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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?

Mary Star
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    This has already been answered here: http://math.stackexchange.com/questions/383601/how-can-i-show-that-if-the-second-fundamental-form-of-a-surface-is-identically-e?rq=1 – Georg Lehner Dec 25 '15 at 12:14
  • I added my question also in my initial post... Could you take a look at it? @GeorgLehner – Mary Star Dec 31 '15 at 17:02
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    You already conluded that the points $\vec{x}$ of your surface satisfy the equation $\vec{x} \cdot N = c$. This is simply the equation of a plane https://en.wikipedia.org/wiki/Plane_%28geometry%29#Point-normal_form_and_general_form_of_the_equation_of_a_plane Hence your surface is a subset of the plane defined by the above equation. Why is it an open subset? Well, by definition $\sigma$, which is defined on an open subset of $\mathbb R^2$, must be a homeomorphism onto its image. – Georg Lehner Dec 31 '15 at 17:33
  • But isn't the equation of plane $\vec{x}\cdot N=0$ and not $\vec{x}\cdot N=c$ ? @GeorgLehner – Mary Star Dec 31 '15 at 18:13
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    That is the equation of a plane through the origin. – Georg Lehner Dec 31 '15 at 18:17
  • Is $\sigma$ always an open set? Or do you mean by definition in this case? @GeorgLehner – Mary Star Jan 03 '16 at 00:43
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    No, $\sigma$ is a function first of all, not a set. I do not know how your textbook defines parametrizations, but the standard one I had in my mind is as a smooth mapping from an open subset of $\mathbb R^2$ into $\mathbb R^3$, being a homeomorphism onto its image and having injective differential. $\sigma$ being a homeomorphism onto its image guaranties you, since you already know that the image lies in a plane (in particular a copy of $\mathbb R^2$), that its image is open. – Georg Lehner Jan 03 '16 at 14:20
  • I see... Thanks a lot for your help!! :-) @GeorgLehner – Mary Star Jan 03 '16 at 14:45

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