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Is every umbilic connected surface $S$ with $0$ curvature cointained in a plane? I know that the answer is "yes" if we also suppose that the surface is orientable. The argument is sketched below:

It's possible to prove that every connected umbilic surface has constant Gaussian curvature $k$. Suppose $k=0$ and let $N$ be a differentiable normal field on $S$. $N$ is constant since $dN_p=0$ for all $p \in S$ and since $S$ is connected. Let $p \in P$ and let $P=\{p+v: v_p \in T_pS\}$. We will show that $S \subset P$. What we must do is to show that given $q \in S$, $q-p$ belongs to $P$. It suffices to show that $\langle q-p, N\rangle=0$ for all $q \in S$. So let $F=\langle Id-p, N\rangle$, where $Id$ is the identity function on $S$. $F$ is differentiable and it is easy to show that $dF_r=0$ for all $r \in s$, therefore $F$ is constant. Notice that $F(p)=0$, therefore $F$ is constantly $0$, as we wanted.

Notice that the argument above used the existence of a normal vector field on $S$. I believe that we may suppress the word "orientable" in the hypothesis by showing that the other parts of the hypothesis implies that $S$ is orientable, however, I don't know how to proceed.

Ivo Terek
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You have the crucial part of the argument. Fix $s_0\in S$, and consider the set $X = \{s\in S: s\in P\}$. By continuity, $X$ is closed, and by your proof, $X$ is open (there is a continuous, hence differentiable normal vector field on any simply connected open set). Thus, by connectedness, $X=S$.

Ted Shifrin
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    Let me add, to allay OP's concern about $N$, that it suffices to say that $N$ is a normal line field rather than vector field; such normal line fields exist even on nonorientable manifolds, and being orthogonal to the normal line field is all that you really need in the proof. (We need, when OP's talking of $F$, to pick a local vector field consistent with the line field, and then show $F$ is locally constant; connectedness then gives that it's globally constant.) – John Hughes Oct 17 '14 at 00:14