How to show a level set isn't a regular submanifold For $F:\mathbb{R}^2\rightarrow \mathbb{R}$ given by $F(x,y)=x^3+xy+y^3$, how do I show that $F^{-1}(0)$ and $F^{-1}(1/27)$ aren't regular submanifolds? I've plotted these on Wolfram alpha:
the first one crosses itself at a point (so it's not a manifold by the standard "remove this point and see it's got more components than it should" argument) 
the second is (edit: the union of) a curve and an isolated point (so it's not a manifold because it doesn't have a well-defined dimension).
But I don't know how to prove these level sets actually look like this. What techniques can I use to work out what they look like?
 A: By the implicit function theorem, $F^{-1}(a)$ will be a regular submanifold if $a$ is a regular value. I.e. $\nabla f\neq 0$ on $F^{-1}(a)$. In this case, if you solve for $\nabla f=0$, you get two critical points $(0,0)$ and $(-1/3,-1/3)$. Thus the two critical values are $0$ and $1/27$. So these are the only two possible places where the preimage can fail to be a manifold. To see that they are not manifolds, use the second derivative test. This will tell you that $(0,0)$ is a saddle and $(-1/3,-1/3)$ is a local minimum. The cross section of a saddle locally looks like two intersecting lines, so is not a manifold. The cross section near a local extremum is a point, so is also not a manifold of the correct dimension.
In general, if the matrix of second-partial derivatives is nonsingular near a critical point, then the preimage fails to be a manifold near that critical point. This follows by the "Morse Lemma," that says that under these conditions, the function locally "looks like" a quadratic function, for which you can make explicit calculations. Depending on how advanced you are, you can check out Milnor's book on Morse Theory.
A: Here (I think) is a more elementary argument that doesn't use the Morse Lemma. Let me know what you guys think. For $\varepsilon < 1/2$, let $C_{\varepsilon} \subset \mathbb{R}^2$ denote the open cube of side length $2\varepsilon$ centered at $0$. Note that, for the smooth curves $\gamma_1, \gamma_2:(-\epsilon, \epsilon) \rightarrow \mathbb{R}^2$ given by $\gamma_1(t) = (t,t)$ and $\gamma_2(t) = (t,-t)$, we have $F(\gamma_1(t)) > 0$ and $F(\gamma_2(t)) < 0$ for $t \neq 0$. Let $R_1 = \{(x,y): -y<x<y, y > 0\}$, $R_2 = \{(x,y): -y<x<y, y < 0\}$, $R_3 = \{(x,y): x<y<-x, x>0\}$, and $R_4 = \{(x,y): x<y<-x, x<0\}$ (the open "quadrants" separated by the lines $y=x$ and $y=-x$). By the intermediate value theorem, $R_i \cap C_{\varepsilon} \cap F^{-1}(0) \neq \varnothing$ for each $i$. Now suppose for contradiction that $F^{-1}(0)$ is an embedded $1$-submanifold of $\mathbb{R}^2$, and let $(V, \varphi)$ be a coordinate chart for $F^{-1}(0)$ centered at $(0,0)$. By shrinking $V$ if necessary, we may assume that $V = C_{\epsilon} \cap F^{-1}(0)$ for some $\epsilon < 1/2$ and $\varphi(V)$ is an open interval. The open sets $R_1, R_2, R_3, R_4$ separate $V \smallsetminus \{(0,0)\}$ into at least four components, but $\varphi(V \smallsetminus \{(0,0)\}) = \varphi(V) \smallsetminus \{\varphi(0,0)\}$ has two components, a contradiction. 
