Find the Christoffel symbols from an expression I have a problem of General Relativity (introduction), I know how to solve it, but I don't know how using the expression. This problem is from here (the first thing) in Spanish, I translated it:

From the definition of covariant derivative, shows that:
$$ V^{\mu}(x^{\alpha}+\Delta x^{\alpha}) = V^{\mu}(x^{\alpha}) - V^{\lambda}(x^{\alpha}) \Gamma^{\mu}_{\lambda\nu}\Delta x^{\nu} + O\big((\Delta x^{\alpha})^2\big)$$
where $V^{\mu}(x^{\alpha}+\Delta x^{\alpha})$ is the parallel transported of $V^{\mu}\,(x^{\alpha})$ over the point $x^{\alpha}+\Delta x^{\alpha}$. Use the expression to find the Christoffel symbols in the flat  space two-dimensional in polar coordinates with the habitual notion of the parallel transport of a plane: the Cartesian components of the vectors don't change.


My attempt:
First, in a parallel transport:
$$ \nabla_{\nu} V^{\mu}(x^{\alpha}) = 0 $$
Developing the covariant derivative:
$$ \partial_{\nu} V^{\mu}(x^{\alpha}) + \Gamma^{\mu}_{\lambda\nu}V^{\lambda}(x^{\alpha}) = 0 $$
$$ \partial_{\nu} V^{\mu}(x^{\alpha}) = - \Gamma^{\mu}_{\lambda\nu}V^{\lambda}(x^{\alpha}) $$
From Taylor:
$$ V^{\mu}(x^{\alpha}+\Delta x^{\alpha}) = V^{\mu}(x^{\alpha}) + \big(\partial_{\lambda}V^{\mu}(x^{\alpha})\big)\Delta x^{\lambda} + O\big((\Delta x^{\alpha})^2\big) $$
Thus:
$$ V^{\mu}(x^{\alpha}+\Delta x^{\alpha}) = V^{\mu}(x^{\alpha}) - V^{\nu}(x^{\alpha})\Gamma^{\mu}_{\lambda\nu}\Delta x^{\lambda} + O\big((\Delta x^{\alpha})^2\big) $$
But I dont' know for why the given expression is relevant, because I know that the Christoffel symbols can be calculated with $\Gamma^{\mu}_{\lambda\nu} = e^{\mu}\cdot\partial_{\nu}e_{\lambda}$.
 A: (continued from ANSWER - Part II)
ANSWER - Part III
$\boldsymbol\S$ F. Parallel Transport on the plane - Geodesics in Polar coordinates
The geodesic curves on smooth two-dimensional surfaces have the following characteristic properties each one of which could be used as equivalent alternate definition

*

*Osculating plane :  Geodesic on a surface is any curve such that at every point its osculating plane is perpendicular to the tangent plane to the surface.


*Shortest path :  Geodesic on a surface is any curve which gives
the shortest path lying on the surface between two given points.


*Autoparallelism :  Geodesic on a surface is any curve with its directions at its various points being all parallel along the geodesic itself.
We will use the autoparallelism as definition of a geodesic. For the curve of equation \eqref{E-14}
\begin{equation}
\boldsymbol\chi\left(\lambda\right)\boldsymbol=
\Bigl(r\left(\lambda\right),\theta\left(\lambda\right)\Bigr) 
\nonumber
\end{equation}
we will parallel transport its tangent vector
\begin{equation}
\boldsymbol\xi\boldsymbol=\left(\xi_r,\xi_\theta\right)\boldsymbol=\dot{\boldsymbol\chi\:}\boldsymbol=
\left(\dot r,\dot{\theta\:}\right)\quad \boldsymbol{\Longrightarrow}\quad \xi_r\boldsymbol{=}\dot{r}\,,\:\xi_\theta\boldsymbol{=}\dot{\theta\:}
\tag{F-01}\label{F-01}  
\end{equation}
and demand this vector to remain tangent to the curve. So, inserting above expressions of $\,\xi_r,\xi_\theta\,$ in equation \eqref{E-16} we have
\begin{equation}
\begin{split}
\ddot r\boldsymbol-r \dot{\theta\:}^{\!_2} &  \boldsymbol=0\\ 
r\ddot{\theta\:}\boldsymbol+2\dot r\dot{\theta\:} & \boldsymbol=0\\
\end{split}
\tag{F-02}\label{F-02}    
\end{equation}
This system of equations must be satisfied by a geodesic. But since the geodesics on the plane are straight lines these equations must represent a straight line. Indeed, in Cartesian coordinates
\begin{equation}
\mathbf x\boldsymbol=\left(r\cos\theta\right)\mathbf e_1\boldsymbol+\left(r\sin\theta\right) \mathbf e_2
\tag{F-03}\label{F-03} 
\end{equation}
so
\begin{equation}
\dot{\mathbf x}\boldsymbol=\left(\dot r\cos\theta\boldsymbol-r\dot{\theta\:}\sin\theta\right)\mathbf e_1\boldsymbol+\left(\dot r\sin\theta\boldsymbol+r\dot{\theta\:}\cos\theta\right) \mathbf e_2
\tag{F-04}\label{F-04} 
\end{equation}
and
\begin{equation}
\begin{split}
\ddot{\mathbf x}\boldsymbol= & \Bigl[\left(\ddot r\boldsymbol-r \dot{\theta\:}^{\!_2}\right)\cos\theta \boldsymbol-\left(r\ddot{\theta\:}\boldsymbol+2\dot r\dot{\theta\:}\right)\sin\theta\Bigr]\mathbf e_1\\
\boldsymbol+ & \Bigl[\left(r\ddot{\theta\:}\boldsymbol+2\dot r\dot{\theta\:}\right)\cos\theta \boldsymbol+ \left(\ddot r\boldsymbol-r \dot{\theta\:}^{\!_2}\right)\sin\theta\Bigr] \mathbf e_2\boldsymbol{=0}\\
\end{split}
\tag{F-05}\label{F-05}   
\end{equation}
that is
\begin{equation}
\mathbf x\boldsymbol=\mathbf a\lambda\boldsymbol+ \mathbf b \qquad \mathbf a,\mathbf b \boldsymbol=\texttt{constant vectors} 
\tag{F-06}\label{F-06}  
\end{equation}
equation of a straight line.
$\boldsymbol\S$ G. Parallel Transport on the plane - Example in Polar Coordinates
As Example we'll try to parallel transport on the plane in polar coordinates a vector $\,\mathbf x_0\,$ along a circular arc of radius $\,R\,$ as shown in Figure-06.
A regular $\,\lambda\boldsymbol-$parametric representation of the arc in polar coordinates is
\begin{equation}
\boldsymbol\chi\left(\lambda\right)\boldsymbol=
\Bigl(r\left(\lambda\right),\theta\left(\lambda\right)\Bigr)\boldsymbol=
\left(R,\lambda\,\pi \right)\,, \quad \lambda \in \mathbb R 
\tag{G-01}\label{G-01}
\end{equation}
so
\begin{equation}
\begin{split}
r\left(\lambda\right)\boldsymbol=R\boldsymbol=\texttt{constant} \quad & \boldsymbol\implies \quad \dot r\boldsymbol=0\\
\theta\left(\lambda\right)
\boldsymbol=\lambda\,\pi \quad & \boldsymbol\implies \quad \dot{\theta\:}\boldsymbol{=}\pi\\
\end{split}
\tag{G-02}\label{G-02}
\end{equation}

The initial vector $\,\mathbf x_0\,$ is supposed to be at the point $\,\texttt P_0\,$ on the cartesian $\,x_1\boldsymbol-$axis. Note the polar and cartesian coordinates of this point $\,\texttt P_0\boldsymbol=\left(R,0\right)_{\texttt{polar}}\boldsymbol=\left(R,0\right)_{\texttt{cartesian}}$. Consider that the polar coordinates of this vector $\,\mathbf x_0\,$ are
\begin{equation}
\mathbf x_0\boldsymbol=
\left(\xi_{0r},\xi_{0\theta}\right)_{\texttt{polar}}
\tag{G-03}\label{G-03}
\end{equation}
Then
\begin{equation}
\mathbf x_0\boldsymbol=\xi_{0r}\mathbf{x}_{0r}\boldsymbol+\xi_{0\theta}\mathbf{x}_{0\theta}
\tag{G-04}\label{G-04}
\end{equation}
where $\,\mathbf{x}_{0r},\mathbf{x}_{0\theta}\,$ the tangents to the $\,r\boldsymbol-,\theta\boldsymbol-$parametric curves at point $\,\texttt P_0\,$ respectively
\begin{equation}
\mathbf{x}_{0r}\boldsymbol=\mathbf e_1\,, \qquad \mathbf{x}_{0\theta}\boldsymbol=R\,\mathbf e_2
\tag{G-05}\label{G-05}
\end{equation}
that is
\begin{equation}
\mathbf x_0\boldsymbol=\xi_{0r}\mathbf e_1\boldsymbol+\xi_{0\theta}R\,\mathbf e_2
\tag{G-06}\label{G-06}
\end{equation}
so for the cartesian coordinates of the initial vector we have
\begin{equation}
\mathbf x_0\boldsymbol=
\left(\xi_{0r},R\,\xi_{0\theta}\right)_{\texttt{cartesian}}
\tag{G-07}\label{G-07}
\end{equation}
Inserting the expressions \eqref{G-02} of $\,r,\dot r, \theta,\dot{\theta\:}\,$ in the equations of parallel transport \eqref{E-16}, repeated here for convenience,
\begin{equation}
\begin{split}
\dot\xi_r\boldsymbol-r \xi_\theta\dot{\theta\:} & \boldsymbol= 0
\nonumber\\ 
r\dot\xi_\theta\boldsymbol+\xi_\theta\dot r\boldsymbol+\xi_r\dot{\theta\:} & \boldsymbol=0\\ 
\end{split}
\nonumber 
\end{equation}
we have
\begin{equation}
\begin{split}
\dot\xi_r\boldsymbol-\pi R\, \xi_\theta &  \boldsymbol=0\\
R\,\dot\xi_\theta\boldsymbol+\pi\xi_r & \boldsymbol=0
\end{split}
\tag{G-08}\label{G-08}
\end{equation}
From the 2nd equation \eqref{G-08}
\begin{equation} 
\xi_r \boldsymbol=\left(\boldsymbol-\dfrac{R}{\pi}\right)\dot\xi_\theta  \quad \boldsymbol\implies\quad \dot\xi_r \boldsymbol=\left(\boldsymbol-\dfrac{R}{\pi}\right)\ddot\xi_\theta 
\tag{G-09}\label{G-09} 
\end{equation}
Inserting this expression of $\dot\xi_r$ in the 1st equation \eqref{G-08} we have
\begin{equation}  
\ddot\xi_\theta \boldsymbol+\pi^2 \xi_\theta \boldsymbol=0
\tag{G-10}\label{G-10} 
\end{equation}
so
\begin{equation} 
 \boxed{\:\:\xi_\theta\left(\lambda\right) \boldsymbol=\mathrm c\sin\left(\lambda\,\pi\boldsymbol+\phi\right)\vphantom{\dfrac{a}{b}}\:\:} \qquad \mathrm c,\phi \in \mathbb R
\tag{G-11}\label{G-11}  
\end{equation}
Now
\begin{equation} 
\eqref{G-09} \:\: :\quad\xi_r\left(\lambda\right) \boldsymbol=\left(\boldsymbol-\dfrac{R}{\pi}\right)\dot\xi_\theta  \quad \boldsymbol\implies \xi_r \boldsymbol{=}\boldsymbol{-}R\,\mathrm c\cos\left(\lambda\,\pi\boldsymbol{+}\phi\right)
\nonumber
\end{equation}
\begin{equation} 
 \boxed{\:\:\xi_r\left(\lambda\right) \boldsymbol{=-}\mathrm cR\cos\left(\lambda\,\pi\boldsymbol+\phi\right)\vphantom{\dfrac{a}{b}}\:\:} \qquad \mathrm c,\phi \in \mathbb R
\tag{G-12}\label{G-12}  
\end{equation}
The constants $\,\mathrm c,\phi\,$ are determined from the initial conditions
\begin{equation}
\begin{split} 
\mathrm c\sin\phi  & \boldsymbol=\xi_\theta\left(0\right)\boldsymbol\equiv\xi_{0\theta}\\ 
\boldsymbol-R\mathrm c\cos\phi  & \boldsymbol=\xi_r\left(0\right)\boldsymbol\equiv\xi_{0r}\\
\end{split}
\tag{G-13}\label{G-13}
\end{equation}
yielding
\begin{equation} 
\tan\phi \boldsymbol=\boldsymbol{-}\dfrac{R\,\xi_{0\theta}}{\xi_{0r}}\,,\qquad \mathrm c\boldsymbol=\boldsymbol-\dfrac{\sqrt{\xi^2_{0r}\boldsymbol+\left(R \xi_{0\theta}\right)^2}}{R}
\tag{G-14}\label{G-14} 
\end{equation}
and finally
\begin{equation}
\begin{split} 
\xi_r\left(\lambda\right) & \boldsymbol{=+}\sqrt{\xi^2_{0r}\boldsymbol+\left(R \xi_{0\theta}\right)^2}\cos\left(\lambda\,\pi\boldsymbol+\phi\right)\\
R\xi_\theta\left(\lambda\right)   &\boldsymbol{=-}\sqrt{\xi^2_{0r}\boldsymbol+\left(R \xi_{0\theta}\right)^2}\sin\left(\lambda\,\pi\boldsymbol+\phi\right)\\
\end{split}
\tag{G-15}\label{G-15}
\end{equation}
So the resulting vector $\,\mathbf x\,$ at point $\,\texttt P\,$ has the following polar coordinates
\begin{equation} 
\mathbf x\boldsymbol= 
\begin{bmatrix}
\xi_r\vphantom{\dfrac{a}{b}}\\
\xi_\theta\vphantom{\dfrac{a}{b}}
\end{bmatrix}_{\texttt{polar}}\!\!\!\!\!\boldsymbol=
\begin{bmatrix}
\hphantom{\boldsymbol-} \sqrt{\xi^2_{0r}\boldsymbol+\left(R \xi_{0\theta}\right)^2} \cos\left(\lambda\,\pi\boldsymbol+\phi\right)\hphantom{/R}\vphantom{\dfrac{a}{b}}\\
\boldsymbol-\sqrt{\xi^2_{0r}\boldsymbol+\left(R \xi_{0\theta}\right)^2} \sin\left(\lambda\,\pi\boldsymbol+\phi\right)/R\vphantom{\dfrac{a}{b}}
\end{bmatrix}_{\texttt{polar}} 
\tag{G-16}\label{G-16}  
\end{equation}
A: Suggestion: don't try to use general expressions throughout. Instead develop your intuition by dealing with two components individually, as follows.

*

*write a two-component vector $(a,b)$ explicitly in polar form (you can write $v^r {\bf e}_r + v^\theta {\bf e}_\theta$ if you like but good old $a$ and $b$ save you the trouble of writing superscript indices)

*apply coordinate transformation to rectangular form.

*add a displacement to the two rectangular coordinates (because we know this is how parallel transport works in 2D flat space charted by rectangular coordinates).

*transform back.

*Look at the result and spot which part or parts has to be the Christoffel symbol(s), by comparing with the expression you were given for parallel transport in general.

This method is perfectly rigorous. (You might like to think about that too).
