Let $M$ be a smooth manifold. For each coordinate chart $(U,\varphi)$ one can form the basis $\partial_1,…,\partial_n$ which are local vector fields defined on $U$. For fixed $x$ this forms a basis of $T_xM$ and $\partial_i|_x$ is obtained as the image of $e_i=(0,…,0,1,0,…,0)$ by the mapping which establishes the isomorphism beetween $\mathbb{R}^n$ and $T_xM$. More concretely, $\partial_i|_x$ is the equivalence class of the curve $\gamma$ where $\gamma(0)=\varphi^{-1}(\varphi(x)+te_i)$.
So far we haven't used the Riemannian structure. Les us now suppose that $M$ is Riemannian manifold.
I have some general questions about the orthogonality of these canonical basis and the domain over which it may hold:
1. In general $\partial_1,…,\partial_n$ need not to be orthogonal. However one can perform Gramm-Schmidt orthogonalization procedure to obtain local orthonormal frame $E_1,…,E_n$. The domain of this frame would be the same as for $\partial_i$'s. Therefore one can consider local orthonormal frames on the domain of each coordinate chart.
When is it possible to choose $E_i$'s in such a wat that the commutator $[E_i,E_j]$ vanish?
And from the other side
- Can we assume without the loss of generality that in a given point $x \in M$ the local frame $\partial_1|_x,…,\partial_n|_x$ is orthogonal?
Best Answer
By the Frobenius theorem, if $[E_i,E_j]$ vanishes for every $i,j$, then there is a parametrization $\varphi$ whose induced $\partial_i$'s are the $E_i$'s. It then follows that $\varphi$ is an isometry, and the curvature must vanish all over the range of $\varphi$.
Yes, just take any parametrization and compose it with an appropriate linear transformation of $\mathbb{R}^n$.