Geodesics of left invariant metrics on matrix Lie groups – Part 1

An elegant derivation of geodesic equations for left invariant metrics has been given by B. Kolev in his paper “Lie groups and mechanics. An introduction”.

Here we will derive these equations using simple tools of matrix algebra and differential geometry, so that at the end we will have formulas ready for applications. We will use
the conservation laws derived in the last post Killing vectors, geodesics, and Noether’s theorem. We will also use the same notation. We consider matrix Lie group G with the Lie algebra Lie(G). The tangent space at a\in G is denoted T_aG.
Thus Lie(G)=T_eG. On Lie(G) we assume nondegenerate scalar product
denoted as g(\xi,\eta),\, \xi,\eta\in Lie(G). We propagate it to the whole group using left translations as in Eqs. (8,9) of Killing vectors, geodesics, and Noether’s theorem

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

which implies for \xi,\eta\in T_bG

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

The metric so constructed is automatically left-invariant, therefore for each \xi\in Lie(G) the vector field \xi(a)=\xi a is a Killing field.

Let a(t) be a geodesic for this metric. We denote by \omega(t)\in Lie(G) the tangent vector left translated to the identity:

(3)   \begin{equation*}\omega(t)=a(t)^{-1}\dot{a}(t).\end{equation*}

Then, from the conservation laws derived in the last post, we know that the scalar product of \dot{a}(t) with \xi a(t) is constant. That is

(4)   \begin{equation*}g_{a(t)}(\xi a(t),\dot{a}(t))=\mbox{const}.\end{equation*}

The metric is left-invariant, therefore g_e(a(t)^{-1}\xi a(t),a(t)^{-1}\dot{a}(t))=\mbox{const}, or

(5)   \begin{equation*}g_e(a(t)^{-1}\xi a(t),\omega(t))=\mbox{const}.\end{equation*}

We will differentiate the last equation with respect to t, but first let us notice that by differentiating the identity a(t) a(t)^{-1}=e we obtain

(6)   \begin{equation*}\frac{d}{dt}a(t)^{-1}=-a(t)^{-1}\dot{a}(t)a(t)^{-1}=-\omega(t)a(t)^{-1}.\end{equation*}

Now, differentiating Eq. (5), and using also \frac{da(t)}{dt}=a(t)\omega(t) we obtain

(7)   \begin{equation*}g_e([a^{-1}\xi a,\omega],\omega)+g_e(a^{-1}\xi a,\dot{\omega})=0.\end{equation*}

We now need a certain bilinear operator on Lie(G) that is defined using the commutator and the scalar product. The commutator [\xi_1,\xi_2] itself is such an operator
from Lie(G)\times Lie(G)\rightarrow Lie(G). But using the scalar product we can define another operator B(\xi_1,\xi_2) by the formula:

(8)   \begin{equation*}g_e(B(\xi_1,\xi_2),\eta)=g_e([\xi_2,\eta],\xi_1),\quad \xi_1,\xi_2,\eta\in Lie(G).  \end{equation*}

The right hand side is linear in \eta, and owing to the nondegeneracy of the scalar product every linear functional is represented by a scalar product with a unique vector. Therefore B(\xi_1,\xi_2) is well defined, and evidently is linear in both arguments.

Let \xi_i be a basis in Lie(G), so that the structure constants are C_{ij}^k

(9)   \begin{equation*}[\xi_i,\xi_j]=C_{ij}^k\,\xi_k.\end{equation*}

We can also write B as

(10)   \begin{equation*}B(\xi_i,\xi_j)=B_{ij}^k\,\xi_k.\end{equation*}

Then Eq. (8) gives

(11)   \begin{equation*} g_e(B(\xi_i,\xi_j),\xi_k)=g_e([\xi_j,\xi_k],\xi_i)\end{equation*}

or

    \[ B_{ij}^l g_{lk}=C_{jk}^lg_{li},\]

which can be solved for B using the inverse metric:

(12)   \begin{equation*}B_{ij}^m=g^{mk}C_{jk}^lg_{li}.\end{equation*}

On the other hand, if we agree to lower the upper index of B and C with the metric, we can write Eq. (11) as

(13)   \begin{equation*}B_{ij,k}=C_{jk,i},\end{equation*}

which is easy to remember.

We can now return to Eq. (7) and rewrite it as

    \[g_e(a^{-1}\xi a,\dot{\omega})=g_e([\omega,a^{-1}\xi a],\omega)=g_e(a^{-1}\xi a,B(\omega,\omega)).\]

Since \xi, and therefore also a^{-1}\xi a is arbitrary, we obtain

(14)   \begin{equation*}\dot{\omega}=B(\omega,\omega),\end{equation*}

or, using a basis and Eq. (13)

(15)   \begin{equation*}g_{kl}\,\dot{\omega}^l=C_{jk,i}\,\omega^{i}\omega^{j}.\end{equation*}

Killing vectors, geodesics, and Noether’s theorem

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.

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

(1)   \begin{equation*} Ad_a: X\mapsto aXa^{-1}.\end{equation*}

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

(2)   \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:

(3)   \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

(4)   \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

(5)   \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

(6)   \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

(7)   \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

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

which implies for \xi,\eta tangent at b

(9)   \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

(10)   \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*}

(11)   \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)

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

(13)   \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.\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. (12)

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

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. (13).

SL2R as anti de Sitter space cont.

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.