Is it always possible to always choose coordinates so that the curvature is locally zero?

Question

I would have thought that this was completely possible as manifolds are so "soft" and the only problems would have been global ones (like Gauss Bonnett etc). But I've never seen the phrase "of course it is always possible to find coordinites in an some open neighbourhood around a point so that the curvature is zero". Also people seem to be very excited about normal coordinites as they have vanishing curvature at a single point, so now I am starting to suspect perhaps it is NOT true. But maybe it is and it's just not very useful so nobody cares. Thoughts?

Edit: I initially thought that vanishing christoffel symbols implied vanishing curvature but this is not true at all. As pointed out in comments it is the christoffel symbols that vanish NOT the curvature.

Answer 1

No, this is not possible. The curvature (tensor and it's derived quantities) are either intrinsic geometric quantities, i.e. they are preserved by coordinate changes, or extrinsic ones (depend on an embedding into an ambient manifold), but here the same statement holds.

Answer 2

The "softness" of the manifolds guarantees that given a smooth manifold $M$ and a point $p \in M$, you can always choose a Riemannian metric $g$ such that $p$ has a flat open neighborhood (and in particular, with respect to any choice of coordinates around $p$, the curvature tensor will be zero). This can't always be done globally just like you said because of topological restrictions.

However, like Thomas wrote, if you start with a Riemannian manifold $(M,g)$ and hold the metric fixed, you can't generally choose local coordinate systems with respect to which the curvature tensor is zero.