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Context My question is about the matrix dimension of the shape operator.

In order to avoid misunderstanding let $S \subset \mathbb{R}^3$ be a regular surface and $$\psi(u,v)=(x(u,v),y(u,v),z(u,v))$$ be a chart on a open subset $W\subset S$. So, we have the induced basis of $T_pM$ given by $\left\{\frac{\partial \psi}{\partial u},\frac{\partial \psi}{\partial v}\right\}$ which allow us to define the Gauss map

$$p \mapsto N(u,v)=\frac{\frac{\partial \psi}{\partial u}\land \frac{\partial \psi}{\partial v}}{\big\vert\big\vert \frac{\partial \psi}{\partial u}\land \frac{\partial \psi}{\partial v}\big\vert\big\vert}$$

Now we can finally define the shape operator $S:\mathfrak X(W)\rightarrow \mathfrak X(W)$; $S(X)=- D_XN$

Question

The matrix that represents the linear operator $D_XN$ lies on the space of $3\times 2$ matrices (here we have a very good derivation about this fact), so by definition of $S$, its matrix representation should also lies on the space of $3\times 2$ matrices.

But at the same time when I'm looking on textbooks I only find that this matrix is a $2\times 2$ matrix whose entries are given by the the one and second fundamental forms coefficients (here gives the formula I'm talking about and here we have an example).

How can I should understand whats going on and how can I interpret this conceptual difference envolved ?

Thank you in advance for any hint about this question :)

Powder
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  • The linked post does not say the shape operator is represented by a $3 \times 2$ matrix: The values of $-D_{X}N$ at a point $p$ are vectors in $\mathbf{R}^3$, but they lie in the tangent plane at $p$ to the surface, so can be expressed as linear combinations of coordinate vectors on the surface. There are two such vectors, so the matrix of the shape operator is $2 \times 2$. – Andrew D. Hwang Dec 30 '21 at 13:27
  • I agree with you, but I did an example with surface $z=xy^2$ and things got weid because the matrix of shape operator $2\times 2$ was exactly the first four entries of matrix $3\times 2$ given by $\left(DN\left(\frac{\partial \psi}{\partial u}\right),DN\left(\frac{\partial \psi}{\partial v}\right)\right)$, this can't be a coincidence but I don't see a reason for that. – Powder Dec 30 '21 at 14:54
  • What is $\mathcal X(W)$ supposed to mean here? If you fix a point $P\in S$, then the shape operator maps the tangent plane $T_PS$ to the tangent plane of the unit sphere at $N(P)$, and this is again $T_PS$. As you vary $P$, you are taking a vector field $X$ on $W$ and getting another vector field on $W$. You must write these vector fields in terms of the basis you've indicated. – Ted Shifrin Dec 30 '21 at 18:49
  • $X\in \mathfrak{X}(W)$ is just a vector field defined on this open set $W\subset S$. The problem is, I have a basis $\left{\frac{\partial \psi}{\partial u},\frac{\partial \psi}{\partial v}\right}$of $T_p S$ induced by the parametrization $\psi=(x(u,v),y(u,v),z(u,v))$. So, they're $3\times 1$ vectors implying that, the matrix of $D_p N$ in this basis is $\left( D_pN\left(\frac{\partial \psi}{\partial u}\right),D_pN\left(\frac{\partial \psi}{\partial u}\right)\right)$ i.e a $3\times 2$ matrix. – Powder Dec 30 '21 at 19:20
  • At the same time, as $T_p S\simeq \mathbb{R}^2$ we sure can representate the matrix of shape operator as a $2\times 2$ matrix. But I really don't understand how this don't contradict what I said above. – Powder Dec 30 '21 at 19:24

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