Can spacetime be curved even in absence of any source? Einstein's equation in absense of any source (i.e., $T_{ab}=0$) $$R_{ab}-\frac{1}{2}g_{ab}R=0$$ has the solution $$R_{ab}=0.$$
But I think $R_{ab}=0$ does not imply that all components of the Riemann-Christoffel curvature tensor $R^c_{dab}$ be zero (or does it?). From this can I conclude that spacetime can be curved even in absence of any source?
 A: What you're asking about is referred to as a vacuum solution to the field equations. This does not mean that there is no mass anywhere, rather that we are considering a region of our curved spacetime in which there is no mass.
The Schwarzschild solution for instance is a "vacuum solution" because we are considering the region outside of the central mass in which there is no matter, but in which the curvature is non-zero.
You are correct that the vanishing of the components of the Ricci tensor does not imply the vanishing of the components of the full Riemann tensor. $R_{\mu\nu}=0$ is a vacuum solution, ${R^\alpha}_{\beta\mu\nu}=0$ is flat spacetime.
A: This is a simple answer:
I would view this in the same light as the following question:
Does
$$ {\bf \nabla \cdot E} = \frac{\rho}{\epsilon_0} $$
imply zero electric field in region with no charge density?
To which the answer is clearly, "No".
And as an example: The astronauts on the moon. They were there in a pretty good vacuum dropping feathers and hammers, which then took off on like geodesics.
A: You are right. $R_{ab}=0$ does not imply $R^{a}_{bcd}=0$. For one thing, $R_{ab}$ has 10 components (in $n=4$ dimensions), whereas $R^{a}_{bcd}$ has $20$ components. The simplest example I can think of is Schwarzschild solution, which has $R_{ab}=0$ everywhere but $R^{a}_{bcd}\neq0$. If you allow the inclusion of a cosmological constant, then the de Sitter metric is an example of an empty solution with non-trivial spacetime curvature. As pointed out here
https://physics.stackexchange.com/a/105336/96768
A spacetime containing gravitational waves is empty but with non-trivial Riemann tensor.
A: That's right. But it doesn't mean that the curvature is from nowhere. The Field equation describes the curvature (locally) at a point only from $T_{\mu \nu}$ at the same point (Since it's all built in a differential manifold and tangent spaces at each points aren't related to each other). If $T_{\mu \nu}$ is zero at a point, then you end up deriving a vacuum solution.
