This is a homework problem from my differential geometry class:
Show that a curve in $\varepsilon^3$ lies on a sphere or plane if and only if the curvatures $\kappa_g$ and $\kappa_n$ of a parallel frame satisfy a linear equation in $\mathbb{R}^2$. How can the radius of the sphere be read from the equation?
$\kappa_g = -\frac{1}{|X'|}(N',T)$ and $\kappa_n = -\frac{1}{|X'|}(T',B)$, parallel frame means $\nabla^TN := N'-(N',T)T=0$.
First I tried $"\implies":$ I know that $N := (X-o)\frac{1}{r}$ (with $X$...curve, $o$...center of the cirle and $r$...radius of the circle) is a unit normal field for a curve on a sphere, which is also parallel. Further it curvatures can be expressed as $\kappa_n = \frac{1}{r}$ and $\kappa_g = \frac{det(X-o,X',X'')}{r}$ which satisfy a linear equation $\kappa_g = \alpha \kappa_n$ by the choice $\alpha = det(X-o,X',X'')$. So far so good. Then any other parallel unit normal field should just be a rotation in the normal plane, which changes the curvatures as follows: $\tilde{\kappa_n} = \kappa_n cos(\phi) - \kappa_g sin(\phi)$ and $\tilde{\kappa_g} = \kappa_n sin(\phi) + \kappa_g cos(\phi)$. But then they no longer satisfy a linear equation as above. Thats the point where I am stuck, maybe somebody could help me?
I actually found the answer to my question already, it is proven in the paper "There is more than one way to frame a curve" by Richard L Bishop. As the paper has just about 10 pages and is freely accessible on his research gate account, I will just share the link here.
https://www.researchgate.net/publication/210222830_There_is_More_than_One_Way_to_Frame_a_Curve