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If I am given a map $f:U\subset \mathbb R^2 \to \mathbb R^3$ where $(x,y)\to (f(x,y),g(x,y),h(x,y))$. Is this necessarily a "parametrized surface"? 

Am I right in thinking that any map of the above form satisfies the "parametrized" bit. Is it necessary a surface though?

Please correct me if I am wrong. Thank you.

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3 Answers 3

up vote 1 down vote accepted

If you are interested in the viewpoint of the classical differential geometry of surfaces, then we say that your map $f$ is a local parametrization of a surface in $\mathbb R^3$ if and only if $f$ is three-times continuously differentiable (i.e. $f \in C^3$) and the cross product $f_x \times f_y$ is nonzero for each $x, y \in U$. Here $f_x$ and $f_y$ represent the partial derivatives of $f$ w.r.t. $x$ and $y$, respectively. Of course there are other contexts as well.

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It depends on your definition of "parametrized surface", but, without any more requirements, your maps include cases like

$$ (x,y) \mapsto (x,x,x) \ , $$

which you would hardly call a "surface", would you? Or take simply

$$ (x,y) \mapsto (0,0,0) \ . $$

If you think that my examples are tricky because the maps don't "truly" depend on two variables, then think of

$$ (x,y) \mapsto (x+y, x+y, x+y ) \ . $$

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You'll want to assume your map is locally one-to-one, I think. This is true if your map is continuously differentiable and the null space of the Jacobian matrix is trivial.

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