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in standard text books on (smooth) manifolds, for example the known series by John M. Lee or Jeffrey Lee, you either deal with continuous manifolds, or with smooth manifolds.

However, neither in these books nor in lectures I have encountered real examples when a manifold may be $C^k$, but not $C^{k+1}$.

Intuitively, I would suppose the $|\cdot|_\infty$-Ball with radius $1$ to be a merely continuous, non-smooth manifold, because smoothness fails at the edges of the cube. In contrast to this, polar coordinates show the $|\cdot|_{2}$ with radius $1$ is in fact a smooth manifold.

I'd be thankful for some examples with clues to basic techniques, how the different degrees of smoothnesses manifest 'in real life'.

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2 Answers 2

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Take a look at Morris Hirsch's text "Differential Topology", in particular Theorem 2.9 on page 51. The upshot is that if $M$ is a $C^k$-manifold for $k \geq 1$, then it admits the structure of a $C^j$-manifold for any $j \geq k$, and that structure is unique up to diffeomorphism. So in that sense the different degrees of smoothness is largely just an artefact of how your manifold is constructed -- you can always "do better" provided the manifold is at least $C^1$.

I finished this as Emerton's answer appeared, anyhow, take this as a reference for the result.

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One can show that any $C^k$-manifold, for $k \geq 1$, has a unique enrichment to a $C^{\infty}$-manifold. (I.e. given $M$ with its $C^k$-atlas, we can find a $C^{\infty}$-atlas on $M$, compatible with the given $C^k$-atlas, and this $C^{\infty}$-atlas is unique up to equivalence; see wikipedia for more details.) So there is not much point in considering $C^k$-manifolds other than for $k = 0$ or $\infty$.

With regard to your unit ball examples, note that the $| \cdot |_{\infty}$-unit ball, although it has corners, is homeomorphic to the $| \cdot |_2$-unit ball; one says that it can be smoothed. There are topological manifolds that cannot be smoothed (in dimension 4 and higher), in the sense that they are not homeomorphic to a smooth manifold. There are also smooth manifolds that are homeomorphic, but not diffeomorphic. (E.g. when $n \geq 7,$ one can find smooth manifolds that are homeomorphic to $S^n$, but not diffeomorphic to it; these are so-called exotic spheres.) Again, the wikipedia entry has more details.

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Interesting. Being $C^0$ is a weaker condition than being PL, right? (A linear transition function is continuous. Or maybe I actually don't have the right definition of PL.) Which should mean that $C^\infty \cong C^1 \subset PL \subset C^0$? And a "$C^0$ manifold" is really just a topological manifold? –  Aaron Mazel-Gee Dec 17 '10 at 8:04

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