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I'm trying to decide whether the following set is a regular surface: $$\{(x^3 - 3xy^2, 3x^2y - y^3, 0) : (x, y) \in \mathbb{R}^2\}$$

I know that the map ${\bf x} : \mathbb{R^2} \to \mathbb{R^3}$ defined by ${\bf x}(x, y) = (x^3 - 3xy^2, 3x^2y - y^3, 0)$ doesn't work as a coordinate map, because the Jacobian determinant $\frac{\partial (x, y)}{\partial (x, y)}$ vanishes at $x = y = 0$. However I guess this doesn't rule out the possibility of other coordinate maps.

Obviously the surface isn't the graph of a function and I can't use the fact that it's a preimage of regular value of a function. So I'm thinking I have to find some coordinate maps which satisfy the appropriate properties, but I can't think what they would be. Can anyone help here? Is this set in fact not a regular surface, and if so how do I prove that?

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The set is $(u,v,0)$ where $u$ and $v$ are the real and imaginary part of $z^3$. –  PAD Oct 9 '12 at 21:50
    
Hm, you're right. How would I use this fact though? –  rt93 Oct 9 '12 at 21:54
    
Perhaps your set is the $xy$ plane in three dimensions? –  PAD Oct 9 '12 at 22:08
    
Of course, thanks. –  rt93 Oct 9 '12 at 23:28

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up vote 1 down vote accepted

The map $$f:(x,y)\in\mathbb R^2\mapsto(x^3-3xy^2,3x^2y-y^3)\in\mathbb R^2$$ is surjective (to see this, you can use the observation made by Pantelis in a comment above and the fact that every complex number has a cubic root) That means that your set is just the $xy$-plane in $\mathbb R^3$, which is surely a regular surface.

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