Proof that the first curvature vector P of a curve C is at right angles to C

First, some notation: write T,a for the (ordinary) partial derivative of the tensor T with respect to xa. This will avoid a lot of messy notation (as if there is not enough already!).

To show the result, we compute

P, T =  P, dx/ds
=  Pi dxjds gij
(Definition of the inner product)
=  d2xids2 + ip q dxpds dxqds dxjds gij
(Definition of Pi)
=  d2xids2 + 12 gli [gql,p + glp,q - gpq,l] dxpds dxqds dxjds gij
=  d2xids2 dxjds gij + 12 [gqj,p + gjp,q - gpq,j] dxpds dxqds dxjds
(since gligij = jl)

Looking at the term on the right, we distribute to get 3 terms of the form

 gqj,p dxpds dxqds dxjds = dgqjds dxqds dxjds by the chain rule,

and we notice that the other two terms are identical to this (except for dummy variables). Thus, we are left with

 d2xids2 dxjds gij + 12 dgqjds dxqds dxjds
=  12 dds dxids dxjds gij
(by symmetry of the gij)
=  12 dds T, T
(definition of the inner product)
=  12 dds ( ±1)
(refer back to the Proof of 5.5 to check this)
= 0,

as asserted.