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A smooth (i.e. $C^{\infty}$) $n$-manifold $M$ can be defined as a topological manifold such that each point has a neighborhood which is diffeomorhic to an open subset of $\mathbb{R}^n$. In particular, every point has a neighborhood which is diffeomorphic to $\mathbb{R}^n$ (see e.g. here).

This definition is unambiguous if $n\neq 4$, since then $\mathbb{R}^n$ has a unique smooth structure. However, in dimension 4, Euclidean space $\mathbb{R}^4$ has uncountably many incompatible smooth structures.

Suppose $M$ is a smooth 4-manifold and $p$ is a point in $M$. Then $p$ has a neighborhood $U$ which is diffeomorphic to $\mathbb{R}^4$, but which smooth $\mathbb{R}^4$? Is the convention that the smooth structure on $\mathbb{R}^4$ is taken to be the standard one?

Are there smooth 4-manifolds such that any point has a neighborhood diffeomorphic to an exotic $\mathbb{R}^4$ (other than the exotic $\mathbb{R}^4$s themselves)? In other words, can one construct exotic 4-manifolds from exotic $\mathbb{R}^4$s? (For example, if you take a quotient $\mathbb{R}^4/\mathbb{Z}^4$ of an exotic $\mathbb{R}^4$, do you get an exotic torus?)

rpf
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    Just a comment: the action of $\mathbb Z^4$ on an exotic $\mathbb R^4$ is almost certainly not a smooth action, so the quotient won't inherit a smooth structure. – Cheerful Parsnip Jun 25 '18 at 17:06

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It is always the standard $\mathbb{R}^4$.

Also notice that the non-standard $\mathbb{R}^4$ are locally diffeomorphic and globally homeomorphic to the standard $\mathbb{R}^4$, but not globally diffeomorphic.

Daniel Robert-Nicoud
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  • I am confused for the following reason: according to the "local diffeomorphism" wikipedia page, a diffeomorphism is a bijective local diffeomorphism. So if an exotic $\mathbb{R}^4$ is locally diffeomorphic and homeomorphic to standard $\mathbb{R}^4$, it must be diffeomorphic to standard $\mathbb{R}^4$. Are you using a different definition of local diffeomorphism? – rpf Jun 25 '18 at 17:07
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    The second comment of this answer is important. If you are locally built on exotic $\mathbb R^4$s, then you are locally built on standard $\mathbb R^4$s, so you won't get a new theory this way. – Cheerful Parsnip Jun 25 '18 at 17:08
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    @rpf: "Locally diffeomorphic" means "around each point $p$ there is a neighborhood $U_p$ of $p$ and a diffeomorphism $\phi_p:U_p\rightarrow V_p$ for some open subset $V_p$ of $\mathbb{R}^4$". A function $\phi$ being a local diffeomorphism means around each point $p$, there is an open neighborhood $U_p$ for which $\phi|_{U_p}:U_p\rightarrow \phi(U_p)$ is a diffeomorphism. So, the real difference between the two notions is that in the first, you get to pick a different $\phi_p$ for each point. For the second, you are stuck using $\phi$ everywhere. – Jason DeVito - on hiatus Jun 25 '18 at 17:56