Geodesics on the upper half-plane – parametrization

The last note ended with the following problem:

Thus: geodesics are circles. Or better: straight lines are circles! In fact: half-circles, because their centers are on the x-axis, and our arena is only upper half-plane.

Except that we have missed some solutions. In Conformally Euclidean geometry of the upper half-plane there was a sentence:

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?

Well, we have these other ones, but how the vertical lines fit our reasoning above?

If \gamma(t)=(x(t),y(t)) is a vertical line, then x(t)=x_0, therefore \dot{x}=0. In Geodesics on the upper half-plane – Part 2 circles we arrived at equations

(1)   \begin{equation*} (\dot{\gamma}(t),K_1(\gamma(t)))=\frac{\dot{x}}{y^2(t)}=\mathrm{const},\end{equation*}

(2)   \begin{equation*} (\dot{\gamma}(t),K_2(\gamma(t)))=\frac{\dot{x}(t)x(t)+\dot{y}(t)y(t)}{y^2(t)}=\mathrm{const},\end{equation*}

and took the ratio of the second to the first. But for vertical lines taking the ratio is not allowed. The first equation is satisfied with the constant on the right hand side equal to zero. The second equation reduces to

(3)   \begin{equation*}(\dot{\gamma}(t),K_2(\gamma(t)))=\frac{\dot{y}(t)}{y(t)}=\mathrm{const}.\end{equation*}

If we now recall Eq. (2) from Conformally Euclidean geometry of the upper half-plane :

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

we see that ds and dt on the vertical line must be proportional:

(5)   \begin{equation*}s=ct+s_0.\end{equation*}

Whenever this last equation holds, one says that t is an “affine parameter”: it is proportional to the arc length, possibly translated. In fact that is part of the definition of the geodesic that enters the “Noether’s theorem” that we are using. Usually we choose the proportionality constant equal to one.

Of course we can use Eq. (3) in order to determine the parameter t. Choosing the constant equal to 1, we have

(6)   \begin{equation*}\frac{dy}{y}=dt,\end{equation*}

therefore

(7)   \begin{equation*}\log y=t+t_0.\end{equation*}

To my surprise Bjab has discovered this all by himself – see the discussion under the previous post

Therefore, when discussing geodesics, we will assume that they are always parametrized by their arc length or, in other words, that the tangent vector \dot{\gamma}(t) is of unit length. In our case that is equivalent to

(8)   \begin{equation*}(\dot{\gamma}(s),\dot{\gamma}(s))_{\gamma(s)}=\frac{\dot{x}(s)^2+\dot{y}(s)^2}{y(s)^2}=1.\end{equation*}

Remark: There are many important examples of pseudo-Riemannian metrics, that are not positive definite. In such a case “length square” along a geodesic line is normalized to +1 or -1 or 0, so that there are three kinds of geodesics. In physics this happens for space-time metrics with Minkowski signature, and in multi-dimensional Kaluza-Klein theories. We will discuss another such case in the following posts.

We can now use these insights also in the case of circular geodesics. Before we have taken the ratio of two “conservation law” equations (1,2). Now, that we know we have a circle, we can write circle equation

(9)   \begin{eqnarray*}x&=&r\cos (\phi)+x_0,\\ y&=&r\sin( \phi),\label{eq:yr}\end{eqnarray*}

where \phi is some function of the arc length parameter s, and substitute into Eqs. (1,2).

We have

(10)   \begin{eqnarray*}\dot{x}&=&-r\sin (\phi)\, \dot{\phi}\\ \dot{y}&=&r\cos (\phi)\,\dot{\phi},\label{eq:dy}\end{eqnarray*}

therefore Eq. (8) reduces to

(11)   \begin{equation*} \frac{\dot{\phi}}{\sin\phi}=\pm 1.\end{equation*}

It is now a straightforward exercise to verify that with Eqs. (911) equations (1,2) are satisfied automatically.

Now that we know geodesics on the upper half-plane, we can draw them for pleasure. There is a famous pattern known as Dedekind’s tesselation

Dedekind tessellation

I was able to reproduce a part of it, namely the part described on p. 3 in the paper on SL(2,Z) by Keith Conrad

SL(2,Z) tessellation

But I would like to be able to reproduce the pretty image from website of Jerzy Kocik from Southern Illinois University

Dedekind tessellation Click on the image to view it full size

And I do not know yet how to do it.

Update: After several hours I managed to produce this:

My poor version – very primitive

Geodesics on the upper half-plane – Part 2 circles

In the last note Geodesics on the upper half-plane – Part 1 Killing vectors we verified that SL(2,R) transformations are isometries of the upper half-plane endowed with the Riemannian metric defined by the line element

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

The corresponding scalar product between tangent vectors being

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

The idea is to use a “classical result” from differential geometry (a version of Noether’s theorem) that we have quoted from the online book by Sean Carroll:

Before applying this theorem to our case let us first see how it works for the standard plane Euclidean geometry. Translations are certainly isometries there. Streamlines of the vector fields generating horizontal and vertical translations form a square grid.

That straight lines intersect these grid lines under a constant angle is evident. In fact, that could serve as an alternative definition of a straight line.

So far so good.

But rotations are also isometries of the Euclidean geometry.

Looking at the image that shows the intersections of a straight lines with streamlines of the vector field generating rotation – nothing is evident. The angle of intersection is certainly not constant. We have to look at the theorem more closely.

The theorem states that the scalar product between Killing vector field (generating isometry) and the tangent vector to the geodesic is constant.

A general straight line \gamma(t)=(x(t),y(t)) on the plane has equation:

(3)   \begin{eqnarray*} x(t)&=&at+x_0,\\ y(t)&=&bt+y_0. \end{eqnarray*}

The tangent vector (\dot{x},\dot{y}) has components (a,b)

Rotation is defined by

(4)   \begin{eqnarray*} x(\phi)&=&\cos(\phi)x-\sin(\phi)y,\\ y(\phi)&=&\sin(\phi)x+\cos(\phi)y. \end{eqnarray*}

The corresponding vector field X(x,y) is obtained by taking derivatives with respect to \phi at \phi=0:

(5)   \begin{equation*} X(x,y)=(-y,x).\end{equation*}

It is streamlines of this vector field that we have drawn in the figure above.

Let us now calculate the scalar product of X and the vector tangent to our straight line at a point (x(t).y(t)) on our line:

(6)   \begin{equation*} (\dot{\gamma}(t),X(\gamma(t))=a(-y(t))+b(x(t))=a(-bt-y_0)+b(at+x_0)=-ay_0+bx_0=\mathrm{const}.\end{equation*}

Somewhat miraculously the non-constant (t-dependent) terms have cancelled out, and the scalar product is indeed constant. Not so evident, but nevertheless true. Perhaps there is some simple argument predicting this result, but I do not know it. Either use Noether’s theorem, or use direct calculation as above.

Let us now move from the Euclidean geometry of the plane to the hyperbolic geometry of upper half-plane. In SL(2,R) generators and vector fields on the half-plane we have focused our attention on two Killing vector fields (now we know that they are “Killing”), we will call them K_1,K_2 here:

(7)   \begin{eqnarray*} K_1(x,y)&=&(1,0),\\ K_2(x,y)&=&(x,y). \end{eqnarray*}

The first one corresponds to horizontal translation, the second one to uniform dilation. Their streamlines are

Suppose now \gamma(t)=(x(t),y(t)) is a geodesic, with tangent vector \dot{\gamma}(t)=(\dot{x}(t),\dot{y}(t)). Its scalar products with K_1 and K_2 should be constant. But now we remember that in calculating the scalar product we must use Eq. (2), so there will be denominator:

(8)   \begin{equation*} (\dot{\gamma}(t),K_1(\gamma(t)))=\frac{\dot{x}}{y^2(t)}=\mathrm{const},\end{equation*}

(9)   \begin{equation*} (\dot{\gamma}(t),K_2(\gamma(t)))=\frac{\dot{x}(t)x(t)+\dot{y}(t)y(t)}{y^2(t)}=\mathrm{const}.\end{equation*}

Their ratio should be therefore constant:

(10)   \begin{equation*}\frac{\dot{x}(t)x(t)+\dot{y}(t)y(t)}{\dot{x}(t)}=\mathrm{const},\end{equation*}

or

(11)   \begin{equation*}\dot{x}(t)x(t)+\dot{y}(t)y(t)=\mathrm{const }\, \dot{x}(t).\end{equation*}

This looks rather simple. In fact it should remind us about the equation of a circle. A circle with center on some point x_0 on x-axis and radius r has equation:

(12)   \begin{equation*}(x-x_0)^2+y^2=r^2.\end{equation*}

Assuming x=x(t),y=y(t) and differentiating with respect to t we get

(13)   \begin{equation*} 2\dot{x}(t)(x(t)-x_0)+2\dot{y}(t)y(t)=0,\end{equation*}

or

(14)   \begin{equation*} \dot{x}(t)x(t)+\dot{y}(t)y(t)=x_0\,\dot{x}(t).\end{equation*}

This is exactly Eq. (11)

Thus: geodesics are circles. Or better: straight lines are circles! In fact: half-circles, because their centers are on the x-axis, and our arena is only upper half-plane.

Except that we have missed some solutions. In Conformally Euclidean geometry of the upper half-plane there was the following sentence:

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?

Well, we have these other ones, but how the vertical lines fit our reasoning above?

This is a homework. It is somewhat tricky though ….

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.