Talk:Geodesic convexity

Latest comment: 1 year ago by Bernanke's Crossbow in topic Disputed

Relation to convex metric space edit

I realize that Riemannian metrics are not the same thing as distance metrics, but the definition of convexity here is almost the same as that in convex metric space (or rather, not the main definition there, but the related one mentioned in the article and used more explicitly e.g. in injective metric space that any pair of points can be connected by the isometric image of a line segment). Should there be a closer connection between these articles? Merge, maybe? Or rename this article geodesic metric space, discuss in it non-Riemannian as well as Riemannian metrics, clarify the difference between that and the notion of convexity discussed in convex metric space, and link articles that need this version of the concept, such as injective metric space, to here instead of there? —David Eppstein 17:09, 8 August 2007 (UTC)Reply

I agree wholeheartedly. Having a page devoted to the property of "Geodesic convexity" rather than about "Spaces with convex geodesics" is the problem. The argument in the "disputed" section below seems to be an artifact of trying to use a particular definition of convexity without having nailed down the space we're talking about (i.e. Riemannian manifold or length metric). Tom3118 (talk) 13:30, 8 September 2010 (UTC)Reply

Disputed edit

Where does the quoted definition come from?

It does not say that it is required for the geodesic to be minimizing the distance between the two given points (i.e. the length of the arc is required to be the distance between the two points). It is not mentioned if other geodesics (minimizing or not) are allowed connecting the two points, and if allowed, if these must also lie completely within C.

The example with the "big" disc A inside S2 only makes sense if one requires the geodesics to be minimizers.

In the book Chavel, Isaac (1993), Riemannian geometry – A modern introduction, Cambridge University Press, three distinct definitions are given:

  • weakly convex: for any two points from C there exists exactly one minimizing geodesic in C connecting them
  • convex: for any two points in C there exists exactly one minimizing geodesic in M connecting them, and that geodesic arc lies completely in C
  • strongly convex: for any two points in C there exists exactly one minimizing geodesic in M connecting them, and that geodesic arc lies completely in C; and furthermore there exists no nonminimizing geodesic inside C connecting the two points.

Consider the open northern hemisphere of S2. It is strongly convex. Now add two antipodal points from the equator. The resulting set is only weakly convex. The closed northern hemisphere (whole equator included) is not even weakly convex.

Other definitions are possible.

/84.238.115.151 (talk) 00:06, 15 April 2009 (UTC)Reply

Sorry, the upper hemispere plus two antipodal equatorial points is not weakly convex in this sense. Consider instead for example half a great circle including endpoints. /84.238.115.151 (talk) 20:50, 15 April 2009 (UTC)Reply
I'm assuming that if A and B are geodesically convex, we want their intersection to be geodesically convex as well. That means that if p and q are points in the intersection of A and B, I would expect the set of geodesics connecting p to q in A to be the same as the set of geodesics connecting p and q in B, so that when we intersect A and B that same set of geodesics is contained in the result, making it geodesically convex. That is not the case with the current definition on the page, so it apparently violates the principle of preservation of convexity under intersection. Probably the best way to fix that is to do as suggested above and require that the geodesics to be length minimizers. In certain cases it might work to define a set as geodesically convex if and only if for every two points it contains, it contains every geodesic between them - that would also lead to a geodesic convexity preserved under intersection. —Preceding unsigned comment added by 128.61.118.127 (talk) 15:11, 23 November 2009 (UTC)Reply
I don't think it is a good idea to require a convex set to contain every geodesic between two of its points. for example, on a sphere it will mean that every convex set with at least two points have to be the whole sphere. 79.178.51.153 (talk) 15:01, 24 December 2010 (UTC)Reply
There are even weaker versions of convexity, only requiring the following:
for any two points from C if there exists exactly one minimizing geodesic between them, it must be contained in C.
As an example where such notion appears, see [1] for the case of the sphere.160.45.109.217 (talk) 17:33, 8 November 2012 (UTC)Reply
I ran a quick survey of some common differential geometry textbooks, to try to see the consensus of the literature. Here's the results:
  • do Carmo, Manfredo (1976). Differential Geometry of Curves and Surfaces. Prentiss-Hall. p. 303. One natural question about Prop. 1 is whether the geodesic of length less than δ which joins two points q1, q2 of W is contained in W. If this is the case for every pair of points in W, we say that W is convex.
  • Petersen, Peter (2016). Riemannian Geometry. GTM 171 (3rd ed.). Cham, Switzerland: Springer. p. 463. doi:10.1007/978-3-319-26654-1. ISBN 978-3-319-26654-1. A subset AM of a Riemannian manifold is said to be totally convex if any geodesic in M joining two points in A also lies in A: There are in fact several different kinds of convexity, but as they are not important for any other developments here we confine ourselves to total convexity.
  • Jost, Jürgen (2017). Riemannian Geometry and Geometric Analysis. Universitext (7th ed.). Cham, Switzerland: Springer. p. 570. doi:10.1007/978-3-319-61860-9. ISBN 978-3-319-61860-9. We call a closed subset A of a Riemannian manifold N convex if any two points in A can be connected by a geodesic arc in A. We call A strictly convex if this geodesic arc is contained in the interior of A with the possible exception of its endpoints. We call A strongly convex if its boundary A is a smooth submanifold (of codimension 1) in N and if all its principal curvatures w.r.t. the normal vector pointing to the interior of A are positive.{{cite book}}: CS1 maint: multiple names: authors list (link)
  • Spivak, Michael (1999). A Comprehensive Introduction to Differential Geometry. Vol. 1 (3rd ed.). Houston: Publish or Perish. p. 363. A set UM is geodesically convex if every pair q,q′∈U has a unique geodesic of minimum length between them, and this geodesic lies completely in U.
  • Shoshichi Kobayashi; Katsumi Nomizu (1963). Foundations of Differential Geometry. Vol. 1. New York, NY: Interscience. p. 149: "U(x0;ρ) is convex in the sense that any two points of U(x0;ρ) can be joined by a geodesic which lies in U(x0;ρ)." p. 166: "We now proceed to prove the existence of a convex neighborhood around each point of a Riemannian manifold in the following form....(1) Any two points of U(x;ρ) can be joined by a unique minimizing geodesic; and it is the unique geodesic joining the two points and lying in U(x;ρ); (2) In U(x;ρ), the square of the distance d(y,z) is a differentiable function of y and z."
  • Nicolaescu, Liviu I (2021). Lectures on the Geometry of Manifolds (3rd ed.). Hackensack, NJ/Singapore: World Scientific. p. 154. ISBN 9789811214820. A subset UM is said to be convex if any two points in U can be joined by a unique minimal geodesic which lies entirely inside U.
  • Hawking, S. W.; Ellis, G. F. R. (1973). The Large-Scale Structure of Space-Time. Cambridge Monographs on Mathematical Physics. Cambridge University Press (published 1994). p. 34. ISBN 0-521-09906-4. Further, one can choose Np to be convex, i.e. to be such that any point q of Np can be joined to any other point r in Np by a unique geodesic starting at q and totally contained in Np.
I conclude that the article does in fact use the correct definition. Bernanke's Crossbow (talk) 21:00, 15 September 2022 (UTC)Reply

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