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How should I show that if every normal section of a surface in $\mathbb{R}^3$ is a geodesic, then for every point of the surface, the curvature is the same in any direction?

I would like some hints. Thank you!

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Sketch: Assume $p=(0,0,0)$ is a point on the surface $S$ and the unit normal vector to $S$ at $p$ is $(0,0,1)$. Take a ball of radius $r>0$ small enough on the tangent plane at $S$ (the $X,Y$-plane). Note that if you go $r$ distance from $p$, along $S$, in any direction, to say $x$, the normal vector at $x$ is in the plane passing through $Z$-axis and $x$ (uniqueness of geodesics given a direction and at a point).

Now consider the curve on $S$ which is the set of all points at Riemannian distance $r$ from $p$. Consider the function which assigns to each point on this curve its euclidean distance from the $X,Y$-plane. Note that the vectors tangent to this curve are perpendicular to planes containing the base point and the $Z$-axis by the observation above. This means that the function considered above is a constant. This is true for any $r$. Check that this gives local symmetry (rotations in the $X,Y$-plane send geodesics from $p$ to geodesics at $p$). The claim follows from this.

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  • $\begingroup$ When considering the curve which is the set of all points at distance $r$ from $p$ (second paragraph), do you mean Euclidean distance? or intrinsic distance? $\endgroup$
    – stoneaa
    Nov 30, 2020 at 6:33
  • $\begingroup$ I made changes to clarify what I wrote. $\endgroup$
    – deb
    Nov 30, 2020 at 8:05

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