Determine all the values $(a,b)$ such that $F^{-1}(a,b)$ is a smooth one-dimensional submanifold of $\mathbb{R^3}$

Question

Consider the map $F:\mathbb{R^3} \to \mathbb{R^2}$ given by $F(x,y,z)=(z^2-xy,x^2+y^2)$

(a) Find all the critical values of $F$

(b) Determine all the values $(a,b)$ such that $F^{-1}(a,b)$ is a smooth one-dimensional submanifold of $\mathbb{R^3}$

(a) $$DF=\begin{bmatrix} -y & -x & 2z\\ 2x & 2y & 0\\ \end{bmatrix}$$

This matrix has full rank except when $x+y=0,z=0$ or $x-y=0,z=0$ or $x=y=0$ for any $z$. Thus, the critical points of $F$ are $(x,-x,0),(x,x,0)$ and $(0,0,z)$.

(b) Now $F(x,-x,0)=(x^2,2x^2), F(x,x,0)=(-x^2,2x^2), F(0,0,z)=(z^2,0)$

I am kind of confused after this. How do I proceed after this??

Thanks for the help!!

Answer 1

You are almost done. You just need to describe your critical values as a set. For example values of the form $(x,y)=(a^2,2a^2)$ are critical. How can they be described? Since $x=a^2$ for some $a\in\mathbb{R}$ you need $x\geq 0$ and using $y=2a^2=2x$ you can say, that $(x,y)=(a^2,2a^2)$ is equivalent to $x\geq 0$ and $y=2x$.

Your set of critical values is $$Crit=\{(x,y)~:~(x\geq 0\wedge y=2x)\vee(x\leq 0\wedge y=-2x)\vee (x\geq 0\wedge y=0)\}$$You can simplify $(x\geq 0\wedge y=2x)\vee(x\leq 0\wedge y=-2x)$ to $y=2|x|$ and you get $$Crit=\{(x,y)\in\mathbb{R}^2~:~y=2|x|\vee(x\geq 0\wedge y=0)\}.$$And the set of regular values is $Reg=\mathbb{R}^2\setminus Crit$ and for $(a,b)\in Reg$ you get, that $F^{-1}(a,b)$ is a smooth one-dimensional submanifold of $\mathbb{R}^3$.