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This is a theorem of Hassler Whitney:

For $0<k<\infty$ and any $n$-dimensional $C^k$ manifold the maximal atlas contains a $C^\infty$ atlas on the same underlying set.

It seems to me that this theorem says every $C^{k}$ manifold can be thought of as if it were a $C^{\infty}$ manifold proceeding like this:

  1. Start with a given a $C^{k}$ atlas $\mathcal{A}$ on a topological manifold $M$

  2. Consider the maximal $C^{k}$ atlas $\overline{\mathcal{A}}$ containing $\mathcal{A}$ (i.e. the atlas containing every $C^{k}$ chart on $M$ compatible with $\mathcal{A}$)

  3. Extract a $C^{\infty}$ subatlas $\mathcal{A}_{\infty}$ of $\overline{\mathcal{A}}$ (it can be done because of Whitney's theorem)

  4. Now make $M$ a $C^{\infty}$ manifold considering the atlas $\mathcal{A}_{\infty}$

After this maneuver we ended up with two differentiable structures over the same underlying topological manifold $M$: one of $C^{k}$ type given by $\mathcal{A}$ and the other of $C^{\infty}$ type given by $\mathcal{A}_{\infty}$.

My questions are:

  1. Up to what extent are this two differentiable manifolds, $(M,\mathcal{A})$ and $(M,\mathcal{A}_{\infty})$, the same?

  2. Is it true that the maximal atlas $\overline{\mathcal{A}_{\infty}}$ generated by $\mathcal{A}_{\infty}$ is the same as $\overline{\mathcal{A}}$ with all the non $C^{\infty}$ charts removed?

  3. Are there examples of $C^{k}$ manifolds for which exists a $C^{k}$ atlas such that none of its charts is $C^{r}$ for some $r>k$?

  4. When studying functions defined on a $C^{k}$ manifold $M$ we can consider differentials up to order $k$, with this limit $k$ imposed by the $C^{k}$ differentiable structure. Is the theorem of Whitney telling us that this restriction is artificial since we can get a $C^{\infty}$ atlas for $M$?

Thanks.

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  • $\begingroup$ You can deform a square into a circle, but the result no longer comprises all details of the square $\endgroup$ Apr 25, 2014 at 18:12
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    $\begingroup$ I did't get it. Could you elaborate the idea? $\endgroup$
    – Pipicito
    Apr 25, 2014 at 18:15

2 Answers 2

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Let $f:\mathbb R^n\to \mathbb R^n$ be a $C^k$ diffeomorphism which is not $C^{k+1}$ and consider the $C^k$ manifold $M$ whose underlying topological space is $\mathbb R^n$ and whose atlas consists of $f$ and the identity map $Id:\mathbb R^n\to \mathbb R^n:x\mapsto x$ .
If you discard one of the maps you obtain two different $C^\infty$ manifolds $M_{Id}$ and $M_{f}$ whose atlas consists of respectively the single homeomorphism $Id$ and the single homeomorphism $f$.
These different manifolds have the same underlying $C^k$ manifold, namely $M$.
This proves that passing from a $C^k$ manifold to a $C^\infty$ manifold is indeed always possible, but not in a canonical way. The same is true for passing from $C^k$ to $C^{k+1}$.

Remark
The existence of a $C^k$ diffeomorphism $f$ which is not $C^{k+1}$ is probably easy to prove but the only reference I could find (by browsing the web for two minutes...) is for the case $n=2$ in this probably much too sophisticated article .

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    $\begingroup$ The function $f : \mathbb R \to \mathbb R$ given by $f(x) = x^n |x|$ is $C^{n}$ but not $C^{n+1}$, and a diffeomorphism provided $n$ is odd. You can construct such diffeomorphisms quite easily like this. $\endgroup$ Apr 25, 2014 at 19:10
  • $\begingroup$ Thanks, Ryan. I was sure that real experts would find this quite easy . What should one do for even $n$ ? $\endgroup$ Apr 25, 2014 at 19:13
  • $\begingroup$ Instead of using $|x|$ in the above definition, you could use a PL function which is linear on $[0,\infty)$ and $(-\infty,0]$ but with different (but positive) slopes. $\endgroup$ Apr 25, 2014 at 19:19
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(1) The identity map from the $C^k$-manifold to its $C^\infty$-refinement is a $C^k$-diffeomorphism.

(2) I don't understand your question.

(3) This question does not make sense to me. What does it mean for a map to be more differentiable than the maximum-definable level of differentiability?

(4) You're asking a vague question here. Could you be more specific.

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  • $\begingroup$ For (3). Two ideas to define it:take the $C^{\infty}$ subatlas and test differentiability of original charts for orders $r>k$ using that. $\endgroup$
    – Pipicito
    Apr 25, 2014 at 18:44
  • $\begingroup$ Another possibility for (3): to find charts $(U,\phi)$ and $(V,\psi)$ in the $C^{k}$ original atlas such that $U \cap V \neq \emptyset$ and $\psi \circ \ \phi^{-1}$ is $C^{r}$ for some $r>k$. $\endgroup$
    – Pipicito
    Apr 25, 2014 at 18:53
  • $\begingroup$ The same for (2). Using the notion of the first comment throw away all the charts that are not $C^{\infty}$ when you give $M$ the differentiable structure induced by $\mathcal{A}_{\infty}$. $\endgroup$
    – Pipicito
    Apr 25, 2014 at 18:56
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    $\begingroup$ I think you need to re-think your questions. There is no notion of differentiability of a single chart in an atlas. You need to perhaps talk about the atlas with a chart removed, but then you're dealing with a different differentiable structure. $\endgroup$ Apr 25, 2014 at 19:12
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    $\begingroup$ @Pipicito: Ryan's last comment was exactly what I was going to say: there is a notion of $C^k$ atlas, in which every overlapping pair of charts has $C^k$ transition functions, but (to my knowledge) one cannot make sense of a single "$C^k$ chart" by itself. (Although, sometimes authors say "$C^k$ chart" to mean "a chart in a $C^k$ atlas.) $\endgroup$ Apr 25, 2014 at 19:15

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