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I'm looking for a continuously differentiable parametrization of $$x^3+y^2-z^2=1$$ but I'm actually totally stuck. If the $x$ term were quadratic instead of cubic, it would be simple: $$(x,y,z)=(\sqrt{t^2+1}\cos\theta, \sqrt{t^2+1}\sin\theta, t)$$ But with the cubic term there, I'm stuck. I naturally thought about $$(x,y,z)=(\sqrt[3]{t^2+1}\cos^{\frac{2}{3}}\theta, \sqrt{t^2+1}\sin\theta, t)$$ but this isn't continuously differentiable in $\theta$.

Hints or suggestions?

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What is your definition of a continuously differentiable parameterization? –  Michael Albanese Oct 17 '12 at 13:08
    
I think OP wants a bunch of maps $U_i \to \mathbb{R}^3$, where $U_i$ is some open subset of $\mathbb{R}^2$, such that each map is a homeomorphism to an open subset of the surface, the images of the maps cover the surface, and each map is $C^1$ when thought of as a map from $\mathbb{R}^2$ to $\mathbb{R}^3$. –  user29743 Oct 17 '12 at 13:23
    
I thought this might be the case, but I wasn't sure if the inverse of each map had to be continuously differentiable or not. –  Michael Albanese Oct 17 '12 at 13:25
    
I don't know! He can probably do it with 5 total maps (just by painstakingly solving for $x$, $y$, $z$, taking both branches of the root) where each map fails to be C^1 along a certain hyperbola and he has to argue that the intersection of these hyperbolae is empty. This, however, is horrible. –  user29743 Oct 17 '12 at 13:30
    
my way of thinking about it is it doesn't make sense to demand that the inverse of the map be continuously differentiable (we are "importing the smooth structure from $\mathbb{R}^2$" along the homeomorphism - the open subset of the surface is not open when thought of as living in $\mathbb{R}^3$ so it doesn't make sense to ask if the inverse is differentiable there.) But I am not an expert - maybe this is wrong. –  user29743 Oct 17 '12 at 13:31

2 Answers 2

up vote 2 down vote accepted

We are given the function $f(x,y,z):=x^3+y^2-z^2-1$ and have to consider the solution set (a "surface") $$S:=\{(x,y,z)\in{\mathbb R}^3\ |\ f(x,y,z)=0\}\ .$$ As $\nabla f(x,y,z)=(3x^2,2y,-2z)$ is $\ =(0,0,0)$ only at the origin $O\notin S$, by the implicit function theorem the set $S$ is a smooth surface in the neighborhood of all of its points. Here is a picture of $S$:

enter image description here

For given $y$ and $z$ the equation $f(x,y,z)=0$ has exactly one solution $x=\phi(y,z)\in{\mathbb R}$ which is commonly written as $\phi(y,z)=\root 3\of {1-y^2+z^2}$. Unfortunately along the hyperbola $y^2-z^2=1$ the function $\phi$ is not differentiable as a function of $y$ and $z$.

If we are allowed to use more than one patch to cover all of $S$ we could use three patches as follows: $$(x,t)\mapsto\bigl(x,-\sqrt{1-x^3}\cosh t,\sqrt{1-x^3}\sinh t\bigr)\qquad(-\infty<x<1, \ -\infty<t<\infty)\ ,$$ $$(x,t)\mapsto\bigl(x,\sqrt{1-x^3}\cosh t,\sqrt{1-x^3}\sinh t\bigr)\qquad(-\infty<x<1, \ -\infty<t<\infty)\ ,$$ $$(y,z)\mapsto\bigl(\root 3\of{1-y^2+z^2}, y, z\bigr)\qquad\bigl(-\infty<z<\infty,\ |y|<\sqrt{1+z^2}\bigr)\ .$$

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Thanks Christian. This problem has gotten me to wonder: is there a nice characterization of the minimum number of coordinate patches needed to cover a given submanifold of Euclidean space? –  symplectomorphic Oct 17 '12 at 21:23
    
@symplectomorphic: As $S$ is diffeomorphic to ${\mathbb R}^2$, in principle you can do it with one patch; but it will be difficult to do it in terms of elementary functions. –  Christian Blatter Oct 18 '12 at 8:18

Have you looked for a polynomial parametrization, with x a quadratic in t and y, z being cubics?

Or even simpler,

$x^3 + y^2 - z^2 = 1$

<=> $1 - x^3 = y^2 - z^2$

<=> $(1 - x)(1 + x + x^2) = (y - z)(y + z)$.

If you assume $1 - x = y - z$ and $1 + x + x^2 = y + z$ you can get a simple parametrization by solving the simultaneous equations for $y$ and $z$.

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This is slick and what I wanted. Thanks. –  symplectomorphic Oct 17 '12 at 20:56
    
wait, wait, that was too hasty on my part -- solving the last two equations for y and z gives y and z as functions of x. the image of that parametrization will only be a curve. but I want a single $C^1$ parametrization that covers the whole surface, if possible. –  symplectomorphic Oct 17 '12 at 21:05

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