$\gamma$ is a geodesic iff $\omega\ne 0$ and it is parallel along $\gamma$?

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.

• For starters, the “Gaussian” is wrong. Commented Jun 6, 2021 at 17:00
• Why or where is it wrong? Commented Jun 6, 2021 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. Commented Jun 6, 2021 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. Commented Jun 6, 2021 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. Commented Jun 6, 2021 at 19:51

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.
• 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? Commented Jun 6, 2021 at 21:30
• 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. Commented Jun 6, 2021 at 21:37