Clarifications about $dN_p$: What does $dN_p$ actually does? I am reading Do Carmo's Differential Geometry book. I have two questions, here:

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*Question 1:





He talks about the tangent vector to the curve with coordinates $(u'(0),v'(0),0)$ and $dN_p$ sends this vector to $(2u'(0),-2v'(0),0)$. But here, he says that $N'(0)=(2u'(0),-2v'(0),0)$$ is the tangent vector. I'm a bit confused: Are they both tangent vectors?





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*Question 2:
What does $dN_p$ does? Does it sends $(x\circ \alpha(t))_t$ to $(N\circ \alpha(t))_t$?


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*Question 3:
How is he computing the eigenvalues? Im my mind, we would have to compute it this way. Ie: Find the matrix of the linear transformation, etc. Is he doing that without showing the details or there is some alternative way to do it?
 A: Below I will call $S$ the surface we are studying and $\Bbb S^2$ the 2 dimensional unit sphere.

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*$N'(0)$ is the tangent vector to the curve $N\circ \alpha \colon I \to \Bbb S^2$, at $t=0$ (this curve itself lives in the 2 dimensional sphere). It is not a vector that is tangent to the surface $S$.


*$N\colon S \to \Bbb S^2$ is a differentiable map from a surface to the two dimensional sphere. It thus has a differential (or a linear tangent map) $dN \colon TS \to T\Bbb S^2$. This is a collection of linear maps $dN_p \colon T_pS \to T_{N(p)}\Bbb S^2$. If $p\in S$ is fixed, $dN_p$ sends a vector $X\in T_pS$ tangent to $S$ at $p$ to a vector $dN_p(X) \in T_{N(p)}\Bbb S^2$ tangent to $\Bbb S^2$ at the point $N(p)$.


*The chain rules says that you can compute $N_p(X)$ by chosing a curve $\alpha\colon I\to S$ satisfying $\alpha(0) = p$ and $\alpha'(0)=X$, because then
$$
(N\circ \alpha)'(0) = dN_{\alpha(0)}(\alpha'(0)) = dN_p(X).
$$
Now fix $p=(0,0,0)$. After identifying $T_pS$ ans $T_{N(p)}\Bbb S^2$ (which surely has been justified earlier in the book), the computations lead by do Carmo show that in the ambient coordinates, $dN_p$ is the linear map
$$
\begin{array}{r|ccc}
dN_p \colon & T_pS & \longrightarrow & T_pS \\
& (x,y,0) & \longmapsto & (2x,-2y,0).
\end{array}
$$
(Recall that $T_pS$ here is the linear subspace of $\Bbb R^3$ given by $z=0$). It is an endomorphism of the 2 dimensional space $T_pS$, and it is clear that its matrix in the basis $\{(1,0,0),(0,1,0)\}$ is given by $\begin{pmatrix} 2 & 0 \\ 0 & -2 \end{pmatrix}$. Hence, its eigenvalues are obviously $2$ and $-2$.
