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When you solve the vacuum field equations of gravity $ R_{ab} = 0$ you come to the conclusion that the spacetime is flat in terms of scalar curvature when you have chosen a suitable metric tensor. However, if you calculate the Kretschmann scalar invariant this is not necessarily zero which indicates your chosen manifold is not flat...

Example - Schwarzschild solution

The Schwarzschild solution in given by

$$(ds)^2 = A(r) dt^2 - dr^2/A(r) - r^2d\Omega^2, $$

where $d\Omega^2$ is the usual line element of the two-sphere. For $ A(r) = 1 + C/r $ the Einstein tensor field equations are satisfied i.e. the spacetime is flat. However, the Kretschmann scalar $K = R_{abcd} R^{abcd}$ goes like

$$ K \propto 1/r^6. $$

Question

What is the difference between Kretschmann and Ricci scalar curvature in laymans terms?

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The Schwarzschild metric is constructed assuming a spherically symmetric spacetime, consequent to a static spherical mass. It is a vacuum solution of the EFE (Einstein field equations), but constrained by the spherical symmetry.
Therefore the Schwarzschild spacetime is not flat, even if the Ricci curvature tensor $R_{\alpha \beta}$ and the Ricci curvature scalar $R$ are zero.
A manifold is flat if the Riemann curvature tensor $R^\alpha_{\beta \gamma \delta}$ vanishes everywhere, which is not in Schwarzschild. If the Riemann tensor is zero, the Ricci tensor and scalar are zero as well, but the opposite does not necessarily hold.

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