Find all values of $t$ for which $S_t$ is a submanifold of $\mathbb{R}^2$.

Question

Let $f : \mathbb{R}^2 \to \mathbb{R}, f(x,y)=x^2+4y^4-4y^2$ and denote $S_t=f^{-1}(t)$. Find all values of $t$ for which $S_t$ is a submanifold of $\mathbb{R}^2$.

So this is asking us to find the regular values of $f$? What we have is that $$D_f=\begin{pmatrix}2x&16y^3-4y\end{pmatrix}$$ and we are trying to find when this matrix does not have full rank. That is when $\operatorname{rank}(D_f)=0$ which happens when $2x = 0$ and $16y^3-4y=0$ i.e. for $(0, \pm 1/2)$. So is $S_t$ a submanifold for all values except $f(0,1/2)=f(0,-1/2)=-3/4$?

Answer 1

Your idea is correct, but it is not sufficient to solve the problem completely.

Let us first observe that $D_f = \begin{pmatrix} 2x & 16y^3 - 8y \end{pmatrix}$. We have $\operatorname{rank}(D_f)=0$ iff the following two conditions are satisfied:

1. $x = 0$
2. $y = 0$ or $y = \pm \sqrt 2 /2$.

Let us now determine the image of $f$.

Let us first study $g(y) = 4y^4 - 4y^2$. It has zeros at $y = 0$ and $y = \pm 1$. Local extrema are attained at $y = 0$ (local maximum since $g''(0) = - 8 < 0$) and $y = \pm \sqrt 2 / 2$ (local minima since $g''(\pm \sqrt 2 / 2) = 16> 0$). We have $g(\pm \sqrt 2 / 2) = -1$. Thus the image of $g$ is $[-1,\infty)$.

Since $x^2 \ge 0$, also the image of $f$ is $[-1,\infty)$.

We conclude that $S_t = \emptyset$ for $t < -1$; this can be regarded as a "trivial submanifold" of $\mathbb R^2$.

The critical points of $f$ are $(0,0)$ and $(0, \pm\sqrt 2 / 2) $. Thus $f$ has the two critical values $0$ and $-1$. All other values of $f$ are regular thus $S_t$ is a $1$-dimensional submanifold for $-1 < t < 0$ and $t > 0$.

By the above considerations we have $S_{-1} = \{ (0,+\sqrt 2 / 2), (0, - \sqrt 2 / 2)$ which is a $0$-dimensional submanifold of $\mathbb R^2$.

$S_0$ is not a submanifold. At the point $(0,0) \in S_0$ four curves are arriving which prevents $S_0$ from being a manifold. To see this, let us understand what $f(x,y) = 0$ means. Since $x^2 > 0$ for $x \ne 0$, we only get solutions $\ne (0,0)$ for $f(x,y) = 0$ when $g(y) = 4y^4 - 4y^2 < 0$. This is the case for $0 < \lvert y \rvert < 1$. We thus get the two curves described by $$x = + \lvert y \rvert \sqrt{y^2-1} \text{ with } -1 < y < 0$$ $$x = - \lvert y \rvert \sqrt{y^2-1} \text{ with } -1 < y < 0$$ arriving at $(0,0)$ from below and the two curves described by $$x = + \lvert y \rvert \sqrt{y^2-1} \text{ with } 0 < y < 1$$ $$x = - \lvert y \rvert \sqrt{y^2-1} \text{ with } 0 < y < 1$$ arriving at $(0,0)$ from above.