1

I have given a function $$f:\mathbb{R}^4 \to \mathbb{R}^3: (x,y,z,u) \mapsto (xz-y^2, yu-z^2,xu-yz)$$ and I want to show that $f^{-1}(0)\setminus\{0\}$ is an embedded surface. If it would be an embedded curve I could verify that $f$ is an submersion to get a 1-dim submanifold. But I don't know how to show that this is an embedded surface.

  • 1st get $f$'s the derivative – janmarqz Jul 18 '18 at 19:44
  • 1
    Are you sure this is right? It seems to consist of the $x$-axis, the $u$-axis, and the curve $(t^2,t,1,1/t)$. Moreover, $f$ has rank $3$ along the non-linear curve. I was expecting all the components of $f$ to be homogeneous. – Ted Shifrin Jul 18 '18 at 19:45
  • sorry there was a mistake on the last component. –  Jul 18 '18 at 19:51
  • The derivative is given by: $$Df(x,y,z,u) = \begin{pmatrix} z & -2y & x & 0\ 0 & u & -2z & y\ u & -z & -y & x\end{pmatrix}$$ which has rank 3 for all $(x,y,z,u) \in \mathbb{R}^4 \setminus {0}$ So, $f^{-1}(0)\setminus{0}$ is not an embedded surface but an 1-dim submanifold of $\mathbb{R}^4$. Is this argumentation right? –  Jul 18 '18 at 20:14
  • @AndreasK... but at $(x,0,0,0)$ and $(0,0,0,u)$ has rank two – janmarqz Jul 18 '18 at 21:03

1 Answers1

1

I saw this problem from Amann's Analysis II, page 257.

Note that $$\begin{aligned} y^2&=xz\\ z^2&=yu\\ yz&=xu. \end{aligned}$$ If $yz\neq 0$, then $$\begin{aligned} y^2\cdot yz=xz\cdot xu&\implies y=x^{2/3}\cdot u^{1/3}\\ z^2\cdot yz=yu\cdot xu&\implies z=x^{1/3}\cdot u^{2/3}. \end{aligned}$$ In the case $yz=0$, then it is easy to see that $xu=0$ and $y=0=z$, where the above relations also hold.

Let $(a,b)=(x^{1/3},u^{1/3})$ and $$g:(a,b)\mapsto (a^3,a^2b,ab^2,b^3).$$ It is clear that $g$ is a homeomorphism. And $$\partial g=\begin{pmatrix} 3a^2& 0\\ 2ab&a^2\\ b^2&2ab\\ 0&3b^2 \end{pmatrix}$$ and $g$ is an embedding when $(a,b)\neq (0,0)$.

It follows that $g(\mathbb R^2\setminus \{(0,0)\})=f^{-1}(0)\setminus \{0\}$ is a embedded curve.

stlinex
  • 782