The Christoffel symbol of a maximally symmetric space What is the form of the Christoffel symbol associated with the curvature tensor of a maximally symmetric space given by $$R_{\mu\nu\alpha\beta}=\frac{R}{d(d-1)}\left(g_{\mu\alpha}g_{\nu\beta}-g_{\mu\beta}g_{\nu\alpha}\right)~?$$
 A: For completion I calculated the Christoffel symbols from Baptiste Bermond's answer using Mathematica.
$$\left.
\begin{array}{cc}
 \Gamma ^t{}_{r,r} & \frac{a(t) a'(t)}{1-k r^2} \\
 \Gamma ^t{}_{\theta ,\theta } & r^2 a(t) a'(t) \\
 \Gamma ^t{}_{\phi ,\phi } & r^2 a(t) \sin ^2(\theta ) a'(t) \\
 \Gamma ^r{}_{t,r} & \frac{a'(t)}{a(t)} \\
 \Gamma ^r{}_{r,r} & \frac{k r}{1-k r^2} \\
 \Gamma ^r{}_{\theta ,\theta } & r \left(k r^2-1\right) \\
 \Gamma ^r{}_{\phi ,\phi } & r \sin ^2(\theta ) \left(k r^2-1\right) \\
 \Gamma ^{\theta }{}_{t,\theta } & \frac{a'(t)}{a(t)} \\
 \Gamma ^{\theta }{}_{r,\theta } & \frac{1}{r} \\
 \Gamma ^{\theta }{}_{\phi ,\phi } & \sin (\theta ) (-\cos (\theta )) \\
 \Gamma ^{\phi }{}_{t,\phi } & \frac{a'(t)}{a(t)} \\
 \Gamma ^{\phi }{}_{r,\phi } & \frac{1}{r} \\
 \Gamma ^{\phi }{}_{\theta ,\phi } & \cot (\theta ) \\
\end{array}
\right.$$
They seem to agree with this website. To calculate these symbols I used the this Mathematica notebook

EDIT: I uploaded this without much thought but after looking some more into it this answer isn't entirely correct. The Robertson-Walker metric has two parts: a time part and a maximally symmetric space part.
$$ ds^2=-dt^2+a^2(t)\gamma_{ij}(u)du_idu_j$$
Here $\gamma_{ij}$ is the maximally symmetric metric for the space part. In this case it is given by
$$ds^2=\frac{dr^2}{1-kr^2}+r^2\left(d\theta^2+\sin^2\theta d\phi^2\right)$$
For this part you can check that it obeys $R_{ijkl}=k(g_{ik}g_{jl}-g_{il}g_{jk})$. The Christoffel symbols are calculated to be 
$$\left.
\begin{array}{cc}
 \Gamma ^r{}_{r,r} & \frac{k r}{1-k r^2} \\
 \Gamma ^r{}_{\theta ,\theta } & r \left(k r^2-1\right) \\
 \Gamma ^r{}_{\phi ,\phi } & r \sin ^2(\theta ) \left(k r^2-1\right) \\
 \Gamma ^{\theta }{}_{r,\theta } & \frac{1}{r} \\
 \Gamma ^{\theta }{}_{\phi ,\phi } & \sin (\theta ) (-\cos (\theta )) \\
 \Gamma ^{\phi }{}_{r,\phi } & \frac{1}{r} \\
 \Gamma ^{\phi }{}_{\theta ,\phi } & \cot (\theta ) \\
\end{array}
\right.$$
In response to your comment: how can the Riemann tensor be a function of the metric itself? Well for the maximally symmetric metric you can calculate the Riemann tensor in two ways. The first way is by calculating the Christoffel symbols and from that construct the Riemann tensor (the painful way). The second way is by using the simple definition, i.e. $R_{ijkl}=k(g_{ik}g_{jl}-g_{il}g_{jk})$. It turns out that the expression you get from the painful way simplifies to the simple definition.
A: Following the paper by Douglas H Laurence, http://www.blazartheory.com/files/notes/grnotes/Maximally_Symmetric_Spaces.pdf ,maximally symmetric spaces correspond (up to coordinates transformation) to the Robertson Walker metric, i.e.
$$ds^2=dt^2-a(t)^2\left(\frac{dr^2}{1-kr^2}+r^2d\Omega^2\right)$$
with $k$ the curvature value which can be equal to -1(hyperbolic geometry), 0(flat geometry), or +1 spheric geometry.
For the christoffel, since these are not tensors, it seems difficult to express them as easily.
