Are $C^{k}$ manifolds the same as $C^{\infty}$ manifolds?

Solution 1:

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 .

Solution 2:

(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.