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Diffeomorphism: surface x4 + y6 + z2 = 1

  1. May 8, 2009 #1
    I'm trying to figure out what the surface x4 + y6 + z2 = 1 looks like.
    I want to say that it is diffeomorphic to the sphere because (x2)2 + (y3)2 + (z)2 = 1
    but i can't seems to actually construct the diffeomorphism (I am having problems with the x2 being invertible).
    Please let me know if i'm on the right track (if i'm even right)
     
  2. jcsd
  3. May 9, 2009 #2
    Re: Diffeomorphism

    I tried (x,y,z)|-->(x2,y3,z) but then only positive x are mapped to (and twice)
    So then I tried (x,y,z)|-->(sgn(x)x2,y3,z) which is bijective but isn't smooth (it doesn't have a second derivative when x=0)
    Any suggestions?
     
  4. May 9, 2009 #3

    Hurkyl

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    Re: Diffeomorphism

    You could just graph it. What are its cross sections in planes parallel to the x-y plane?

    Or, you could just use your mapping to understand the points away from the points where where x=0, and find some other means of understanding the subspace of points of small x.

    What is the application? Is it really not enough to simply have a homoeomorphism?

    P.S. my instinct is to map radially from the origin. Descartes rule of signs proves this is well defined for all points not lying in a coordinate plane.,,,
     
    Last edited: May 9, 2009
  5. May 9, 2009 #4
    Re: Diffeomorphism

    I was hoping to find the euler characteristic for it, so i suppose a homeomorphism would be sufficient. In this case my second construction would be suitable correct?
    I graphed the level curves and they are rounded off squares which is much like i anticipated. (also it suggests your idea of mapping out radially would probably work)
     
    Last edited: May 9, 2009
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