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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.

GreekCorpse
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  • For starters, the “Gaussian” is wrong. – Ted Shifrin Jun 06 '21 at 17:00
  • Why or where is it wrong? – GreekCorpse Jun 06 '21 at 17:28
  • 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. – Ted Shifrin Jun 06 '21 at 19:26
  • 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. – GreekCorpse Jun 06 '21 at 19:47
  • 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. – John Hughes Jun 06 '21 at 19:51
  • I disagree with that computation. The first term of your cross product does not disappear, does it? ($(N\times \gamma')\times \gamma'$ is just $-N$, isn't it?) – Ted Shifrin Jun 06 '21 at 19:52
  • @JohnHughes I think that's a red herring. The ansatz is certainly that $k\ne 0$, so the Frenet and Darboux frames are well-defined. – Ted Shifrin Jun 06 '21 at 19:53
  • @TedShifrin I didn't say that the first term vanishes. $\omega$ refered to the tangential component of the binormal, so I only took the side that had to do with the tangent. My apologies if it wasn't clear enough. Thanks for the reply, though! – GreekCorpse Jun 06 '21 at 21:21
  • Oh right, of course. I did it a slicker way which I wouldn't expect a student to do. :) I used a determinant formula for $(A\times B)\cdot (A\times C)$. – Ted Shifrin Jun 06 '21 at 21:42
  • Oh, you expect right hahahaha. But it is cool to have as a tool in the future. – GreekCorpse Jun 06 '21 at 21:47
  • @TedShifrin: I don't know what "ansatz" is (and reading the definition didn't enlighten me much), but I personally think that when you state a theorem, it's best to put in all the hypotheses. It's just possible that this particular OP didn't actually realize that the setup omitted this special (but important!) case, even though it's not at all central to the ideas of the question. – John Hughes Jun 06 '21 at 23:11
  • @JohnHughes Didn't the OP state explicitly that curvature was everywhere nonzero? Before any iff statements were uttered? Come on. – Ted Shifrin Jun 06 '21 at 23:12
  • Yes ... sort of. I was misled by "Gaussian curvature not zero" thinking OP was talking about the surface, which could then have been a hyperboloid or other ruled surface with nonzero Gaussian curvature. I suppose a more charitable reading (and more careful reading of the comment chain) would have been a Good Thing. – John Hughes Jun 06 '21 at 23:29

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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.

Ted Shifrin
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  • 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? – GreekCorpse Jun 06 '21 at 21:30
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    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. – Ted Shifrin Jun 06 '21 at 21:37
  • That's right! I only looked at the implication to the right. Thanks again! – GreekCorpse Jun 06 '21 at 21:42
  • 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. – Ted Shifrin Jun 06 '21 at 21:45
  • Sure, it might be helpful for my finals. Thanks again! – GreekCorpse Jun 06 '21 at 21:48