# Using Geodesics to show that every direction is an eigenvector of Shape Operator

Suppose that every geodesic of a connected surface $M \subset \mathbb{R}^3$ has positive curvature $\kappa > 0$ and is contained in a plane.

a. Prove that every direction $T_p M$ is an eigenvector of for the shape operator $S_p$, for all $p \in M$
b. Prove that $M$ is contained in a sphere.

My starting work is that I know $\alpha$ is geodesic iff $\alpha''(s)$ is colinear to $U(\alpha(s))$ for each $s$ in the domain of $\alpha$ and $U$ is the unit normal to $M$. This then implies that $\alpha''(s) \cdot \alpha'(s) = 0$ and $\alpha''(s) \cdot U'(\alpha'(s)) =0$ as $\alpha'(s), \ U'(\alpha'(s)) \in T_{\alpha(s)} M$. Finally, (I think) $\alpha$ contained in a plane implies that $U(\alpha(s))$ is constant. I would then jump to say that $U'(\alpha'(s)) = 0$. Is this a correct path to take?

For a., I know that I want to show $k_1 = k_2$, where $k_1$ and $k_2$ are the principal directions of $M$ and are the eigenvalues of $S_p$. Is this the

Geodesics have an accelerator (curvature) vector perpendicular to the surface. So the plane must be perpendicular to the surface. Otherwise suppose it is not perpendicular at $p$, the intersection curve has a curvature vector at $p$ that is inside the plane thus not perpendicular to the surface, a contradiction.
• If the geodesic is contained in a plane, is the unit normal $U$ on $M$ not constant along the geodesic? – swygerts Jan 5 '17 at 1:48