# 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?

-
How about something like... $\nabla F = \langle 3x^2+y, 3y^2+x \rangle$. Notice that $\nabla F(0,0) = \langle 0,0 \rangle$ and $\nabla F(-1/2,1/2) \not= \langle 0,0 \rangle$. So what should be the "tangent space" (vectors perpendicular to $\nabla F$) is 2-dimensional at $(0,0)$ and is 1-dimensional at $(-1/2,1/2)$. Both of these points are on the 0 level surface. Hence it is not a manifold (the tangent space at each point must have the same dimension). The other level set probably will need a different argument. Are you sure it's a single point? –  Bill Cook Nov 9 '11 at 21:57

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.