# Pushing a geodesic

1. Apr 5, 2008

### CompuChip

Hello.

Suppose that $\sigma: (M, g) \to (N, h)$ is an isometric diffeomorphism between two Riemannian manifolds M and N and let $\gamma: [0, 1] \to M$ be a geodesic on M.
Because $\sigma$ preserves distances, and geodesics are locally length minimizing, it is intuitively clear that $\sigma_* \gamma = \sigma \circ \gamma$ is a geodesic on N, but I'm having some trouble proving this.
In particular, I don't see which characterization of geodesics is the most convenient (I suppose it is the locally length minimizing property; but I don't really see how to express that formally).
Any help is greatly appreciated.

2. Apr 5, 2008

### genneth

Perhaps the fact that geodesics parallel transport their own tangent vectors? The local isometry pretty much by definition does so, so geodesics are preserved.

3. Apr 5, 2008

### CompuChip

That's definitely a possibility, but I think it's quite elaborate. It involves dealing with the two different Levi-Civita connections
$$\nabla^M_{\dot\gamma} \dot\gamma = \nabla^N_{(\sigma \circ \gamma)^\cdot} (\sigma \circ \gamma)^\cdot$$
with $\nabla^{M,N}$ denoting the connection on the respective manifolds, and $(\sigma \circ \gamma)^\cdot$ is the derivative w.r.t. t.

Since there is such an intuitive meaning to the statement (geodesics have to do with minimal length curves, and isometries are precisely those maps that preserve lengths) I was sort of hoping for a proof in that direction.

Last edited: Apr 5, 2008
4. Apr 5, 2008

### genneth

I would have thought that the latter is more difficult actually. Isometry is an intrinsically local concept, and finite lengths are not.

5. Apr 8, 2008

### wofsy

Your idea works. A geodesic minimizes length locally. Under an isometry this must be true in both manifolds because curve length is preserved.

I think it is instructive though to prove this directly from the definition of a geodesic rather than appeal to this difficult theorem.