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I'm trying to fill in some details in Do Carmo's curves and surfaces proof on p.149,

If all points of a connected surface $S$ are umbilical points, then $S$ is either contained in a sphere or in a plane.

also discussed here. We have proven the local version, i.e., for every point in the surface $S$ there is a connected coordinate neighborhood $V_p$ such that every point in $V_p$ is contained in either a plane or a sphere. I am trying to understand how to conclude the global version, i.e., that all points in $S$ belong to the same plane or sphere. The author first proves that we can connect each point in $S$ to any other point $r$ with a curve $\alpha$, and that we can cover the curve with a finite number of $V_p$ neighborhoods satisfying the above. Then he just states that

If the points of one of these neighborhoods are on a plane, all the others will be on the same plane. Since $r$ is arbitrary, all the points of $S$ belong to this plane. And similarly, if the points of one of these neighborhoods are on a sphere.

I can intuitively see that should be the case, but I am not sure how to prove it. I think we have to rely on overlaps between pairs of neighborhoods (how to prove none of them have $\emptyset$ overlap?), and perhaps on some fact about sufficient conditions for overlap of some open subset of a plane with another plane implying that they are the same plane (and same for a sphere). But I am not sure I am going the right way, or if I am overcomplicating this.

MathPhysForFun89
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If $N$ is the surface normal, then in the planar case we know $N=N_0$ is constant. Now consider the quantity $N_0\cdot x$ (where $x$ is the position vector of the surface). Take a curve $\alpha(t)$ in the surface joining any two points (here we use connectedness). Then $(N_0\cdot\alpha)’(t)=N_0\cdot\alpha’(t)=0$ for all $t$. Then $N_0\cdot x = N_0\cdot\alpha(t_0)=c$ for every $x$ in the surface. This says the surface is contained in a single plane.

The spherical case is analogous.

Ted Shifrin
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  • Apologies if I’m missing something obvious, but how do we know that N=$N_0$ everywhere? I.e., that the same normal applies outside of the individual coordinate neighborhood? I think that’s the point of my question. Aren’t you assuming that in your answer? – MathPhysForFun89 Mar 14 '24 at 05:12
  • If the principal curvatures are $0$ everywhere, then the derivative of $N$ is zero along any curve. As before, connectedness tells us that $N$ is constant on the surface. No mention of coordinate neighborhoods (other than to define principal curvatures, but truly those are defined independent of local coordinates. – Ted Shifrin Mar 14 '24 at 05:24
  • At this point in the proof we haven’t proved that the principal curvatures are 0 everywhere. We know that every point is umbilical, and we know that each point has a neighborhood such that within the neighborhood the principal curvatures are constant. But I think proving that there’s no going from planes to spheres is also still to be proved. – MathPhysForFun89 Mar 14 '24 at 13:16