3.1 Theorem (O'neill) If its shape operator is identically zero, them M is part of a plane in $R^3$
The first part of Proof.
By the definition of shape operator, $S = 0$ means that any unit normal vector field $E_3$ on $M$ is Euclidean parallel, and hence can be identified with a point of $R^3$.
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Question. I don't know why S = 0 implies a parallel normal vector field.
I have tried the following: Let $M$ be a orientable path-connected surface. Since M is orientable, let Z denote the derivatable unit-normal vector field defined in M.
Claim. If $dZ = 0 $ (shape operator is identically zero), Z is parallel normal vector field. (Z is Gaussian mapping)
Try.
Since $dZ = 0$, Normal curvature is zero. On the other hand, there is coordinate patch $\mathtt x : D \rightarrow M$ for each point $p\in M$. Since $R^3$ is locally path-connected space, there is path- connected open set $D'$. Thus, coordiante path $\mathtt x : D' \rightarrow M$ is a well - defined coordinate function. Z$\circ \mathtt x (u,v)$ = $(x, y , z)$. Since $Z_u = Z_u = 0$, $x_u = x_v = 0$. $D'$ is domain, Because $D'$ is path- connected open set. Since gradient $x$ = 0, $Z$ is identically constant on $\mathtt x(D')$.
In short, Z is a constant function in the appropriate neighborhood of each point. I can not go on since then. Maybe I should use path- connecte surface M, but I don't like it. Please help.
If $Z$ - (the Gauss map assigning to each point $p$ on the surface the (outer) unit normal at that point) - is constant in a neighborhood of $p$, then for every point $q$ in a neighborhood of $p$ on the surface, the tangent space is one and the same - the one that is orthogonal to the constant value of $Z$. This means that in a neighborhood of $p$, all points of the surface are orthogonal to the constant value of $Z$, hence belong to a plane.