Does the existence of exotic smooth structure in $\mathbb{R}^4$ imply the existence of an atlas which has a $C^0$ mapping to the Cartesian atlas, but not a $C^k$ mapping (for some finite $k$)? Does the nonexistence of exotic smooth structure in $\mathbb{R}^n$, $n\neq 4$ imply that all atlases therein have smooth mappings to the Cartesian atlas?
[Math] exotic smooth structure clarification
differential-topologygt.geometric-topologysmooth-manifolds
Best Answer
Regarding your 1st question, perhaps you meant to ask something else? Any atlas can be composed with a non-smooth homeomorphism to produce an atlas that isn't smooth in the standard sense. For example, $\mathbb R \to \mathbb R$ defined by $t \longmapsto t^{1/3}$ is an atlas on $\mathbb R$ but it's not $C^1$. This answers your 2nd question in the negative.
Alternatively, some exotic smooth $\mathbb R^4$'s are diffeomorphic to open subsets of the standard $\mathbb R^4$, so even for exotic smooth $\mathbb R^4$'s you could potentially have only a one-map atlas, which is smooth in the standard sense.
You might like to read this article: http://en.wikipedia.org/wiki/Exotic_R4