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Let $f$ be a parametrized surface $f: \Omega \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3$ and $N : \Omega \rightarrow Tf$ the Gauß map. Then the shape operator is defined as $L = -DN \circ Df^{-1}.$ Now the thing is that $Df$ is a $3 \times 2$ matrix, so I cannot invert this matrix easily. So how do I get a matrix representation for my shape operator?

If anything is unclear, please let me know.

  • The shape operator $L$ is just negative of the derivative $DN$ of the sphere map $$p\to N(p)=\frac{\partial_1\times\partial_2}{||\partial_1\times\partial_2||},$$ where $\partial_1,\partial_2$ are the coordinated tangent basis – janmarqz Jan 26 '15 at 16:33
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    @janmarqz well, i guess not, cause $L$ is supposed to be a $3 \times 3 $ matrix, so that things like $\langle L f_u ,f_v \rangle$ are well-defined –  Jan 26 '15 at 19:13
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    http://en.wikipedia.org/wiki/Differential_geometry_of_surfaces#Shape_operator – Xipan Xiao Jan 26 '15 at 19:52

1 Answers1

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The Gauss' map is $\Omega\to\Bbb R^3$ given by $p\longmapsto \dfrac{\partial_1\times\partial_2}{||\partial_1\times\partial_2||}$, where $\partial_1=\frac{\partial f}{\partial v}$ and $\partial_2=\frac{\partial f}{\partial w}$.

We can see this map as a vector field $N:\Sigma\to\Bbb R^3$, i.e. an assignation $ \left(\begin{array}{c} x\\ y\\ z \end{array}\right) \longmapsto \left(\begin{array}{c} N_1\\ N_2\\ N_3 \end{array}\right)$. So its Jacobian is $DN= \left(\begin{array}{ccc} \dfrac{\partial N_1}{\partial x}&\dfrac{\partial N_1}{\partial y}&\dfrac{\partial N_1}{\partial z}\\ \\ \dfrac{\partial N_2}{\partial x}&\dfrac{\partial N_2}{\partial y}&\dfrac{\partial N_2}{\partial z}\\ \\ \dfrac{\partial N_3}{\partial x}&\dfrac{\partial N_3}{\partial y}&\dfrac{\partial N_3}{\partial z} \end{array}\right)$.

Now observe that -by the use of the Chain's Rule- one gets $$DN(\partial_1)= \left(\begin{array}{ccc} \dfrac{\partial N_1}{\partial x}&\dfrac{\partial N_1}{\partial y}&\dfrac{\partial N_1}{\partial z}\\ \\ \dfrac{\partial N_2}{\partial x}&\dfrac{\partial N_2}{\partial y}&\dfrac{\partial N_2}{\partial z}\\ \\ \dfrac{\partial N_3}{\partial x}&\dfrac{\partial N_3}{\partial y}&\dfrac{\partial N_3}{\partial z} \end{array}\right) \left(\begin{array}{c} \dfrac{\partial x}{\partial v}\\ \\ \dfrac{\partial y}{\partial v}\\ \\ \dfrac{\partial z}{\partial v} \end{array}\right) = \left(\begin{array}{c} \dfrac{\partial N_1}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial N_1}{\partial y}\dfrac{\partial y}{\partial v}+\dfrac{\partial N_1}{\partial z}\dfrac{\partial z}{\partial v}\\ \\ \dfrac{\partial N_2}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial N_2}{\partial y}\dfrac{\partial y}{\partial v}+\dfrac{\partial N_2}{\partial z}\dfrac{\partial z}{\partial v}\\ \\ \dfrac{\partial N_3}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial N_3}{\partial y}\dfrac{\partial y}{\partial v}+\dfrac{\partial N_3}{\partial z}\dfrac{\partial z}{\partial v}\ \end{array}\right) = \left(\begin{array}{c} \dfrac{\partial N_1}{\partial v}\\ \\ \dfrac{\partial N_2}{\partial v}\\ \\ \dfrac{\partial N_3}{\partial v} \end{array}\right) = \dfrac{\partial N}{\partial v} $$

and similarly

$$DN(\partial_2) = \left(\begin{array}{c} \dfrac{\partial N_1}{\partial w}\\ \\ \dfrac{\partial N_2}{\partial w}\\ \\ \dfrac{\partial N_3}{\partial w} \end{array}\right) = \dfrac{\partial N}{\partial w} $$

In another hand we have $||N||=1$, then $DN\bullet N=0$, that is, $DN$ is orthogonal to $T_p\Sigma$.

Hence we can construct $$L:T_p\Sigma\longrightarrow T_p\Sigma$$ $$\partial_1\longmapsto L(\partial_1)=\frac{\partial N}{\partial v}$$ $$\partial_2\longmapsto L(\partial_2)= \frac{\partial N}{\partial w}$$ which is a basis change in case that $\frac{\partial N}{\partial v}$ and $\frac{\partial N}{\partial w}$ are linearly independent.

janmarqz
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    This is a great answer! I wish understood all of it though. What are $\Omega$ and $\Sigma$? Also, if we are dealing with surfaces, wouldn't $N$ map from a two-dimensional space into $\mathbb{R}^3$? (I.e. shouldn't $DN$ be a $3 \times 2$ matrix, and shouldn't the argument be of the form $( x y)$ instead of $(x y z)$. – Chill2Macht Aug 03 '16 at 15:15
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    @William: $\Omega$ is an open subset of $\Bbb R^2$ and it is also the domain of the parametererization $\Phi:\Omega\to\Bbb R^3$ of the surface $\Sigma={\rm im}\Phi$. The normal field $N$ maps $\Sigma\subset\Bbb R^3\to\Bbb R^3$. Then $DN_p$ would map $\Bbb R^3\to \Bbb R^3$. We use $(v,w)$ as the coordinates for $\Bbb R^2$ and $(x,y,z)$ for coordinates of $\Bbb R^3$, "ambient" for the surface $\Sigma$. – janmarqz Aug 03 '16 at 22:46
  • So the shape operator can be seen as a change of basis? Thanks for this post man it was great! and $\partial_1$ can be seen as the velocity of the curve when holding the $u$ variable constant in the paramterization... I think I'm getting the hang of this stuff a little bit haha –  May 07 '20 at 16:39
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    @HaKuNaMaTaTa derivatives are interpreted as velocities? yeah. But, here the geometry for $\partial_1$ is a tangential stuff, where $w$ is kept constant and for $\partial_2$ is $u$ constant. The shape operator relates $\partial_1,\partial_2$ to a new pair of tangent vectors $D_{\partial_1}N$ and $D_{\partial_2}N$ which maybe couldn't be linear independent: that happens when, in such a point on the surface, the curvature is zero. Keep on this happy "struggles". – janmarqz May 07 '20 at 17:32
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    Hm.. If the curvature is zero at a point then the shape operator might mess up the linear independence? Yeah, I really want to develop a better understanding of this stuff, I'm just now opening my eyes to how cool with differential geometry is. –  May 07 '20 at 21:15
  • go for it, there's a bunch of amusement ahead. Let me invite you to see this collection of selected contributions I did manufacture: https://math.stackexchange.com/search?tab=votes&q=user%3a74166%20%5bdifferential-geometry%5d about differential geometry. It si also this another https://math.stackexchange.com/search?tab=votes&q=user%3a74166%20%5btensors%5d on tensors – janmarqz May 07 '20 at 22:30