Why the set $g^{-1}(\{0\}) $ is not a differentiable manifold?

Question

Let $g:\mathbb{R}^2 \to \mathbb{R}$ given by $g(x,y) = x^2 - y^2$. Then I am triying to figure out why this function is not a differentiable manifold , I was trying to give an explicit coordinate system, but I can't figure out how.

Then I tried to use the result that says the following: Let $A \subset \mathbb{R} $ open, $p<n$, $g:A \to \mathbb{R}^p$ of class $C^1$ such that $g'(x)$ has rank $p$ for all $x \in A$ such that $g(x)=0$ therefore $g^{-1}(\{0\})$ is a differentiable manifold in $\mathbb{R}^n$ of dimension $n-p$.

So I computed the Jacobian matrix and it gives me the following: $(2x \;\; -2y )$ and it has rank 1 so I don't know what is the problem with this manifold. Can someone help me to find where is the property that does not hold please? Thanks a lot.

Definition: A subset $M \subset \mathbb{R}^n$ is a differentiable manifold of dimension $k$ if, for each $x \in M$, exists open sets $U \subset \mathbb{R}^k$ and $V \subset \mathbb{R}^n$ and a class $C^1$ function $f:U \to V$ such that:

1) $x \in V$

2) $f(U)=V \cap M$, $f$ is an homeomorfism

3) for each $y \in U$ the jacobian matrix has rank k

Answer 1

This is an elaboration of Ayman’s comment above. The set $ {g^{\leftarrow}}[\{ 0 \}] $ is simply $$ \{ (x,y) \in \Bbb{R}^{2} \mid y = \pm x \}. $$ This is not a topological submanifold of $ \Bbb{R}^{2} $. Assume the contrary. Then $ (0,0) $ would possess an open neighborhood in $ {g^{\leftarrow}}[\{ 0 \}] $ that is homeomorphic to an open ball in $ \Bbb{R}^{n} $ for some $ n \in \{ 1,2 \} $. We can rule out $ n = 2 $ because the removal of any point in an open ball does not take away its connectivity, but removing $ (0,0) $ from $ {g^{\leftarrow}}[\{ 0 \}] $ produces four distinct connected components. We can also rule out $ n = 1 $ because the removal of any point in an open interval yields exactly two connected components, not four. We therefore have a contradiction.

Note: We do not consider $ n \geq 3 $ due to Invariance of Domain. We also do not consider $ n = 0 $ because $ {g^{\leftarrow}}[\{ 0 \}] $ is not discrete.

Answer 2

Property 2. fails at the origin. If $V$ is an arbitrary neighborhood of the origin in $\mathbf{R}^{2}$, there does not exist an open set in the real numbers and a smooth map $f:U \to V$ whose derivative has rank one everywhere and whose image is $M = V \cap g^{-1}(0)$.

Intuitive reasons are given in the comments. One approach is, $g^{-1}(0)$ is the union of the lines $y = x$ and $y = -x$. By hypothesis, some number $t_{0}$ maps to the origin, i.e., $f(t_{0}) = (0, 0)$. If $f$ were a homeomorphism from $U$ to $M = V \cap g^{-1}(0)$, then $f$ would induce a homeomorphism from $U \setminus\{t_{0}\}$ to $M \setminus\{(0, 0)\}$. But the former has precisely two connected components and the latter has at least four.