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

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.

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 $$\lambda \equiv r^2$$.