# 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”.
[latexpage]
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
$$g_a(\xi,\eta)=g_e(a^{-1}\xi,a^{-1}\eta),$$
which implies for $\xi,\eta\in T_bG$
$$g_{ab}(a\xi,a\eta)=g_b(\xi,\eta),\,a,b\in G.$$
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:
$$\omega(t)=a(t)^{-1}\dot{a}(t).$$

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
$$g_{a(t)}(\xi a(t),\dot{a}(t))=\mbox{const}.$$
The metric is left-invariant, therefore $g_e(a(t)^{-1}\xi a(t),a(t)^{-1}\dot{a}(t))=\mbox{const}$, or
$$g_e(a(t)^{-1}\xi a(t),\omega(t))=\mbox{const}.\label{eq:tod}$$
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
$$\frac{d}{dt}a(t)^{-1}=-a(t)^{-1}\dot{a}(t)a(t)^{-1}=-\omega(t)a(t)^{-1}.$$
Now, differentiating Eq. (\ref{eq:tod}), and using also $\frac{da(t)}{dt}=a(t)\omega(t)$ we obtain
$$g_e([a^{-1}\xi a,\omega],\omega)+g_e(a^{-1}\xi a,\dot{\omega})=0.\label{eq:a1}$$
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:
$$g_e(B(\xi_1,\xi_2),\eta)=g_e([\xi_2,\eta],\xi_1),\quad \xi_1,\xi_2,\eta\in Lie(G). \label{eq:b1}$$
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 $C_{ij}^k$
$$[\xi_i,\xi_j]=C_{ij}^k\,\xi_k.$$
We can also write $B$ as
$$B(\xi_i,\xi_j)=B_{ij}^k\,\xi_k.$$
Then Eq. (\ref{eq:b1}) gives

g_e(B(\xi_i,\xi_j),\xi_k)=g_e([\xi_j,\xi_k],\xi_i)\label{eq:b2}
or
$B_{ij}^l g_{lk}=C_{jk}^lg_{li},$
which can be solved for $B$ using the inverse metric:
$$B_{ij}^m=g^{mk}C_{jk}^lg_{li}.$$
On the other hand, if we agree to lower the upper index of $B$ and $C$ with the metric, we can write Eq. (\ref{eq:b2}) as
$$B_{ij,k}=C_{jk,i},\label{eq:b3}$$
which is easy to remember.

$g_e(a^{-1}\xi a,\dot{\omega})=g_e([\omega,a^{-1}\xi a],\omega)=g_e(a^{-1}\xi a,B(\omega,\omega)).$
Since $\xi$, and therefore also $a^{-1}\xi a$ is arbitrary, we obtain
$$\dot{\omega}=B(\omega,\omega),$$
or, using a basis and Eq. (13)
$$g_{kl}\,\dot{\omega}^l=C_{jk,i}\,\omega^{i}\omega^{j}.$$

## 7 thoughts on “Geodesics of left invariant metrics on matrix Lie groups – Part 1”

1. [latexpage]
$LHS=g_e(B(\xi_i,\xi_j),\xi_k)=g_e(B_{ij}^l\xi_l,\xi_k)=B_{ij}^lg_e(\xi_l,\xi_k)=B_{ij}^lg_{kl}.$
$RHS=g_e(C_{jk}^l\xi_l,\xi_i)=C_{jk}^lg_{li}.$

1. dakunin says:

To jest ostatnia notka tutaj – czy coś skasowano?

Y..nie wiem dlaczego nie mogę się zalogować i komentować tutaj z mojego konta blogowego (blog.cassiopaea.pl) – może dlatego, że tamto jest na polskim wordpressie i tak nie można (?) Muszę to sprawdzić.
Wpisałem się na salonie24. pozdrawiam m