Does knowledge of all the geodesics of a Riemannian manifold suffice to determine the metric up to a scaling factor?

The metric completely characterizes the shape of a Riemannian manifold. However, knowledge of all the geodesics on a Riemannian manifold does not uniquely determine the metric.

This problem is a case of obtaining local properties from global ones. The metric is a local quantity; it depends only on the chosen point of the manifold and a local neighborhood around it. Geodesics, on the other hand, connect points on the manifold, appealing to a larger global structure of the manifold.

There are a variety of theorems dealing with the relationship between local and global properties in Riemannian geometry, a prolific example of which is the Hopf-Rinow theorem. This tells you that a Riemannian manifold is a complete metric space if and only if it is geodesically complete - that is, if at any point you can extend a geodesic infinitely far in any direction, the metric on a space is such that the manifold is complete. We can deduce a few interesting properties like this, but we cannot fully determine the metric from information about the geodesics.

So we can't fully determine the metric from information about the geodesics, but can we determine the metric up to a scaling factor from the geodesics?


The answer to your question is "no". Consider any constant metric on $\mathbb{R}^n$. The geodesics associated to this metric are just straight lines and thus, any two such metrics indistinguishable by just their geodesics. But it is not necessarily the case that two such metrics are conformal to one another.

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    $\begingroup$ What is a constant metric? $\endgroup$ – Steve D Aug 18 '18 at 23:19
  • $\begingroup$ A metric which does not vary from point to point. This just amounts to choosing a positive definite matrix $A$ and saying that at every point, the metric $g$ is defined by $g(v,w):=v^TAw$. Implicitly here, we are identifying all of the tangent spaces for $\mathbb{R}^n$ with $\mathbb{R}^n$ in the standard, canonical way. Since the Levi-Civita symbols identically vanish for this metric, we get that the Levi-Civita connection for this metric is just the ordinary derivative. Thus, the geodesics are just lines. $\endgroup$ – Or Eisenberg Aug 19 '18 at 18:11
  • $\begingroup$ Doesn't that argument only work for constant metrics $\endgroup$ – Ultradark Aug 20 '18 at 21:52
  • $\begingroup$ I'm not entirely sure what you mean. In any event, we can generalize this to nonconstant metrics by noting that if two metrics $g$ and $g'$ on $\mathbb{R}^n$ satisfy the condition that $g-g' \in C^{\infty}\big(\mathbb{R}^n,\mathrm{Mat}_{n \times n}(\mathbb{R})\big)$ is constant, then $g$ and $g'$ have precisely the same geodesics. In particular, then, for $any$ metric $g$ on $\mathbb{R}^n$, one has some other metric $g'$ such that the geodesics associated to $g$ are the same as those associated to $g'$, yet $g \neq \lambda g'$ for all $\lambda \in \mathbb{R}$. $\endgroup$ – Or Eisenberg Aug 20 '18 at 21:59

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