Constant orthonormal frame

Question

I'm reading a geometry article, and some doubts arose, and I hope someone can help me. At a certain point in the work, the author says: Choosing a point wise constant local orthonormal frame $\{e_1, ....., e_n\}$ on the Riemannian manifold $(M, g)$ we have...

What does that mean? It would be something like $\nabla_X e_i=0$ for all index $i$ and $X$ in $M$? Given any Riemannian manifold, there is always this kind of frame?

Thanks!

Answer 1

What is true is the following:

If $M$ is a Riemannian manifold with Levi-Civita connection $\nabla$, then for all $p\in M$ there is a local orthonormal frame $e_1,\dots,e_n$ with $\nabla_{X_p}e_i=0$ for all $X_p\in T_pM$.

Take for example an orthonormal basis $e_1(p),\dots,e_n(p)$ of $T_pM$ and parallel transport it in a normal neighborhood of $p$ along radial geodesics. The frame will stay orthonormal since $\nabla$ is a metric connection and $\nabla_{X}e_i=0$ for all $X\in T_pM$ is true by construction. The crucial point maybe is smoothness of the frame, this is explained here.

The condition that for each point, there is a local parallel frame (i.e. with vanishing derivative everywhere in this neighboorhood) is equivalent to $M$ having zero curvature, $R=0$ and is not true in general.