Suppose that the second fundamental form of a surface patch $\sigma$ is zero everywhere.
How can we prove that $\sigma$ is an open subset of a plane?
The second fundamental form of a surface patch $\sigma$ is $$Ldu^2+2Mdudv+Ndv^2$$ where $L=\sigma_{uu}\cdot \textbf{N}, \ M=\sigma_{uv}\cdot \textbf{N}, \ N=\sigma_{vv}\cdot \textbf{N}$.
What does it mean that the second fundamental form of a surface patch $\sigma$ is zero everywhere? That $L=M=N=0$ ?
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EDIT:
We have that $\sigma_u\times\sigma_v$ is perpendicular to both $\sigma_u$ and $\sigma_v$. So $N=\frac{\sigma_u\times \sigma_v}{\|\sigma_u\times\sigma_v\|}$ is perpendicular to both $\sigma_u$ and $\sigma_v$. Therefore, $ \sigma_u \cdot N=0$ and $\sigma_v\cdot N=0$.
We also have that $$\sigma_{uu}\cdot N=0, \sigma_{uv}\cdot N=0, \sigma_{vv}\cdot N=0$$
So, $$( \sigma_u\cdot N)_u=0 \Rightarrow \sigma_{uu}\cdot N+\sigma_u \cdot N_u=0 \Rightarrow \sigma_u \cdot N_u=0$$ $$( \sigma_u\cdot N)_v=0 \Rightarrow \sigma_{uv}\cdot N+\sigma_u \cdot N_v=0 \Rightarrow \sigma_u \cdot N_v=0$$ $$( \sigma_v\cdot N)_u=0 \Rightarrow \sigma_{vu}\cdot N+\sigma_v \cdot N_u=0 \Rightarrow \sigma_v \cdot N_u=0$$ $$( \sigma_v\cdot N)_v=0 \Rightarrow \sigma_{vv}\cdot N+\sigma_v \cdot N_v=0 \Rightarrow \sigma_v \cdot N_v=0$$
So, $N_u$ and $N_v$ are perpendicular to $\sigma_u$ and $\sigma_v$.
We have that $N_u$ and $N_v$ are perpendicular to $N$. $N$ is perpendicular to the tangent plane. So, this implies that $N_u$ and $N_v$ lie in the tangent plane of the surface, which is spanned by $\sigma_u$ and $\sigma_v$.
We have that $N_u$ lies in the tangent plane that is spanned by $\sigma_u$ and $\sigma_v$, and that $\sigma_u\cdot N_u=0$ and $ \sigma_v\cdot N_u=0$, which means that $N_u$ is perpendicular to $\sigma_u$ and $\sigma_v$. This can only hold when $N_u=0$. Respectively, we get $N_v=0$. Is this correct?
Since $N_u=N_v=0$, we conclude that $N$ is constant.
Since $N$ is perpendicular to the tangent plane, we have that $\sigma_u \cdot N=\sigma_v\cdot N=0$. So,
$(\sigma\cdot N)_u =\sigma_u\cdot N=0$
$(\sigma\cdot N)_v =\sigma_v\cdot N=0$
So, we have that $\sigma\cdot N=c$, where $c$ is a constant.
How co we conclude from here that $\sigma$ is an open subset of a plane?