Covariant derivatives and curvature

Let $\nabla$ be a covariant derivative on a manifold $M$ with $T$ and $R$ the torsion and curvature respectively. Suppose that $T=R=0$, could we ensure that the Christoffel symbols are 0 respect to a specific local chart?

In the case of the answer is affirmative, does somebody know a reference to see the proof?

Thank you very much.

• math.stackexchange.com/a/718147/565 – AnonymousCoward Aug 24 '17 at 14:08
• If you look at Mike Spivak's 5-volume differential geometry opus, Volume 2, you'll see that he refers to this as the Test Case. He gives various proofs, depending on your favorite tools. – Ted Shifrin Aug 25 '17 at 0:54
• In this book he only proves the Riemannian case, i.e., $\nabla$ is a Levi-Civita connection. – V.Jiménez Oct 20 '17 at 12:12

Choose a point $p \in M$ and a small neighborhood $U$ of $p$ diffeomorphic to a ball. Since the curvature vanishes, you know that parallel transport doesn't depend on the path in $U$ (see any book which discusses holonomy). Choose some basis $(e_1,\dots,e_n)$ of $T_pM$ and using parallel transport extend it to a local frame $(E_1,\dots,E_n)$ on $U$. The vector fields $E_i$ satisfy $\nabla E_i = 0$. Since $\nabla$ is torsion-free, we have
$$0 = T(E_i, E_j) = \nabla_{E_i} E_j - \nabla_{E_j} E_i - [E_i, E_j] = -[E_i, E_j]$$
so $[E_i,E_j] = 0$ for all $1 \leq i, j \leq n$. By the theorem describing the canonical form for commuting vector fields (Theorem 9.46 in "Introduction to Smooth Manifolds" by Lee), this implies that we can find a coordinate neighborhood around $p$ such that $E_i = \frac{\partial}{\partial x^i}$ in that neighborhood. Since $\nabla_{E_i} E_j = \nabla_{\partial_i}{\partial_j} = 0$, the Christoffel symbols with respect to the chart $(x^1,\dots,x^n)$ will vanish.