# Recovering a frame field from its connection forms

1. Apr 6, 2014

### underflow

Hi,

I have a faced a research problem where I would need to recover a frame field given its connection forms. More precisely, I begin with an orthonormal frame field (given by data) in $\Re^3$ written as
$$\mathbf F=\begin{pmatrix}\vec f_1\\\vec f_2\\\vec f_3\end{pmatrix}$$
where $\vec f_i:\Re^3\rightarrow\Re^3$ are vectors fields with $\vec f_i\cdot\vec f_j=\delta_{ij}$, and $\delta_{ij}$ is the Kronecker delta. I then obtain the connection forms
$$\omega_{ij}=\text{d}\vec f_i\cdot\vec f_j,$$
which yields
$$\text{d}\mathbf F=\Omega\mathbf F$$
where $\Omega=[\omega_{ij}]\in\Re^{3\times3}$ is the skew-symmetric matrix of connection forms.

In my application, I then proceed by computing the interior product of the 1-forms $\omega_{ij}$ onto the frame fields themselves (the Christoffel symbols), i.e., I compute
$$\omega_{ijk}\equiv\omega_{ij}\langle\vec f_k\rangle\in\Re$$
where $\langle\cdot\rangle$ denotes the standard interior product for forms. I thus obtain 9 different measurements at each point, i.e., $\omega_{121},\omega_{122},\omega_{123},\omega_{131},\omega_{132},\omega_{133},\omega_{231},\omega_{232},\omega_{232}$, each of which has a very precise meaning in the application at hand.

Now, I would like to do the converse, i.e., **I would like to solve for $\mathbf F$ given the list of interior contractions** $\omega_{121},\omega_{122},\omega_{123},\omega_{131},\omega_{132},\omega_{133},\omega_{231},\omega_{232},\omega_{232}$.

The Frobenius theorem states the unique existence of $\mathbf F$ in the neighborhood of $0$ if we set $\mathbf F(0)=I$ and if the following are satisfied:
$$\Omega=\text{d}(F)F^{-1}$$
$$\text{d}\Omega-\Omega^2=0.$$
but I'm unsure about where to start. Would anybody have a suggestion on how to approach this problem? Is there a formal name to what I'm trying to do?

Perhaps a better way to formulate the problem is to enumerate the frame axis differentials directly:
\begin{align}
\text{d}\vec f_1&=\omega_{12}\vec f_2+\omega_{13}\vec f_3\\
\text{d}\vec f_2&=-\omega_{12}\vec f_1+\omega_{23}\vec f_3\\
\text{d}\vec f_3&=-\omega_{13}\vec f_1-\omega_{23}\vec f_2.
\end{align}

Writing the direction of contraction as
$$\vec v=\sum_i^3(\vec v\cdot\vec f_i)\vec f_i=\sum_i^3v_i\vec f_i$$
we obtain
\begin{aligned}
\text{d}\vec f_1\langle\vec v\rangle&=\left(\sum_i^3v_i\omega_{12i}\right)\vec f_2+\left(\sum_i^3v_i\omega_{13i}\right)\vec f_3\\
\text{d}\vec f_2\langle\vec v\rangle&=-\left(\sum_i^3v_i\omega_{12i}\right)\vec f_1+\left(\sum_i^3v_i\omega_{13i}\right)\vec f_3\\
\text{d}\vec f_3\langle\vec v\rangle&=-\left(\sum_i^3v_i\omega_{13i}\right)\vec f_1-\left(\sum_i^3v_i\omega_{23i}\right)\vec f_2.
\end{aligned}

Which would give us three coupled differential equations to solve...

Thanks in advance for any help!

2. May 4, 2014