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Why is the cusp not a submanifold?

  1. Oct 3, 2008 #1

    quasar987

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    My book says that the cusp y=x^2/3 is not an embedded submanifold of R². Why is that?
     
  2. jcsd
  3. Oct 4, 2008 #2
    According to my notes the embedding is defined like this:

    Let M and N be differentiable manifolds, and [tex]f:M\to N[/tex] a smooth ([tex]C^{\infty}[/tex]) mapping. If for all [tex]p\in M[/tex] the tangent space mapping [tex]f_{*p}:T_p(M)\to T_{f(p)}(N)[/tex] is injective, and [tex]f:M\to f(M)[/tex] is a homeomorphism when [tex]f(M)[/tex] has the induced topology from N, then [tex]f[/tex] is an embedding of M in N.

    If we set the natural differentiable structures on [tex]\mathbb{R}[/tex] and [tex]\mathbb{R}^2[/tex], then a mapping

    [tex]f:\mathbb{R}\to\mathbb{R}^2,\quad\quad f(x)=(x, (x^2)^{1/3})[/tex]

    is not an embedding, because it is not smooth at origo.
     
  4. Oct 4, 2008 #3

    quasar987

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    True, but an embedded submanifold is by definition (or characterisation) the image of a smooth embedding. Couldn't there be a pair (M, f) other than M=R and f given in your post such that f(M) = the cusp, and such that f is an smooth embedding?
     
  5. Oct 4, 2008 #4
    So is the real question this: We give [tex]\mathbb{R}^2[/tex] the natural differentiable structure, define [tex]f:\mathbb{R}\to\mathbb{R}^2[/tex] like in my post, and ask that does this mapping somehow induce a differentiable structure on [tex]\mathbb{R}[/tex] so that [tex]f[/tex] becomes smooth?

    Or perhaps the question is, that why cannot this [tex]f[/tex] induce differentiable structure in such way, assuming that the book's claim is right?
     
  6. Oct 4, 2008 #5

    quasar987

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    The real question is this:
    Consider R² with the natural differentiable structure, and C the subset of R² defined by the equation y=x^2/3. Is there a smooth manifold M and an embedding f:M-->R² whose image is C.
     
  7. Oct 4, 2008 #6

    mathwonk

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    every line through the origin of that set has intersection number ≥ 2 with the set. for a manifold, the generic intersection number will be one.
     
  8. Oct 5, 2008 #7
    Isn't the content of the posts #4 and #5 the same?
     
  9. Oct 5, 2008 #8

    quasar987

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    Well, the cusp is a smooth manifold, with smooth atlas consisting of the unique chart "projection onto the x coordinate".

    Well, it seems to me that the question in post #5 is more general than any of the 2 questions of post #4. Or perhaps they are equivalent?
     
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