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I have a bits-and-pieces understanding on how to solve this problem and just a very rough intuition of the path to solve the problem but very much struggling to get to show $dL=L$.

This is part of a question if Do Carmo's book that reads as follows:

Prove that if $L:R^3\to R^3$ is a linear map and $S\subset R^3$ is a regular surface invariant under $L$, i.e., $L(S)\subset S$, then the restriction $L|S$ is a differentiable map and

$$dL_p(w)=L(w), \text{ }p\in S, w\in T_p(S).$$

I could show that the restriction $L|S$ is a differentiable map but I am struggling to show that $dL_p(w)=L(w)$.

So let's try this:

Let $p$ be a point on the surface $S$, $x:U\subset \mathbb R^2\rightarrow S$ be a parametrization s.t. $x(0)=p$ and $y:V\subset \mathbb R^2\rightarrow S$ be another parametrization s.t. $L(p)=y(0)$.

Let $f(u,v)=y^{-1}\circ L \circ x(u,v)$. By using chain rule, we get $df_0=d(y^{-1})_{L(p)}\circ dL \circ dx_0$.

My question is:
1. What the map does is $\mathbb R^2 \underset{dx}{\longrightarrow} T_pS \underset{dL}{\longrightarrow} T_{L(p)}S\underset{dy^{-1}}{\longrightarrow} \mathbb R^2$. From here, how can we conclude that $dL=L$?

I have seen some similar problems from some other sources, but I am still very much struggling to connect all the dots, everything for me is still bits and pieces.

Helps are greatly appreciated! Hopefully after making those points clear I can finish the problem by myself. Thanks.

user71346
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    Roughly, the differential of a map is the best linear approximation of this map in the neighbourhood of a point. What is the best linear approximation of a linear map? – Bernard Dec 28 '15 at 11:57
  • @Bernard. Is it the linear map itself? But I still fail to see how do we get dL=L from the chain rule? – user71346 Dec 28 '15 at 12:22
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    I think maybe you're confused because in high school, you learn the derivative of $ax$ is $a$. Note the differential itself in this case is $h\mapsto ah$ (the differential is not the derivative: the latter is a number or a vector, the differential is a linear map). – Bernard Dec 28 '15 at 12:38
  • @Bernard the differential is a linear map between the tangent space right? I think I know that. But I am still struggling to see how do we get from $T_pS \underset{L}{\longrightarrow} T_{L(p)}S$ to $dL=L$? Thanks. – user71346 Dec 28 '15 at 12:51
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    You're considering the wrong map here: you have $L\colon \Bbb{R}^3 \to \Bbb{R}^3$, which is then restricted to $S$, and $dL \colon T_pS \to T_{L(p)}S$. You have to prove that $dL$ coincides with $L$ (restricted to $S$), after the natural isomorphisms $T_pS \simeq \Bbb{R}^2 \simeq T_{L(p)}S$ induced by the coordinate system. By the way, you may have to use the coordinate system induced by the embedding $S \to \Bbb{R}^3$ instead of a preferred chart, though I'm not sure. – A.P. Dec 28 '15 at 14:03

1 Answers1

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Do Carmo defines a tangent vector $w \in T_p S$ as a velocity vector $w = \alpha'(0)$ for some curve $\alpha:(-\epsilon,\epsilon) \to S$ with $\alpha(0) = p$.

Now by definition $dL_p(w) = \left.\frac{d}{dt}\right\rvert_{t=0} L(\alpha(t)) = L(\alpha'(0)) = L(w)$, where we used the chain rule and the fact that $L$ is linear. Indeed, note that as a (linear) map $L:\mathbb{R}^3 \to \mathbb{R}^3$, $L$ coincides with its differential at any point (in particular, at $p$): if, say, $$L(x,y,z) = (a_1x + a_2y + a_3z, b_1x+b_2y+b_3z,c_1x+c_2y+c_3z),$$ then the Jacobian of $L$ at any point is $$\begin{pmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{pmatrix}.$$

Alex Provost
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