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Recovering a frame field from its connection forms

  1. Apr 6, 2014 #1

    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 [itex]\Re^3[/itex] written as
    \mathbf F=\begin{pmatrix}\vec f_1\\\vec f_2\\\vec f_3\end{pmatrix}
    where [itex]\vec f_i:\Re^3\rightarrow\Re^3[/itex] are vectors fields with [itex]\vec f_i\cdot\vec f_j=\delta_{ij}[/itex], and [itex]\delta_{ij}[/itex] 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 [itex]\Omega=[\omega_{ij}]\in\Re^{3\times3}[/itex] is the skew-symmetric matrix of connection forms.

    In my application, I then proceed by computing the interior product of the 1-forms [itex]\omega_{ij}[/itex] 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 [itex]\langle\cdot\rangle[/itex] denotes the standard interior product for forms. I thus obtain 9 different measurements at each point, i.e., [itex]\omega_{121},\omega_{122},\omega_{123},\omega_{131},\omega_{132},\omega_{133},\omega_{231},\omega_{232},\omega_{232}[/itex], 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 [itex]\mathbf F[/itex] given the list of interior contractions** [itex]\omega_{121},\omega_{122},\omega_{123},\omega_{131},\omega_{132},\omega_{133},\omega_{231},\omega_{232},\omega_{232}[/itex].

    The Frobenius theorem states the unique existence of [itex]\mathbf F[/itex] in the neighborhood of [itex]0[/itex] if we set [itex]\mathbf F(0)=I[/itex] and if the following are satisfied:
    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:
    \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.

    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
    \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.

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

    Thanks in advance for any help!
  2. jcsd
  3. May 4, 2014 #2
    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
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