# I Taylor expansion and parallel transport

1. Jun 15, 2016

### 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

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2. Jun 15, 2016

### mertcan

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

3. Jul 19, 2016

### 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.