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Othonormal moving frame on ${\bf R}^3$ is a set $\{ e_i\}$ such that for any $p\in {\bf R}^3$, $$ e_i(p)\cdot e_j(p) = \delta_{ij}$$

When $M^2\rightarrow {\bf R}^3$, a frame $\{ e_i\}$ is adapted on $U=V\cap M$, $V\subset {\bf R}^3$, if for any $q\in U$, $e_1$ and $e_2$ are tangent to $M$.

In otherwords, trajectories of $e_1$ and $e_2$ are in $M$.

If $M$ is regular we can easily construct adapted frame by using a parametrization of $M$.

Here I have a question : The converse hold ?

For given frame $\{ e_i\}$ on ${\bf R}^3$, given $p\in {\bf R}^3$, and $\{e_1, e_2\}$,

there exists $f : U\subset {\bf R}^2\rightarrow {\bf R}^3$ such that $p\in f(U)$ and $ \{ e_1, e_2\}$ is tangent to $f(U)$.

Thank you in advance.

HK Lee
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    I think that you will have to look at the distribution of two planes spanned by $e_1$ and $e_2$. The Frobenius Theorem should be what you are looking for. – THW Aug 27 '13 at 15:08

1 Answers1

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In the language of differential geometry, your question is whether the distribution generated by $\{e_1, e_2\}$ is integrable. This is not true for general frames, but there is a nice characterisation of those frames it is true for: the Frobenius Theorem.

In this case the Frobenius theorem reduces to the necessary and sufficient condition that the Lie bracket of the two vectors lies in their span:

$$ [e_1, e_2] = D_{e_1} e_2 - D_{e_2} e_1 \in \operatorname{span}\{e_1, e_2\} $$

where $D_u v$ is the derivative of $v$ in the direction $u$.

Anthony Carapetis
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