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Given a metric $g_{\mu\nu}$ it is possible to find the equations of the geodesic on the Riemannian manifold $M$ defined by the metric itself:

$$\frac{d^2x^a}{ds^2} + \Gamma^{a}_{bc}\frac{dx^b}{ds}\frac{dx^c}{ds} = 0$$ where: $$\Gamma^a_{bc} = \frac{1}{2} g^{ad} \left( g_{cd,b} + g_{bd,c} - g_{bc,d} \right)$$ are the Christoffel symbols and $$g_{ab,c} = \frac{\partial {g_{ab}}}{\partial {x^c}}$$ Now, given a parametric equation of a curve, is it possible to find the metric of a Riemannian manifold which gives that curve as a geodesic? If the answer is 'Yes', is there a bijective correspondence between the curve and the metric? Or are there many metrics giving the same geodesic? Thanks in advance.

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In general the answer is of course no. Take any curve $\gamma : [a,b] \to M$ where $\gamma(t_1) = \gamma(t_2)$ and $\gamma'(t_1) = \gamma'(t_2)$ with $t_2 \neq t_1$. Provided it's not true that $\gamma(t+t_2-t_1) = \gamma(t)$ then you've got a counter-example, because geodesics are determined by their velocity vectors at a point. A good counter-example would be a curve that is tangent to itself, but with different curvatures at the point of tangency. –  Ryan Budney Feb 17 '12 at 23:06
The title doesn't ask the same thing as the question. –  Ben Crowell Feb 18 '12 at 1:48
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2 Answers

up vote 6 down vote accepted

I'm not sure what happens for general curves, but I think I can prove the following:

Let $\gamma:[0,1]\rightarrow M$ be any injective curve segement. Then there is a Riemannian metric for which $\gamma$ is a geodesic. If instead $\gamma$ is a simple closed curve and $\gamma'(0) = \gamma'(1)$, the conclusion still holds.

I'm not sure what happens in the other cases.

Here's the idea of the proof in the (slightly harder) second case:

Pick a background Riemannian metric once and for all. The normal bundle of $\gamma$ embeds into $M$ via the exponential map (for a suitably short time). Call the image of this embedding $W$. Choose an open set $V$ with the property that $V\subseteq \overline{V}\subseteq W$ and let $U = M-\overline{V}$. Notice that $W\cup U = M$, so we can find partition of unity $\{\lambda_U,\lambda_W\}$ subordinate to $\{U,W\}$.

Now, the classification of vector bundles over circles is easy: There are precisely 2 of any rank - the trivial bundle of rank $k$ and the Möbius bundle + trivial bundle of rank $k-1$. The point is that both of these have (flat) metrics where the $0$ section ($\gamma$) is a geodesic.

Since $W$ is diffeomorphic to a vector bundle over the circle, we can assume it has a metric $g_W$ for which $\gamma$ is a geodesic. Now, pick any Riemannian metric $g_U$ on $U$. Finally, define the metric $g_M$ on $M$ by $\lambda_W g_W + \lambda_U g_U$. This is a convex sum of metrics, and hence is a metric. Near $\gamma$, $\lambda_U \equiv 0$ and $\lambda_W\equiv 1$, so the metric near $\gamma$ looks just like $g_W$, so $\gamma$ is a geodesic in $M$.

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I expect that if $M$ is a (connected!) differentiable manifold and $\gamma_1, \gamma_2: S^1 \rightarrow M$ are any two smooth embeddings, there is a diffeomorphism $\Phi: M \rightarrow M$ such that $\gamma_2 = \Phi \circ \gamma_1$. If so, this gives a positive answer to your question restricted to smoothly embedded loops. And something similar should work for smooth embeddings of $\mathbb{R}$ with closed image.

Added: The above is certainly not generally valid: I seem to have forgotten about the fundamental group. It seems like it might still have a chance to hold in the simply connected case. (Also, in the case of surfaces, if you take a metric of constant curvature, I seem to recall that every homotopy class has a unique geodesic representative, so this obstruction is not a problem at least in that case.)

As for the second question: of course there are going to be many Riemannian metrics than geodesic curves: changing the metric in an open set bounded away from the geodesic will certainly not disturb that curve's being a geodesic. As for changes of metric which preserve all geodesic curves rather than just a given one, that's a more interesting question, but at least you can uniformly rescale the metric without affecting any of the geodesics.

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I don't think your expectation is correct. Consider $M = \mathbb{R}P^2$ and choose as $\gamma_1$ something representing a nontrivial element in $\pi_1$ (this can be chosen to be embedded) and $\gamma_2$ a small contractible loop (which can also be chosen to be embedded). Then no diffeomorphism of $\mathbb{R}P^2$ maps $\gamma_1$ to $\gamma_2$ because no group isomorphism of $\pi_1$ takes the nontrivial element to the trivial element. –  Jason DeVito Feb 17 '12 at 22:26
In fact, I think if $\dim M \geq 3$, and $M$ is not simply connected, you can find such a $\gamma_1$ and $\gamma_2$. (In $\dim M = 2$, you can't always assume your curves are embedded). –  Jason DeVito Feb 17 '12 at 22:29
@Jason: You're right. This was a rather naive thought, in retrospect. I wonder whether it's true in the simply connected case? –  Pete L. Clark Feb 18 '12 at 0:19
If you take a nontrivial knot $K$ in $S^3$, then I believe the knot complement is not homeomorphic to the knot complement of an unknot. Thus, there can be no homeomorphism of $S^3$ taking $K$ to the unknot, since the homeomorphism restricted to the knot complements is still a homeomorphism. I don't know what happens if $\dim M \geq 4$ so that there are no knots. –  Jason DeVito Feb 18 '12 at 1:21
@Riccardo: I don't think this should be the accepted answer. –  Pete L. Clark Feb 18 '12 at 7:37
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