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Bending of manifold and coordinate change

  1. Jun 20, 2010 #1
    I am new to this subject of topology. I want to know if bending and stretching of a manifold is same as a general transformation of a coordinate system drawn on the manifold. Or the mathematical definition of bending and stretching shall equally help.
     
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  3. Jun 20, 2010 #2

    quasar987

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    Deforming a local coordinate system of a manifold, you only get a new coordinate system for the same manifold.

    Mathematically, I would describe the bending or stretching of a manifold M embedded in some R^n as an homotopy h:[0,1] x M --> R^n which is an embedding at all times.

    Given a metric on the manifold, there may be a more intrinsic description of bending\stretching.
     
    Last edited: Jun 20, 2010
  4. Jun 20, 2010 #3
    There is then another result you may be able to use here:

    Say M is your manifold in its initial state, and M' is the "final"

    state of the manifold, after doing some bending and stretching.

    If the bending and stretching are done homeomorphically, i.e.,

    if there is a homeo. h between M and M' , then you can use h

    to pullback the charts of M into those of M'. If the bending and

    stretching are not a homeomorphisms, i.e., if the homotopy that

    Quasar described is not an isotopy, then you are dealing with a

    manifold M' that is topologically different from M, i.e., there may

    not be atlases for M' that are compatible with those for M.
     
  5. Jun 24, 2010 #4

    lavinia

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    coordinate transformations have nothing to do with the shape of the manifold. Stretching is a change of shape and must be represented as a change of metric not as a coordinate transformation.

    Bending has meaning only for an embedded manifold. Here the bending does not change the metric but rather changes the embedded curvature - which is an extrinsic geometrical property of the embedding. Again this has nothing to do with coordinate transformations.
     
    Last edited: Jun 24, 2010
  6. Jun 24, 2010 #5
    "quasar" and "lavinia"
    Here is my real problem. Consider a n-plane. And take two parallel lines on it. Any deformation of this plane will not affect the relation between lines- that they never meet.
    The riemann curvature tensor (or some tensors of raychaudhari equation), which governs this relation and is zero in n-plane, should vanish even in deformed n-surface. So vanishing of tensors can be obtained by any coordinate transform of the coordinate on plane (except the case of bending). Unless there is some other transform too, after which the tensors vanish, I am having quite a difficulty in understanding that deformation is not a coordinate transform.
     
  7. Jun 24, 2010 #6
    "lavinia"
    Given a new metric g'ab and an old one gab, both infinitesimally different, we can define lie derivative of a vector field 'Zi' with respect to the metric g to be (g'-g)/e , where e ->0, and transform coordinates by xi -> xi + eZi , e->0. With this transform we can get the original metric, for deformed manifold, thus identifying stretching again with coordinate change.
    ( I am not sure if i am correct or not. Expert's comments required).
     
  8. Jun 24, 2010 #7

    lavinia

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    I am not sure about this. Can you elaborate with more detail what you are talking about? Maybe we can just figure the whole thing out together.
     
  9. Jun 25, 2010 #8
    "lavinia"
    Take a manifold, and a vector field 'f' on it. Moving the existing coordinate system along the field changes the tensors on the manifold(transformation). Given a desired change of a tensor field( eg. metric tensor), a vector field 'f' can always be found on it. So if stretching is a change in metric tensor, then a coordinate change can be identified which does the same.
    Now bending of an embedded manifold does not change local properties of the manifold (at least when embedding is done in euclidean space).

    "lavinia, bacle, quasar"
    Again, consider gravity. It is simply the ability of space to converge (or diverge?) to initially parallel geodesics. So there could be a classification of spaces, based on what they do with geodesics. Since this property is not affected by stretching or bending (bending case is clear), and as i have heard, such deformations preserve the topology of the space, is this classification the topological classification? Example, plane and sphere. We know what they do with their geodesics.
    And since converging of geodesics is determined by tensors, which preserve their properties upto arbitrary coordinate change, so I thought coordinate change and stretching might be related.
     
  10. Jun 28, 2010 #9

    lavinia

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    I still need more detail. How do you move a tensor along a vector field? I do not know what that means. Can you give an example?
     
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