Calculating the derivative of a mapping $\varphi: S^2 \rightarrow S^2$

Question

I have recently learned about tangent spaces and derivatives in the context of manifolds and I am having a hard time solving the following exercise:

Let A be a $3\times3$ orthogonal matrix. Consider the map $\varphi: S^2 \rightarrow S^2$ defined by $\varphi(x) = Ax$.
($S^2$ means the 2 sphere)

Calculate the mapping $D\varphi(x) = T_x S^2 \rightarrow T_{\varphi(x)}S^2$

If I am not mistaking I need to find bases in $T_x S^2$ and $T_{\varphi(x)}S^2$ and then calculate $D(f_2^{-1}\circ\varphi \circ f_1)(x)$ where $f_1$ and $f_2$ are local charts but I do not know how to do this in practice.

Could you help me?

Answer 1

$S^2$ is a smooth submanifold of $\mathbb{R}^3$. Let $i : S^2 \to \mathbb{R}^3$ denote the (smooth) embedding and let $\phi : \mathbb{R}^3 \to \mathbb{R}^3, \phi(x) = Ax$. Then $$(*) \phantom{x} Di(\varphi(x)) D\varphi(x) = D\phi(x) Di(x) .$$ But $T_y \mathbb{R}^3 = \mathbb{R}^3$ and $D\phi(y) = \phi$ for all $y \in \mathbb{R}^3$. The tangent spaces $T_yS^2$, $y \in S^2$, can be identified via $Di(y)$ with two-dimensional linear subspaces of $\mathbb{R}^3$. Doing so, $(*)$ shows that $\phi(T_x S^2) = T_{\varphi(x)} S^2$ and $D\varphi(x)$ is the restriction of $\phi$.

Concerning the tangent spaces of a smooth submanifold $M \subset \mathbb{R}^n$ also see my answer to The motivation for a tangent space.