Consider Lie groups of matrices: SO(3) or SO(2,1). Their double covering groups are SU(2) and SU(1,1) (or, after Cayley transform, SL(2,R)). We prefer to use these covering groups as they have simpler topologies. SU(2) is topologically a three-sphere, SL(2,R) is an open solid torus. Our discussion will be quite general, and applicable to other Lie groups as well.

[latexpage]

We denote by $Lie(G)$ the Lie algebra of $G$. It is a vector space, the set of all tangent vectors at the identity $e$ of the group. It is also an algebra with respect to the commutator.

$G$ acts on its Lie algebra by the adjoint representation. If $X\in Lie(G)$ and $a\in G,$ then

\begin{equation}

Ad_a: X\mapsto aXa^{-1}.\end{equation}

We define the scalar product $(X,Y)$ on $Lie(G)$ using the trace

\begin{equation}(X,Y)=\mbox{const}\frac{1}{2}\mbox{Re}(\mbox{Tr}(XY)).\end{equation}.

In each particular case we will choose the constant so that the formulas are simple.

Due to trace properties this scalar product is invariant with respect to the adjoint representation:

\begin{equation}(aXa^{-1},aYa^{-1})=(X,Y).\end{equation}

We will assume that this scalar product is indeed a scalar product, that is we assume it being non-degenerate. For SO(3) and SO(2,1) it certainly is. Lie groups with this property are called semisimple.

Let $X_i$ be a basis in $Lie(G).$ The structure constants $C^{i}_{jk}$ are then defined through

\begin{equation}[X_i,X_j]=C_{ij}^k\,X_k.\end{equation}

We denote by $\mathring{g}_{ij}$ the matrix of the metric tensor in the basis $X_i$

\begin{equation}\mathring{g}_{ij}=(X_i,X_j).\end{equation}

The inverse matrix is denoted $\mathring{g}^{ij}$ so that $\mathring{g}_{ij}\mathring{g}^{jk}=\delta^k_i.$

For SU(2) the Lie algebra consists of anti-Hermitian $2\times 2$ matrices of zero trace. For the basis we can take

\begin{equation}X_1=\frac{1}{2}\begin{bmatrix}0&i\\i&0\end{bmatrix},\,X_2=\frac{1}{2}\begin{bmatrix}0&1\\-1&0\end{bmatrix}, \, X_3=\frac{1}{2}\begin{bmatrix}i&0\\0&-i\end{bmatrix}.\end{equation}

For the constant $\mbox{const}$ we chose $\mbox{const}=-2$. Then $\mathring{g}_{ij}=\mathring{g}^{ij}=\mbox{diag}(1,1,1).$

The structure constants are

\begin{equation}C_{ij}^k=\mathring{g}^{kl}\epsilon_{ijl}.\end{equation}

In this case, since $\mathring{g}_{ij}$ is the identity matrix, there is no point to distinguish between lower and upper indices. But in the case of SU(1,1) it will be important.

We will now consider a general left-invariant metric on the group $G.$ The discussion below is a continuation of the discussion in Riemannian metrics – left, right and bi-invariant.

That is we have now two scalar products on $Lie(G)$ – the Ad-invariant scalar product with metric $\mathring{g},$ and another one, with metric $g.$ We propagate the scalar products from the identity $e$ to other points in the group using left translations (see Eq. (1) in Riemannian metrics – left, right and bi-invariant). We have a small notational problem here, because the letter $g$ often denotes a group element, but here it also denotes the metric. Moreover, we have two scalar products and we need to distinguish between them. We will write $g_a(\xi,\eta)$ for the scalar product with respect to the metric $g$ of two vectors tangent at $a\in G.$ Then left invariance means

\begin{equation}g_a(\xi,\eta)=g_e(a^{-1}\xi,a^{-1}\eta),\end{equation}

which implies for $\xi,\eta$ tangent at $b$

\begin{equation}g_{ab}(a\xi,a\eta)=g_b(\xi,\eta),\,a,b\in G\end{equation}

Infinitesimal formulation of left invariance is that the vector fields $\xi(a)=\xi a$ are “Killing vector fields for the metric” – Lie derivatives of the metric (cf. SL(2,R) Killing vector fields in coordinates, Eq.(13)) with respect to these vector fields vanish. What we need is a very important result from differential geometry: scalar products of Killing vector fields with vectors tangent to geodesics are constant along each geodesic. For the convenience of the reader we provide the definitions and a proof of the above mentioned result (a version of Noether’s theorem). Here we will assume that there are coordinates $x^1,…,x^n$ on $G.$ Later on we will get rid of these coordinates, but right now we will follow the standard routine of differential geometry with coordinates.

We define the Christoffel symbols of the Levi-Civita connection

\begin{equation}\Gamma_{kl,m}=\frac{1}{2}\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}

\begin{equation}\Gamma^{i}_{kl}=g^{im}\Gamma_{kl,m}=\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}

The geodesic equations are then (in Geodesics on upper half-plane factory direct we have already touched this subject)

\begin{equation}\frac{d^2 x^i}{ds^2}= -\Gamma^{i}_{jk}\frac{dx^j}{ds} \frac{dx^k}{ds}.\label{eq:geo}\end{equation}

A vector field $\xi$ is a Killing vector field for $g_{ij}$ if the Lie derivative of $g_{ij}$ with respect to $\xi$ vanishes, i.e.

\begin{equation}0=(L_\xi g)_{îj}=\xi^k\partial_k g_{ij}+g_{ik}\partial_j \xi^k+g_{jk}\partial_i\xi^k.\label{eq:kil}\end{equation}

The scalar product of the Killing vector field and the tangent vector to a geodesic is constant. That is the “conservation law”. A short proof can be found online in Sean Carroll online book “Lecture notes in General Relativity”. The discussion of the proof can be found on physics forums. But the result is a simple consequence of the definitions. What one needs is differentiating composite functions and renaming indices. Just for fun of it let us do the direct, non-elegant, brute force proof.

Suppose $x^{i}(t)$ is a geodesic, and $\xi$ is a Killing field. The statement is that along geodesic the scalar product is constant. That means we have to show that

\[ g_{ij}(x(t))\,\dot{x}^{i}(t)\,\xi^{j}(x(t))=\mbox{const}.\]

We differentiate with respect to $t$, and we are supposed to get zero. So, let’s do it. We have derivative of a product of three terms, so we will get three terms $t_1,t_2,t_3$:

\[t_1=\frac{d}{dt}(g_{ij}(x(t)))\,\dot{x}^{i}(t)\,\xi^{j}(x(t)),\]

\[t_2=g_{ij}(x(t))\,\frac{d}{dt}(\dot{x}^{i}(t))\,\xi^j(x(t)),\]

\[t_3=g_{ij}(x(t))\,\dot{x}^{i}(t)\,\frac{d}{dt}(\xi^j(x(t))).\]

Let us calculate the derivatives. After we are done, in order to simplify the notation, we will skip the arguments.

\[\frac{d}{dt}(g_{ij}(x(t)))=\partial_k\,g_{ij}\dot{x}^k.\]

Thus

\[t_1=\partial_k\,g_{ij}\dot{x}^{i} \dot{x}^{k}(t)\,\xi^{j}(x(t)).\]

Then, from Eq. (\ref{eq:geo})

\[\frac{d}{dt}(\dot{x}^{i}(t))=-\Gamma^{i}_{kl}\dot{x}^k\dot{x}^l,\]

therefore

\[t_2=-\Gamma_{kl,j}\dot{x}^k\dot{x}^l\xi^j=-\frac{1}{2}\partial_k g_{lj}\dot{x}^k\dot{x}^l\xi^j-\frac{1}{2}\partial_l g_{kj}\dot{x}^k\dot{x}^l\xi^j+\frac{1}{2}\partial_jg_{kl}\dot{x}^k\dot{x}^l\xi^j.\]

Renaming the dummy summation indices $k,l$ we see that the two first terms of $t_2$ are identical, therefore

\[t_2=-\partial_k g_{lj}\dot{x}^k\dot{x}^l\xi^j+\frac{1}{2}\partial_jg_{kl}\dot{x}^k\dot{x}^l\xi^j.\]

Again, renaming the dummy summation indices we see that the first term of $t_2$ cancels out with $t_1,$ therefore

\[t_1+t_2=\frac{1}{2}\partial_jg_{kl}\,\dot{x}^l\dot{x}^k\xi^j.\]

For $t_3$ we have

\[t_3=g_{ij}\,\dot{x}^{i}\,\partial_k\xi^j\dot{x}^k.\]

Owing to the symmetry of $\dot{x}^{i}\dot{x}^k=\dot{x}^{k}\dot{x}^i$, we can write it as

\[t_3=\frac{1}{2}g_{ij}\,\partial_k\xi^j\,\dot{x}^{i}\,\dot{x}^k+\frac{1}{2}g_{kj}\,\partial_i\xi^j\,\dot{x}^{i}\,\dot{x}^k.\]

Therefore

\[t_1+t_2+t_3=\frac{1}{2}\left(\partial_jg_{kl}\,\dot{x}^l\dot{x}^k\xi^j+g_{ij}\,\partial_k\xi^j\,\dot{x}^{i}\,\dot{x}^k+g_{kj}\,\partial_i\xi^j\,\dot{x}^{i}\,\dot{x}^k\right)\]

We rename the indices to get

\[t_1+t_2+t_3=\frac{1}{2}\left(\xi^j\partial_jg_{ik}+g_{ij}\,\partial_k\xi^j+g_{kj}\,\partial_i\xi^j\right)\dot{x}^{i}\,\dot{x}^k\]

But the expression in parenthesis vanishes owing to Eq. (\ref{eq:kil}).