5
$\begingroup$

Maybe this question is so clear or maybe it is not exact. It is because of my very little knowledge of differential geometry. I am reading some material in this field and I got a question which seems has some meaning, but I am not sure.

Is it true that any simple curve $\gamma$ in the plane is geodesy of a finite volume Riemannian manifold $M$ in which its curvature is not zero? If this is true, is there any strategy to construct at least one manifold for that simple curve?

$\text{Added later by @Anton Petrunin suggestions:}$

$\gamma$ is orthogonal projection to the plane and $M$ is a closed surface.

I think the answer is positive, but I do not have any idea (except imagination) to prove it.

Actually, I did many searches and I did not find any answer in papers or books.

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ Related: Is every closed curve in 3D a geodesic on a genus-0 surface?. $\endgroup$
    – Joseph O'Rourke
    Commented Apr 8, 2019 at 13:40
  • 1
    $\begingroup$ You can take your simple curve (if it is embedded) to a circle by a diffeomorphism of the plane, which is the identity outside a compact set, and then bend your plane into a light bulb shape. Is that what you are looking for? The details of such an argument would be long, I suppose. $\endgroup$
    – Ben McKay
    Commented Apr 8, 2019 at 16:00
  • 1
    $\begingroup$ The formulation is very unclear. Did you want say that $\gamma$ is orthogonal projection to the plane and $M$ is a closed surface? $\endgroup$
    – Anton Petrunin
    Commented Apr 8, 2019 at 18:06
  • 2
    $\begingroup$ I believe that it is known (probably from Grayson's work on shrinking of curves) that, for any embedded closed smooth curve in the plane, there is a diffeomorphism of the plane taking that curve to a circle. From there you can be quite explicit, mapping the plane to, for example, a punctured sphere, so that the circle is mapped to a great circle, by projection from the north pole of the sphere (Ptolemaic projection). $\endgroup$
    – Ben McKay
    Commented Apr 8, 2019 at 19:47
  • 1
    $\begingroup$ Well, then you should change the question (others do not understand it still). $\endgroup$
    – Anton Petrunin
    Commented Apr 9, 2019 at 23:56

1 Answer 1

Reset to default
1
$\begingroup$

So $\gamma$ is a simple plane curve; let us assume it is smooth and regular (otherwise the question has no sense).

Note that the curve $\gamma$ is a geodesic on the cylindrical surface $\mathbb{R}\times \gamma$.

This surface is has infinite area, but it is easy to fix, cut it by planes $z=\pm1$ and smooth the corners (it can be done explicitely in a tubular neighborhood of the corner).

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .