Flat manifold and metric 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"?
 A: 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.
A: No. In particular, the second statement is always false. For if $X$ is a smooth manifold, then in local coordinates (diffeomorphic to $\mathbb R^n$, say) on $X$ choose a non-flat metric and then extend it arbitrarily to $X$. Then there does not exist a coordinate change (on the chosen local coordinates) that makes $g$ identically constant.
The correct statement is "$(X,g)$ is a flat Riemannian manifold if and only if about each point $x \in X$ there exists a neighborhood $U \ni x$ and local coordinates on  $U$ in which $g$ is the standard Euclidean metric." (The correct statement is always more of a mouthful, isn't it?)
The point is that saying $X$ is a Riemannian manifold implies that $X$ carries with it a fixed Riemannian metric $g$. You do not get to choose $g$ as in your initial statement.
From the standpoint of definitions, saying that $(X,g)$ is flat means precisely that the Riemann curvature tensor is identically zero, and then from this one proves the statement in my second paragraph (see Petersen's or Do Carmo's Riemannian Geometry books, for instance).
