I'm trying to do excercise 4.8 in "Riemannian manifolds" by John Lee. (It's about showing that the geodesics of [itex]\mathbb R^n[/itex] are straight lines).(adsbygoogle = window.adsbygoogle || []).push({});

The result I'm getting is that the definition of a geodesic implies the well-known identity 0=0, which isn't very useful. I must have made a mistake somewhere, but I don't see it.Edit:I do now. See my edit in #3.

Lemma 4.9 defines an operator that Lee writes as D_{t}, but I don't see the point of that t, so I'll just call it D. Lee defines a geodesic as a curve [itex]\gamma:I\rightarrow M[/itex] (where I is some intervale) such that [itex]D\dot\gamma(t)=0[/itex] for all t. The Euclidean connection is defined by [itex]\nabla_XY=XY^i\partial_i[/itex]. The velocity vector field along [itex]\gamma[/itex] is defined by [itex]\dot\gamma(t)=\gamma_*D_t[/itex], where D_{t}(not to be confused with the other D) is the operator that takes a function to its derivative at t. So the components of the velocity vector field in a coordinate system x are

[tex]\dot\gamma(t)x^i=\gamma_*Dx^i=D(x^i\circ\gamma)=(x^i\circ\gamma)'(t)[/tex]

Lee writes this as [itex]\dot\gamma^i(t)[/itex], so I will too. Note that since the manifold we're going to be dealing with is [itex]\mathbb R^n[/itex], we can take the coordinate system to be the identity map. The obvious definition [itex]\gamma^i=x^i\circ\gamma[/itex] implies that [itex]\gamma^i'(t)=\dot\gamma(t)[/itex].

Let V be an extension of [itex]\dot\gamma[/itex] to a neigborhood of the image of the curve. This means that we have [itex]V_{\gamma(t)}=\dot\gamma(t)[/itex].

[tex]0=D\dot\gamma(t)=\nabla_{\dot\gamma(t)} V=(\nabla_V V)_{\gamma(t)}=(VV^i\partial_i)_{\gamma(t)}=V_{\gamma(t)}V^i\partial_i|_{\gamma(t)}[/tex]

The vector is only zero if its components are, so this implies

[tex]0=V_{\gamma(t)}V^i=V^j(\gamma(t))\partial_j|_{\gamma(t)}V^i[/tex]

To continue we need to know that

[tex]V^i(\gamma(t))=V_{\gamma(t)}x^i=\dot\gamma(t)x^i=\dot\gamma^i(t)[/tex]

and that this implies that

[tex]V^i=\dot\gamma^i\circ\gamma^{-1}[/tex]

Let's continue with the main calculation. We have

[tex]0=V^j(\gamma(t))\partial_j|_{\gamma(t)}V^i=\dot\gamma^j(t)(V^i\circ x^{-1})_{,j}(x(\gamma(t))[/tex]

where ",j" means the partial derivative with respect to the jth variable. We choose x to be the identity map, so the above is

[tex]=\dot\gamma^j(t)(\dot\gamma^i\circ\gamma^{-1})_{,j}(\gamma(t))=\dot\gamma^j(t)\ddot\gamma^i(t)(\gamma^{-1})_{,j}(\gamma(t))=\ddot\gamma^i(t)(\gamma^{-1}\circ\gamma)'(t)=\ddot\gamma^i(t)\cdot 0=0[/tex]

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# Geodesic implies the well-known identity 0=0

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