# Geodesic implies the well-known identity 0=0

1. Feb 11, 2010

### Fredrik

Staff Emeritus
I'm trying to do excercise 4.8 in "Riemannian manifolds" by John Lee. (It's about showing that the geodesics of $\mathbb R^n$ are straight lines).

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 Dt, but I don't see the point of that t, so I'll just call it D. Lee defines a geodesic as a curve $\gamma:I\rightarrow M$ (where I is some intervale) such that $D\dot\gamma(t)=0$ for all t. The Euclidean connection is defined by $\nabla_XY=XY^i\partial_i$. The velocity vector field along $\gamma$ is defined by $\dot\gamma(t)=\gamma_*D_t$, where Dt (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

$$\dot\gamma(t)x^i=\gamma_*Dx^i=D(x^i\circ\gamma)=(x^i\circ\gamma)'(t)$$

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

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

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

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

$$0=V_{\gamma(t)}V^i=V^j(\gamma(t))\partial_j|_{\gamma(t)}V^i$$

To continue we need to know that

$$V^i(\gamma(t))=V_{\gamma(t)}x^i=\dot\gamma(t)x^i=\dot\gamma^i(t)$$

and that this implies that

$$V^i=\dot\gamma^i\circ\gamma^{-1}$$

Let's continue with the main calculation. We have

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

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

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

Last edited: Feb 11, 2010
2. Feb 11, 2010

### espen180

Re: Geodesics

Just a thought: If you work in euclidean n-space the Christoffel symbols are zero and the indicial geodesic equation simplifies to

$$\frac{d^2x^m}{ds^2}=0$$

ruling out any change of direction or speed, therefore straight lines.

Last edited: Feb 12, 2010
3. Feb 11, 2010

### Fredrik

Staff Emeritus
Re: Geodesics

Thanks. That inspired me to try to calculate

$$\frac{d}{dt}V^i(\gamma(t))$$

to see what I get, and the result is

$$=D_t(V^i\circ\gamma)=\gamma_*D_tV^i=\dot\gamma(t)V^i=V_{\gamma(t)}V^i$$

which I know is =0, if the first few steps in my calculation in #1 are correct. And I also showed in #1 that $V^i=\dot\gamma^i\circ\gamma^{-1}$, so

$$0=\frac{d}{dt}V^i(\gamma(t))=\frac{d}{dt}\dot\gamma^i(t)$$

which implies constant velocity. I guess that completes the solution of the excercise, but I still don't see what I did wrong in #1. Hm, looks like the mistake has to be in the last line, but I still don't see it.

Edit: Found it. The derivative of the identity map $t\mapsto t$ is 1, not 0.

Last edited: Feb 11, 2010