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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
<br /> \mathbf F=\begin{pmatrix}\vec f_1\\\vec f_2\\\vec f_3\end{pmatrix}<br />
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!
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
<br /> \mathbf F=\begin{pmatrix}\vec f_1\\\vec f_2\\\vec f_3\end{pmatrix}<br />
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!