1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Patches and Surfaces (Differential Geometry)

  1. Feb 23, 2010 #1
    I'm completely confused with patches, which were introduced to us very briefly (we were just given pictures in class). I am using the textbook Elementary Differential Geometry by O'Neill which I can't read for the life of me. I'm here with a simple question and a somewhat harder one.

    1. The problem statement, all variables and given/known data
    Is the following mapping x:R^2 to R^3 a patch?

    x(u,v)=(u, uv, v)?

    2. Relevant equations

    For a mapping to be a patch, it must be one-to-one (injective) and regular (smooth).

    3. The attempt at a solution

    I understand how to show that it is regular; for any arbitrary direction, either the directional derivative of the x component or the directional derivative of the y component is non-zero. Now, I don't know how to prove that it is injective. The book gives a hint: x is one-to-one iff x(u,v) = x(u_1, v_1) implies (u,v)=(u_1,v_1).

    So my attempt was to just let x(u_1,v_1) = (u_1, u_1v_1, v_1) so that
    (u_1, u_1v_1, v_1)=(u, uv, v). Is this the correct way of going about it? I feel like I didn't show anything.

    Can someone also point me toward a better book or online notes where I can try to understand some of this material?

    Thank you, any help or suggestions will be appreciated.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 23, 2010 #2
    Suppose that x(u,v)=x(w,z). Then (u,uv,v)=(w,wz,z), and this implies (u,v)=(w,z).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook