Showing that $f:S_1 \to S_2$ is a conformal mapping.

Question

Exercise :

Given the surfaces $S_1 = \{ x \in \mathbb R^3 : \|x\| = 1\}$ and $S_2 = \{x \in \mathbb R^3 : \|x\| = r\}$ where $r >0$, check if the mapping $f:S_1 \to S_2$ with the formula $f(x,y,z) = (rx,ry,-rz)$ is conformal.

Thoughts :

According to my Differential Geometry notes, I have the following equivalent definitions :

1 : The topical diffeomorphism $f : S_1 \to S_2$ is conformal, if and only if $f_p^* = \lambda(p)\langle \cdot, \cdot\rangle_p, \; \forall p$ where $\lambda : S_1 \to \mathbb R$ is a function.

2 : The topical diffeomorphism $f:S_1 \to S_2$ is conformal, if and only if $\forall r$ patches around $p$, $f \circ r$ is also a patch around $f(p)$ with : $$E_p = \lambda(p)E_{f(p)}, \; F_p = \lambda(p)E_{f(p)}, \; G_p = \lambda(p)G_{f(p)}$$

I assume that the second equivalent definition is or more use, as we have an explicit formula for $f$ that can help finding the fundamental form. But, I am kind of stuck in decoding it. What is $E_p$ and what $E_{f(p)}$ in that case ? Also, I first need to show that $f$ is a Diffeomorphism, which I found to have many equivalent remarks here.

Any explanation or thorough elaboration will be much appreciated as I am still a beginner in differential geometry regarding my experience on the subject.

Answer 1

Let's do this with your first definition. Consider the map $F \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ given by $F(x,y,z) = (rx,ry,-rz)$. The map $F$ is linear and so for any $p \in \mathbb{R}^3$ and $(a,b,c) \in \mathbb{R}^3$ we have

$$ dF|_{p}(a,b,c) = (ra,rb,-rc). $$

Now, the map $f$ is just the restriction of the map $F$ to $S_1$ in the domain and $S_2$ in the codomain. You can verify that for $p \in S_1$, the map $df|_p \colon T_p(S_1) \rightarrow T_{f(p)} S_2$ is just the restriction of the map $dF|_{p} \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ to $T_p(S_1)$ in the domain and $T_{f(p)} S_2$ in the codomain. Then, for $p \in S_1$ and $v = (a,b,c), w = (a',b',c') \in T_p(S_1)$ we have

$$ \left< df|_{p}(v), df|_p(w) \right>^{S_2}_{f(p)} = \left< (ra, rb, -rc), (ra', rb', -rc') \right> = r^2 \left< (a,b,c), (a',b',c') \right> = r^2 \left< v, w \right>^{S_1}_{p}.$$

Also, it is clear that $f$ is bijective, smooth and with invertible differential so $f$ is a conformal diffeomorphism with $\lambda \equiv r^2$.