SL(2,R) Killing vector fields in coordinates

In Parametrization of SL(2,R) we introduced global coordinates 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

(1)   \begin{equation*} 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}. \end{equation*}

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

(2)   \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:

(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}. \end{eqnarray*}

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

(4)   \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

(5)   \begin{equation*}\xi_{1,L}^{i}=\frac{dx^{i}(U_1(t)A)}{dt}|_{t=0}.\end{equation*}

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

(6)   \begin{equation*}\xi_{1,L}=(r^2,-ur,1+r^4-u^2),\end{equation*}

The same way we get

(7)   \begin{equation*}\xi_{2,L}=(0,-r,-2u),\end{equation*}

(8)   \begin{equation*}\xi_{3,L}=(0,-ur,-1+r^4-u^2),\end{equation*}

Then we can calculate vector fields of right shifts

(9)   \begin{equation*}\xi_{j,R}^{i}=\frac{dx^{i}(AU_j(t))}{dt}|_{t=0},\end{equation*}

to obtain:

(10)   \begin{equation*}\xi_{1,R}=(\cos (2 \theta ),-r \sin (2 \theta ),2 r^2 \cos (2 \theta )),\end{equation*}

(11)   \begin{equation*}\xi_{2,R}=(-2 \sin (\theta ) \cos (\theta ),-r \cos (2 \theta ),-4 r^2 \sin (\theta ) \cos (\theta )),\end{equation*}

(12)   \begin{equation*}\xi_{3,R}=(1,0,0),\end{equation*}

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

(13)   \begin{equation*} (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}. \end{equation*}

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.