Computation 1. Let m be a positive integer and ϕ:Rm→Cm be the standard parametrization of the m-dimensional torus Tm in Cm given by ϕ(x1.…,xm)=(eix1,…,eixm). Prove that ϕ is isometric (w.r.t. the standard Riemannian metric on Rm and Cm).Solution. The first thought would be to consider the map C:(z1,…,zm)↦(Rez1,Imz1,…,Imzm)∈R2m and then to consider the "local" representation of ϕ, the C∘ϕ, but then it would be found that this is not of any help to compute dϕp, p∈Rm.
We can only find dϕp pointwise. Namely, let v∈Rm and let γ′(0)=v, with γ(0)=p, then
dϕp(v)=dϕp(γ′(0))=ddtϕ∘γ(t)|t=0. From this we readily see that dϕp is computable! And in fact, for (v1,…,vm)∈Rm, dϕp(v1,…,vm)=(v1eip1,…,vmeipm). Therefore ϕ is an isometry because
⟨dϕpv,dϕpv′⟩Cm=m∑j=1v′je−ipjvjeipj=m∑j=1vjv′j=⟨v,v′⟩Rm.◼
Computation 2. Let m be a positive integer andSolution. As before for p∈Sm∖{(1,0,…,0)} and v∈TpSm=(Rp)⊥, we have
πm:(Sm∖{(1,0,…,0)},⟨,⟩Rm+1)→(Rm,4(1+|x|2)2⟨,⟩Rm). be the stereographic projection given by
πm(x0,…,cm)=11−x0(x1,…,xm). Prove that πm is an isometry.
d(πm)(p0,…,pm)(v0,…,vm)=11−p0(v1,…,vm)+v0(1−p0)2(p1,…,pm), therefore
=⟨d(πm)p(v),d(πm)p(v′)⟩πm(p),(Rm,4(1+|x|2)2⟨,⟩Rm)=4(1+|πm(p)|2)2⋅⟨v,v′⟩Rm+1(1−p0)2=⟨v,v′⟩.◼
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