TITLE: Flat manifold and metric
QUESTION [6 upvotes]: Is "X is a flat Riemannian manifold" equivalent to "for any metric g on X, there is a change of coordinates that can transform g to a tensor with only constants on its diagonal"?

REPLY [6 votes]: From the comments: "I found that in a seemingly reputable astronomy book. "If it is possible to find a frame in which all the components of g are constant, the space is 'flat'"." --Frank
This question makes sense, but it is phrased in a new-user-hostile way. We have a space $(M,g)$ on our hands. Now consider an open neighborhood $U$ of some point and start picking local frames $(\xi_1, \dots, \xi_n)$ of the tangent space $T_M$ on $U$. The statement in the book says that if you can find a frame such that $g(\xi_j,\xi_k)$ are constant functions on $U$ for all $j,k$, then $g$ is flat on $U$.
The condition says that the $\xi_j$ are parallel with respect to $g$, or that $\nabla \xi_j = 0$, where $\nabla$ is the Levi-Civita connection of $g$. This implies that
$$
R(\xi_j,\xi_k) \xi_l = \nabla_{\xi_j} \nabla_{\xi_k} \xi_l - 
\nabla_{\xi_k} \nabla_{\xi_j} \xi_l - \nabla_{[\xi_j,\xi_k]} \xi_l = 0
$$
for all $j, k, l$. Here $R$ is the curvature tensor of $g$. Since $R$ is a tensor on the space of tangent fields and $(\xi_1, \dots, \xi_n)$ is a frame on $U$, this implies that $R = 0$ on $U$, so $g$ is flat.