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I believe one can fill $\mathbb{R}^3$ with horizontal translates of the helix $(\cos t, \sin t, t) \;,\; t \in \mathbb{R}$, so that every point of $\mathbb{R}^3$ lies in exactly one helix. I am wondering if it is possible to assign a metric to $\mathbb{R}^3$ so that these helices are all geodesics? (I am imagining these as world lines of particles stationary in the plane.)

                helices

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Giving (only) a metric such that the helices are geodesics is easy. For example you can define one by setting the distance between two different points on the $(x,y,0)$-plane to be 1, distance between two points in a helix to be the distance along the helix and the distance between other points to be the distance you get by combining the previous two. Also, it is easy to give a norm so that helices are geodesics. For example set $||(x,y,z)|| = \max\\{\sqrt{x^2+y^2},2\pi|z|\\}$. Based on the tags you have chosen (and the imagined world lines of particles) you probably want more from the metric. – Tapio Rajala Mar 26 2011 at 17:21

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Consider the following reparametrization of $\mathbb R^3$: $$(x,y,z)\mapsto (x-\cos z,y-\sin z,z).$$ Note that the horizontal translations of your helix go to the vertical lines. So the pullback of the canonical metric on $\mathbb R^3$ is the metric you want.

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Euclidean metric is indeed more natural than the metrics which I pointed out. – Tapio Rajala Mar 27 2011 at 8:20
@Anton: Thank you! I was not sufficiently familiar with the notion of a pullback metric. I appreciate being enlightened! – Joseph O'Rourke Mar 27 2011 at 13:30
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I'm pretty sure that Nil geometry works, but I don't know a reference. I seem to remember that one may think of Nil geometry as fibering over the plane, and that geodesics connecting different points in the same fiber had projections to circles. I think then these give helices.

A more general criterion was given by Dennis Sullivan for when a foliation may be realized as the geodesics of a Riemannian metric.

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Thank you for these leads. I will have to look into Nil geometry. (The Wikipedia article is thin: en.wikipedia.org/wiki/Nil_geometry#Nil_geometry). – Joseph O'Rourke Mar 28 2011 at 11:24

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