Q: convexity of level sets

January 29th, 2008 ~ Posted in: Mathematics

When are the levels sets of an arbitrary function the boundary of some convex set? I.e. if I give you the level set \{ x\,:\, \phi(x) = C \} , what kind of conditions on \phi and C guarantee this?

I’ve been spending a lot of time looking at the equation y^2 = 2(C + \omega^2 \cos(x)) lately. From Mathematica’s implicit plot, it sure looks like an ellipse (locally). So I’m wondering what kind of results you can get about level sets.

Thinking out loud, for one dimensional level sets in 2d, one condition that looks like it might be necessary and sufficient if you assume the level set is a smooth manifold is that the gradient on the level set has to be a one to one function onto the unit circle. I say this because it looks like you could stretch the unit circle onto the boundary of an arbitrary convex set by just growing it at different rates in the direction of its outward normal.

On a related note, are there higher dimensional analogs to the Jordan Curve Theorem: if it’s hard to show that a simple closed curve separates the plane into two disjoint regions, is it harder to show compact submanifolds do the same to space? I’ve not actually seen the proof of the JCT, so I don’t know if the arguments used are intrinsic to the plane, or more generally applicable.

Mathematics ~

5 Responses to “Q: convexity of level sets”

  • 1. ObsessiveMathsFreak
    January 30th, 2008 at 8:47 am

    Thinking out loud, for one dimensional level sets in 2d, one condition that looks like it might be necessary and sufficient if you assume the level set is a smooth manifold is that the gradient on the level set has to be a one to one function onto the unit circle.

    You may be right I think. I recall problems with “slowness surfaces” in ray theory in which, as you move around the surface, you trace out the normal vectors of the slowness curve/surface, to get the normal curve/slowness.

    Essentially, when you trace out these normal vectors, any concavity in the slowness surface leads to cusps/overlap in the normal surface.

    Going on a differential geometry type route, I think a sufficient condition for concavity could be that the curvature never becomes zero, but this is not a necessary condition I think.

    A necessary and sufficient condition might be that the tangent, normal and binormal triplet (Fresnet frame), never changes orientation as you traverse the curve. I think it can still become zero though. But how would you deal with polygons in this case?

    Perhaps for 2d curves, you could set the binormal to be (0,0,1) throughout. Then I think the orientation condition will work, with the orientation being zero for straight lines.

  • 2. Fanfan
    January 31st, 2008 at 2:32 am

    The definition you are looking for is “quasi-convexity”.

  • 3. Alex
    February 1st, 2008 at 1:16 am

    OMF:

    Frenet frame orientation doesn’t change orientation? Isn’t that always true, by definition?

    As for the curvature idea, you’re correct: looking through Thorpe’s differential geometry book, in the chapter on convex surfaces, there’s this theorem:

    Let S be a compact connected oriented n-surface in R^(n+1) whose Gauss-Kronecker (?) curvature is nowhere zero. Then
    1) the Gauss map is a bijection
    2) S is strictly convex

  • 4. ObsessiveMathsFreak
    February 1st, 2008 at 11:16 am

    If the curvature becomes zero then there is no set fresnet frame. Think of a straight line. The tangent is set, but you can whirl around the normal and binormal vectors as much as you like, and even change their orientation.

  • 5. Captain Strychnine
    February 6th, 2008 at 3:56 am

    The analog of the Jordan Curve Theorem is the Jordan-Brouwer Separation Theorem. A submanifold M of R^n homeomorphic to S^(n-1) separates R^n\M into two connected components, one bounded and one unbounded, each having M as their boundary.

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