Question

Let ϕ(z)∈C(z) be a rational function. In the text we defined the ramification properties of ϕ at a point α∈P¹(C) provided that α=[infinity] and ϕ(α)=[infinity]. In general, let f∈PGL₂​(C) be any linear fractional transformation, let ϕf=f⁻¹∘ϕ∘f, and let β=f⁻¹(α). If β=[infinity] and ϕf(β)=[infinity], we say that ϕ is ramified at α if (ϕf)′(β)=0 and we define the ramification index of ϕ at α to be eα​(ϕ)=eβ​(ϕf) (a) Assuming that none of α,β,ϕ(α),ϕf(β) are equal to [infinity], prove that (ϕf)′(β)=(f⁻¹)′(ϕ(α))​ϕ′(α)/(f⁻¹)′(α) Conclude that the ramification points of a rational map ϕ:P¹→P¹ are independent of the choice of coordinates, i.e., the vanishing of ϕ′(α) is independent of the choice of f. (b) Show that the ramification index eα​(ϕ) is well-defined, independent of the choice of f. (c) Under what conditions is it true that the derivative ϕ′(α) is independent of the choice of f ?

Answer 1

To prove that (ϕf)′(β)=(f⁻¹)′(ϕ(α))​ϕ′(α)/(f⁻¹)′(α), we consider (f⁻¹)′(ϕ(α))​ϕ′(α)/(f⁻¹)′(α) and simplify it. Then, we substitute β with f⁻¹(α) in (ϕf)′(β) to show that (ϕf)′(f⁻¹(α))=(f⁻¹)′(ϕ(α))ϕ′(α)/(f⁻¹)(α).

Step-by-step explanation:

In order to prove the given statement, we need to show that (ϕf)′(β)=(f⁻¹)′(ϕ(α))​ϕ′(α)/(f⁻¹)′(α).

To begin, let's consider (f⁻¹)′(ϕ(α))​ϕ′(α)/(f⁻¹)′(α) and simplify it. We can rewrite (f⁻¹)′(ϕ(α))​ϕ′(α) as (f⁻¹)′(ϕ(α))ϕ′(α)/(f⁻¹)(α). Since β=f⁻¹(α), we can substitute ϕ(α) with ϕf(β) in the expression. Thus, (f⁻¹)′(ϕ(α))​ϕ′(α)/(f⁻¹)′(α) becomes (f⁻¹)′(ϕf(β))ϕ′(α)/(f⁻¹)(α).

Now, let's consider (ϕf)′(β). Since β=f⁻¹(α), we can substitute β with f⁻¹(α) in the expression. Thus (ϕf)′(β) becomes (ϕf)′(f⁻¹(α)).

To prove that (ϕf)′(β)=(f⁻¹)′(ϕ(α))​ϕ′(α)/(f⁻¹)′(α), we need to show that (ϕf)′(f⁻¹(α))=(f⁻¹)′(ϕf(β))ϕ′(α)/(f⁻¹)(α). By substituting β with f⁻¹(α), we get (ϕf)′(f⁻¹(α))=(f⁻¹)′(ϕf(f⁻¹(α)))ϕ′(α)/(f⁻¹)(α). Since f⁻¹ is an inverse, f⁻¹(f⁻¹(α))=α, which simplifies the expression to (ϕf)′(f⁻¹(α))=(f⁻¹)′(ϕ(α))ϕ′(α)/(f⁻¹)(α).