Is it possible to define a vector field of non-coordinate basis vectors?

Question

I'm studying non-coordinate basis of (pseudo-)riemannian manifolds and I found the following definition from Nakahara - Geometry, topology and physics: a non-coordinate basis $\{\hat{e}_\alpha\}$ is the frame of basis vectors which is obtained by a $GL(n,\mathbb{R})$-rotation of the coordinate basis $\{e_\mu\}\in T_PM$ preserving the orientation,

$$\hat{e}_\alpha=e_\alpha^{\ \mu}e_\mu$$

with $(e_\alpha^{\ \mu})\in GL(n,\mathbb{R})$ and $\det(e_\alpha^{\ \mu})>0 $. We also require that $\{\hat{e}_\alpha\}$ be orthonormal with respect to the metric tensor

$$g(\hat{e}_\alpha,\hat{e}_\beta)=e_\alpha^{\ \mu}e_\beta^{\ \nu}g_{\mu\nu}=\delta_{\alpha\beta}$$

So this means that for each point $P\in M$ I can find a non-coordinate basis, but I was wondering if it is possible to build a vector field with all the non-coordinate bases. I think that the answer should be no, unless the manifold is Euclidean. My problem is that I (think I) know that this is the answer because I know that a generic manifold is Euclidean locally and (of course) not globally, so if this vector field exists, this means that the manifold is Euclidean, because in this case I have found a global basis in which the metric is Euclidean. But I don't know how to prove it.

---

I tried to dissect this topic with some examples:

1. $M=\mathbb{R}^2$ with Euclidean metric $\delta_{\mu\nu}$ and the coordinate basis $\{\partial_\mu\}$ $\Rightarrow$ in polar coordinate I have another coordinate basis (I also give a look at this question) $$x=r\cos \phi \ \ \ \ y=r\sin\phi$$ $$\partial_r=\cos \phi\partial_x+\sin\phi\partial_y \ \ \ \ \partial_\phi=-r\sin\phi\partial_x+r\cos\phi\partial_y$$ and from this basis, I can find a non-coordinate basis $$\hat{e}_r=\partial_r \ \ \ \ \hat{e}_\phi=\frac{1}{r}\partial_\phi$$ The metric tensor in polar coordinates is $g_{\mu\nu}=\operatorname{diag}(1,r^2)$, so it's degenerate for $r=0$ and indeed the non-coordinate basis is not well defined in that point. This is a Euclidean manifold, so in this case should the Euclidean non-coordinate basis be global (except for $r=0$)? How do I see this?

2. So I did the exact same thing with a non-flat manifold for comparison, $M=S^2$ with $r=1$, metric tensor $g_{\mu\nu}=\operatorname{diag}(1,\sin^2\theta)$ and the coordinate basis $\{\partial_\theta, \partial_\phi\}$. I found the non-coordinate basis $$\hat{e}_\theta=\partial_\theta \ \ \ \ \hat{e}_\phi=\frac{1}{\sin\theta}\partial_\phi$$ and also in this case I have that the metric tensor is degenerate when $\sin\theta=0$ and indeed the non-coordinate basis is not well defined in the north and in the south poles. It seems to me quite similar to the previous case, so what all this means? Is it possible that this means that the 2-sphere is Euclidean in all the chart $S^2-\{$north pole, south pole$\}$?

Answer 1

I was wondering if it is possible to build a vector field with all the non-coordinate bases. I think that the answer should be no, unless the manifold is Euclidean.

This is incorrect; you can always (locally) find an orthonormal frame, which is a system of vector fields that forms an orthonormal basis at each point, as you did for $\mathbb S^2 - \{N, S\}.$

The existence of an orthonormal coordinate frame implies the metric is Euclidean (i.e. isometric to $\mathbb R^2$ with the standard metric); but the presence of the word coordinate here is crucial. You seem to have some confusion about the common phrase "manifolds are locally Euclidean" - this is only a topological claim, not a geometric one. Anywhere a metric has non-zero curvature, it is not locally Euclidean in the geometric sense.