Roberto, you got the math right, but were slightly off the mark in actually proving the argument. In particular, you can't

use the result of [14.26], because that's the very thing we're trying to demonstrate!

Instead, look at the equation at the top of your page 2 (of the attachment).

As you later point out, if

u is chosen to be an affine parameter,

R is a constant along the curve and thus the RHS of this equation vanishes, leaving you with an equation defining

(=

dt/du) in terms of

t,

g, and derivatives of

g.

Now consider that along the affine geodesic

is zero. Expanding the expression for the covariant derivative (connection derivative) of a vector

t along

t yeilds:

and since that is zero, you can substitute your

g expression to replace the partial derivative term, yielding the

definition of

, just as it is in [14.26], as required!

[Actually, there's a little subtlety here: What you actually get is an expression for

that's true for any vector

t at any point (since you can always construct an affine geodesic through any point in any direction).

It's a simple matter to show, however, that this is sufficient to completely define

when it's symmetric in it's two bottom indices: Just note that due to the symmetry, given

any two arbitrary vectors

s and

t,

st =

uu-

vv where

s = u + v,

t = u - v.]