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When can we extend the minimal geodesics indefinitely?

  1. Jun 12, 2013 #1
    What sort of structure must a manifold possess in order to talk about minimal geodesics between two points on it? When can we extend the minimal geodesics indefinitely?
     
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  3. Jun 12, 2013 #2

    lavinia

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    The answer is the Hopf Rinow theorem. I think it suffices that the manifold is a complete metric space.

    It is clear that incomplete implies that there are not always length minimizing geodesics. For instance in a Euclidean disk with its center removed two radially opposite points can not be connected by any geodesic.
     
  4. Jun 12, 2013 #3
    Given a minimal geodesic between two points, is there a unique way to extend it past the points? Indefinitely?
     
  5. Jun 13, 2013 #4

    lavinia

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    no. Think about the example I gave above.
     
  6. Jun 13, 2013 #5
    I meant on a closed manifold.
     
  7. Jun 19, 2013 #6

    mathwonk

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    I'm not sure what you mean by a closed manifold (compact without boundary?), but here is the statement which seems to answer your question. I.e. part 3 seems to say then you can extend geodesics indefinitely.

    http://en.wikipedia.org/wiki/Hopf–Rinow_theorem
     
  8. Jun 20, 2013 #7
    OK, thanks to both of you.

    Now, given a closed (compact without boundary) Riemannian manifold M, a point P on it and a vector V in P's tangent space, how do we decide whether the geodesic which extends indefinitely from P in the direction of V fills the entire manifold? That is, how do we know whether the domain of this "maximal geodesic" is just M?

    For example, on a sphere, you always just get great circles; however, on a torus, I'm pretty sure sometimes you get the entire torus.
     
  9. Jun 20, 2013 #8

    WannabeNewton

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    Are you asking for when the exponential map is a global diffeomorphism?
     
  10. Jun 20, 2013 #9
    I'm not sure. What does it mean for the exponential map to be a global diffeomorphism?
     
  11. Jun 21, 2013 #10
    I'm asking that given a point and a direction, are there any non-trivial criteria which decides whether the maximal geodesic in that direction is closed?
     
  12. Jun 22, 2013 #11

    mathwonk

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    First, it seems impossible for the geodesic to fill the manifold for dimension reasons, although it could be dense. Second I cannot think of any way to determine whether the geodesic is closed from local data like the direction, since it seems to depend on global properties of the manifold. I.e. it seems two manifolds could be isometrically isomorphic near a point and have different global geodesics.

    With a torus, if it had zero curvature, maybe if the local tangent space were considered as the universal covering space, and one knew the lattice of pre-images of each point, one could compare the angle of the direction vector to the angle made by the generating vectors of the lattice to determine the image of the geodesic, but that seems a very special case.

    I guess any time the exponential map is surjective one onto a compact manifold, could look at the shape of a "fundamental polygon" in the tangent space, that maps almost isomorphically onto the manifold, and it may become a question related to the billiard table problem of when a ball struck returns to its original position, or whether the course of the ball is closed or not. But the billiard table here is a non euclidean multidimensional polyhedron.

    ????
     
  13. Jun 22, 2013 #12
    Can a geodesic really not fill a manifold for dimension reasons? What about space-filling curves? Like Peano's space filling curve. A geodesic can self-intersect, and having it fill a space doesn't make it a homeomorphism.

    The billiard ball problem is actually what motivated me to ask this. But I am still too uncomfortable with manifolds to fully understand your answers. I'm going to work on learning the basics some more.
     
  14. Jun 22, 2013 #13

    micromass

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  15. Jun 22, 2013 #14
    Ah OK, gotcha.
     
  16. Jun 24, 2013 #15

    lavinia

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    As Micromass said, a space filling curve can not be smooth. an interval completely onto an area of a volume.

    With smooth curves the best you can do is a curve whose image is dense in the manifold. This can be done even without self-intersections. The standard example with a geodesic is this.

    The surface is a flat torus that is the quotient of the unit square in the plane. The geodesic is the projection of a straight line that intersects the x-axis at an irrational angle.
     
  17. Jun 30, 2013 #16

    lavinia

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    It would be interesting to study this. In the case of a sphere the fundamental polygon is an n-1 sphere and all of the points on it are identified to a single point. The exponential map is singular on the fundamental n-1 sphere. On the flat torus, the exponential map is non-singular - also on a Riemann surface of constant negative curvature.

    I think there are two types of points on the fundamental polygon: points where the exponential map is singular and point where it is not but beyond which the radial geodesic through it is no longer length minimizing.
     
  18. Jun 30, 2013 #17

    lavinia

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    So here is a related question. Can a curve like a loxodrome be a geodesic in some metric on the sphere? This is the curve that cuts latitude lines at a constant angle. It winds around the poles infinitely many times but has finite length in the standard metric on the sphere.
     
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