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A celebrated theorem of Whitney says that every $C^1$-manifold can be made a smooth manifold by considering an appropriate differentiable structure.

On the other hand, a set $M\subseteq \mathbb R^n$ (with the subspace topology) has the property of being a $C^k$-manifold means that, for every $x\in M$, there are an open set $V\subseteq \mathbb R^m$ containing $x$ and an injective $C^k$-function $f:U\to\mathbb R^m$ on some open set $U\subseteq\mathbb R^n$ such that $M\cap V=f(U)$, $f^{-1}:M\cap V\to U$ is continuous, and $f'(t):\mathbb R^n\to\mathbb R^m$ is injective for all $t\in U$.

I am quite sure that Whitney's theorem does not say that the properties of being a $C^k$- or a $C^p$-manifold are equivalent for different $k,p\in\mathbb N$.

What are simple examples?

If $g:\mathbb R \to \mathbb R$ is $C^1$ but not $C^2$ (e.g., $g(t)=0$ for $t\le 0$ and $g(t)=t^2$ for $t>0$), the graph $\{(t,g(t)):t\in\mathbb R\}$ is clearly a $C^1$-manifold. I am very embarrassed that I don't see if it can be $C^2$.

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  • $\begingroup$ One thing you could perhaps clarify: in your first paragraph, I suspect you are asking about abstract manifolds defined in terms of an atlas; whereas in the second paragraph you are asking about submanifolds of $\mathbb R^n$. Is this so? $\endgroup$
    – Lee Mosher
    Jul 5 at 15:18
  • $\begingroup$ Yes, that's exactly what I mean. $\endgroup$
    – Jochen
    Jul 5 at 16:27
  • $\begingroup$ How are you defining your $C^1$-atlas on the graph of $g$? Also I recall you that the graph of a continuous map $f$ is homeomorphic to the domain, via this homeomorphism we can endow the graph with the pullback of any atlas on the domain. In particular you can endow the graph of your $g$ with the pull back of the smooth atlas of $\mathbb R$ which makes it into a smooth manifold diffeomorphic to $\mathbb R$. $\endgroup$
    – Overflowian
    Jul 5 at 17:37
  • $\begingroup$ Of course, the graph is $C^1$-isomorphic to the line, but this does not help to answer the question if it can be locally parametrized by $C^2$-embedings. $\endgroup$
    – Jochen
    Jul 5 at 20:33

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The curve you mentioned cannot be $C^2$ because its curvature function is discontinuous at the origin. If you approach the origin from the left, the curvature is zero. If you approach from the right, the curvature tends to 2.

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  • $\begingroup$ I understand the spirit of the argument. What exactly is the curvature function of a 1-dimensional $C^2$-submanifold of $\mathbb R^2$? $\endgroup$
    – Jochen
    Jul 5 at 17:18
  • $\begingroup$ en.wikipedia.org/wiki/Curvature gives enough information. $\endgroup$
    – Jochen
    Jul 6 at 7:37

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