Let $C_n = C_n(\mathbb{R}^3)$ denote the configuration space of $n$ distinct points in $\mathbb{R}^3$. Is there a Riemannian metric $g$ on $C_n$ such that given any two configurations in $C_n$, there is a unique geodesic joining them?

In addition, it would be nice if $g$ was also geodesically complete, and if $g$ came from natural considerations in Physics (for instance if it is the kinetic term of some naturally occuring Lagrangian etc.).

Edit 1: I accepted Andy Putman's answer below, because it does answer negatively my question (thank you!). However, could someone please indicate whether or not there exists a complete Riemannian metric $g$ on $C_n$? Is it more appropriate to create another post perhaps? I just found out that Nomizu and Ozeki proved that any connected smooth (second countable) manifold admits a complete Riemannian metric. This is nice. However, is there a known explicit such complete Riemannian metric $g$ on $C_n$? If two of the points say are going towards each other and seem about to collide, there has to be a repulsive force that forbids collision (in physical terms).

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    $\begingroup$ What about bijectively mapping the configuration of $n$ points to a sum of Dirac measures and then using the optimal transport (Wasserstein-2) metric? Can this yield a Riemannian metric on $C_n(\mathbb{R}^3)$ by pullback? $\endgroup$ – S.Surace Jun 5 at 7:56
  • $\begingroup$ See the related question mathoverflow.net/questions/301908/… $\endgroup$ – S.Surace Jun 5 at 7:58
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    $\begingroup$ Due to Mather's avoidance principle, the Wasserstein-2 metric will indeed induce a metric on $C_n(\mathbb{R}^3)$ (this is the repulsive force you want) . The geodesic between two configurations will not necessarily be unique, but the metric will be complete. Also, you can compute the distances reasonably quickly using linear programming and it should be feasible to determine when two configurations are in the cut locus. $\endgroup$ – Gabe K Jun 5 at 9:13
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    $\begingroup$ I would suggest to take a look at Villani's books. More details can be found in Ambrosio, Gigli and Savaré's book from 2008. $\endgroup$ – S.Surace Jun 5 at 12:43
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    $\begingroup$ Ah. I realized I made a mistake last night, which I should correct. The metric is not complete, but it is geodesically complete in that the shortest path between any two configurations remains a configuration. If you follow a geodesic too long though, points can collide. $\endgroup$ – Gabe K Jun 5 at 13:06

The answer is no. This relies on two things:

  1. A uniquely geodesic proper metric space is contractible; see here for a proof.

  2. $C_n$ is not contractible. Indeed, it has many nontrivial homology groups (there is a huge literature on this).

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  • $\begingroup$ This answers the questions under the assumption that the metric is complete, correct? Using the hypothesis that “it would be nice if $g$ was also geodesically complete”? $\endgroup$ – Pierre PC Jun 5 at 15:35

This is just a long comment, and a pretty speculative one at that. However it might perhaps be of interest to you since:

  • there is a natural connection to physics,
  • the construction only works in three dimensions,
  • the construction is equivariant with respect to the action of the permutations.

I have in mind the (conjectured) map described by Atiyah in [1] which maps configurations of points to the complex flag manifold: $$ C_n(\mathbb{R}^3) \to U(n) / T^n. $$

Since the flag manifold is homogeneous, this map would provide a metric on $C_n(\mathbb{R}^3)$ if we could spot a natural metric on the fibres. I don't know if this is possible but for $n=2$, the fibres of the map are pairs of distinct points defining the same direction (first point looking at second point) and so are naturally parameterised by their midpoint $m$ and distance apart $t$. It's a bit of a stretch but if we give this fibre the metric of $dm^2 + (dt/t)^2$ then we get something you might regard as "nice".

[1] Atiyah, M., "Configurations of Points", R. Soc. Lond. Philos. Trans. Ser. A Math. Phys. Eng. Sci. 359 (2001), no. 1784, 1375-1387.

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  • $\begingroup$ Thank you for your comments. I am actually motivated by this specific problem you have mentioned. A small correction. For $n=2$, the map factors through the sphere, which is the relative direction from $x_1$ to $x_2$, and the factor map from the sphere to $U(2)/T^2$ is actually a diffeomorphism. Permuting $x_1$ and $x_2$ corresponds to the antipodal map on the sphere, which corresponds to interchanging the two columns of the corresponding two by two matrix. $\endgroup$ – Malkoun Jun 16 at 12:08
  • $\begingroup$ Please note though, that the fibers of those maps (assuming linear independence) are not so easy to describe (in part because these maps involve the polar decomposition, at the end, which is a Weyl equivariant orthogonalization process). $\endgroup$ – Malkoun Jun 16 at 12:11
  • $\begingroup$ I knew I had heard of your name before. I think I have read part of your thesis a while back. I think you had worked on the problem: "what is the natural geometry on the moduli space of hyperbolic monopoles?" and related problems. $\endgroup$ – Malkoun Jun 16 at 12:27
  • $\begingroup$ You have a good memory! I did indeed write my thesis on that problem, though it has been many years since I've thought about it. I have just looked at your personal website and I see that you have already published results on Atiyah's configurations conjecture so I now realise that you were aware of all that I said. (Nice work btw!) I agree with your remarks except I am confused by the correction: note that I claimed that the (conjectured) map is equivariant, not invariant. Anyway, very good luck with your interesting research. $\endgroup$ – Oliver Nash Jun 16 at 12:53
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    $\begingroup$ Thank you. Oh, it is just a minor point. I just meant that the fiber over a point in the image of the map from the sphere of relative directions to $U(2)/T^2$, for $n=2$, consists of just a single point on the sphere, rather than a pair of antipodal points. Maybe I misunderstood what you had written. It is just a minor point. Thank you for your nice comments. $\endgroup$ – Malkoun Jun 16 at 12:59

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