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.
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.