 Solution

At a chosen event $p$, local inertial coordinates have $g_{ab}(p)={\eta }_{ab}$, ${\partial }_c{g}_{ab}(p)=0$ and ${\Gamma }^a{}_{bc}(p)=0$. The metric agrees with Minkowski space to first order and freely falling trajectories are momentarily straight, expressing the local equivalence principle. Second derivatives generally remain, recording curvature and tidal effects.

Differentiate the Levi-Civita connection. At $p$, the terms from derivatives of the inverse metric multiply first derivatives of $g$ and vanish, givingFor the convention $R^a{}_{bcd}={\partial }_c{\Gamma }^a{}_{bd}-{\partial }_d{\Gamma }^a{}_{bc}+\Gamma \Gamma -\Gamma \Gamma$, the connection products vanish at $p$. Lowering the first index and subtracting these derivatives cancels the ${\partial }_c{\partial }_d{g}_{ab}$ terms, yieldingSymmetry of $g$ and commutation of second derivatives show $R_{abcd}=R_{cdab}$ in these coordinates. Both sides are tensors, so equality in an inertial chart at each event implies equality at any point in every chart.

For the specified conformal metric, let $s=L^{-2}{\eta }_{ef}{x}^e{x}^f$. Its expansion is $g_{ab}={\eta }_{ab}(1-2s+O(|x{|}^4))$. Thus the original coordinates are already inertial at zero, andSubstitution gives $R_{abcd}(0)=4({\eta }_{ac}{\eta }_{bd}-{\eta }_{ad}{\eta }_{bc})/L^2$. In four dimensions the indicated contraction isThe sign is fixed by the Riemann-tensor formula in the question; reversing the Riemann convention would reverse the scalar as well.