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SL2R as anti de Sitter space cont.

Posted on by arkajad

We continue Becoming anti de Sitter.

Every matrix \Xi in the Lie algebra o(2,2) generates one-parameter group e^{\Xi t} of linear transformations of \mathbf{R}^4. Vectors tangent to orbits of this group form a vector field. Let us find the formula for the vector field generated by \Xi. The orbit through y\in \mathbf{R}^4 is

(1)   \begin{equation*}y(t)=e^{\Xi t}y.\end{equation*}

Differentiating at t=0 we find the vector field \Xi(y)

(2)   \begin{equation*}\Xi(y)=\Xi y.\end{equation*}

If \Xi is a matrix with components \Xi^{\mu}_{\phantom{\mu}\nu}, then \Xi(y) has components

(3)   \begin{equation*}\Xi^{\mu}(y)=\Xi^{\mu}_{\phantom{\mu}\nu}y^{\nu}.\end{equation*}

Vectors tangent to coordinate lines are often denoted as \partial_\mu. Therefore we can write the last formula as:

(4)   \begin{equation*}\Xi(y)=\Xi^{\mu}_{\phantom{\mu}\nu}y^{\nu}\partial_\mu.\end{equation*}

In the last post Becoming anti de Sitter we have constructed six generators \Xi_{(\mu\nu))}. Their vector fields now become

(5)   \begin{equation*}\Xi_{(1,2)}=y^2\partial_1-y^1\partial_2,\Xi_{(1,3)}=y^3\partial_1+y^1\partial_3,\Xi_{(1,4)}=y^4\partial_1+y^1\partial_4,\end{equation*}

(6)   \begin{equation*}\Xi_{(2,3)}=y^3\partial_2+y^2\partial_3,\Xi_{(2,4)}=y^4\partial_2+y^2\partial_4,\Xi_{(3,4)}=-y^4\partial_3+y^3\partial_4.\end{equation*}

Bengtsson and Sandin in their paper “Anti de Sitter space, squashed and stretched” discussed in the previous note use coordinates y^1=X,y^2=Y,y^3=U,y^4=V. Our vector field \Xi_{(1,2)} is the same as their J_{XY}, our \Xi_{(1,3)} is the same as their J_{XU} etc.

In SL(2,R) Killing vector fields in coordinates we introduced six Killing vector fields acting on the group manifold SL(2,R). How they relate to the above six generators of the group O(2,2)?

Vectors from the fields \xi_{iL},\xi_{iR} are tangent to SL(2,R). We have expressed them in coordinates of the group SL(2,R) x^1=\theta,x^2=r,x^3=u. The manifold of SL(2,R) is a hipersurface of dimension 3 in \mathbf{R}^4 endowed with coordinates y^1,y^2,y^3,y^4. What is the relation between components of the same vector in different coordinate systems? The formula is easy to derive and is very simple. If \xi^{i}, (i=1,2,3) are coordinates of the vector in SL(2,R) and \xi^{\mu},\, (\mu=1,2,3,4) are coordinates of the same vector in \mathbf{R}^4, then

(7)   \begin{equation*}\xi^\mu=\frac{\partial y^\mu}{\partial x^{i}}\xi^{i}.\end{equation*}

How y^\mu depend on x^{i}? That is simple. In SL(2,R) vector fields in coordinates we have represented each matrix A from SL(2,R) as

(8)   \begin{equation*} A=\begin{bmatrix} r \cos (\theta )+\frac{u \sin (\theta )}{r} & \frac{\cos (\theta ) u}{r}-r \sin (\theta ) \\ \frac{\sin (\theta )}{r} & \frac{\cos (\theta )}{r}\end{bmatrix}. \end{equation*}

On the other hand, Becoming Anti-de Sitter, we represented it as

(9)   \begin{equation*}A=\begin{bmatrix} V+X & Y+U \\ Y-U & V-X \end{bmatrix}.\end{equation*}

Therefore coordinates y^\mu are easily expressed in terms of x^{i}. It remains to do the calculations. I have used computer algebra software to make these calculations for me. My Mathematica notebook doing all calculations can be downloaded from here. The result of all these calculations is the expression of vector fields \xi_{iL},\xi_{iR} in terms of the generators of O(2,2) used in the paper on anti de Sitter spaces. Here is what I have obtained:

(10)   \begin{eqnarray*} \xi_{1R}&=&-J_1=J_{XU}+J_{YV},\\ \xi_{2R}&=&J_2=J_{YU}-J_{XV},\\ \xi_{3R}&=&J_0=-J_{XY}-J_{UV},\\ \xi_{1L}&=&\tilde{J}_1=J_{YV}-J_{XU},\\ \xi_{2L}&=&\tilde{J}_2=-J_{XV}-J_{YU},\\ \xi_{3L}&=&\tilde{J}_0=J_{XY}-J_{UV}. \end{eqnarray*}

Bengtsson and Sandin introduce then their own parametrization of SL(2,R) and study the invariant metric on the group. We will find the connection between ours and their approaches in the next posts. We came to our problems starting from T-handles spinning freely in zero gravity. They are studying spinning black holes. It is interesting to see and to research similarities.

Geodesics on upper half-plane factory direct

Posted on by arkajad

This is a continuation of Einstein the Stubborn.
We have calculated the Christoffel symbols of the Levi-Civita connection of the SL(2,R) invariant metric on the upper half-plane.

We have used the standard formula, the same that physicists and astronomers are using in their calculations of Black Holes, White Holes, Big Bangs and Small Bangs:

(1)   \begin{equation*}\Gamma^{i}_{kl}=\frac{1}{2}g^{im}\left(\frac{\partial g_{mk}}{\partial x^{l}}+\frac{\partial g_{ml}}{\partial x^{k}}-\frac{\partial g_{kl}}{\partial x^{m}}\right).\end{equation*}

With the metric of upper half-plane hyperbolic geometry given by

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

which is a very simple kind of metric, only four of the six Christoffel symbols are non-zero. They are:

(3)   \begin{eqnarray*} \Gamma^1_{21}&=&\Gamma^1_{12}=-\frac{1}{y},\\ \Gamma^2_{11}&=&\frac{1}{y},\\ \Gamma^2_{22}&=&-\frac{1}{y}. \end{eqnarray*}

In the previous post I have originally written “only three of the six Christoffel symbols are non-zero”, but I have forgotten about the symmetry, as in the first line above.

The Christoffel symbols, that is “the coefficients of the torsion free metric affine connection” serve as the tools for defining “parallel transport” of geometric objects along curves. The transport is, in general, path dependent, when there is a non-vanishing “curvature”.

Curvature is expressed in terms of Christoffel symbols and their derivatives. But “geodesics” are expressed directly in terms of the Christoffel symbols. Here are their equations:

(4)   \begin{equation*}\frac{d^2 x^i}{ds^2}= -\Gamma^{i}_{jk}\frac{dx^j}{ds} \frac{dx^k}{ds}.\end{equation*}

One has to remember that the Einstein convention is being used, so that in the above formula summation over the dummy indices j,k is implied.

Let us apply this general formula to our case, with x^1=x,x^2=y. Let us calculate the right hand side for i=1. With i=1 we have \Gamma^1_{12}=\Gamma^1_{21}=-\frac{1}{y}, therefore

(5)   \begin{equation*} \frac{d^2x}{ds^2}=\frac{2\frac{dx}{ds}\frac{dy}{ds}}{y}.\end{equation*}

With i=2 we have \Gamma^2_{11}=1/y, \Gamma^2_{22}=-1/y. Therefore

(6)   \begin{equation*}\frac{d^2y}{ds^2}=-\frac{\left(\frac{dx}{ds}\right)^2-\left(\frac{dy}{ds}\right)^2}{y}.\end{equation*}

These are the geodesic equations as they come directly from the factory, in their original shape.

In Geodesics on the upper half-plane – Part 2 circles we have derived the formula for geodesics from conservation laws, that is “second-hand”. We have obtained the following formulas:

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

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

They are simpler than Eqs. (5,6). But they are consequences of (5,6). Taking derivatives of the left hand sides of Eqs. (7,8) we can easily check that they are automatically zero if Eqs. (5,6) are satisfied! By using conservation laws we have simply taken a short way.

Once we got into the main objects of Riemannian differential geometry, in the next post we will calculate the curvature of our metric.

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

Posted on by arkajad

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.