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
\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

\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
\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:

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

The same way we get

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

Then we can calculate vector fields of right shifts

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

to obtain:
\begin{equation}\xi_{1,R}=(\cos (2 \theta ),-r \sin (2 \theta ),2 r^2 \cos (2 \theta )),\end{equation}
\begin{equation}\xi_{2,R}=(-2 \sin (\theta ) \cos (\theta ),-r \cos (2 \theta ),-4 r^2 \sin (\theta ) \cos (\theta )),\end{equation}
\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
\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.

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