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Silly question about the Theorema Egregium

  1. Mar 19, 2013 #1

    quasar987

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    So the Theorema Egregium for a surface is remarkable because it expresses the Gaussian curvature K (a priori extrinsic) in terms of the metric tensor, which is intrinsic to the surface hence K is intrinsic as well after all. Here "intrinsic" means that a 2d creature living in the surface can compute it by making measurements solely in the surface.

    So how does a creature computes the metric? I thought that it does so by computing the lengths of curves and then use:
    [tex]
    \left.\frac{d}{dt} L(\gamma)|_{[-\epsilon,t)}\right|_{t=0}=\frac{d}{dt}\int_{-\epsilon}^t||\gamma'(\tau)||d\tau=||\gamma'(0)||
    [/tex]
    But how does a creature in the surface measure lengths? It chooses some small stick lying in the 2d forest behind his 2d house to use as "unit of length", and then he measures. But then the length of curves he gets, and hence also the derivative of this length will in general be some multiple of the same derivative computed in the units of the 3d space in which the surface is embedded (i.e. the LHS above). So a creature can only measure the metric up to a constant (homethety). Am I wrong?
     
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  3. Mar 20, 2013 #2
    The theorema allows you to forget about the ambient space, because the curvature K of the surface will be the same whether you obtain it extrinsically or intrinsically. So there is no more valid measure in the ambient space than in the surface, the metric is intrinsic, the measure of a curve in the surface by the bug living in 2d and the measure of that curve constrained to a surface in 3d ambient space must coincide, that is what's "egregium" about the theorem.
     
  4. Mar 20, 2013 #3

    quasar987

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    This is what I'm trying to understand. Why is this so? What would be a procedure that can be carried out by a 2d being living in the surface that will allow him to compute the induced metric g.
     
  5. Mar 20, 2013 #4
    Because Carl and Bernhard said so. :tongue2:

    Measure angles of triangles and see if they add up to, more or less than 180º, measure length of circunferences and see if they measure more or less than 2πr...
     
  6. Mar 20, 2013 #5

    Ben Niehoff

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    All you're doing here is observing that length is a dimensionful quantity. If we scale our units of measurement by ##a##, then arc lengths (in these units) scale by ##1/a## and curvatures scale by ##a^2##.

    We could just as well ask how to measure distances in the ambient 3d space, if we are not given the metric tensor a priori. You will find the same ambiguity. For the intrinsic Gauss curvature to agree with the product of the extrinsic principle curvatures, the 2d beings and the 3d beings need to agree on a standard measuring stick.
     
  7. Mar 21, 2013 #6

    lavinia

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    The creature living on the surface has a measuring stick already. He is also able to measure angles between two sticks.

    Just as in the Theory of Relativity, where one finds curvature by comparing the circumference of a sphere to its geodesic radiius, one does the same thing on a surface for the circumference of a circle. (here as usual, a sphere or circle is a set of points equidistant from a central point.)

    So curvature on a surface is intrinsic to measurement on the surface itself.

    The Theorem Egregium shows that the determinant of the Gauss mapping can be computed from the metric on the surface instead of from the differential of the Gauss map. One then shows that the creature would compute it from the ratios of circles to their radii.


    ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

    Suppose the surface is embedded in 4 space rather than 3 space. Then there is no unit normal but rather a normal plane. How would you compute the Gauss curvature extrinsically in this case?

    Suppose instead of a surface one has a hypersurface of n dimensional Euclidean space. A hypersurface has a unit normal so there is a Gauss map onto the n-1 dimensional unit sphere.
    Is the determinant of the Gauss map still intrinsic?
     
    Last edited: Mar 21, 2013
  8. Mar 21, 2013 #7

    quasar987

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    I see. So saying g is intrinsic is a tautology. When we say something is intrinsic to S, we really mean "to (S,g)".
     
  9. Mar 21, 2013 #8

    lavinia

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    Right. The metric is intrinsic.

    When a manifold is embedded in another manifold it inherits a metric from the ambient manifold. But this inherited metric is intrinsic in itself and is the same when the embedding is changed by an isometry, e.g a bending of the manifold.

    Historically it seems that surfaces were seen as subspaces of 3 space and their shape derived intuituvely from the way they curved around. People measured this curving from the derivative of the unit normal vector field much in the same way that the curving of curves in the plane were measured from the derivative of the unit tangent or equivalently the unit normal.It took time to realize that the Gauss curvature could be measured without the unit normal.
     
  10. Mar 21, 2013 #9

    lavinia

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    It is interesting that the principal curvatures, the maximum and minimum curvatures of curves passing through a point, are not intrinsic yet there product which is the Gauss curvature is intrinsic.

    The curvature of a curve can not be derived from the curve itself. And geodesics which may be curved in 3 space will appear straight to the surface creature.
     
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