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Consider a function $f:M\rightarrow\mathbb{R}$, where $M$ is a $C^{\infty}$ manifold. Recall that a function f is smooth if $\forall$ $p$ $\in M$, $\exists$ a smooth chart $(U,\phi)$ that contains $p$ such that $f o \phi^{-1}$ is a smooth map from $\phi(U)$ to $\mathbb{R}$.

Suppose now I set $M=\mathbb{R}^{n}$. Does there exist a function $f$ and a smooth atlas for $M$ such that the function $f$ is smooth in the sense that I have just defined, but it is not smooth in the sense of ordinary calculus (Note: smooth in the sense of ordinary calculus means that the function has continuous partial derivative of all order)?

KnobbyWan
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    The answer should be yes, and the smooth atlas should be given by any of the exotic differential structures on $\mathbb{R}^4$ (https://en.wikipedia.org/wiki/Exotic_R4), whose existence is proven using a good deal of beautiful, but pretty advanced mathematics. – Daniel Robert-Nicoud Mar 10 '16 at 15:50

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It is very easy to change the smooth structure on $\mathbb{R}^n$: simply push forward the usual smooth structure using a non-smooth homeomorphism.

For example, consider the following homeomorphism which is not a diffeomorphism $$h(x) = g(|x|) \cdot x $$ where $g : [0,\infty) \to [0,\infty)$ is some terrible homeomorphism, maybe non differentiable at a dense set and its inverse is similarly terrible. The push forward atlas under $h$ is the set of all $(U,\phi)$ such that $U$ is an open set, $\phi : U \to V$ is a homeomorphism onto some open subset $V \subset \mathbb{R}^n$, and $\phi \circ h^{-1} : h(U) \to V$ is a diffeomorphism.

A function $f : \mathbb{R}^n \to \mathbb{R}$ is smooth with respect to this push forward atlas if and only if $f \circ h^{-1}$ is smooth in the usual old atlas. But for almost any example $f$ you choose (such as projection onto a coordinate axis), if $f \circ h^{-1}$ is smooth then $f$ is not.

Lee Mosher
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It sounds like you're asking for exotic $\mathbb{R}^n$s: differential structures on $\mathbb{R}^n$ that are not equivalent to the usual one. No such manifolds exist for $n\not = 4$: any manifold homeomorphic to $\mathbb{R}^n$ (in particular, one that's just $\mathbb{R}^n$ with a different underlying differential structure) is actually diffeomorphic to it. For $n = 4$, there are actually infinitely many such $\mathbb{R}^n$, no two of which are diffeomorphic. Their construction, and even a proof of their existence, is nontrivial. The one-sentence answer is that both phenomena above are controlled by the $h$-cobordism theorem, but that's sweeping quite a bit of math under the rug. A great reference for exotic $\mathbb{R}^4$ is Scorpan's "Wild World of $4$-manifolds", but it assumes a significant familiarity with algebraic and differential topology (say, at the level of Hatcher's and Lee's books, plus a background on characteristic classes).

anomaly
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  • Yes, but take a smooth map from exotic $\mathbb{R}^4$ to $\mathbb{R}$. How do you show that the same function, seen as a map on standard $\mathbb{R}^4$, is possibly non-smooth? I.e. how do you construct a map which is smooth on the exotic Euclidean space, but not on the classical one? – Daniel Robert-Nicoud Mar 10 '16 at 15:53
  • In at least one exotic $X = \mathbb{R}^4$, there aren't arbitrarily large smoothly embedded $3$-spheres. That should imply that the usual norm $X\to \mathbb{R}$ isn't smooth. – anomaly Mar 10 '16 at 16:00