Christoffel Symbols of a parallel shift Regarding the transformation $p=3x+5y;q=x+y$ (where $x,y$ are cartesian coordinates) I want to look at the christoffel symbols for the covariant coordinate system for a parallel shift (using ruler and triangle).
Therefore I calculated the covariant basis vectors and then looked at coordinates of a vector $\overrightarrow{A}=A^x\overrightarrow{e}_x+A^y\overrightarrow{e}_y$ in the p-q coordinates system (calculating $A^p$ and $A^q$), because I then need to shift it parallel to another point. The coordinates of $A$ then become $A^p+\delta A^p,A^q+\delta A^q$
I now need to write $\delta A^p$ and $\delta A^q$ as a total differential of the variables $p,q$ and find a way to put the parallel shift in mathematic language. This is where I am stuck.
So far I have $\vec{e}_p=\frac{1}{2}(-\vec{e}_x+\vec{e}_y)$ and $\vec{e}_p=\frac{1}{2}(5\vec{e}_x-3\vec{e}_y)$.
I then also want to compare the solution to $\delta A^i=\Gamma^i_{ml}\cdot A^m\delta x^l $.
Can somebody help me understand what I need to do here?
 A: This case is not a proper one to understand the parallel transport and the role of the Christoffel Symbols. This is due to the fact that the latter are all identically zero.
For the parallel transport of a vector $\:V^\alpha\:$ along a curve with parametric equation
\begin{equation}
\boldsymbol{p}\left(\lambda\right) \boldsymbol{=}\left[p^1 \left(\lambda\right), p^2 \left(\lambda\right)\right]\qquad \left(p^1\boldsymbol{\equiv}p,\:p^2\boldsymbol{\equiv}q\right)
\tag{01}\label{01} 
\end{equation}
we use the equation
\begin{equation}
\frac{\mathrm dV^\alpha}{\mathrm d\lambda} \boldsymbol{=} \boldsymbol{-}\Gamma_{\beta\nu}^{\alpha}V^\nu\frac{\mathrm dp^{\,\beta}}{\mathrm d\lambda} \qquad \left(\alpha,\beta,\nu \boldsymbol{=}1,2\right)
\tag{02}\label{02} 
\end{equation}
or equivalently
\begin{equation}
\mathrm dV^\alpha \boldsymbol{=} \boldsymbol{-}\Gamma_{\beta\nu}^{\alpha}V^\nu \mathrm dp^{\,\beta} \qquad \left(\alpha,\beta,\nu \boldsymbol{=}1,2\right)
\tag{03}\label{03} 
\end{equation}
The Christoffer symbols are expressed through the components of the metric tensor $\mathbf{g}$
\begin{equation}
\Gamma_{\beta\nu}^{\alpha}\boldsymbol{=} \frac{1}{2}g^{\alpha k} \left(\frac{\partial g_{k\nu}}{\partial p^{\,\beta }}\boldsymbol{+}\frac{\partial g_{\beta k}}{\partial p^{\,\nu}}\boldsymbol{-}\frac{\partial g_{\beta \nu}}{\partial p^{\,k}}\right) 
\tag{04}\label{04} 
\end{equation}
In all equations we make use of the Einstein summation convention.
The metric tensor is determined from the expression of the infinitesimal displacement
\begin{equation}
\mathrm ds^2 \boldsymbol{=}g_{\rho\sigma}\mathrm dp^{\,\rho}\mathrm dp^{\,\sigma} 
\tag{05}\label{05} 
\end{equation}
In our case
\begin{equation}
\mathrm ds^2 \boldsymbol{=}\mathrm d\mathbf r\boldsymbol{\cdot}\mathrm d\mathbf r\boldsymbol{=}\left(\mathbf e_1\mathrm dp\boldsymbol{+}\mathbf e_2\mathrm dq\right)\boldsymbol{\cdot}\left(\mathbf e_1\mathrm dp\boldsymbol{+}\mathbf e_2\mathrm dq\right) \boldsymbol{=}g_{11}\mathrm dp^{2}\boldsymbol{+}2g_{12}\mathrm dp\mathrm dq\boldsymbol{+}g_{22}\mathrm dq^{2}
\tag{06}\label{06} 
\end{equation}
explicitly
\begin{equation}
\mathrm ds^2 \boldsymbol{=}\overbrace{\underbrace{\left(\mathbf e_1\boldsymbol{\cdot}\mathbf e_1\right)}_{g_{11}}}^{\tfrac{1}{2}}\mathrm dp^{2}\boldsymbol{+}2\overbrace{\underbrace{\left(\mathbf e_1\boldsymbol{\cdot}\mathbf e_2\right)}_{g_{12}}}^{\boldsymbol{-}2}\mathrm dp\mathrm dq\boldsymbol{+}\overbrace{\underbrace{\left(\mathbf e_2\boldsymbol{\cdot}\mathbf e_2\right)}_{g_{22}}}^{\tfrac{17}{2}}\mathrm dq^{2}
\tag{07}\label{07}  
\end{equation}
where as the OP gives
\begin{equation}
\mathbf e_1 \boldsymbol{=}\frac12\left(\boldsymbol{-}\mathbf e_x\boldsymbol{+}\mathbf e_y\right)\,,\qquad \mathbf e_2 \boldsymbol{=}\frac12\left(5\mathbf e_x\boldsymbol{-}3\mathbf e_y\right)
\tag{08}\label{08}  
\end{equation}
So the metric tensor is
\begin{equation}
\mathbf g\boldsymbol{=}g_{\rho\sigma}\boldsymbol{=}
\begin{bmatrix}
g_{11} & g_{12} \vphantom{\dfrac{a}{b}}\\
g_{21} & g_{22} \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\hphantom{\boldsymbol{-}}\tfrac{1}{2} & \boldsymbol{-}2 \vphantom{\dfrac{a}{b}}\\
\boldsymbol{-}2 & \hphantom{\boldsymbol{-}}\tfrac{17}{2} \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\tag{09}\label{09}    
\end{equation}
The metric tensor is independent of the $''$curvilinear$''$  coordinates $p,q$ so all Christoffel Symbols are identically zero
\begin{equation}
\frac{\partial g_{\rho\sigma}}{\partial p^{\tau }}\boldsymbol{=}0\quad \boldsymbol{\implies} \quad \Gamma_{\beta\nu}^{\alpha}\boldsymbol{=} \frac{1}{2}g^{\alpha k} \left(\frac{\partial g_{k\nu}}{\partial p^{\,\beta }}\boldsymbol{+}\frac{\partial g_{\beta k}}{\partial p^{\,\nu}}\boldsymbol{-}\frac{\partial g_{\beta \nu}}{\partial p^{\,k}}\right)\boldsymbol{=}0 
\tag{10}\label{10}  
\end{equation}
and from equation \eqref{03}
\begin{equation}
\mathrm dV^\alpha \boldsymbol{=} 0  \qquad \left(\alpha \boldsymbol{=}1,2\right)
\tag{11}\label{11} 
\end{equation}
A: Location $(x,y,z)$ isn't a vector in most spaces. $(dx/dt,dy/dt,dz/dt)$ is a contravariant vector.
$dp = (dp/dx) dx + (dp/dy) dy$
I believe the basis vectors transform "opposite" to that.
$e_p = (dx/dp) e_x + (dy/dp) e_y =(dx/dp) \hat{x} + (dy/dp) \hat{y}$
As they said, these are all constants so the derivatives are going to be zero. The basis vectors are constant.
You should try a more interesting transformation. Your christoffel symbols will be zero and the parallel transformed vector will be identical to the original vector.
