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Math Help - Polemic problem in differential geometry

  1. #1
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    Polemic problem in differential geometry

    Hello

    I was asked in a test to do an integration of an exact form on a manifold. The first thing you think is to apply the Stokes theorem. The problem here is to determine the boundary of the manifold. Let S=\{(x,y,z,w):x^2+y^2+z^2+w^2=2\} and let T^2=\{(x,y,z,w):x^2+y^2=z^2+w^2=1\}. Now, is the boundary of the component of S-T^2, such that x^2+y^2<1, empty?

    My teacher says it's not empty, but a fellow PhD of mine says it's empty. I don't know who's right. Can anyone find a way to decide this without any doubt remaining?

    Thanks in advance.
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  2. #2
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    Re: Polemic problem in differential geometry

    T^2 is closed. So S-T^2 is open. What is the definition of the boundary of an open manifold? It must be empty.
    Last edited by xxp9; December 1st 2012 at 09:03 AM.
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    Re: Polemic problem in differential geometry

    S is compact in R^4 and thus closed, so you can consider S, as a space itself with the subspace topology. In it, S is open and T^2 is closed, thus S-T^2 is open. However, it's open in the subspace topology, not necessarily in R^4. For example, I^2=[0,1]X[0,1] is open as the whole space. Take [1/2,1]X[0,1] out of [0,1]X[0,1]. The result is an open set in I^2, but not in R^2. A similar example with disks can be made to fit in the context of smooth manifolds.

    It's tricky. Help please!
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    Re: Polemic problem in differential geometry

    You don't need to( or you should not always) consider a manifold as an embedded sub-manifold of R^n. So my argument from the first reply works well.
    For your example, what is the boundary of the result?
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    Re: Polemic problem in differential geometry

    In my example, the manifold isn't smooth. But consider this example: [0,1) is a smooth manifold. [0,1/2) is open in [0,1). Its boundary is {0}.
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    Re: Polemic problem in differential geometry

    This is because that [0,1) itself has a boundary. So an open sub-manifold of [0,1) can have a boundary.
    However S itself has no boundary.
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    Re: Polemic problem in differential geometry

    I'm sorry, but you haven't proved what was asked so far.
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    Re: Polemic problem in differential geometry

    Let me prove the statement: an open sub-manifold M of a (boundaryless) manifold N has no boundary.
    Suppose M has a boundary and p is a point on the boundary, there is a chart (U, \phi) of N around p such that \phi: U\cap M \rightarrow R^n_+=\{(x^1, ..., x^n) \in R^n | x^n \ge 0 \} is a diffeomorphism, and \phi(p)=0. However since M is open in N, there is an open set V of N around p such that V \subset M\cap U . The image \phi(V) is an open set around 0 in R^n, so must contain a point with x^n < 0, which contractdicts the fact that \phi(V) \subset R^n_+
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  9. #9
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    Re: Polemic problem in differential geometry

    Thank you very much for your effort and help.
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