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How to show that set: $M = \left\{(x,y,z): x^2+y^2+3z^3 = xy + 6z^{\frac{1}{3}}, z \neq 0\right\} \cup \left\{(0,0,0)\right\}$ is not a manifold?

I know the problem is with point $(0,0,0)$. I think i can even show that you cannot define $x$ as $y$ and $z$ function, I mean $x=x(y,z)$. And the same with $y=y(x,z)$. But how to show that you cannot define $z=z(x,y)$?

Thank you for help.

Prold
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2 Answers2

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I think that $M$ IS a manifold.

The map $z \mapsto z^3$ is a bijection from $\mathbb{R}$ to itself. Hence, doing the change of variables $z' = z^{1/3}$, we get another equation for $M$, that is:

$$ M = \{(x,y,z'): x^2 + y^2 + 3z'^9 = xy + 6z', z' \neq 0\} \cup \{(0,0,0)\}.$$

But we have in fact the equality

$$ M = \{(x,y,z') : x^2 + y^2 + 3z'^9 = xy + 6z'\},$$

since the only real solution of $x^2 + y^2 = xy$ is the couple $(0,0)$.

With this equation, we see that $M$ is a manifold since the map $f(x,y,z') = x^2 + y^2 + 3z'^9 - xy - 6z'$ is a submersion at all points of $M$.

Edit: As remarked in a comment, this only shows that $M$ is a topological submanifold of $\mathbb{R}^3$. But I think that $M$ is nevertheless a smooth manifold.

Denote by $M'$ the set $\{(x,y,z) : x^2 + y^2 + 3z^9 = xy + 6z\}$ (this was written $M$ before but now I want to make explicit the homeomorphism between $M$ and $M'$). Since $\partial_z f$ is not zero at $(0,0,0)$, there is a smooth function $g: U \rightarrow V$, with $U$ an open set in $\mathbb{R}^2$ containing $(0,0)$, $V$ is an open set in $\mathbb{R}$ containing $0$ and an open set $W$ in $\mathbb{R}^3$ such that $M' \cap W = \{(x,y,g(x,y)): (x,y) \in U\}$.

Let $\psi: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be the smooth homeomorphism $(x,y,z) \mapsto (x,y,z^3)$. One has $M = \psi(M')$ and $M \cap \psi(W) = \{(x,y,g^3(x,y)): (x,y) \in U\}$.

Edit 2: This is much more easier. Let $g(x,y,z) = (x^2 + y^2 + 3z^3 - xy)^3 - 216z$. Then $M$ is $g^{-1}(0)$ and $\partial_z g$ does not vanish at $(0,0,0)$.

Jeremy Daniel
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  • I really like this idea. It looks ok, but it'd be great if somebody could confirm. Thank you! – Prold Dec 14 '13 at 19:44
  • This proves that the set is a topological sub manifold, not smooth one. – Moishe Kohan Dec 15 '13 at 17:37
  • I don't understand ; I have explained that $M$ is the set $f^{-1}(0)$, with $f$ a smooth submersion. This shows that $M$ is a smooth submanifold, hence a smooth manifold. – Jeremy Daniel Dec 15 '13 at 21:41
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    @JeremyDaniel: Yes, it is smooth after a topological change of coordinates, not differentiable one. Try the same method with the cuspidal curve $x^2=y^3$ (use the change of variables $y'=y^3$). You get the parabola $x^2=y'$ which is of course smooth. But the original curve is not a smooth submanifold. – Moishe Kohan Dec 19 '13 at 06:03
  • @studiosus, so, do you know how to show it's (not) differentiable manifold? – Prold Dec 19 '13 at 11:38
  • @Prold: I am not sure, I am thinking about it. – Moishe Kohan Dec 19 '13 at 12:09
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Use implicit differentiation to see that there is no tangent space of the right dimension at $(0, 0, 0).$

Igor Rivin
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  • Could you help me a little bit more? I know how it works on easy examples, but here I cannot do it. – Prold Dec 14 '13 at 17:41