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I'm having some trouble trying to prove the following:

If we consider $\gamma:I\to S$ a differentiable curve parametrized along the length of arc $s\in I$, with its curvature different from 0, and $\omega(s)$ is the tangential component of $b(s)$, where $b$ is the binormal of $\gamma$, then these statements are equivalent:

  • $\gamma$ is a geodesic.
  • $\omega(s)\ne0$ and it is parallel along $\gamma$.

I have computed beforehand $\omega(s) = -\frac{k_n(s)}{k(s)}(N(s)\times\gamma'(s))$, where $N$ is the normal field of the surface S.

In order to get the implication to the left, I tried to see that since $\omega$ is parallel along $\gamma$, then we would have that it only happens if $N\times\gamma'' = 0$, which determines that they are proportional (since $\omega(s)\ne 0$), so $\gamma$ is a geodesic.

For the implication to the right, if $\gamma$ is geodesic, then since $k\ne0$, by $k^2 = k_g^2+k_n^2$ we know that $k_n\ne0$. Since $N\times\gamma'$ is a generator of the Darboux trihedron, $N\times\gamma'$ is not zero. So $\omega(s)\ne0$. I still have to prove that $\omega(s)$ is a parallel along $\gamma$.

Is my reasoning correct, or is there a mistake somewhere? Also, could anyone please help with the rest of the proof, even if it is just a hint? Thanks in advance!

P.S.: I've seen this question around, but this is about the tangential component of the normal. I don't know if the computations there help somehow in here.

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  • $\begingroup$ For starters, the “Gaussian” is wrong. $\endgroup$ Commented Jun 6, 2021 at 17:00
  • $\begingroup$ Why or where is it wrong? $\endgroup$ Commented Jun 6, 2021 at 17:28
  • $\begingroup$ Gaussian curvature is never used in reference to curves. It is used commonly for surfaces and for hypersurfaces in general. OK, I have started to think about your question. Where does the $k$ in the denominator of your formula for $\omega$ come from? I disagree with that. $\endgroup$ Commented Jun 6, 2021 at 19:26
  • $\begingroup$ Oh, I see now the problem. I'll edit it quickly, since I was refering to the curvature of the curve. My apologies! As for where the $k$ comes from, I know that $k(s)n(s) = k_g(s)(N(s)\times\gamma'(s)) + k_n(s)N(s)$, so by the cross-product of both sides of the equality with $\gamma'(s)$, we get that $k(s)\omega(s) = -k_n(s)(N(s)\times\gamma'(s))$, and since $k(s)\ne 0$, then we get to the expression from above. $\endgroup$ Commented Jun 6, 2021 at 19:47
  • $\begingroup$ Here's another difficulty: for a straight line $\gamma$ in a plane $S$ in 3-space, the normal and binormal are either undefined or zero, so that idea that $\omega(s) \ne 0$ is pretty much a non-starter, because $\gamma$ is certainly a geodesic. But Ted's gonna help you get things straightened out, so I can probably drop out of this discussion. $\endgroup$ Commented Jun 6, 2021 at 19:51

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Assume $\gamma$ is a geodesic. Then $k_g=0$ and $\gamma''\times N = 0$, as you commented. On the other hand, if we differentiate $\omega$, we get (omitting all the evaluations at $s$) $$-\omega' = \left(\frac{k_n}k\right)'(N\times\gamma') + \left(\frac{k_n}k\right)(N'\times\gamma' + N\times\gamma'').\tag{$\star$}$$ Since $k_g=0$, we have $k_n/k = \pm1$, and the derivative vanishes. As we said, $N\times\gamma'' = 0$, and so we're left only with the $N'\times\gamma'$ term, which is normal to the surface (why?). Thus, $\omega$ is parallel.

Conversely, if $\omega$ is parallel, the tangential component of $-\omega'$ must vanish. Note that $N\times\gamma'$ and $N\times\gamma''$ will be orthogonal and $N\times\gamma'$ is nonzero. Therefore we must have $k_n/k$ constant and either $k_n = 0$ or $N\times\gamma''=0$. But if $k_n=0$, then $\omega$ vanishes, and so we conclude that $N\times\gamma''=0$, which says precisely that $\gamma$ is a geodesic.

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  • $\begingroup$ Thank you so much for the answer! The idea was almost there, but you gave a much better solution. I imagine that $N'\times\gamma'$ is normal to the surface, since we have that $n'(s) = k t(s) + \tau b(s)$, $n(s)$ is proportional to $N(s)$, and $\gamma' = T$, so $N'\times\gamma' = \tau b(s)\times t(s) = \tau n(s)\propto\tau N(s)$. Is it correct? $\endgroup$ Commented Jun 6, 2021 at 21:30
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    $\begingroup$ This argument presupposes a geodesic, and you need it for both directions. The key fact is that $N'$ is always in the tangent plane (because the derivative of $N$ maps the tangent plane to the tangent plane), so you have the cross-product of two tangent vectors. $\endgroup$ Commented Jun 6, 2021 at 21:37
  • $\begingroup$ That's right! I only looked at the implication to the right. Thanks again! $\endgroup$ Commented Jun 6, 2021 at 21:42
  • $\begingroup$ You're welcome. If you're interested, you might check out my differential geometry text — freely downloadable at the link in my profile. There are a number of unusual and interesting exercises in it. $\endgroup$ Commented Jun 6, 2021 at 21:45
  • $\begingroup$ Sure, it might be helpful for my finals. Thanks again! $\endgroup$ Commented Jun 6, 2021 at 21:48

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