are great circles (like the equator). The geodesics in a space depend on the Riemannian metric, which affects the notions of distance and acceleration.

It is time for us to study geodesics on the upper half-plane, and do it in a semi-rigorous way. That would require a rigorous definition of geodesics, which would take us into differential geometry and variational calculus. That would certainly not be a length-minimizing and straight way of achieving our goal. For us, interested mainly in Lie groups and their actions, there is a shorter way. It is like in classical mechanics, where in many important cases we do not have to solve complicated Newton’s differential equations, it is enough to use the law of conservation of energy, or momentum.

That is what we will do. We will use conservation laws. Usually these come as theorems in courses of differential geometry (Noether’s theorem). For instance Sean Carroll in his online book Lecture Notes on General Relativity, in Chpater 5, More geometry has this piece:

And that is what we will use. And how to use it, in detail, we will see below.

Let us start with this sentence:

If a one-parameter family of isometries is generated by a vector field V^{\mu}(x), then V^{\mu} is known as a Killing vector field.

We have our candidates for Killing vector fields. We were plotting some of their streamlines in SL(2,R) generators and vector fields on the half-plane .

But are we sure that they generate “isometries”? Till now we have only a roundabout argument: metric on the upper half-plane comes from the metric on the disk, metric on the disk comes from geometry on the hyperboloid, metric on the hyperboloid comes from flat space-time metric of signature (2,1) and the SL(2,R) group comes from the SO(2,1) group of linear transformations preserving the flat space-time metric. That could be enough for a while, but can’t we check directly if indeed we have isometries?

Yes, we can check, and that is, in fact, quite easy. Generators of the SL(2,R) group form the Lie algebra sl(2,R) of real 2\times 2 matrices of trace zero. After exponentiation they generate one-parameter groups of SL(2,R) matrices. SL(2,R) acts on the upper-half plane \mathbb{H} by linear fractional transformations. If A is in SL(2,R)

(1)   \begin{equation*}A=\begin{bmatrix}\alpha&\beta\\ \gamma&\delta,\end{bmatrix}\end{equation*}

with \det A=\alpha\delta-\beta\gamma=1, then A acts on \mathbb{H} through

(2)   \begin{equation*}z\mapsto \tilde{z}=A\cdot z=\frac{\delta z+\gamma}{\beta z+\alpha}\end{equation*}

Is the transformation defined in Eq. (2) an isometry? The formula looks relatively simple when written in terms of complex variables. But if we write z=x+iy, \tilde{z}=\tilde{x}+i\tilde{y}, then the coordinates (\tilde{x},\tilde{y}) of the transformed point become not that simple functions of the coordinates (x,y) of the original point:

(3)   \begin{equation*}\tilde{x}=\frac{\alpha  \gamma +\alpha  \delta  x+\beta  \gamma  x+\beta  \delta  x^2+\beta  \delta  y^2}{\alpha^2+2 \alpha  \beta  x+\beta^2 x^2+\beta^2 y^2},\end{equation*}

(4)   \begin{equation*}\tilde{y}=\frac{y (\alpha  \delta -\beta  \gamma )}{\alpha^2+2 \alpha  \beta  x+\beta^2 x^2+\beta^2 y^2}.\end{equation*}

Is it an isometry? And what is isometry?

In Conformally Euclidean geometry of the upper half-plane we have derived the formula for calculating the length of a given curve:

(5)   \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*}

Now, suppose, we transform our curve using the matrix A. The length of the transformed curve is then given by the formula

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

where the relation between (\tilde{x},\tilde{y}) and (x,y) is given by Eqs. (3,4).

The transformation is an isometry if \tilde{s}((t_0,t_1)=s(t_0,t_1) for any segment of any curve. Is true in our case? In order to verify it some little calculations are needed. If we listen to Leibniz:

“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

we use our computer. I used Mathematica. Here is the result:

To summarize: after somewhat lengthy calculation we end up with

(7)   \begin{equation*}\frac{\dot{\tilde{x}}(t)^2+\dot{\tilde{y}}(t)^2}{\tilde{y}(t)^2}=\frac{\dot{{x}}(t)^2+\dot{{y}}(t)^2}{{y}(t)^2},\end{equation*}

even without using the \det A=1 condition. Therefore SL(2,R) transformations are indeed isometries (for our metric). Therefore our vector fields are “Killing vector fields”. Therefore we can use their properties in our derivation of geodesic equations. Which we will continue in the following post.

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!