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I am looking at the following exercise:

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Could you give me some hints how we could show that?

Do we use the matrix of the Weingarten map with respect to the basis $\{\sigma_u,\sigma_v\}$ of the tangent plane? But how exactly?

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EDIT1:

The matrix of the Weingarten map with respect to the basis $\{\sigma_u,\sigma_v\}$ of the tangent plane is $$\mathcal{F}_I^{-1}\mathcal{F}_{II}= \begin{pmatrix} \cos^2 v & 0 \\ 0 & 1 \end{pmatrix}^{-1}\begin{pmatrix} -\cos^2 v & 0 \\ 0 & -1 \end{pmatrix}=\frac{1}{\cos^2 v}\begin{pmatrix} -\cos^2 v & 0 \\ 0 & -\cos^2 v \end{pmatrix}=\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$$

We have that $$\mathcal{F}_I^{-1}\mathcal{F}_{II}= \begin{pmatrix} a & c \\ b & d \end{pmatrix}$$ where $-\textbf{N}_u=a\sigma_u+b\sigma_v$ and $-\textbf{N}_v=c\sigma_u+d\sigma_v$.

Therefore, we get $-\textbf{N}_u=-\sigma_u\Rightarrow \textbf{N}_u=\sigma_u \Rightarrow \textbf{N}=\sigma+A(v)$ and $-\textbf{N}_v=-\sigma_v \Rightarrow \textbf{N}_v=\sigma_v \Rightarrow \textbf{N}=\sigma+B(u)$.

From the relations $\textbf{N}=\sigma+A(v)$ and $\textbf{N}=\sigma+B(u)$, we see that $A(v)=B(u)$ must be a constant. So $\textbf{N}=\sigma+\tilde{C}$.

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EDIT2:

Could you give me some hints how we could find a parametrization of $S^2$ with the above first and second fundamental forms?

Mary Star
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    I don't know if this is helpful, but the curvature of your patch is $\det II / \det I = 1$, so by Minding's theorem it is locally isometric to $S^2$. – A.P. Jan 05 '16 at 23:41
  • When we have that the surface is locally isometric to $S^2$, can we conclude that it is an open subset of $S^2$ ? @A.P. – Mary Star Jan 06 '16 at 11:32
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    "Is" is a strong word. "Looks like" is better, albeit fuzzy. Isometric is the word you want to use. Locally isometric is defined as: Around each point of the surface there exists an open neighbourhood of that point that is isometric to an open surface of the sphere. Since you were only talking about a surface patch anyway, it amounts to the same thing. – Georg Lehner Jan 07 '16 at 21:46
  • Ah ok... Can we justify it only using Minding's theorem? @GeorgLehner – Mary Star Jan 07 '16 at 22:00
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    To get back to your explicit question: The Weingarten map in this case is $-Id$ (I denote with Id the identity). Remember, the Weingarten map is the differential of the Gauss map, hence the Gauss map is (locally) an isometry (from your surface to the sphere, of course). – Georg Lehner Jan 07 '16 at 22:00
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    Regarding your edit: If you know that $N = \sigma + \tilde{C}$ you can deduce that $| \sigma - \tilde{C} |^2_u = 0 = | \sigma - \tilde{C} |^2_v$ hence the term $| \sigma - \tilde{C} |^2$ is constant, which is the equation for a sphere. – Georg Lehner Jan 08 '16 at 16:48
  • Couldn't we take also the norm of $N=\sigma +\tilde{C}$ to deduce that the term $| \sigma - \tilde{C} |^2$ is constant? @GeorgLehner – Mary Star Jan 08 '16 at 17:28
  • We have that $|N|^2=|\sigma+\tilde{C}|^2\Rightarrow 1=|\sigma+\tilde{C}}^2$, right? So, this is the equation for th unit sphere, or not? Have we shown in that way that the surface is an open subset of unit sphere? @GeorgLehner – Mary Star Jan 08 '16 at 20:03
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    Yes, that is also how you can conclude your statement. (Your equation is the equation for a sphere centered at $-\tilde{C}$.) – Georg Lehner Jan 09 '16 at 18:49
  • Since this is the equation for the unit sphere, why is the surface an open subset of the unit sphere? @GeorgLehner – Mary Star Jan 09 '16 at 19:51
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    By exactly the same reasoning as in your other question: http://math.stackexchange.com/questions/1586043/open-subset-of-a-plane?rq=1

    With the important difference that it isn't an open subset of 'the' unit-sphere, but of a unit sphere, in your case centered at $-\tilde{C}$.

     – Georg Lehner Jan 09 '16 at 19:53
  • Ahaa... Ok... $$$$ Do you maybe have also an idea for the second edit of my initial post? @GeorgLehner – Mary Star Jan 09 '16 at 20:06
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    @Edit2: Have you tried sphercial coordinates? ;) Btw: To avoid this comment section to get any longer, ask further questions in the chat, please. – Georg Lehner Jan 09 '16 at 20:18

1 Answers1

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Let me give an explicit answer: Write $M$ for your surface.

First, the $2-$nd fundamental form is typically defined as

$II(u,v) := - I(d \nu (u) , v)$

where $\nu : M \rightarrow S^2$ is the Gauss map, i.e. the map that assigns to each point $p$ the unit normal vector to the tangent plane of $p$. (Note that I had a different sign in my comment. I was used to a different convention, but let us stick with how it is on wikipedia)

The first and second fundamental forms in your example fulfill the relationship $II = - cos(v)du^2 - dv^2 = - I$, hence $d \nu$ must be the identity map. This means in particular, that $\nu$ is a local isometry, i.e. around each point $p$ there exists an open neighbourhood $U$ of $p$, such that $\nu |_U$ is an isometry.

PS: A small remark is that the proof of Mindings theorem for surfaces of positive curvature goes in much the same way.

PPS: What I call $\nu$, you denoted by $N$ in your edit. It's the same function.

Georg Lehner
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  • I haven't really understood the definition $II(u,v) := - I(d \nu (u) , v)$. How do we see that in this case we have that $II = - I$ ? $$$$ I added in my initial post what I have tried using the Weingarten map... Could you take a look at it and tell if we also could use this to get to the desired result? – Mary Star Jan 07 '16 at 22:31
  • I added why $II = - I$. Could you elaborate what definition for the second fundamental form you use? – Georg Lehner Jan 08 '16 at 16:40
  • I added the definition of my book for the second fundamental form in my initial post. $$$$ When $d \nu$ is the identity map, why is $\nu$ then a local isometry? – Mary Star Jan 08 '16 at 17:26
  • https://en.wikipedia.org/wiki/Isometry_%28Riemannian_geometry%29 In the notation of this article, the differential $d \nu$ is written $\nu_*$, and $g$ is in our example $I$. Clearly, the identity map fullfils $I( Id(u) , Id(v) ) = I(u,v)$. – Georg Lehner Jan 09 '16 at 18:48