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A Opposite "sides" of a surface - Differential Geometry.

  1. Dec 24, 2015 #1
    How, if at all, would differential geometry differ between the opposite "sides" of the surface in question. Simplest example: suppose you look at vectors etc on the outside of a sphere as opposed to the inside. Or a flat plane? Wouldn't one of the coordinates be essentially a mirror while the other remained the same? Would the affine connection change at all? Is this at all related to the differences between a covariant and contra variant vector?
     
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  3. Dec 24, 2015 #2

    WWGD

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  4. Dec 25, 2015 #3

    WWGD

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    Ultimately, as I understand it, a surface S embedded in a manifold M is one-sided if S has a trivial normal bundle, which is equivalent to being co-orientable, meaning sort of being orientable within the ambient space. A coorientation is a smooth assignment of a (unit) normal vector field at each point of S .I believe this is equivalent to having a continuous non-tangent vector field . I don't understand what the question about the connection or the coordinates is about. Of course, normality is itself a function of the choice of Riemannian metric, but I don't get the question. As I understand it, a "side" is given by a choice of coorientation, tho maybe someone can chime in and add something here.
     
    Last edited: Dec 25, 2015
  5. Dec 25, 2015 #4

    lavinia

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    Trivial normal bundle is correct. For a smooth compact hypersurface,##S##, with trivial normal bundle, a tubular neighborhood is diffeomorphic to ##S##x##[-1,1]##. The two pieces, ##S##x##[-1,0]## and ##S##x##[0,1]## define the two sides.

    On the other hand, if the normal bundle is not trivial, then the hypersurface is not two sided. So trivial normal bundle is both necessary and sufficient.

    A hypersurface of Euclidean space is always two sided because it separates Euclidean space into two disjoint pieces. One of these pieces is a bounded domain and intuitively can be thought of as the interior region defined by the hypersurface. One might generalize this to compact manifolds ,##S##, that are boundaries of one higher dimensional manifolds,##M##. In this case, normal bundle to ##S## is automatically trivial and ##M## can be thought of as the interior region defined by ##S##. (This does not mean that the boundary manifold has to be orientable. For instance, the Klein bottle is the boundary of a 3 dimensional manifold. ). This construction does not work in general, because there are many manifolds that are not boundaries e.g. the projective plane.
     
    Last edited: Dec 25, 2015
  6. Dec 25, 2015 #5

    lavinia

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    A compact surface in 3 space divides 3 space into two disjoint regions, one bounded the other unbounded. These two regions are the two sides that you were thinking of. A compact hypersurface of a higher dimensional Euclidean space will also divide the Euclidean space into two disjoint regions.

    AS WWGD has rightfully pointed out, two sidedness really has to do with whether there are well defined opposite directions normal to the hypersurface (trivial normal bundle) and this condition generalizes the case of hypersurfaces of Euclidean space. A hypersurface of a general manifold even if it is two sided may not separate the manifold into two disjoint pieces.

    - Two sidedness has nothing to do with affine connections on the manifold. It is a question of Differential Topology, not of Differential Geometry. Why do you think this has anything to do with affine connections?
    - The question about whether two sidedness has anything to do with the difference between covariant and contravariant vectors doesn't seem to make sense. Can you explain what you were thinking of?
     
    Last edited: Dec 25, 2015
  7. Dec 25, 2015 #6

    berkeman

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    Thread closed for Moderation...

    EDIT: an impolite exchange has been removed and the thread will remain closed.
     
    Last edited by a moderator: Dec 25, 2015
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