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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?
K.defaoite
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Cathy
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2 Answers2

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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.

C. Falcon
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  • Thank you for taking the time to answer my questions. As you mentioned, in the case $n < 1$ (and $n=1$), the superellipse is not a $C^1$ submanifold of $\mathbb{R}^2$, and I am interested in the case $n > 1$. Thanks to your answer, I understand, regarding 1., that the superellipse (with $n > 1$) is a $C^1$ submanifold of $\mathbb{R}^2$. Regarding 2., I understand that there is still a possibility that the superellipse is a smooth submanifold of $\mathbb{R}^2$ even when $f$ is not smooth. Do you have some idea about this? I appreciate your other comments as well. They are very informative. – Cathy Dec 25 '21 at 17:13
  • To put it more clearly, the question I still have is as follows: Is the superellipse smooth (not only $C^1$) submanifold of $\mathbb{R}^2$? If $n$ is an even integer, the answer is clearly yes. How about the case of general real number $n > 1$ (odd integer, rational number, or irrational number)? For example, if $n=3$, the curve is a $C^2$ submanifold of $\mathbb{R}^2$ since $f$ is $C^2$ in this case, but is it a $C^3$ (or $C^\infty$) submanifold of $\mathbb{R}^2$? – Cathy Dec 25 '21 at 17:26
  • I get what your remaining question is, but I have no answer to it right now, as I don't immediately see how the equation defining the superellipse can be smoothed. – C. Falcon Dec 25 '21 at 17:30
  • Thank you for your kind answer. – Cathy Dec 25 '21 at 17:35
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Take the case $n=3$. $$ f(x) = \big(1 - |x|^3\big)^{1/3} $$ This is the graph of $f'''(x)$
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$f'''(x)$ has a jump discontinuity at $x=0$.

GEdgar
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    Thank you for your answer and the nice figure. I also first explicitly expressed the part of the superellipse for $n = 3$, i.e., the curve $y = (1-|x|^3)^{1/3}$. However, my question still remains: Even though $f(x) := (1-|x|^3)^{1/3}$ is not $C^3$, can we conclude that the curve is not a $C^3$ submanifold of $\mathbb{R}^2$? – Cathy Dec 25 '21 at 17:46
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    I have posted this several times before: A $C^k$ submanifold must be the graph of a $C^k$ function over one of the standard coordinate axes (or, more generally, coordinate planes of the appropriate dimension). This is a consequence of the $C^k$ inverse function theorem. – Ted Shifrin Dec 25 '21 at 17:51
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    Thank you for your comment. So, regarding C. Falcon's example, $y = \pm\sqrt{x}$ are not smooth but $x = y^2$ is smooth (with respect to the right-hand sides). In contrast to this, the curve defined by $|x|^3+|y|^3 = 1$ is expressed as $y = \pm(1-|x|^3)^{1/3}$ and $x = \pm(1-|y|^3)^{1/3}$, both of which are not $C^3$. Therefore, the curve cannot be a $C^3$ submanifold of $\mathbb{R}^2$. Am I correct? I'll try to rigorously prove by myself the "consequence of the $C^k$ inverse function theorem" you mentioned. But if this fact is famous and there is literature, could you introduce some? – Cathy Dec 25 '21 at 18:09
  • @Cathy I believe Ted Shifrin is saying that GEdgar's answer does not show that the superellipse for $n=3$ is not $C^3$. He is saying that smooth $k$-dimensional submanifolds are locally graphs of smooth maps over a $k$-tuple of the coordinates, but this $k$-tuple needs not to be constituted of the first $k$ coordinates. Going back to my example, $y=\sqrt{x}$ is smooth, but around the origin, it is not a smooth graph of the $x$-coordinate, but rather a smooth graph of the $y$-coordinate. – C. Falcon Dec 25 '21 at 18:40
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    Yes, Cathy, your argument is correct. You have only to check $\binom Nn$ possible functions for an $n$-dimensional submanifold of $\Bbb R^N$. I agree that this is a result that is not sufficiently well-known. It appears, for example, as an exercise in Guillemin and Pollack's Differential Topology (#9 on p. 19). I always assigned it!! – Ted Shifrin Dec 25 '21 at 19:07
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    Here is another application of this approach. – Ted Shifrin Dec 25 '21 at 19:13
  • Many thanks to all of you for writing so many helpful comments! From what I understand, an answer to my original question, "Do superellipses provide examples of submanifolds of $\mathbb{R}^2$ that are not smooth?" is "Yes, consider the curve defined by $|x|^3+|y|^3=1$, for example. It's a $C^2$ submanifold of $\mathbb{R}^2$, but is not a $C^3$ (nor smooth) submanifold of $\mathbb{R}^2$." – Cathy Dec 26 '21 at 07:31