Proving a level set is not an embedded submanifold of $\mathbb R^2$

Question

I was trying to prove the following:

Let $F(x,y)=x^3+xy+y^3$ be defined on $\mathbb R^2$. Show that $F^{-1}[\{0\}]$ is not an embedded submanifold of $\mathbb R^2$.

I have verified that $(0,0)$ is not a regular point of $F$ and it is a saddle point. I have read in another question that this implies $F^{-1}[\{0\}]$ is the image of a smooth curve with self-intersections.

I would like to solve this problem without using this "result", since I don't know a proof of it and haven't found one anywhere, or else I'd like a reference to such a proof.

Answer 1

Suppose by contradiction $M=F^{-1}[\{0\}]$ is an embedded submanifold of $\mathbb R^2$. Then it is locally a graph of a function around $(0, 0)$, that is, there exists an open neighborhood $I\subset \mathbb R$ of $0$, $h:\rightarrow \mathbb R$ smooth and an open neighborhood $G\subset M$ of $(0, 0)$ such that $G=\{(x, h(x)): x \in I\}$ or $G=\{(h(x), x): x \in I\}$. Suppose it is the first case (since $F$ is symmetric, the second case is analogous). Notice that $h(0)=0$.

For every $x \in I$, it follows that:

$$x^3+xh(x)+h(x)^3=0$$ $$\Rightarrow 3x^2+h(x)+xh'(x)+3h(x)^2h'(x)=0$$ $$\Rightarrow 6x+2h'(x)+xh''(x)+6h(x)h'(x)^2+3h(x)^2h''(x)=0$$ $$\Rightarrow 6+3h''(x)+xh'''(x)+6h'(x)^3+12h(x)h'(x)h''(x)+6h(x)h''(x)h'(x)+3h(x)^2h'''(x)=0$$

The third line tells us that $h'(0)=0$, and the last line tells us that $h''(0)<0$. Therefore by the second derivative test there exists a open nhood $J_0\subset I$ of $0$ such that $h(x)<h(0)=0$ for every $x \in J_0\setminus\{0\}$.

Since $G$ is open in the subspace topology, there exists a nhood $J_1$ of $0$ such that $J_1\subset J_0$ and $(0, 0)\in J_1\times J_1 \cap M\subset G$. Since $h(0)=0$ and $h$ is continuous, $J_3=J_2\cap h^{-1}[J_2]$ is an open nhood of $0$. There exists $\epsilon>0$ such that $(-\epsilon, \epsilon)\subset J_2$. Since $J_2\subset J_0$, by the intermediate value theorem there exists $a, b$ such that $-\epsilon<a<0<b<\epsilon$ and $h(a)=h(b)$. Since $a, b \in J_2$, then $a, b, h(a), h(b) \in J_1$. Since $(a, h(a))$ and $(b, h(b))$ in M and $F$ is symmetric, it follows that $(h(a), a), (h(b), b)\in M\cap J_1\times J_1\subset G$. So there exists $x, y \in I$ such that $(h(a), a)=(x, h(x))$ and $(h(b), b)=(y, h(y))$, so:

$$a=h(h(a))=h(h(b))=b$$

a contradiction.

Answer 2

As you noticed, $p:=(0,0)$ is a non-degenerated critical point of $F$, which means that $\nabla_pF=0$ and $H_pF$, the hessian matrix of $F$ at $p$ is invertible, indeed, one has the following equality: $$H_pF=\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$ Therefore, using Morse's lemma (Taylor's formula+inverse function theorem), up to a change of coordinates sending the origin on the origin, in a neighborhood of $(0,0)$, one has that: $$F(x,y)=x^2-y^2.$$ From there it is clear that $S:=F^{-1}(\{0\})$ is not a manifold. For example, you can notice that $S\setminus\{(0,0)\}$ has $4$ connected-components, exactly as the set $\{x=y\}\cup\{x=-y\}$ minus the origin.

---

Here is a sketch of $S$:

As you can see, in a neighborhood of $(0,0)$, $S$ can be straightened onto $\{x=y\}\cup\{x=-y\}$, which is exactly what we have done.