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Why the Gaussian curvature is extrinsic ?

  1. Nov 21, 2011 #1
    Hi all!
    Let's start from the begin to see where I get lost.

    Extrinsic curvature defines the way an object relates to the radius of curvature of circles that touch the object (a couple of further nicer definitions come from physics, for the moments I am not mentioning them), and intrinsic curvature defines each point in a Riemannian manifold.

    A Mean curvature locally describes the curvature of an embedded surface.
    [itex]1/2 (k_1 + k_2)[/itex] as extrinsic.

    A Gaussian curvature is defined as an intrinsic curvature as [itex] k_1 \times k_2 [/itex], the reason of this is just that Gaussian curvature involves surfaces ? Then why cannot we define the mean curvature as well as intrinsic
    Last edited: Nov 21, 2011
  2. jcsd
  3. Nov 21, 2011 #2
    The curvatures k1 and k2 ARE extrinsic curvatures.

    It's not at all obvious that their product is intrinsic. It took the genius of Gauss to figure that out. For a good explanation, you can consult Geometry and the Imagination by Hilbert and Cohn-Vassen. I forget where in the book it is, but it's not hard to find. There's a section on differential geometry.
  4. Nov 21, 2011 #3
    I was wondering that a layman explanation can be related to the fact that the principal curvature and their mean are not describing very well intrinsic behaviour of the surface while a product of them can :

    If the product is positive we are in a parabolic point, if the product is negative we are in a hyperbolic point
  5. Nov 21, 2011 #4
    Basically, what you want to show is that the mean curvature is not invariant under bending.

    For this, you only need to curl a piece of paper so that it becomes cylindrical. One of the principal curvatures will increase and the other will stay equal to zero, so the mean curvature changes, but the Gaussian curvature stays the same.
  6. Nov 22, 2011 #5


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    what about the product of the principal curvatures for hypersurfaces in higher dimensions?
    Is it intrinsic? Say a 3 manifold in R^4 or a 10 manifold in R^11.
    Last edited: Nov 22, 2011
  7. Nov 22, 2011 #6
    Damn it! Nice question :biggrin:

    You destroyed our observations, at this point I have no idea. Or we assume my and homeomorphic's observation valid only on 3D representations
  8. Nov 22, 2011 #7
    Yes, from context, I interpreted the question as being about surfaces embedded in R^3.
  9. Nov 22, 2011 #8
    Ok, ok ok ... I have "Geometry and the Immagination" in front of me.
    (I summarized a bit)
    Pag. 178: arc length does not make any reference to a particular system of coordinate. For this reason arc length is considered "natural" or intrinsic coordinate of a curve

    Pag. 205 The Gaussian curvature is invariant to bending and must depend on the arclength and angle of the curve lying on it. For this reason Gaussian curvature, and its analogue in higher dimension are "intrinsic" on the surface.

    Therefore, according to Hilbert-Cohn-Vossen, the definition of invariant to bending should hold in embedding in R^n also with n>3.
  10. Nov 22, 2011 #9
    Invariant to bending is sort of the way to think of it, but more precisely, the idea is that it is invariant under isometry, which is a map that preserves the metric.

    But the idea of principal curvatures is a little different in higher dimensions. If you have a 2-d surface embedded in R^n, you can cut it up along many different planes, each of which will intersect the surface in an arc (generically) with some value of curvature. The definition of Gaussian curvature as the product of the principal curvatures will only work for a surface embedded in R^3.

    If it's embedded in a bigger R^n, it will still have a Gaussian curvature that is intrinsic, but you can't compute it the same way.

    It's a good exercise to think about what will happen.

    When they say analogue of Gaussian curvature in higher dimensions, I assume they mean Riemannian curvature, which is a more complicated beast than Gaussian curvature.
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