Normal coordinates and the metric matrix While trying to follow and check the proof of Theorem 1 in this work on manifold averaging I reached the notion of normal coordinates. An important property is that the metric tensor at a point $\mathbf{p}$ is the classic Euclidean one, i.e. 
\begin{equation}
g_{ij}(\mathbf{p}) = \delta_{ij}.
\end{equation}
Where does this apparently trivial/obvious implication come from? 
This answer states that one can diagonalize the metric tensor at a point, but, almost obviously, not over a neighborhood. So, the question is equivalent to what allows one to perform this diagonalization?
 A: Around any point $p$ in a Riemannian manifold $M$ you can find a normal neighbourhood; that is neighbourhoods $\mathcal U$ of $0\in T_pM$  and $\mathcal {\tilde U}$ of $p \in M$ such that $\exp : \mathcal U \to \mathcal{ \tilde U}$ is a diffeomorphism. Riemannian normal coordinates are then obtained by taking Cartesian coordinates on $\mathcal U \subset T_pM \simeq \mathbb R^n$ and composing with $\exp^{-1}$. 
The most intuitive picture is probably geodesic polar coordinates - if a point $x$ is close enough to $p$, we can draw a unique minimizing geodesic from it to $p$. The geodesic polar coordinates $(r,\omega)$ of $x$ are then determined thus: $r \ge 0$ is simply the length of the geodesic drawn, and $\omega \in S^{n-1}$ is the direction in which it strikes $p$ (relative to the Cartesian coordinates chosen on the tangent space). From there we can convert from ((hyper)spherical) polar coordinates back to Cartesian in the usual way to get the normal coordinates.
This gives $g_{ij}(p) = \delta_{ij}$ but more interestingly $\Gamma^k_{ij}(p)=0$. You can find a proof of this (and the existence of normal neighbourhoods) in any text on Riemannian geometry - see e.g. Chapter 3 of O'Neill's Semi-Riemannian Geometry.
Note that if all you want is the first condition $g_{ij}(p) = \delta_{ij}$, you don't need to get nearly as involved - just take the symmetric matrix $g_{ij}$ in whatever coordinates you currently have and diagonalize it, then use the eigenvector matrix you attain to transform your whole coordinate system.
