I'm looking for an easily-checked, local condition on an $n$-dimensional Riemannian manifold to determine whether small neighborhoods are isometric to neighborhoods in $\mathbb R^n$. For example, for $n=1$, all Riemannian manifolds are modeled on $\mathbb R$. When $n=2$, I believe that it suffices for the scalar curvature to vanish everywhere (this is certainly necessary). But my intuition is poor for higher-dimensional structures.


Put another way: given a Riemannian structure $g$ on a smooth manifold, when can I find coordinates $x^1,\dots,x^n$ so that $g\_{ij}(x) = \delta\_{ij}$?



When is a Riemannian metric equivalent to the flat metric on $\mathbb R^n$?
If the Riemannian metric is twice differentiable in some co-ordinate system, then this holds in any dimension if and only if the Riemann curvature tensor vanishes identically.


In dimension 2, it suffices for the scalar curvature to vanish.
In dimension 3, it suffices for the Ricci curvature to vanish.
In higher dimensions, you need to have the full Riemann curvature vanish.


