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 with the Lie algebra The tangent space at is denoted .

Thus On we assume nondegenerate scalar product

denoted as We propagate it to the whole group using left translations as in Eqs. (8,9) of Killing vectors, geodesics, and Noether’s theorem

(1)

which implies for

(2)

The metric so constructed is automatically left-invariant, therefore for each the vector field is a Killing field.

Let be a geodesic for this metric. We denote by the tangent vector left translated to the identity:

(3)

Then, from the conservation laws derived in the last post, we know that the scalar product of with is constant. That is

(4)

The metric is left-invariant, therefore , or

(5)

We will differentiate the last equation with respect to but first let us notice that by differentiating the identity we obtain

(6)

Now, differentiating Eq. (5), and using also we obtain

(7)

We now need a certain bilinear operator on Lie(G) that is defined using the commutator and the scalar product. The commutator itself is such an operator

from But using the scalar product we can define another operator by the formula:

(8)

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

Let be a basis in so that the structure constants are

(9)

We can also write as

(10)

Then Eq. (8) gives

(11)

or

which can be solved for using the inverse metric:

(12)

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

(13)

which is easy to remember.

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

Since , and therefore also is arbitrary, we obtain

(14)

or, using a basis and Eq. (13)

(15)

functional i represented ->

?

Fixed. Thanks.

How do we get eq.(11a) from eq.(11) ?

Thanks.