Taylor expansion and parallel transport

[mertcan]

hi, first of all in this image there is a fact that we have parallel transported vector, and covariant derivative is zero along the "pr"path as you can see at the top of the image. I consider that p, and r is a point and in the GREEN box we try to make a taylor expansion of the contravariant component of Z vector at the "p" and "r" points. But in the RED box we try to make a taylor expansion of the contravariant component of Z vector at the "u" and "r" points. Here is my confusion: Although there is not a "u" point, instead is just "p" and "r" points, How can we use such a point? What does "u"point stand for?

I quoted from Differential Geometry Fionn Fitzmaurice page 36

 

[mertcan]

I have seen so many views, but I have not been replied for a long time... I am waiting your valuable return

 

[Max Renn]

Hi.

The point $u = h_{ds}(q)$, i.e. transporting from $p$ to $q$ by $dt$ and then from $q$ to $u$ by $ds$. It's the same construction as section 6.3.