I would like to give a general expression to the Riemann tensor of a metric
\begin{equation}\text{d}s^2=r^2(-\text{d}t^2+\text{d}\Omega^2_{d-1}).\end{equation}
This metric is actually decomposed in $R\times \mathbb{S}_{d-1}$ and the Riemann tensor of $\mathbb{S}_{d-1}$ satisfies $R_{abcd}=g_{ac}g_{bd}-g_{ad}g_{bc}$. I am looking for a relation of this type. I already know that the Ricci tensor and the Ricci scalar of $R\times \mathbb{S}_{d-1}$ read, respectively, $R_i^j=\frac{d-2}{r^2}\delta_i^j$ and $R=\frac{(d-1)(d-1)}{r^2}$, with $a,b=1,…d-1$ and $i,j=2,…,d-1$
Best Answer
I found that the answer it is actually, simpler than I expected. As the submanifold is maximally symmetric, the Riemann tensor is given by
$$R_{ikjl}=\frac{1}{r^2}(g_{ij}g_{kl}-g_{il}g_{kj}).$$