conservation of dot product with parallel transport I have a question about the parallel transport of a vector :
Why does one say that parallel transport preserves the value of dot product (scalar product) between the transported vector and the tangent vector ?
Is it due to the fact that angle between the tangent vector and transported vector is always the same during the operation of transport (which is the definition of parallel transport) ?
So If I take 2 different points, the dot product is the same since angle is the same ??
How to demonstrate it or translate this statement from a mathematic point of view ?
Thanks for your help
UPDATE 1 :
Thanks for your quick answer. Unfortunately, I am not an expert in tensor calculus but I know some basics like the definition of covariant derivative of a vector $V$ along a geodesic - like with this notation :
$\nabla_{i}V^{j}=\partial_{i}V^{j}+V^{k}\Gamma_{ik}^{j}\quad\quad(1)$
and the absolute derivative : $D\,V^{j}=(\nabla_{i}V^{j})dx^{i}\quad\quad(2)$
Could give me the link between your equation (
$Z \langle X,Y \rangle = \langle \nabla_Z X, Y\rangle + \langle X, \nabla_Z Y\rangle$ ) and the equation (1) or (2).
Moreover, you define $Z$ like $$\text{d}\gamma/\text{d}t$$ but after, you only take $\text{d}/\text{d}t$ in :
$\frac{d}{dt}\langle X,Y \rangle = \langle \nabla_{\overset{\cdot}{\gamma}} X, Y\rangle + \langle X, \nabla_{\overset{\cdot}{\gamma}} Y\rangle = 0$
You say that $Z$ is a vector field : is it an operator or a vector field ?
And what about $\langle X,Y\rangle$ ? 
Can one write :
$\langle X,Y\rangle=g_{ij}X^{i}Y^{j}$
with $g_{ij}$ the metrics ???
Regards
UPDATE 2 : 
I think there is a little error on index for the third term in first factor, this sould be :
$\Big(\dfrac{\partial g_{ij}}{\partial x^p} - g_{kj}\Gamma_{pi}^k - g_{ik}\Gamma_{pj}^k\Big)\xi^i\eta^j\mathrm{d}x^p = 0.$
and not 
$\Big(\dfrac{\partial g_{ij}}{\partial x^p} - g_{kj}\Gamma_{pi}^k - g_{ik}\Gamma_{pi}^k\Big)\xi^i\eta^j\mathrm{d}x^p = 0.$
UPDATE 3 : 
You say "Why ?" above "equal symbol" but you don't give the reason :
$\mathrm{D}\langle\xi\eta\rangle \stackrel{why?}{=} \mathrm{d}\langle\xi\eta\rangle = \mathrm{d}(g_{ij}\xi^i\eta^j) = 0.$
Could you justify please this expression : $\mathrm{d}(g_{ij}\xi^i\eta^j) = 0$
I know that norm of 2 vectors defined by $\text{d}x^{i}$ and $\text{d}x^{j}$ is constant (I mean indepedently of basis used) because :
$\text{d}s^2=g_{ij} \text{d}x^{i}\text{d}x^{j}=\text{length}=\text{constant}$
Is this the same justification with $\xi^i\eta^j$ vectors ? 
 A: I can prove the following statement:

In Riemannian Space (Riemannian manifold with metric) there exist unique connection $\Gamma_{ij}^k(x)$ such that
   1. $\Gamma_{ij}^k = \Gamma_{ji}^k$
   2. Scalar product of two parallel transported vectors is constant along path.

Preliminaries
Let vectors $\xi$ and $\eta$ be transported along path $P$. Also, let us denote metric tensor as $g_{ij}(x)$.
In Euclidean Space (where permanency of scalar product is obvious (why?) ) $\mathbb{R}^n$  it is possible to introduce curvilinear coordinate system which produces affine connection $\Gamma_{ij}^k$ so as metric tensor (or Gram matrix) $g_{ij}$ (inverse metric we will denote as $g^{kl}$). These two object have the following connection (sorry for ambiguity):
$$\Gamma_{ij}^{k} = \frac12g^{kl}\Big(\frac{\partial g_{li}}{\partial x^j} + \frac{\partial g_{lj}}{\partial x^i} - \frac{\partial g_{ij}}{\partial x^l}\Big).$$
Proof of the statement
$\rhd$ Starting with the definition of scalar product we can write:
$$\langle\xi,\eta\rangle = g_{ij}\xi^i\eta^j.$$
Requiring scalar product to be constant we write:
$$\mathrm{D}\langle\xi,\eta\rangle \stackrel{why?}{=} \mathrm{d}\langle\xi,\eta\rangle = \mathrm{d}(g_{ij}\xi^i\eta^j) = 0.$$
By the chain rule we obtain:
$$\mathrm{d}g_{ij}\xi^i\eta^j + g_{ij}\mathrm{d}\xi^{i}\eta^{j} + g_{ij}\xi^{i}\mathrm{d}\eta^{j} = 0$$ 
But vectors $\xi,\eta$ are transported parallel and then
$$\mathrm{d}\xi^i = -\Gamma_{pi}^{k}\xi^i\mathrm{d}x^{p},$$
where $\mathrm{d}x^{p}$ - differential along path $P$, $\eta$ is transported similarly. More over,
$$\mathrm{d}g_{ij} = \frac{\partial g_{ij}}{\partial x^p}\mathrm{d}x^p.$$ 
So we have (renaming summation indexes)
$$\Big(\frac{\partial g_{ij}}{\partial x^p} - g_{kj}\Gamma_{pi}^k - g_{ik}\Gamma_{pj}^k\Big)\xi^i\eta^j\mathrm{d}x^p = 0.$$
On the other hand, vectors $\xi, \eta$ and path $P$ can be arbitrary, so we have to require:
$$\Big(\frac{\partial g_{ij}}{\partial x^p} - g_{kj}\Gamma_{pi}^k - g_{ik}\Gamma_{pj}^k\Big) = 0.$$
This equation has the same form as in case of Euclidean Space so we solve this in the same way.
From this and 1st condition of the statement we are able to find desired $\Gamma_{ij}^{k} \lhd.$ 

About ''why's''


*

*Why in Euclidean Space it is obvious that scalar product of two vectors remains constant while parallel transported along path $P$.
This is because in Euclidean Space you can choose Cartesian coordinate system where coordinates of a vector remain constant during parallel transport.

*Why $$\mathrm{D}\langle\xi,\eta\rangle = \mathrm{d}\langle\xi,\eta\rangle = 0.$$
I.e. why in this case absolute diferential ($\mathrm{D}$) is  equal to ordinary ($\mathrm{d}$).
This is because scalar product is a scalar i.e. has no indexes (but no every quantity that has no indexes is invariant).


Now about the same form as in Euclidean Space. Our purpose here is to express connection $\Gamma$ through metric tensor $g$. In ES we introduce 
$\Gamma_{ij}^{k}$ from the following expression:
$$\frac{\partial^2 \vec{x}}{\partial x^i \partial x^j} = \Gamma_{ij}^{k}\frac{\partial \vec{x}}{\partial x^k},$$
where $\frac{\partial \vec{x}}{\partial x^k} = \vec{x}_k$ - local basis in $\mathbb{R}^n$. Then having a scalar product we have
$$\langle x_l, x_{ij}\rangle = \Gamma_{ij}^k g_{lk},$$
where we used $\langle x_l, x_k\rangle = g_{lk}$. Next it is convenient to denote:
$$\Gamma _{l, ij} = g_{lk}\Gamma_{ij}^k.$$
Next, we obtain
$$\frac{\partial}{\partial x^m}\langle x_l, x_k \rangle = \frac{\partial g_{lk}}{\partial x^m},$$
or by the chain rule
$$\langle x_{lm}, x_k \rangle + \langle x_{l}, x_{km} \rangle = \Gamma_{k,lm} + \Gamma_{l, km} = \frac{\partial g_{kl}}{\partial x^m}.$$
Here we have one equation with two unknown quantities, but if we permute indices cyclically, we obtain two more equations with only one new unknown. Solving we have
$$\Gamma_{l, km} = \frac12 \Big( \frac{\partial g_{lk}}{x^m} + \frac{\partial g_{lm}}{x^k} - \frac{\partial g_{km}}{x^l}\Big).$$ Raising index $l$ we get desirable result. 

P.S.you can find this proof in 
Rashevsky's ''Riemannian geometry and tensor analysis.'' Though I do not know if there exists English version of this book.
A: By definition, the parallel transport of $e \in T{p}M$ along a path $\gamma(t), \gamma(0) = p$ is the unique vector fields $X_t$ with $X_t \in T_{\gamma(t)}M$ such that $\nabla_{\overset{\cdot}{\gamma}}X = 0 $ and $X_0 = e$.
Now, by definiton your connection is compatible with the metric, i.e $Z \langle X,Y \rangle = \langle \nabla_Z X, Y\rangle + \langle X, \nabla_Z Y\rangle$ for any vector field $Z$.
Thus taking $Z = d\gamma/dt$, we obtain that $\frac{d}{dt}\langle X,Y \rangle = \langle \nabla_{\overset{\cdot}{\gamma}} X, Y\rangle + \langle X, \nabla_{\overset{\cdot}{\gamma}} Y\rangle = 0$ since $X,Y$ are parallel vector. Thus $\langle X, Y \rangle = \langle X_0, Y_0 \rangle$ as wished.
