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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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up vote 7 down vote accepted

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, 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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Why is the following not a proof that the Lie brackets are trivial?: $[X_i,X_j]=\nabla_{X_i}X_j-\nabla_{X_j}X_i$, with $\nabla_{X_i}X_j=0$ for all $i,j$. – Zach Blumenstein Mar 18 at 6:13
@ZachBlumenstein For $X_i,X_j$ two vector fields in an orthonormal frame, in general $\nabla_{X_i}X_j \neq 0$. Integrability, orthonormality, local framing: choose two. – Neal May 9 at 2:11

Another way to think about this:

1.) if the manifold is flat then you can find coordinate vector fields which are locally orthogonal.

2.) if the manifold is not flat then you can find non-coordinate fields which are locally orthogonal. In physics, in general relativity, these non-coordinate vector fields are sometimes taken to be locally orthogonal to the minkowski metric. The curvature of spacetime is then encoded in the details of how this non-coordinate frame varies from point to point.

Some of the formulas you have at your disposal are likely tacitly assuming a coordinate basis. If you forget that and assume the non-coordinate basis is appropriate for such formulas then you get contradictions like you mention.

A vaguely similar problem: we cannot find a set of orthonormal eigenvectors for a non-symmetric real matrix. Even if the matrix is diagonalizable the eigenbasis will not remain an eigenbasis after it is put through Gram-Schmidt.

To summarize: one way to understand the other answers thus far given is that they serve to show you why the Gram-Schmidt process need not produce coordinate vector fields.

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