# Ricci Scalar as Curvature

So I understand that the Ricci scalar represents the curvature of the space. Since any manifold can be considered locally flat, is Ricci scalar always zero locally for any manifold? On one hand it being defined as curvature makes sense it should be. But mathematically it involves the second order derivative of metric tensor which should be zero locally unless the space is globally flat. So I am kinds confused.

The locally flatness means that we can choose a coordinate system at $$P$$ such that $$g_{\alpha\beta}(\mathcal{P}) = \eta_{\alpha\beta} = {\rm diag}(-1,1,1,1)$$ and $$g_{\alpha\beta,\gamma}(\mathcal{P}) = 0$$. In this way, the Christoffel symbols vanishes at $$\mathcal{P}$$ and so there is no gravitational acceleration. However, we can not make $$g_{\alpha\beta,\gamma\delta}(\mathcal{P})$$ vanishes generally. Actually, Riemann curvature tensor is a true tensor. If it vanishes in one coordinate system, it will vanish in all coordinate system. In physics, we can eliminate gravitational acceleration, but we can not eliminate the tidal force generated by the inhomogeneity of gravitation field. And this is the difference between gravitation and inertial force.
1. On one hand, locally flat appears here to be a layman's characterization that a tangent space $$T_pM$$ locally approximate a manifold $$M$$ at a point $$p$$.
3. Example: The (idealized) surface of the Earth looks flat to an earthling, but it still have a radius of curvature given by $$R_E\approx 6\times 10^6 m$$, so that that the 2D Ricci scalar (which is twice the Gaussian curvature) is $$2/R_E^2$$. Indeed small, but technically speaking not locally flat :)