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This is a repost of a question I posted at MSE.

Mark L. Irons' paper The Curvature and Geodesics of the Torus gives a concise overview of the geodesics on the torus:

  • There are five clear-cut families of geodesics.
  • Most of the geodesics are "ergodic": aperiodic and covering either the entire surface - by spiraling endlessly around - or substantial parts of it.
  • Some of the geodesics are "boring": the meridians, the inner and the outer equator
  • A few of them are "æsthetically pleasing": returning to their starting point after just a few circuits.

Does the structure of geodesics change when twisting the "hose" before gluing its ends?

alt text

E.g., there might be no equators anymore because after twisting the (two) equators lost their "ends".

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I did not understand the question but want to say something. $$ $$ If the torus can be is fibered by closed geodesics then any other geodesic has to go transverally to the fibers. I think that is the only idea behind any statement in this direction; i.e., if it does not help then nothing will help. – Anton Petrunin Oct 23 '12 at 22:39
BTW, the equator will move but it will survive. – Anton Petrunin Oct 23 '12 at 22:40
@Anton: Where do the equators go? How do they survive? – Hans Stricker Oct 23 '12 at 22:49
@Anton: To be honest, I have nothing overly specific in mind when asking for "the structure of geodesics". It's partly about the classification of geodesics - does it still hold after a twist? -, and it's partly about something comparable to the cycle space of a graph ( – Hans Stricker Oct 23 '12 at 23:09
Take a marker, draw a geodesic curve on the torus, then do a Dehn twist. The resulting curve is a geodesic as the geodesic equation is local. With this you can easily see how the families above map to one-another. – Kelly Davis Oct 24 '12 at 16:06

It is the same submanifold of $\mathbb R^3$ with same geodesics. You are just describing another chart for it which is open and dense. Another way to say this and to visualize the geodesics in this other chart is:
Lift each geodesic to the universal cover $\mathbb R^2$ and use a parallelogram instead of a square as fundamental domain.

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This is not a flat torus, it has curvature. – Will Sawin Oct 27 '12 at 19:31
Sure, the geodesics will not be straight lines in the universal cover. Nevertheless you can lift them. – Peter Michor Oct 29 '12 at 7:55

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