# orthonormal vector fields

In a Riemannian manifold $(M^n,g)$, is it true that given a point $p$ we can find $n$ vector fields $X_i$ on a neighborhood $U$ of $p$ such that $g(X_i,X_j)(x)=\delta_{ij}$ with $x \in U$? It seems that if I do Gram-Schmidt to the coordinate basis would work, but then the sectional cuvature $K$ will be $0$ and as it is intrinsic, it will imply that the surface (in the case $n=2$) is locally isometric to a plane which obviously is not true in general

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Yes, you can do this. Take an orthonormal frame at a fixed central point and extend it in all directions by parallel transport along geodesics leaving the central point. What you cannot also do is say anything about the pairwise Lie brackets of the resulting fields. Chern did most of his calculations in this fashion.

EDDIITT: I am not seeing a proof anywhere, so...one of the defining relations of the Levi-Civita connection of a Riemannian metric, with smooth vector fields $U,V,W$ is $$U \langle V,W \rangle = \langle \nabla_UV, W\rangle + \langle V, \nabla_U W \rangle.$$ If $U$ is the velocity field of a geodesic, we do get (by a little extension trickery that requires a full lesson) $$\nabla_U U = 0,$$ which is nice. But parallel transport of the two vector fields $V,W$ along the geodesic means precisely that $$\nabla_U V = 0, \; \; \nabla_U W = 0.$$ In particular, $$U \langle V,W \rangle = \langle 0, W\rangle + \langle V, 0 \rangle = 0.$$ The inner products of the fields remain constant. So, if $V=W = e_i$ at the central point is one of the orthonormal basis vectors, we find that $|V| =1$ always. Next, if $V= e_i$ and $W = e_j$ at the central point with $i \neq j,$ we find that $V,W$ remain orthogonal. Finally, all the transported vector fields are defined on a fairly large ball, up to the injectivity radius actually.

Well and good. As I said, no control of the pairwise Lie brackets, as those being $0$ really would require flatness.

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