Christoffel symbol for parabolic coordinates I am wondering why am I getting all the Christoffel symbol for the parabolic co-ordinates 
$$p(x, y) =x \qquad q(x, y) = y - x^2/2$$ ZERO ? The metric for this co-ordinate is $$ g^{\alpha\beta} = \left( \begin{array}{cc}
1 & -p \\
-p & 1 + p^2 \end{array} \right)$$
or
$$ g_{\alpha\beta} = \left( \begin{array}{cc}
1 + p^2 & p \\
p & 1  \end{array} \right).$$
 A: While the OP made a mistake in calculation and not all $\Gamma^\mu_{\alpha\beta}$ are zero, one can ask what are the conditions on the metric for all $\Gamma^\mu_{\alpha\beta}$ to be zero.
In order to answer that question, one must first see up to which equivalence class does connection determine the metric. Well, if the connection $\Gamma^\mu_{\alpha\beta}$ is compatible with the metric $g_{\mu\nu}$, then, by definition, the following differential equation must hold:
\begin{equation}
\nabla_\rho g_{\mu\nu}=0
\end{equation}
This is the first order differential equation, so it has one constant of integration - if you know the metric in one point, then the connection determines it uniquely.
Now, for the case $\Gamma^\mu_{\alpha\beta}=0$,the above equation reduces to
\begin{equation}
\partial_\rho g_{\mu\nu}=0.
\end{equation}
The conclusion is: Connection vanishes if and only if the metric tensor is constant.
Therefore, if you get your Christoffels equal to zero for non-constant metric, then it is easy to see that you made a mistake in your calculation.
