Why do isometries preserve geodesics (locally minimizing paths)? Under the isometry Ο•:(M,g,βˆ‡)β†’(M~,g~,βˆ‡~), we have that βˆ‡=Ο•βˆ—βˆ‡~. In other words, the only Levi-Civita connection on M is the pullback connection. How to prove this? The approach is to prove that the right hand side satisfies the properties of a connection on M, which is torsion-free and compatible with the metric g. Uniqueness of the connection on M satisfying these properties then gives βˆ‡=Ο•βˆ—βˆ‡~. These aspects can be shown entirely within the framework tensor fields, making the results automatically global. Once we understand why βˆ‡=Ο•βˆ—βˆ‡~ we can answer our question.

  • First note that Ο•βˆ—βˆ‡~ is a map Ξ“(TM)Γ—Ξ“(TM)β†’Ξ“(TM). Then we want to check linearity in the first entry and the characteristic Leibniz rule in the second. Recall that Ο•βˆ—(fX)=(fβˆ˜Ο•βˆ’1) Ο•βˆ—X. Let for notation’s sake ψ:=Ο•βˆ’1:Nβ†’M, D:=Ο•βˆ—βˆ‡~ and the application of a derivation X on a function h be X.h=Xh. Then for f∈C∞(M)
(1)DfXY=Οˆβˆ—βˆ‡~Ο•βˆ—(fX)Ο•βˆ—Y=Οˆβˆ—βˆ‡~f∘ψ Ο•βˆ—XΟ•βˆ—Y(2)=Οˆβˆ—(f∘ψ βˆ‡~Ο•βˆ—XΟ•βˆ—Y)=(fβˆ˜Οˆβˆ˜Ο•) Οˆβˆ—βˆ‡~Ο•βˆ—XΟ•βˆ—Y(3)=fDXY

and recalling for g∈C∞(N), (Ο•βˆ—X).g=(X.(fβˆ˜Ο•))∘ψ as functions on N,

(4)DX(fY)=Οˆβˆ—βˆ‡~Ο•βˆ—XΟ•βˆ—(fY)=Οˆβˆ—βˆ‡~Ο•βˆ—X(f∘ψ)Ο•βˆ—Y(5)=Οˆβˆ—((Ο•βˆ—X).(f∘ψ)Ο•βˆ—Y+f∘ψ βˆ‡~Ο•βˆ—XΟ•βˆ—Y)(6)=Οˆβˆ—((X.(fβˆ˜Οˆβˆ˜Ο•))∘ψ)Ο•βˆ—Y+fΟˆβˆ—βˆ‡~Ο•βˆ—XΟ•βˆ—Y(7)=Οˆβˆ—((X.f)∘ψ)⏟∈C∞(N)Ο•βˆ—Y+fDXY(8)=(X.f)Y+fDXY
  • Next we show that it is torsion free, using the fact that [Ο•βˆ—X,Ο•βˆ—Y]=Ο•βˆ—[X,Y]. We have from torsion-freeness of βˆ‡~ that βˆ‡~Ο•βˆ—XΟ•βˆ—Yβˆ’βˆ‡~Ο•βˆ—YΟ•βˆ—X=[Ο•βˆ—X,Ο•βˆ—Y]=Ο•βˆ—[X,Y], whereby DXYβˆ’DYX=Οˆβˆ—Ο•βˆ—[X,Y]=[X,Y] since pushforwards are isomorphisms.
  • Lastly we use compatibility of βˆ‡~ with g~ to get it for D with g. Recall g(X,Y)=gId(X,Y)=g~Ο•(Ο•βˆ—X,Ο•βˆ—Y) since Ο•:Mβ†’N is an isometry. Let Ξ“(TM)βˆ‹Z=Οˆβˆ—ΞΎ for ΞΎβˆˆΞ“(TN). Then since the expressions need to be equal as functions on M: (9)g(DXY,Z)=g(Οˆβˆ—βˆ‡~Ο•βˆ—XΟ•βˆ—Y,Οˆβˆ—ΞΎ)=g~Ο•(βˆ‡~Ο•βˆ—XΟ•βˆ—Y,ΞΎ)=g~Ο•(βˆ‡~Ο•βˆ—XΟ•βˆ—Y,Ο•βˆ—Z)(10)=((Ο•βˆ—X).g~Id(Ο•βˆ—Y,Ο•βˆ—Z))βˆ˜Οˆβˆ’g~Ο•(Ο•βˆ—Y,βˆ‡~Ο•βˆ—XΟ•βˆ—Z)(11)=X.g~Ο•(Ο•βˆ—Y,Ο•βˆ—Z)βˆ’g(Y,Οˆβˆ—βˆ‡~Ο•βˆ—XΟ•βˆ—Z)(12)=X.g(Y,Z)βˆ’g(Y,DXZ).

Thus βˆ‡=Ο•βˆ—βˆ‡~. As a bonus we can nicely see from this that geodesics get mapped by an isometry Ο•:Mβ†’N. If c:Iβ†’M is a geodesic then the corresponding curve Ξ³:=Ο•βˆ˜c:Iβ†’N is as well. (This curve is the flow ΞΈ~t(n)=Ξ³(t) with Ξ³(0)=n from the vector field Ο•βˆ—cβ€²(t) with c(0)=ψ(n).) (13)Οˆβˆ—βˆ‡~Ξ³β€²Ξ³β€²=Οˆβˆ—βˆ‡~Ο•βˆ—cβ€²Ο•βˆ—cβ€²=Dcβ€²cβ€²=βˆ‡cβ€²cβ€²=0 so βˆ‡~Ξ³β€²Ξ³β€²=0 since Οˆβˆ— is an isomorphism. (Note Ξ³β€²(t)=ddt(Ο•βˆ˜c(t))=Tc(t)Ο•cβ€²(t)=(Ο•βˆ—cβ€²)(t) where Tc(t)Ο•:Tc(t)Mβ†’TΞ³(t)N is the tangent map or differential.)