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Example of differentiable manifold of class C^1

  1. Nov 3, 2013 #1

    I read from several sources the statement that the set of points M[itex]\in[/itex]ℝ2 given by [itex](t, \, |t|^2)[/itex] is an example of differentiable manifold of class C1 but not C2.

    Is that true?
    To be honest, that statement does not convince me completely, because in order for M to be a manifold, we should be able to find an atlas of charts [itex]x_i:U_i \rightarrow M[/itex] such that all the points of M are covered by the atlas.
    So how do we cover the point (0,0) ?

    Do we need to use 3 charts as follows?

    [tex]x_1(t) = (t,\, t^2) \quad \, t\in(0,+\infty)[/tex]
    [tex]x_2(t) = (t,\, -t^2) \quad \, t\in(-\infty, 0)[/tex]
    [tex]x_3(t) = (t,t^2) \quad \, t\in(-1,1)[/tex]
    Last edited: Nov 3, 2013
  2. jcsd
  3. Nov 3, 2013 #2
    I don't understand what you're doing. For real numbers ##t##, ##|t|^2=t^2##, right?
  4. Nov 3, 2013 #3
    Sorry, I reported the wrong equation for the curve . What I meant was that the points of M are given by [itex] (t, \, \mathrm{sign}(t)t^2)[/itex]
  5. Nov 3, 2013 #4


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    Why don't you use a ##C^1## function f that is not ##C^2## and try to find a manifold whose transition functions ae given by f ?

    Only one I can think of now is $$f(x)= \frac {-x^2}{2}; x\leq 0$$ and

    $$f(x) =\frac{x^2}{2} ; x>0$$

    We have f'(x)=|x| , which is not differentiable. I'm not sure this will work; just an idea. Seems like an interesting question: given a (finite) collection of functions, can I find a manifold for which the transition functions are precisely those functions? I'm thinking this is the way one can construct bundles by choosing cocycles satisfying certain properties; can we do something similar for manifolds?
  6. Nov 4, 2013 #5
    Hi WWGD,

    I think the function you suggested is the same as the one I am considering, up to a scalar factor of 1/2.
    In my first post I gave three coordinate charts x1, x2, x3 that I regarded as possibly correct candidates to form an atlas.

    What I missed were the transition maps.
    In this case, it seems that the transition maps are always the identity function on some domain: [tex]\tau_{1,3}(t)=\tau_{3,1}(t)=t \quad ; \, t\in(0,1)[/tex]
    [tex]\tau_{2,3}(t)=\tau_{3,2}(t)=t \quad ; \, t\in(-1,0)[/tex]

    If this last step is correct, it should prove that the curve is indeed a [itex]C^1[/itex]-differentiable manifold (but not [itex]C^2[/itex]).
    Last edited: Nov 4, 2013
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