Kaluza-Klein Christoffel Symbols I have a question regarding the connection coefficients as they pertain to the following paper: http://www.weylmann.com/kaluza.pdf . When I try to calculate the 4D Christoffel symbols from the 4D part of the 5D metric $\tilde{g}_{\mu \nu}=g_{\mu \nu}+kA_\mu A_\nu$ I get an extra term that shouldn't be there. I get:
$$\tilde{\Gamma}^\lambda_{\mu \nu }=\frac{1}{2}\tilde{g}^{\lambda \sigma}(\partial_\mu \tilde{g}_{\nu \sigma}+\partial_\nu \tilde{g}_{\mu \sigma}-\partial_\sigma \tilde{g}_{\mu \nu})$$
$$=\Gamma^\lambda_{\mu \nu }+\frac{k}{2} g^{\lambda \sigma}(\partial_\mu (A_\nu A_\sigma)+\partial_\nu (A_\mu A_\sigma)-\partial_\sigma (A_\mu A_\nu))$$
$$=\Gamma^\lambda_{\mu \nu }+\frac{k}{2} g^{\lambda \sigma}(A_\mu (\partial_\nu A_\sigma -\partial_\sigma A_\nu)+A_\nu (\partial_\mu A_\sigma -\partial_\sigma A_\mu)+A_\sigma (\partial_\mu A_\nu +\partial_\nu A_\mu ))$$
$$=\Gamma^\lambda_{\mu \nu }+\frac{k}{2} g^{\lambda \sigma}(A_\mu F_{\nu \sigma }+A_\nu F_{\mu \sigma}+A_\sigma (\partial_\mu A_\nu +\partial_\nu A_\mu ))$$
According to the paper the $A_\sigma$ term shouldn't be there, but I can't figure out how to make it go away. Anybody have any ideas?
 A: You missed a term in expanding the upper-indexed metric. The full version is below:
\begin{align}
\tilde{\Gamma}^\lambda_{\mu\nu} & = \frac{1}{2} \tilde{g}^{\lambda X} \left(\partial_\mu \tilde{g}_{\nu X} + \partial_\nu \tilde{g}_{\mu X} - \partial_X \tilde{g}_{\mu\nu}\right) \\
& =\frac{1}{2} \tilde{g}^{\lambda\sigma} \left(\partial_\mu \tilde{g}_{\nu\sigma} + \partial_\nu \tilde{g}_{\mu\sigma} - \partial_\sigma \tilde{g}_{\mu\nu}\right) + \frac{1}{2} \tilde{g}^{\lambda5} \left(\partial_\mu \tilde{g}_{\nu5} + \partial_\nu \tilde{g}_{\nu5} - \partial_5 \tilde{g}_{\mu\nu}\right) \\
& = \frac{1}{2} g^{\lambda\sigma} \left(\partial_\mu \left(g_{\nu\sigma} + k A_\nu A_\sigma\right) + \partial_\nu \left(g_{\mu\sigma} + k A_\mu A_\sigma\right) - \partial_\sigma \left(g_{\mu\nu} + k A_\mu A_\nu\right)\right) \\
& \quad \qquad + \frac{1}{2} \left(-A^\lambda\right) \left(\partial_\mu \left(k A_\nu\right) + \partial_\nu \left(k A_\mu\right) - \partial_5 \left(g_{\mu\nu} + k A_\mu A_\nu\right)\right) \\
& = \frac{1}{2} g^{\lambda\sigma} \left(\partial_\mu g_{\nu\sigma} + \partial_\nu g_{\mu\sigma} - \partial_\sigma g_{\mu\nu}\right) + \frac{k}{2} g^{\lambda\sigma} \left(\partial_\mu \left(A_\nu A_\sigma\right) \partial_\nu \left(A_\mu A_\sigma\right) - \partial_\sigma \left(A_\mu A_\nu\right)\right) \\
& \quad \qquad - \frac{k}{2} A^\lambda \left(\partial_\mu A_\nu + \partial_\nu A_\mu\right) \\
& = \Gamma^\lambda_{\mu\nu} + \frac{k}{2} g^{\lambda\sigma} \left(A_\mu \left(\partial_\nu A_\sigma - \partial_\sigma A_\nu\right) + A_\nu \left(\partial_\mu A_\sigma - \partial_\sigma A_\mu\right) + A_\sigma \left(\partial_\mu A_\nu + \partial_\nu A_\mu\right)\right) \\
& \quad \qquad - \frac{k}{2} g^{\lambda\sigma} A_\sigma \left(\partial_\mu A_\nu + \partial_\nu A_\mu\right) \\
& = \Gamma^\lambda_{\mu\nu} + \frac{k}{2} \left(A_\mu F_{\nu\sigma} + A_\nu F_{\mu\sigma}\right).
\end{align}
This is because we have
\begin{cases}
\tilde{g}_{\mu\nu} = g_{\mu\nu} + k A_\mu A_\nu \\
\tilde{g}_{\mu5} = k A_\mu \\
\tilde{g}_{55} = k
\end{cases}
and also
\begin{cases}
\tilde{g}^{\mu\nu} = g^{\mu\nu} \\
\tilde{g}^{\mu5} = -A_\mu \\
\tilde{g}^{55} = \frac{1}{k} + A_\mu A^\mu.
\end{cases}
A: As I understand in the above answer the coefficient $k$ is taken to be constant. Since I am doing very similar calculations right now, let me just put the results for the full Christoffel symbols, using the metric with all coefficients depending on the base space. I am using my own notations, but everything is defined below.
The metric and its inverse are given by
\begin{equation}
\begin{aligned}
& g^{\mu\nu}=
\begin{bmatrix}
f^{-2}(-1+f^3\omega^p\omega_p) & -f \omega^n \\
- f \omega^m  & f h^{mn}
\end{bmatrix}, && 
g_{\mu\nu}=
\begin{bmatrix}
-f^2 & -f^2 \omega_m\\
-f^2 \omega_m & f^{-1}h_{mn}-f^2 \omega_m\omega_n
\end{bmatrix}
\end{aligned}
\end{equation}
The corresponding vielbein takes the following form
\begin{equation}
\begin{aligned}
\hat{e}^0&=f(dt+\omega),\\
\hat{e}^a&=f^{-1/2}e^a,
\end{aligned}
\end{equation}
where $\hat{e}^A\hat{e}^B\eta_{AB}=ds^2$ and $e_m^{\bar{m}}e_n^{\bar{n}}\eta_{\bar{m}\bar{n}}=h_{mn}$.
All small Latin indices are raised and lowered by the space metric $h_{mn}$. The Christoffel symbols $\hat{\Gamma}_{\mu\nu}{}^\rho$ of the metric $g_{\mu\nu}$ then read
\begin{equation}
\begin{aligned}
\hat{\Gamma}_{00}{}^\mu&=fg^{\mu m}\partial_mf,\\
\hat{\Gamma}_{0n}{}^0&=f^{-1}\partial_n f-f^2\omega_n \omega^m\partial_m f-\frac12f^3\omega^mF_{mn},\\
\hat{\Gamma}_{0m}{}^k&=f^2\partial^kf \omega_m+\frac12f^3 F^k{}_{m},\\
\hat{\Gamma}_{mn}{}^0&=\nabla_{(n}\omega_{n)}+3f^{-1}\omega_{(m}\partial_{n)}f-\frac12f^{-1}\partial_\omega f h_{mn}+f^3\omega^k\omega_{(m}F_{n)k}-f^2\partial_\omega f \omega_m\omega_n,\\
\hat{\Gamma}_{mn}{}^k&=\Gamma_{mn}{}^k+f^3F^k{}_{(m}\omega_{n)}+f^2\partial^kf\omega_m\omega_n-f^{-1}\partial_{(m}f\delta^k_{n)}+\frac12f^{-1}\partial^kfh_{mn},
\end{aligned}
\end{equation}
where we define as usual the field strength $F_{mn}=2\partial_{[m}\omega_{n]}$ and denote $\omega^k\partial_k=:\partial_\omega$. Note the $g^{\mu m}$ in the first line, that actually contains some powers of $f$ as well.
Decomposition of the spin connection then reads
\begin{equation}
\begin{aligned}
\hat{\omega}^{\bar{m}\bar{n}}&=\omega^{\bar{m}\bar{n}}-f^{-1}\partial^{[\bar{m}}fe^{\bar{n}]}-\frac12f^3 F^{\bar{m}\bar{n}}(dt+\omega),\\
\hat{\omega}^{\bar{m}\bar{0}}&=f^{\frac12}\partial^{\bar{m}}f \omega+\frac{1}{2}f^{\frac 32}F^{\bar{m}}{}_{\bar{n}}e^{\bar{n}},
\end{aligned}
\end{equation}
where $e^{\bar{m}}=e^{\bar{m}}_mdx^m$ is the  vielbein generating the metric $h_{mn}$. My definition (in all dimensions) for the spin connection is:
\begin{equation}
\omega_m{}^{\bar{m}\bar{n}}=-\Gamma_{mn}{}^ke^{n\bar{n}}e_k^{\bar{m}}+e^{k\bar{n}}\partial_m e_k^{\bar{m}}.
\end{equation}
This follows from the vielbein postulate 
\begin{equation}
0=\nabla_m[\Gamma,\omega]e^{\bar{m}}_n=\partial_m e^{\bar{m}}_n-\Gamma_{mn}{}^ke^{\bar{m}}_k- \omega_m{}^{\bar{m}\bar{n}}e_{\bar{m}n}
\end{equation}
