Curvature of Einstein manifold Is there a general expression for the Riemann curvature tensor $R_{ijkl}$ on an Einstein manifold (i.e., where $R_{ij} = kg_{ij}$ for some constant $k$)?
 A: There is no formula that expresses the full curvature tensor in terms of the Ricci tensor, if that's what you're looking for.  The Ricci decomposition of the curvature tensor can be written
$$
\operatorname{Riemann} = \operatorname{Weyl} +\frac{1}{n-2} \operatorname{Ricci}  \bigcirc
\kern-2.5ex\wedge  g - \frac{ S}{2(n-1)(n-2)} g \bigcirc
\kern-2.5ex\wedge \ g,
$$
where $\operatorname{Weyl} $ is the Weyl tensor and $\bigcirc
\kern-2.5ex\wedge\ $ is the Kulkarni-Nomizu product. If the metric is Einstein, the last two terms can be combined into 
$$
\operatorname{Riemann} = \operatorname{Weyl} + \frac{ S} {2n(n-1)} g \bigcirc
\kern-2.5ex\wedge \ g.
$$
But the Weyl tensor can be any trace-free $4$-tensor with the symmetries of the curvature tensor (at a given point, at least).  So unless you have some additional information about the Weyl tensor, knowing that the metric is Einstein doesn't tell you very much about its full curvature tensor.
EDIT: I just realized I should have added a remark that dimension 3 is different: The Weyl tensor is identically zero in that case, so when $g$ is Einstein, the metric has constant sectional curvature, and the entire Riemann curvature tensor is determined by the scalar curvature:
$$
\operatorname{Riemann} = \frac{ S} {12} g \bigcirc
\kern-2.5ex\wedge \ g.
$$
(And, of course, dimension 2 is different too, because the entire curvature tensor is determined by the Gaussian curvature in that case.)
