Riemann normal coordinates

Riemann manifolds

Also known as Gaussian normal coordinates.
It is possible, at least in Riemannian manifolds with the Levi-Civita connection (I'm not sure if this is more general), to find local coordinate systems for every point $P$ , such that the coordinates curves are geodesics. This is done thanks to the exponential map for Riemannian manifolds. In general relativity it corresponds to choosing a local inertial frame (the observer moves along a geodesic...).

In this coordinate system, called Riemann normal coordinates, the Christoffel symbols are all null, and the Riemannian metric tensor corresponds to the identity matrix (keep an eye: only in $P$ ).

This can be extended to a neighborhood if and only if the manifold is flat. See Schuller GR-Parallel transport and curvature.

Related: geodesic polar coordinates.

Manifolds with an affine connection

Schuller GR-connections.
Let $(M, O, A, \nabla)$ be an arbitrary affine manifold and let $p \in M$ . Then one can construct a chart $(U, x) \in A$ with $p \in U$ such that

Γ_{(x) (j k)}^{i} (p) = 0.

This says that we can make the symmetric part of the $Γ$ s vanish at the point $p \in M$ , not that we can necessarily make them vanish in some neighborhood of $p$ . Most connections are symmetric (torsion-free), so we can "flatten the connection" in a particular point...

Proof. Let $(V, y) \in A$ be any chart with $p \in V$ . Thus, in general, the $Γ_{(y) (j k)}^{i} \neq 0$ . Then consider a new chart $(U, x)$ to which one transits by virtue of

(x \circ y^{- 1}) (α^{1}, \dots, α^{d}) := α^{i} - \frac{1}{2} α^{j} α^{k} Γ_{(y) (j k)}^{i} (p) .

Then

{(\begin{matrix} \frac{\partial x^{i}}{\partial y^{j}} \end{matrix})}_{p} := \partial_{j} (x^{i} \circ y^{- 1}) |_{(α^{1}, \dots, α^{p})} = δ_{j}^{i} - α^{m} Γ_{(y) (j m)}^{i} (p)

⟹ {(\frac{\partial^{2} x^{i}}{\partial y^{k} \partial y^{j}})}_{p} = - Γ_{(y) (j k)}^{i} (p) .

Taking into account the expression for the change of chart of "the gammas", we can choose, w.l.o.g., the chart $(V, y)$ such that $y (p) = (0, \dots, 0)$ , then we have

Γ_{(x) (j k)}^{i} (p) = Γ_{(y) (j k)}^{i} (p) - Γ_{(y) (j k)}^{i} (p) = Γ_{(y) (j k)}^{i} (p) - Γ_{(y) (j k)}^{i} (p) = 0,

and so we only have an antisymmetric contribution, therefore the symmetric part vanishes.