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Differential Topology

  1. Jan 6, 2007 #1
    I had to ask myself two simple problems in differential topology:

    1) Why is the rank of a diffeomorphism (on a manifold of dimension m) of rank m?

    2) Why is a chart on a manifold an embedding?

    These are actually quite obvious so textbooks don't even bother proving it. So I've attached my own proofs (note that the Inverse Function Theorem goes only one way, so I couldn't use it to prove #1). My proofs probably seem longer than they need to be, but I wanted to be completely rigourous. Are they correct?
     
    Last edited: Jan 22, 2008
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  3. Jan 6, 2007 #2

    matt grime

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    A diffeomorphism is by definition invertible. Why would you want to invoke the inverse function theorem?

    A chart on is an open set U<M, and a homeomorphism f:U-->R^n. Are you asking why f is an embedding, or why the inclusion of U in M is an embedding?
     
    Last edited: Jan 6, 2007
  4. Jan 6, 2007 #3
    Inverse function theorem: If f:R^m -> R^m has rank m, then f is a local diffeomorphism.

    The theorem does not state the converse, so I only considered using it, but was not allowed to and so didn't.

    The former. I just didn't want to type all that mess in my intro. My pdf file is more clear. You got me thinking about the latter question as well, and I worked out a quick proof in minutes and am sure I got it right (the coordinate expression of the inclusion map turns out to be just the identity and so the Jacobian matrix is just the identity matrix, making the inclusion map an immersion).
     
    Last edited: Jan 6, 2007
  5. Jan 7, 2007 #4

    matt grime

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    I still don't see what you're getting at in 1). If f is a diffeo then it is invertible with differentiable inverese. The Jacobian of the inverse is the inverse of the Jacobian, so it had better have rank m otherwise it isn't invertible. (Just linear algebra).

    I mistyped in the other part - I shouldn't have said homemoprhism - f is just a smooth map onto some assigned open subspace of R^n, not necessarily all of R^n itself - could be the open ball or something, but that is diffeomorphic with R^n so it doesn't matter.

    Anyway, the point is that f is a diffeo onto its image (that f is invertible on its image is part of the definition of f), so that's an embedding.
     
    Last edited: Jan 7, 2007
  6. Jan 7, 2007 #5
    Yes, I did precisely everything you said in my proofs, but I had to use charts because f is acting on a manifold, and not on R^n. You are bypassing the use of charts and giving the intuitive proof. I was just being more rigorous (perhaps somewhat superfluous as well, but that's my style). By the way, in the second question, I never said "differentiable manifold" but only "manifold", so f could not be assumed to be a diffeomorphism onto its image but only a homeomorphism onto its image, so I had to prove separately that f was an immersion to prove that it is an embedding.

    Thanks for your alternative quick approach.
     
    Last edited: Jan 7, 2007
  7. Jan 7, 2007 #6

    matt grime

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    For 1, how have I bypassed charts? The use of charts is implicit because rank is a purely local definition.

    In 2, what precisely are the definitions you're using?
     
    Last edited: Jan 7, 2007
  8. Jan 7, 2007 #7
    Let (U,f) be a chart on a topological manifold M. f is by definition a homemorphism onto a subset of R^m.
    embedding = homeomorphism onto its image that is an immersion (i.e. has rank m).

    Actually, f has to be differentiable in order for it to be an immersion, so you are right that f is differentiable, but I could not assume that its inverse is differentiable (a differentiable homeomorphism is not necessarily a diffeomorphism). So I had to use the homeomorphism property of f to prove that it has rank m in order to prove f is an embedding. Anyways, it was trivial to do that so the problem is finished.
     
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