a $C^r$ manifold that is not $C^{r+1}$ manifold

Question

For a given $r>0$, Is there any example of a $C^r$ manifold which is not $C^{r+1}$? This is an exercise of This book.

Answer 1

Yes, namely consider $f:(a,b)\to \Bbb R$ a function such that:

• it is $C^r$
• it is not $C^{r+1}$, and specifically such that either $f^{(r+1)}(c)$ doesn't exist or $f^{(r+1)}$ is discontinuous at $c$ (your choice).
• it satisfies $f'(x)>0$ for all $x\in (a,b)$.

Then, the atlas $\mathcal A=\{((a,c+\varepsilon), (a,c+\varepsilon), id); ((c-\varepsilon,b), f[(c-\varepsilon,b)], f)\}$ induces a $C^r$ manifold structure on $(a,b)$ which is not $C^{r+1}$. Of course, though, the induced structure is $C^r$-diffeomorphic to the tautological structure induced by the $C^\infty$ atlas $\{((a,b),(a,b),id)\}$.

Added: More generally, if you believe in the existence of $C^r$-diffeomorphisms of open subsets $U\to V$ which are not $C^{r+1}$, you can devise a non-$C^{r+1}$ manifold structure on any manifold in such fashion: let $(X,\mathcal A)$ be a smooth manifold with atlas $\mathcal A$ and consider $(Y,U,\phi)$ any chart, with $U$ open subset of $\Bbb R^n$. Consider $\Psi:U\to V$ a $C^r$-non-$C^{r+1}$-diffeomorphism onto some open set $V\subseteq \Bbb R^n$. It is then clear that $\mathcal A'=\mathcal A\cup\{(Y,V,\Psi\circ \phi)\}$ is a new atlas on $X$, that $(X,\mathcal A')$ is a $C^r$ manifold which is not $C^{r+1}$, and that $(X,\mathcal A')$ is $C^{r}$-diffeomorphic to $(X,\mathcal A)$.