When is the dual coframe of an orthonormal frame closed?

Question

Let $E_1,E_2$ be a smooth orthonormal frame of vector fields on a simply connected open domain $U \subseteq\mathbb{R}^2$.

Let $E^i$ be the associated dual coframe of one-forms, and suppose that $dE^i=0$.

It is true that the $E_i$ are constants? (do not change from point to point).

I know that $E^i=du_i$ or equivalently that $E_i=\nabla u_i$ where the $u_i$ are smooth functions on $U$, but I am not sure how to proceed from here.

Answer 1

Consider the map $u:U\to\mathbb{R}^2$ whose coordinates are the $u_i$. The assumption that the $E_i$ are orthonormal says that $u$ is a local isometry. But an isometry between two open subsets of Euclidean space must be given by a rigid motion, and the connectedness of $U$ then implies that $u$ is globally the restriction of some rigid motion to $U$. In particular $u$ is an affine map, and so its derivative is constant.

(Note that the assumption that $U$ is simply connected is not necessary--you can just apply this argument locally to conclude that the $E_i$ are locally constant and hence constant if $U$ is connected.)

Answer 2

The coframe $1$-forms are both closed if and only if the connection form is $0$. If the Riemannian metric is flat (has zero curvature), you can choose an orthonormal frame so that this will be the case, but it need not be. (Consider the case of polar coordinates on an appropriate open subset of $\Bbb R^2-\{0\}$.)