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Math Help - Showing f is an isometry

  1. #1
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    Showing f is an isometry

    I have shown the second function is a isometry and if I can show the first function is as well, then that gurantess f is a local isometry right? The other thing is I'm not sure what showing f being well defined involves.

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  2. #2
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    Re: Showing f is an isometry

    1. To show that f is well-defined you need to show that \phi is one to one, so that f = \psi \circ \phi^{-1}. This is easy since sinh is one to one.
    2. To show that f is locally isometric, show that first fundamental forms of the two surfaces are the same:  I = \cosh^2v({du}^2+{dv}^2)
    3. To show that f is not globally isometric, choose the two points (u,v)=(0,0) and (2\pi, 0)
    The two points on H is (0,0,0) and (0,0,2\pi). The distance is 2\pi since the z-axis lies on H.
    The two points on C is (1,0,0) and (1,0,0), the same point. So the distance is 0.
    Thanks from Plato13
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  3. #3
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    Re: Showing f is an isometry

    Thanks kindly.
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