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I've been reading Manfredo Do Carmo's Differential Geometry of Curves and Surfaces and was wondering what are the conditions that need to hold for a surface parameterization as this is not defined in the book. For example, one of the exercises asks whether the parameterization:

$\textbf{x}(u,v)=(u+v,u+v,uv)$ defined on the set $U=\{(u,v)\in R^2|u>v\}$ is a parameterization of the plane $P=\{(x,y,z)\in R^3|x=y\}$. Apparently, this is not a parameterization since $\textbf{x}$ is not one-to-one. I don't see why this should be a condition (i.e. being one-to-one) in order to be a parameterization. Is this a condition because the plane is a regular surface?

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There are two definitions in do Carmo: regular surface in section 2.2, page 52, and parametrized surface in section 2.3, page 78. Which one do you mean? –  lhf Feb 3 '13 at 0:08
    
The question I'm referring to is in section 2.2 (Exercise 5) –  Simon Sehayek Feb 3 '13 at 0:12
    
Obviously you can define "parameterization" however you want. And Do Carmo includes being one-to-one as part of his definition. His choice. Actually, in my sloppy field, we would not require a parameterization to be one-to-one, because, for us, the one-to-one property is not very important. Also, we need to be able to "parameterize" surfaces like spheres, which (I think) is impossible if you require one-to-one. Mathematicians would probably say we should use a different term, but it's historically ingrained. It does make it hard to have a discussion with a "serious" mathematician, though. –  bubba Mar 13 '13 at 5:39

1 Answer 1

Isn't the map x(u,v)=(u+v,u+v,uv) one-to-one? Here u and v are the unique (real) roots of the polynomial $z^2-(u+v)z+(uv)$ (provided that the roots are real). The condition u>v then uniquely determines u and v. Am I right?

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