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The geodesics on a cylinder (a cylinder infinite in both directions) are either (1) simple (non-self-intersecting) closed geodesics, or (2) simple infinitely long geodesics (infinite in both directions).
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           (Image from John Oprea.)

Q1. Are there any other surfaces in $\mathbb{R}^3$ all of whose geodesics are either class (1) or class (2) above (and there are geodesics in both classes)?

In other words, does this classification of geodesics characterize the cylinder? Thanks for your thoughts!

Answered. Anton Petrunin's example shows that the answer to my question above is No. However, Paul Reynolds raised a possibly more interesting variant of my question:

Q2. Are there any other surfaces in $\mathbb{R}^3$, topologically distinct from a cylinder, all of whose geodesics are either class (1) or class (2) above (and there are geodesics in both classes)?

In other words, does this classification of geodesics determine that such a surface must have the topology of a cylinder?

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You should at least add that both types of geodesics must occur, otherwise any plane (or any surface of zero curvature, isomorphic to the plane) would also satisfy this. –  Jaap Eldering Jul 7 '12 at 13:24
    
@Jaap: Excellent point! Corrected. Thanks! –  Joseph O'Rourke Jul 7 '12 at 13:55
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Could we not graft a cylinder (with $z$-axis) to the $xy$-plane (with disk removed) in such a way that no geodesic can self-intersect? –  Ramsay Jul 7 '12 at 15:03
    
Dear Joseph O'Rourke: Opera$\to$Oprea. –  Giuseppe Tortorella Jul 7 '12 at 15:47
    
@Giuseppe: Thanks, corrected. –  Joseph O'Rourke Jul 7 '12 at 16:35
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3 Answers 3

up vote 6 down vote accepted

Take the surface of tubular $\varepsilon$-neighborhood $S_\varepsilon$ of a curve $\gamma$(*) for some $\varepsilon < r$. It will have exactly one closed geodesic through any point and the rest will be simple infinitely long geodesics.

Indeed, any geodesic is either a meridian of $S_\varepsilon$ or it intersects transversely any meridian and therefore it has no self-intersections.

$(*)$ I assume that $\gamma$ is an infinite curve such that for any point $p$ on distance $ < r $ from $\gamma$ there is unique point $\bar p\in\gamma$ which minimize the distance $|p-\bar p|$. In particular the curvature of $\gamma$ has to be less $\tfrac1r$.

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Assuming the curve isn't itself periodic, e.g. a circle. –  Paul Reynolds Jul 7 '12 at 16:42
    
@Paul. Now it is corrected. –  Anton Petrunin Jul 13 '12 at 19:42
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There are negatively curved, complete cylinders whose ends are asymptotic to parallel planes; I've seen lots of pictures of these but don't have one available.

Or even more simply, a one-sheeted hyperboloid, $x^2 + y^2 - z^2 = 1$, also negatively curved, with ends asymptotic to cones instead of planes.

These have the same properties, with the interesting additional properties that exactly one geodesic of type (1) occurs, and certain geodesics of type (2), albeit having infinite length as required, are not closed sets but instead limit on the unique type (1) geodesic by spiralling around it (I am interpreting "infinite in both directions" to be about length, not diameter).

EDIT: Oh No! This is example is wrong!

There are bi-infinite geodesics on these surfaces that cross themselves. To see one, start with one of the bi-infinite geodesics $g$ that spirals around the type (1) closed geodesic. Take a tangent vector $v$ on $g$ that points in the direction of the spiralling end. Suppose, for example, that $g$ spirals into the closed geodesic from below, so the vector $v$ has an upward component. Now change the angle of $v$ every so slightly to get a new tangent vector $v'$ based at the same point that has a slightly smaller upward component. The new geodesic $g'$ will appear to be spiralling in around the type (1) closed geodesic, but after only a finite number of spirals its tangent vector will become momentarily horizontal and then it will start spiralling downward away from the type (1) closed geodesic. Once this happens, the downward spirals of $g'$ will cross the upward spirals transversely.

Another way to say this is that the type (2) geodesics that do not cross the type (1) geodesics have a unique point whose tangent vector is horizontal. This point can be arbitrarily close to the type (1) geodesic, and as it gets closer and closer, the type (2) geodesic starts crossing itself transversely, more and more times.

[Added by JORourke: a hyperboloid, and a catenoid:]
      Hyperboloid
    Hyperboloid

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The catenoid $\lbrace x^2+y^2=cosh^2 z\rbrace$ is an example of a negatively curved cylinder with ends asympotic to parallel planes. –  Rbega Jul 7 '12 at 16:46
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I wanted to show that this property of geodesics doesn't even characterise the topology of the surface, and I think this is an example.

This is a bunch of cylinders of the kind Rbega mentions, deformed slightly so as to be joined smoothly together. The cut off edges should be asymptotic to parallel planes.

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Ah but as soon as the surface has more topology than a cylinder --- positive genus, or zero genus and $\ge 3$ ends, then it has self-intersecting closed geodesics. On this surface, there is a self-intersecting figure eight closed geodesic on the visible gold portion: one lobe of the figure eight goes around the downward hole, the other goes around the upward pillar. –  Lee Mosher Jul 8 '12 at 0:46
    
You're right of course, I see it now. The slight deformations needed to patch the cylinders together spoil the party. I guess I'll leave it here as your comment is useful, thanks. –  Paul Reynolds Jul 8 '12 at 0:54
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But Paul has raised a deeper question than my original: Do the cylinder's geodesics properties determine the topology of all surfaces with those properties? [Paul, you should feel free to pose this yourself, if so inclined.] –  Joseph O'Rourke Jul 8 '12 at 1:59
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