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Let $a,b,c\in\mathbb{R}^3$ with $a\wedge b\wedge c\ne 0$ and $a^2>b^2$.

What is the form of this equation $p(\theta,t)=e^{a\wedge c\theta}(a+be^{a\wedge bt})e^{-a\wedge c\theta}$ in standard vector coordinates, where $\theta,t\in\mathbb{R}$.

I am not really sure at all how to go about converting it. I have read some basics on rotors, but cannot find anything on expressions involving exponentials as rotors.

My ultimate goal is to examine the features of this surface; i.e. whether or not it is compact or regular.

Karambwan
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  • Are you working in 3D and have $a,b,c$ vectors and $t,\theta$ scalars? – Mark S. Apr 21 '19 at 17:49
  • Yes. I will update the post. – Karambwan Apr 21 '19 at 17:54
  • This is fairly ugly/unnatural in general. Are you certain this is the expression you want, and that there are no other conditions? For instance, if $\mathbf a,\mathbf b,\mathbf c$ were orthonormal, then this would almost be the very nice "take the vector in a $\mathbf {ab}$-plane that's at an angle $t$ from $\mathbf a$ to $\mathbf b$ and then rotate it an angle of $-2\theta$ around $\mathbf a\times\mathbf c$. – Mark S. Apr 21 '19 at 18:59
  • The additional constraint is $a\wedge b\wedge c\ne0$, so they are indeed orthogonal. So the rotors are rotating the vector in parentheses by -2$\theta$ around $a\times c$, and the vector in parentheses is $a$ rotated an angle $t$ toward $b$? – Karambwan Apr 21 '19 at 19:07
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    $\mathbf a\wedge\mathbf b\wedge\mathbf c\ne0$ does not imply they're orthogonal, just that they're linearly independent. If you meant for them to be orthogonal, that changes things considerably (but you still have the issue of their lengths). Even if they were orthonormal, the vector in parentheses looks to me like a typo for $\mathbf ae^{\mathbf a\wedge\mathbf bt}$ or $\mathbf a\cos t+\mathbf b\sin t$. I think only with that correction would you get exactly the sort of elegant geometric interpretation I alluded to. – Mark S. Apr 21 '19 at 19:10
  • Ah okay. Must be an error here. The only constraint I've neglected to mention is $a^2>b^2$. – Karambwan Apr 21 '19 at 19:16

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