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Question: Given an integer $n \in \mathbb{N}$, then what is an example of a $C^n$-manifold which is not a $C^{n+1}$-manifold?

Minor question: I know everything in theory, but I can not do it in practice. For instance, how can I check that the graph of the function $f\colon \mathbb R^2 \to \mathbb R^3$, $f(x,y)=(\sin x,e^y\cos x, xy)$, is a differentiable manifold.

Please do not use theorems, if it is possible. I want to see the whole procedure for at least one time. Any link and simple references are welcome.

I know the definition of a differentiable (or ~smooth or ~$C^{\infty}$) manifold, which is a Hausdorff, second countable topological space, with a differentiable atlas on it. If an atlas was given, then I feel comfortable with the definition of transition maps, and diffeomorphism (invertible smooth functions, with the smooth inverse), etc.

But I can not realize what is the Atlas in that example? Even if it was given to me, I don't have any idea how can I check all the infinitely many transition maps and diffeomorphism in practice.

Tireless and hardworking
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    My statement about $C^k$-manifolds was not really precise, but there have been questions on this topic where you can find the relevant information (for example: https://math.stackexchange.com/questions/769159/are-ck-manifolds-the-same-as-c-infty-manifolds) – Jake28 Oct 13 '20 at 17:00
  • @Jake28 I have an embarrassing question: what is that unique chart in this case? Consider the Stereographic projection; even in this case, I can not verify that the transition maps are diffeomorphisms. Could you please give a hint in these two cases? – Tireless and hardworking Oct 13 '20 at 17:05
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    The chart I meant is defined as follows: Let $G={(x,y,f(x,y))}$ be the graph of $f$ and let $u:G\to \mathbb{R}^2$ be the projection on the two first coordinates, i.e. $u(x,y,f(x,y))=(x,y)$ – Jake28 Oct 13 '20 at 17:05
  • I know you don't want a theorem, but the answer is indeed a theorem. See https://math.stackexchange.com/questions/1747752/every-mathcalc1-manifold-can-be-made-smooth – Deane Oct 15 '20 at 22:27
  • Your minor question has nothing to do with the main question. You therefore should ask it as a separate question. – Paul Frost Oct 17 '20 at 10:19

1 Answers1

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I interpret your question in the following sense:

A $C^n$-manifold $M$ is determined by a $C^n$-differentiable structure which we may regard as an atlas such that all transition maps are $C^n$ (let us call it $C^n$-atlas). Of course we could assume that the atlas is maximal with respect to this property, but it is irrelevant. Then the question is

Is every $C^n$-atlas also a $C^{n+1}$-atlas?

Variants of the question are

  1. If $M$ has a $C^n$-atlas, does it also have a $C^{n+1}$-atlas (which may be unrelated to the first atlas)?

  2. If $M$ has a $C^n$-atlas, does it always have a $C^{n+1}$-subatlas?

The answer to 2. (and a fortiori also to 1.) is "yes" if $n > 0$. See Every $\mathcal{C}^1$ manifold can be made smooth? For $n= 0$ the answer to 1. (and a fortiori also to 2.) is "no". See Manifold that is NOT smooth

Let $$h : \mathbb R \to \mathbb R, h(x) = \begin{cases} x & x \le 1 \\ x + (x-1)^{n+1} & x \ge 1\end{cases}$$

  1. $h$ is a homeorphism, thus $h^{-1} : \mathbb R \to \mathbb R$ exists.

  2. $h$ and $h^{-1}$ are $C^n$. This is trivial for $n = 0$. For $n > 0$ we get $$h'(x) = \begin{cases} 1 & x \le 1 \\ 1 + (n+1)(x - 1)^n & x \ge 1\end{cases}$$ $$h''(x) = \begin{cases} 0 & x \le 1 \\ (n+1)n (x - 1)^{n-1} & x \ge 1\end{cases}$$ $$h^{(n)}(x) = \begin{cases} 0 & x \le 1 \\ (n+1)!(x-1) & x \ge 1\end{cases}$$ Since $h'(x) > 0$ for all $x$, the inverse function theorem shows that $h^{-1}$ is $C^1$. But then $h^{-1}$ is even $C^n$. See the answer to Example of a $C^k$-homeomorphism $h : \mathbb R \to \mathbb R$ whose inverse is $C^{k-1}$, but not $C^k$

  3. $h$ is not $C^{n+1}$ because $h^{(n+1)}(1)$ does not exist.

Now let $\mathfrak A$ be the atlas on $\mathbb R$ having the two charts $id, h$. Transition functions are $h$ and $h^{-1}$ which are $C^n$. Thus $(\mathbb R, \mathfrak A)$ is a $C^n$-manifold. It is not a $C^{n+1}$-manifold by 3. However, the two subatlases $\{id\}$ and $\{h\}$ are $C^\infty$.

You can generalize this example to $\mathbb R^k$ by considering the map $h_k(x_1,\ldots,x_k) =(h(x_1), \ldots, h(x_k))$.

Paul Frost
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