### SL(2,R) Killing vector fields in coordinates

In Parametrization of SL(2,R) we introduced global coordinates [latexpage] $x^1=\theta,x^2=r,x^3=u$ on the group SL(2,R). Any matrix $A$ in SL(2,R) can be uniquely written as

A(\theta,r,u)=\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}.

If $A$ is the matrix with components $A_{ij},$ then its coordinates can be expressed as functions of the matrix components as follows

\begin{eqnarray}
x^1(A)&=&\mathrm{atan2}(A_{21},A_{22}),\\
x^2(A)&=&\frac{1}{\sqrt{A_{21}^2+A_{22}^2}},\\
x^3(A)&=&\frac{A_{11}A_{21}+A_{12}A_{22}}{A_{21}^2+A_{22}^2}.
\end{eqnarray}
The function $\mathrm{atan2}(y,x)$ returns the angle $\theta$, $0\leq\theta<2\pi$ of the complex number $x+iy=\sqrt{x^2+y^2}e^{i\theta}.$ Once we have coordinates, it is easy to calculate components of tangent vectors to any given path $A(t)$ - they are given by derivatives of the coordinates $x^{i}(A(t)).$ We will calculate now the vector fields resulting from left and right actions of one-parameter subgroups of SL(2,R) that we have already met in SL(2,R) generators and vector fields on the half-plane

We have introduced there the one-parameter groups, that we will denote now as $U_1,U_2,U_3$, generated by $X_1,X_2,X_3$:

\begin{eqnarray}
X_1&=&\begin{bmatrix}0&1\\1&0\end{bmatrix},\\
X_2&=&\begin{bmatrix}-1&0\\0&1\end{bmatrix},\\
X_3&=&\begin{bmatrix}0&-1\\1&0\end{bmatrix}.
\label{eq:x123}\end{eqnarray}

We can take exponentials of the generators and construct one-parameter subgroups

\begin{eqnarray}
U_1(t)&=&\exp tX_1=\begin{bmatrix}\cosh t&\sinh t\\ \sinh t&\cosh t\end{bmatrix},\\
U_2(t)&=&\exp tX_2=\begin{bmatrix}\exp -t&0\\ 0&\exp t\end{bmatrix},\\
U_3(t)&=&\exp tX_3=\begin{bmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{bmatrix}.
\end{eqnarray}

In SL(2,R) generators and vector fields on the half-plane we were acting with these transformations on the upper half-plane using fractional linear representation. Now we will be acting on SL(2,R) itself via group multiplication, either left or right.

Let us start with $U_1$ acting from the left. At $A$ we have the trajectory $U_1(t)A$. Its coordinates are $x^{i}(t)=x^{i}(U_1(t)A).$ Let us denote by $\xi_{1,L}$ the tangent vector field, with components $\xi_{1,L}^{i}, \,(i=1,2,3)$ Then
$$\xi_{1,L}^{i}=\frac{dx^{i}(U_1(t)A)}{dt}|_{t=0}.$$

We can calculate it easily using algebra software. The result is:

$$\xi_{1,L}=(r^2,-ur,1+r^4-u^2),$$

The same way we get

$$\xi_{2,L}=(0,-r,-2u),$$
$$\xi_{3,L}=(0,-ur,-1+r^4-u^2),$$

Then we can calculate vector fields of right shifts

$$\xi_{j,R}^{i}=\frac{dx^{i}(AU_j(t))}{dt}|_{t=0},$$

to obtain:
$$\xi_{1,R}=(\cos (2 \theta ),-r \sin (2 \theta ),2 r^2 \cos (2 \theta )),$$
$$\xi_{2,R}=(-2 \sin (\theta ) \cos (\theta ),-r \cos (2 \theta ),-4 r^2 \sin (\theta ) \cos (\theta )),$$
$$\xi_{3,R}=(1,0,0),$$

We know that our metric is bi-invariant. That means the vector fields of the left and right shifts generate one-parameter group of isometries. They are called Killing fields of the metric. In differential geometry one shows that a vector field $\xi$ is a Killing vector field for metric $g$ if and only if the Lie derivative $L_\xi$ of the metric vanishes. Lie derivative of any symmetric tensor $T_{ab}$ is defined as

(L_\xi T)_{ab}=\xi^c\frac{\partial T_{ab}}{\partial x^c}+\frac{\partial \xi^c}{\partial x^{a}}T_{cb}+\frac{\partial \xi^c}{\partial x^{b}}T_{ac}.

Using any computer algebra software it is easy to verify that Lie derivatives of our metric with respect to all six vector fields $\xi_{j,L},\xi_{j,R}$ indeed vanish. Mathematica notebook verifying this property can be downloaded here.

### Curvature of the upper half-plane

In Geodesics on upper half-plane factory direct we used the Christoffel symbols and identified geodesics on the upper half plane endowed with the hyperbolic geometry metric. The formulas for Christoffel symbols contain derivatives of the metric tensor components:
[latexpage]
$$\Gamma^{i}_{kl}=\frac{1}{2}g^{im}\left(\frac{\partial g_{mk}}{\partial x^{l}}+\frac{\partial g_{ml}}{\partial x^{k}}-\frac{\partial g_{kl}}{\partial x^{m}}\right).$$

From these connection coefficients one constructs the Riemann curvature tensor. Let us take the expression from Wikipedia:

R^{i}_{\phantom{i}jkl}=\frac{\partial \Gamma^{i}_{lj}}{\partial x^k}-\frac{\partial \Gamma^{i}_{kj}}{\partial x^l}+\Gamma^{i}_{km}\Gamma^m_{lj}-\Gamma^{i}_{lm}\Gamma^m_{kj}.

From its construction several symmetries follow, so that only $n^2(n^2-1)/12$ really independent components remain. For $n=4$ one has to calculate 20 components. But for $n=2$ it is only one. But even this one, why should we calculate it? We better listen to Leibniz.

It is unworthy of excellent men to lose hours like slaves in the labour of calculation which could safely be relegated to anyone else if machines were used.

Nowadays software does all that. Right now I ma using Mathematica, so I type

riemann = Simplify[Table[
D[christ[[i, j, l]], coord[[k]] ] –
D[christ[[i, j, k]], coord[[l]] ] +
Sum[christ[[s, j, l]] affine[[i, k, s]] –
christ[[s, j, k]] affine[[i, l, s]],
{s, 1, n}],
{i, 1, n}, {j, 1, n}, {k, 1, n}, {l, 1, n}] ]

With the result that

R^1_{\phantom{1}221}=-R^2_{\phantom{2}121}=\frac{1}{y^2}.$From the curvature we usually calculate the Ricci tensor$R_{jl}$$$R_{jl}=R^{i}_{\phantom{i}jil}.$$ The Ricci tensor is obtained by contraction (summation) over two indices of the Riemann tensor. The Ricci tensor is always automatically symmetric. In our case we get that Ricci is simply the negative of the metric $$R_{ij}=-g_{ij}.$$ Spaces in which this happens are known under a special name: Einstein spaces (or “Einstein manifolds“). From Ricci tensor by contraction with the inverse metric one calculates the scalar curvature. In our case the result is -2: $$R=g^{ij}R_{ij}=-2$$ While Riemann and Ricci are “tensors”, they may have different components in different coordinate systems, the scalar curvature is an invariant. Its value is independent of the coordinates. In our case the result -2 would be the same if we have calculated it with the coordinates and metric that we have used on the Poincare disk. From two dimensions of the Poincare disk and the upper half-plane we will now move to three-dimensions of the group SL(2,R) itself. Below is the view of the Mathematica notebook doing the calculations described in this post. ### Einstein the Stubborn Before developing his 1915 General Theory of Relativity, Einstein held the “Entwurf” theory. Tullio Levi-Civita from Padua, one of the founders of tensor calculus, objected to a major problematic element in this theory, which reflected its global problem: its field equations were restricted to an adapted coordinate system. Einstein proved that his gravitational tensor was a covariant tensor for adapted coordinate systems. In an exchange of letters and postcards that began in March 1915 and ended in May 1915, Levi-Civita presented his objections to Einstein’s above proof. Einstein tried to find ways to save his proof, and found it hard to give it up. Finally Levi-Civita convinced Einstein about a fault in his arguments. However, only in spring 1916, long after Einstein had abandoned the 1914 theory, did he finally understand the main problem with his 1914 gravitational tensor. In autumn 1915 the Göttingen brilliant mathematician David Hilbert found the central flaw in Einstein’s 1914 derivation. On March 30, 1916, Einstein sent to Hilbert a letter admitting, “The error you found in my paper of 1914 has now become completely clear to me”. That is what Weinstein writes about Einstein in Einstein the Stubborn: Correspondence between Einstein and Levi-Civita Finally Einstein learned what he needed to learn and worked out his “General Relativity Theory” – the theory of gravitation based on the mathematics of (pseudo) Riemannian metric tensor. From metric tensor one calculates the “Levi-Civita connection”, encoded in what are called “Christoffel symbols“. From them one calculates the curvature: ‘Matter tells space how to curve, space tells matter how to move.’ That is the essence of Einstein’s theory of gravitation. Einstein was, for a while, happy with this picture. But only for a while. Many physicists are happy with it even today. But that is not the subject of my post today. My plan is simply to calculate the Christoffel symbols for the SL(2,R) invariant metric that we have discussed in Conformally Euclidean geometry of the upper half-plane [latexpage] Let me recall the metric: $$g=\frac{1}{y^2}\begin{bmatrix}1&0\\0&1\end{bmatrix}.\label{eq:g1}$$ So, it is a$2\times 2$symmetric matrix that depends on the coordinates of a point$z=x+iy$in the upper half-plane of complex numbers$z$with positive imaginary part$y.$We usually write$g$as a matrix with entries$g_{ij}$where$i,j=1,2.$We have two coordinates, we may call$x=x^1,y=x^2.$It is customary in differential geometry to use the coordinate index as the upper index. It is a convention, but a useful convention. Usually it is accompanied with another convention, so called “Einstein summation convention”. This Einstein convention is that whenever we see a term that contains one symbol with index down and another symbol with index up – it means that this is an unwritten sum from$i=1$to$n,$where$n$is the number of dimensions. We are dealing with$n=2,$so whenever we see something like$x^i y_i$or$x_ky^k$it means:$x^1y_1+x^2y_2$or$x_1y^1+x_2y^2.$The name of the repeated index does not matter, it can be any letter, just different from other letters present in the given term. We call it a “dummy index”. The metric is written with indices$i,j$as lower indices. We call them “covariant”. Upper indices are usually called contravariant. The metric should always be invertible, otherwise terrible things can happen, the universe may cease to have its ordinary meaning. The inverse metric is usually written as a contravariant tensor$g^{ij}.$In our case the inverse metric exists because we assume that$y>0:$$$g^{-1}=y^2\begin{bmatrix}1&0\\0&1\end{bmatrix}.\label{eq:g2}$$ For$y=0$it would become zero, and the different parts of the Universe would not know what to do. There would be a confusion, the door to paranormal could get opened. And I am not joking. I wrote about it in an unpublished paper. The content of this paper was too scary for the referees of Physics Letters. The paper is available here: Vanishing Vierbein in Gauge Theories of Gravitation. As I said – it was never published – but it is being cited by others, for instance here: Phys.Rev. D62 (2000) 044004 DOI: 10.1103/PhysRevD.62.044004 \url{}, or here: Found.Phys.38:7-37,2008 DOI: 10.1007/s10701-007-9190-0 \url{} Let us now calculate the Christoffel symbols for our metric. We use the standard formulas from differential geometry, they can be found in Wikipedia at Christoffel symbols of the second kind $$\Gamma^{i}_{kl}=\frac{1}{2}g^{im}\left(\frac{\partial g_{mk}}{\partial x^{l}}+\frac{\partial g_{ml}}{\partial x^{k}}-\frac{\partial g_{kl}}{\partial x^{m}}\right).$$ The expression for$\Gamma^{i}_{kl}$is symmetric in$kl$– the theory has no torsion. There are$2\times 3 =6 $symbols that must be computed. If four dimensions one needs to calculate$4\times 10=40$of these symbols – lot of calculation, as each of these 40 symbols is a sum of four terms (sum over$m\$). In old times people were not as lazy as they are today. They were calculating. Today I am using Mathematica (or Maple, or whatever). Using Mathematica, for instance, it goes as follows:

Only four of the six symbols are different from zero, and they are very simple. Christoffel symbols enter geodesic equations, when they are different from zero – the geodesic lines curve. They may curve because coordinates are curved, or because the space is curved. We will discuss it next in our toy model.