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In Lee's Introduction to Riemannian Geometry he stats that the existence of local orthonormal frames does not imply the existence of a local orthonormal coordinate frame.

I am struggling to understand this, so I must be confused about some of the definitions here.

Using the 2-sphere as an example, if we have a chart function for the open set, $U$ where $x\gt0$ $$\varphi(x,y,z) = (y, z)$$ $$\varphi^{-1}(y,z) = (1 - \sqrt{y^2+z^2}, y, z)$$

Then we can define our metric, $g$, on this chart using the typical Euclidean metric on $\mathbb{R}^2$. Doesn't that automatically make the coordinate vector fields, $\partial_y$ and $\partial_z$ an orthonormal frame?

I am obvioiusly missing something in the definitions here. Any help would be appreciated in clarifying.

Jeff
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    You have to fix a metric before looking for coordinates with orthonormal coordinate frame, so you're not at liberty to take $g$ to be the Euclidean metric if your sphere is round. On the other hand, those coordinates do not define an orthonormal frame with respect to the round metric. (That's not a complete answer, just a guide to why your proposed counterexample doesn't work.) – Andrew D. Hwang Apr 05 '21 at 18:13
  • I don't quite follow, why I can't define a metric on this particular coordinate chart as I am proposing. Then glue it with other metrics using a partition of unity. – Jeff Apr 05 '21 at 18:21

1 Answers1

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Let $(M,g)$ be a Riemannian manifold. Fix $p \in M$ and choose a smooth local frame around $p$. By the Gram-Schmidt orthonormalization process, one can construct from the previous local frame a local orthonormal frame (by applying pointwisely the process). From the formula of the Gram-Schmidt process, this local orthonormal frame is smooth. Call it $(E_1,\ldots,E_n)$.

However, in order to be an orthonormaal coordinate frame, there must exist a coordinate patch $(x^1,\ldots,x^n)$ such that $E_j = \partial/\partial x^j$. As for a coordinate patch, the partial derivatives commute (that is, $[\partial/\partial x^i , \partial/\partial x^j]=0$, it follows that if $(E_1,\ldots,E_n)$ is an orthonormal coordinate frame, then $[E_i,E_j]=0$, which is a pretty rigid condition.

Moreover, if $(x^1,\ldots,x^n)$ is an orthonormal coordinate frame on $(M,g)$, it follows that the curvature tensor of $(M,g)$ vanishes on the neighbourhood where $(x^1,\ldots,x^n)$ is defined: this is because the metric in these coordinates is simply $(g_{i,j}) = (\delta_{i,j})$, that is constant, and hence, the Christoffel symbols are zero.

To conclude, I think you are misinterpreting what J. Lee is saying in his book. Given a fixed Riemannian manifold, there may not exist any orthonormal coordinate frame around a point. But of course, given any manifold with a coordinate system, one can construct a local metric for which the coordinate system is orthonormal.

Didier
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    I think I am starting to get it. I may be able to construct a global metric using a partition of unity that for some points the metric is locally flat. I just won't be able to construct one that has that property at every point. – Jeff Apr 05 '21 at 19:21
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    Yes. There exist some topological obstructions to the existence of a globally flat metric. For compact (orientable, without boundary) surfaces, the Gauss-Bonnet formula shows that only the torus admits a gobally flat metric. – Didier Apr 05 '21 at 19:33