Proof a set is a submanifold

Question

Let $M=\{x\in \mathbb{R}^3\colon f(x)=g(x)=0\}$ and

$f\colon \mathbb{R}^3\longrightarrow \mathbb{R}, \ \ f(x_1,x_2,x_3)=-x_1^2-2x_1x_2+2x_2+2x_3\\ g\colon \mathbb{R}^3\longrightarrow \mathbb{R}, \ \ g(x_1,x_2,x_3)=\frac{3}{2}x_1^2-x_1x_2-3x_2+x_3$

be given.

I want to show that $M$ is a onedimensional submanifold of $\mathbb{R}^3$.

The only reasonable approach from what I have learned so far, would be the regular value theorem. But to apply the theorem I would need an implicit function describing the manifold. Not sure if I miss something obvious.

Answer 1

I imagine that what you call the "regular value theorem" is that if $\Phi : \mathbb{R}^n \to \mathbb{R}^m$ is smooth and $y \in \mathbb{R}^m$ is a regular value (i.e. for all $x \in f^{-1}(y)$, the differential $D_x\Phi$ is surjective) then $\Phi^{-1}(y)$ is a submanifold of dimension $n-m$.

A standard thing to do in your case is to consider the map $\Phi : \mathbb{R}^3 \to \mathbb{R}^2$, given by $\Phi(x) = (f(x), g(x))$. Then you're looking at $\Phi^{-1}(0)$ and you want to apply the regular value theorem to that.

So you're left with computing the differential $D_x\Phi$ at some point $x = (x_1,x_2,x_3) \in \Phi^{-1}(0)$. The matrix of the differential $D_x\Phi$ is given by the Jacobian of $\Phi$: $$\operatorname{Jac}_{x}(\Phi) = \begin{pmatrix} -2x_1-x_2 & -2x_1+2 & 2 \\ 3x_1-x_2 & -x_1-3 & 1 \end{pmatrix}$$ And what you now need to check is that if $\Phi(x) = 0$ then this matrix has full range, i.e. the linear map it represents is surjective. Using the rank-nullity theorem you may alternatively check that the null space of this matrix is of dimension $1$, whatever you prefer.

Answer 2

Apply the implicit function theorem to $$F=\left(\array{f \\ g}\right):\mathbb{R}^3 \rightarrow \mathbb{R}^2 $$

Answer 3

The implicit function that you need is already in the question that you wrote. Let me describe it more precisely: Let $F:\mathbb{R}^3 \rightarrow \mathbb{R}^2$ be a the implicit function defined as $$ F(x_1,x_2,x_3) =(f(x_1,x_2,x_3),g(x_1,x_2,x_3))=(x_1^2-2x_1x_2+2x_2+2x_3, \frac{3}{2}x_1^2-x_1x_2-3x_2+x_3) $$ The Jacobian matrix of $F$ is, $$ J_{F} = \left[ \begin{array}{cc} \frac{\partial f}{\partial x_1} & \frac{\partial f}{\partial x_2} & \frac{\partial f}{\partial x_3} \\ \frac{\partial g}{\partial x_1} & \frac{\partial g}{\partial x_2} & \frac{\partial g}{\partial x_3} \end{array}\right]= \left[ \begin{array}{cc} -2x_1-2x_2 & -2x_1+2 & 2\\ 3x_1-x_2 & -x_1-3 & 1 \end{array}\right] $$ To prove that $M$ is submanifold of dimension $1$ is sufficient to prove that it exists a submatrix $2 \times 2$ of $J_F$ with a non-zero determinant. Thus $$ \left| \begin{array}{cc} -2x_1+2 & 2\\ -x_1-3 & 1 \end{array} \right| = 8. $$