Differential Geometry – Can Every Curve on a Riemannian Manifold be a Geodesic?

differential-geometryriemannian-geometry

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.

Best Answer

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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