Take the 2-minute tour ×

At the risk of asking an uninformed question...

Imagine an ant on a compact two-dimensional surface embedded in 3-space. The ant is placed at a point on the surface with random orientation. Once placed and oriented, the ant will walk along a local geodesic path.

Are there examples of such 2D surfaces where we are guaranteed the ant will never return to some starting position?

(Thanks to Henry Wilton for requesting clarification as per his comment below.)

share|improve this question
I very strongly suspect the answer is 'no' for arbitrary positioning or arbitrary orientation. I should have thought a bit more before asking this... –  Mensen Nov 6 '09 at 5:01
I presume you want your surface to be compact? Also, when you say "the ant will walk along the curvature of the surface", do you mean that it follows a local geodesic? –  HJRW Nov 6 '09 at 5:40
Henry - yes, and yes. This was poorly written and thought out on my end. –  Mensen Nov 6 '09 at 7:57
Now I'm just confused about the quantifiers. Is the question "does every surface in R^3 have a closed geodesic?" or "does every surface in R^3 have an infinite geodesic?", or even "does every surface in R^3 have an infinite non-self-intersecting geodesic?"? –  HJRW Nov 6 '09 at 17:33

5 Answers 5

up vote 6 down vote accepted

Silly answer: any plane in 3-space will work (so the answer is "yes").

You probably want a compact surface. These always admit geodesic cycles (so the answer in this case is "no").

share|improve this answer
Yes, I realized shortly after asking this question that I had 'egged' myself. =) Thank you nevertheless for your response. –  Mensen Nov 6 '09 at 7:56

By Poincare recurrence theorem, on a compact manifold, almost every geodesic returns arbitrarily close to its starting point, infinitely many times.

So if you accept a slight error in the measure of the position (and direction) of the ant (say less than 1 picometer, and $10^{-9}$ degree), you are almost guaranted (w.r.t Lebesgue measure) that the ant returns to its initial position if you wait long enough.

The case of a non-compact manifold is even more interesting. If the volume is finite, there is always a geodesic that goes to infinity, yet the result still holds: a.e geodesics are recurrent.

There are also examples of infinite volume surfaces for which a.e. geodesics are recurrent. This is the case for a tubular neighborhood of the Cayley graph of $\mathbb{Z}^2\times \lbrace 0\rbrace$ in $\mathbb{R}^3$.

Finally, there are examples of infinite volume surfaces for which there is a dense set of recurrent geodesics, yet almost all geodesics go to infinity. This is the case for a tubular neighborhood of the Cayley graph of $\mathbb{Z}^3$.

share|improve this answer
...and if you agree that the space is discrete, than this is just Polya's theorem. –  Delio Mugnolo Feb 21 '13 at 8:40

Every smooth two-manifold with nonzero genus has a locally geodesic closed loop: the smallest noncontractible cycle. But that doesn't rule out the existence of a point that is not on a geodesic loop, nor does it rule out the possibility that the closed loops through a point have measure zero among the random choices of direction.

If you just want a single surface, a single point on that surface, and probability one of not returning home, take a standard torus in 3d. If you pick any point on that surface, and choose an orientation randomly, you will (with probability one) get a curve such that it has an irrational change in longitude when it first returns to the same latitude as its starting point.

share|improve this answer
Is it easy to prove this? Forgetting about the "embeddable in R^3" condition, I figured the easiest thing to do was work on the flat torus... –  Aaron Mazel-Gee Nov 7 '09 at 3:02
Let f(x) map a starting angle to the difference in longitude (amount a curve wraps the long way around the torus) when the curve first returns to the same latitude (wraps once the short way around). Then f is two-to-one (if two curves return to the same latitude while wrapping the same amount the long way around the torus, they must go in opposite directions around the short way). So the set of x at which it is rational is countable. Ok, all of these qualitative descriptions are not a rigorous proof, but I think that's not needed to see that this example works. –  David Eppstein Nov 7 '09 at 4:17

An easy case is the flat torus (that is, a torus with the flat metric -- the one induced from its representation as a square in the plane with edges identified). There, the ant will return home if and only if the slope of its walk is rational. I think this means that with probability 1, the ant will not return home!

share|improve this answer
Sorry, I think I may be confused about the order of your quantifiers. Does this have to be true for any point, or does such a point just have to exist? It seems like you're clear on the fact that you want the ant not to return for <i>any</i> initial orientation... –  Aaron Mazel-Gee Nov 6 '09 at 8:35
Dear Aaron, I am happy with the existence of a point! –  Mensen Nov 6 '09 at 8:43

A flat torus doesn't work because it can't be embedded into 3-space (at least, I'm assuming by "3-space" you mean R^3, not, say, T^3, in which case it can be embedded).

share|improve this answer
Beware that a flat torus can be $C^1$ embedded into $\mathbb{R}^3$. Since one can define geodesics as locally minimizing curves, the notion still makes sense in the $C^1$ regularity. –  Benoît Kloeckner Apr 8 '10 at 13:59

Your Answer


