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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!

Elismar Dias
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    Manifolds famously don't always have global coordinate systems or "directions", which is why tangent spaces are defined (and why connections are constructed). All the author is saying in more words is choose an orthogonal basis for the tangent space at point $p$, denoted $T_pM$. The "point wise" part is redundant, unless the author is referring to a sequence of tangent spaces, like say on a curve, but this is not clear as the rest of the sentence is cut off. – Ninad Munshi Jan 30 '21 at 05:29
  • Hard to say without seeing the full sentence and maybe a few before and after. – Deane Jan 30 '21 at 16:44
  • @NinadMunshi my doubt is mainly with the word constant in the paper. In case you want more information, I had previously asked a question here: https://math.stackexchange.com/questions/3992388/trace-of-the-riemman-operator – Elismar Dias Jan 31 '21 at 05:06

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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.

Claire
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