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Suppose that $V$ is a connected open subset of $M$, and, furthermore, suppose that the shape operator vanishes throughout $V$, i.e, for every $p \in M$ and $v \in T_pM$, $S_p(v)=0$. Show then that $V$ must be flat, i.e., it's part of a plane.

So far, I know it's enough to show that the normal vector is the same for any two points $p,q$ in $M$. Since $V$ is open and connected, there exists a smooth curve $\gamma:[0,1]\rightarrow M$ with $\gamma(0)=p$ and $\gamma(1)=q$. So we'll define $f:[0,1] \rightarrow R$ by $f(t)=n(\gamma(t))$ and differentiate to get $f'(t)=dn_\gamma(\gamma'(t))$. Now, if we can justify that this last bit is $0$, we're done. My question is, is it $0$ and why?

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If the shape operator $S_p: T_pM \rightarrow T_pM$ is defined by $S_p(v)=-(\nabla_v N)(p)$ where $N$ is the normal vector field to $M$ and $\nabla$ denotes the covariant derivative in $\mathbb{R}^3$ then the condition that the shape operator vanishes identically gives you $\nabla_v N=0$ for all vector fields $v$. Recall, if the vector field $W$ has cartesian component functions $w^1,w^2,w^3$ then $\nabla_v W = \sum_{i=1}^3 v[w^i] \frac{\partial}{\partial x^i}$. Thus, setting $N = \sum_{i=1}^3 N^i \frac{\partial}{\partial x^i}$ we find: $$ \nabla_v N = \sum_{i=1}^3 v[N^i] \frac{\partial}{\partial x^i} $$ Consider $v=\frac{\partial}{\partial x^1} = \partial_1$, $v=\frac{\partial}{\partial x^2}= \partial_2$ and $v=\frac{\partial}{\partial x^3}= \partial_3$. These give: $$ \nabla_{\partial_1} \, N = \sum_{i=1}^3 \frac{\partial N^i}{\partial x^1} \frac{\partial}{\partial x^i}, \qquad \nabla_{\partial_2} \, N = \sum_{i=1}^3 \frac{\partial N^i}{\partial x^2} \frac{\partial}{\partial x^i}, \qquad \nabla_{\partial_3} \, N = \sum_{i=1}^3 \frac{\partial N^i}{\partial x^3} \frac{\partial}{\partial x^i} $$ Therefore, the triviality of the shape operator implies $\frac{\partial N^i}{\partial x^j}=0$ for all $i,j$. Then, as $M$ is connected, this implies $N$ is constant.

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