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Math Help - simple question on mapping from R^n -> R^m

  1. #1
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    simple question on mapping from R^n -> R^m

    I'm studying surface where I came along the following definition:

    a coordinate patch \mathbf{x}: D \mapsto \mathbb{R}^3 is one-to-one regular mapping of an open set D of \mathbb{R}^2 into \mathbb{R}^3

    Now given this definition how would one go about checking if the given map would constitute a patch?

    I'm thinking that in order to check if it's 1-1 we simply take the Jacobian and if one of the solutions is not 0 then we have a 1-1 relationship. So for instance if I have:
    x,y) = (x^2,y,y^3-y)" alt="\mathbf{x}x,y) = (x^2,y,y^3-y)" /> and I define the first set as (u,v) and the second set as (x,y,z) coordinate then the Jacobians would be:

    \frac{\partial(x,y)}{\partial(u.v)} = \det \begin{pmatrix}<br />
x_u & x_v\\<br />
y_u & y_v<br />
\end{pmatrix}<br />
 = 2x \neq 0

    \frac{\partial(x,z)}{\partial(u.v)}=2x*(3y^2-1)
    \frac{\partial(y,z)}{\partial(u.v)}=0

    so now that at least one of the Jacobians is not 0 then it's 1-1.

    is this correct?
    In this particular case the last one is 0 does that make any difference?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by lllll View Post
    I'm studying surface where I came along the following definition:

    a coordinate patch \mathbf{x}: D \mapsto \mathbb{R}^3 is one-to-one regular mapping of an open set D of \mathbb{R}^2 into \mathbb{R}^3

    Now given this definition how would one go about checking if the given map would constitute a patch?

    I'm thinking that in order to check if it's 1-1 we simply take the Jacobian and if one of the solutions is not 0 then we have a 1-1 relationship. So for instance if I have:
    x,y) = (x^2,y,y^3-y)" alt="\mathbf{x}x,y) = (x^2,y,y^3-y)" /> and I define the first set as (u,v) and the second set as (x,y,z) coordinate then the Jacobians would be:

    \frac{\partial(x,y)}{\partial(u.v)} = \det \begin{pmatrix}<br />
x_u & x_v\\<br />
y_u & y_v<br />
\end{pmatrix}<br />
 = 2x \neq 0

    \frac{\partial(x,z)}{\partial(u.v)}=2x*(3y^2-1)
    \frac{\partial(y,z)}{\partial(u.v)}=0

    so now that at least one of the Jacobians is not 0 then it's 1-1.

    is this correct?
    In this particular case the last one is 0 does that make any difference?
    Are you attempting to appeal to the inverse function theorem?
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    It's not correct, as the example you give shows. The points (x,y) and (-x,y) have the same image.

    I think it is true if the open set U is a small enough neighbourhood of a point where the condition you describe holds. There certainly needs to be a restriction on U.
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  4. #4
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    Quote Originally Posted by Bruno J. View Post
    It's not correct, as the example you give shows. The points (x,y) and (-x,y) have the same image.
    I didn't really see that. Is there a way to verify injectivity of a given map as a increase/decrease the number of dimensions. I don't think verifying it on a case by case basis is feasible.
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by lllll View Post
    I didn't really see that. Is there a way to verify injectivity of a given map as a increase/decrease the number of dimensions. I don't think verifying it on a case by case basis is feasible.
    I don't know of a general method except for the inverse function theorem stated by Drexel, which only applies to differentiable functions f: \mathbb{R}^n \to \mathbb{R}^n. There probably exists a generalization of this theorem which would answer your question; I'll think about it.
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