Do superellipses provide examples of submanifolds of $\Bbb{R}^2$ that are not smooth?

Question

Consider the curve (a kind of Lamé curve or superellipse (https://en.wikipedia.org/wiki/Superellipse)) in $\mathbb{R}^2$ defined by the equation \begin{equation} |x|^n + |y|^n = 1, \end{equation} where $n > 1$ is a real number. Obviously, the function $f(x,y) := |x|^n + |y|^n$ is $C^1$, and $1$ is a regular value of $f$.

1. From this observation, is it true that the curve (the preimage $f^{-1}(1)$) is a $C^1$ submanifold of $\mathbb{R}^2$ by the preimage theorem? For example, the Wikipedia page https://en.wikipedia.org/wiki/Preimage_theorem or Tu's book cited there state the preimage theorem for $C^\infty$ maps. Is a similar discussion applied to $C^1$ maps?
2. In general, $f$ may not be $C^\infty$. When $f$ is not $C^\infty$, we cannot say that the curve is a $C^\infty$ submanifold of $\mathbb{R}^2$ from the preimage theorem, but this does not directly mean that the curve cannot be a $C^\infty$ submanifold of $\mathbb{R}^2$. I'm curious about the possibility that the curve can be a $C^\infty$ submanifold of $\mathbb{R}^2$. Clearly, if $n$ is an even integer, the curve is a $C^\infty$ submanifold of $\mathbb{R}^2$ (from the preimage theorem because $f$ is $C^\infty$). How about the cases that $n$ is an odd integer, rational number, or general real value?

Answer 1

When $n\leqslant 1$, $f$ is not even a differentiable function. It is true that $f$ is smooth outside the set $(\mathbf{R}\times\{0\})\cup(\{0\}\times\mathbf{R})$, but $f^{-1}(1)$ crosses both of these lines, so it makes no sense saying that $1$ is a regular value of $f$.

1. The preimage theorem you are talking about admits $C^k$ versions for all $k\in\mathbf{N}\cup\{\infty\}$, so your reasoning is true when $f$ is a $C^1$-map. The proof of the preimage theorem only requires the implicit function theorem to work, whose proof basically boils down to a Newton's method. So there is nothing specific about smooth maps to it.

2. A critical level set of a function can be a smooth submanifold, but in this case the equation is said to be non-regular and the manifold is said to be non-transversally cut out. The easiest counterexample would be $f^{-1}(0)$ for $f(x,y)=x^2$.

It should be clear from the pictures drawn on Wikipedia that when $n<1$, the corresponding superellipse is not a smooth (nor $C^1$) submanifold of $\mathbf{R}^2$. However, your line of reasoning applies to the $n>1$ case.

Aside remarks.

• This a theorem by Kervaire and Milnor that all topological manifolds of dimension less or equal than $3$ can be endowed with a smooth structure (they are counterexamples in higher dimensions). However, this theorem is not saying that a topological submanifold of $\mathbf{R}^n$ of dimension less or equal than $3$ is a smooth manifold for the smooth structure induced by the one of $\mathbf{R}^n$.

• I should also be saying that if a topological space admits a $C^1$-structure, then it admits a smooth structure. Again, it is not saying that it is a smooth manifold for the same differentiable structure.

Answer 2

Take the case $n=3$. $$ f(x) = \big(1 - |x|^3\big)^{1/3} $$ This is the graph of $f'''(x)$

$f'''(x)$ has a jump discontinuity at $x=0$.