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Differential Geometry Theorem on Surfaces

  1. Jun 18, 2009 #1

    I am having difficulty understanding the proof of the following theorem from Differential Geometry


    [tex]S\subset \mathbb{R}^3[/tex] and assume [tex]\forall p\in S \exists p\in V\subset\mathbb{R}^3[/tex] [tex]V[/tex] open such that

    [tex] f:V\rightarrow\mathbb{R}^3[/tex] is [tex]C^1[/tex]

    [tex] V\cap S=f^{-1}(0)[/tex]

    [tex]\forall\in V\cap S,\quad \nabla_x f\neq 0[/tex]

    [tex]\Rightarrow f[/tex] is a surface

    2. Relevant equations

    3. The attempt at a solution

    The proof that I’m trying to understand goes as follows

    Write out [tex] \nabla_p f[/tex] in coordinates, since [tex]\nabla_p f[/tex] is non-zero at least one of the components of the gradient is non-zero. W.L.O.G take it to be the z coord

    [tex] \frac{\partial f}{\partial x_3}(p)\neq 0[/tex] so that

    [tex]p\in \left\{ q | \frac{\partial f}{\partial x_3}(q)\neq 0\}\subset V[/tex]

    [tex]V[/tex] open

    We then Construct a function

    [tex]\psi :V_1\rightarrow \mathbb{R}^3[/tex]


    Calculating the Jacobian [tex]d_p\psi[/tex] we find that its determinant is non-zero, so that we can apply the Inverse function theorem to [tex]\psi[/tex] but I don’t understand the rest of the proof below

    [tex]\exists p\in V_2\subset V_1,\quad \psi |_{V_2}\rightarrow W_2[/tex]

    Choose [tex]\epsilon[/tex] so that [tex]\psi(p)+(-\epsilon,\epsilon)^3=W_3\subset W_2[/tex]

    What might this mean here?

    [tex]\left(\psi_{V_2}\right)^{-1}\left|_{W_3}\rightarrow V_2[/tex]

    Then we construct another map

    [tex] W_3\cap\left(\mathbb{R}^2\times 0\right)\rightarrow S[/tex]

    [tex] W_3\cap(\mathbb{R}^2\times 0)=\psi(p) +(-\epsilon,\epsilon)^2\times 0[/tex]

    Construct another map
    [tex] \phi : (-\epsilon, \epsilon)^2\rightarrow S[/tex]

    [tex] (u,v)\rightarrow \phi (\psi(p)+(u,v,0))[/tex]
    is a parameterization around p

    I can’t seem to understand what is going on here, maybe there is another way to prove the theorem. I have an intuitive proof using Taylor’s theorem and showing that f is like a plane locally but its not a rigorous proof
  2. jcsd

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