3
$\begingroup$

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?)

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Cheerful Parsnip Jun 25 '18 at 17:06
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – rpf Jun 25 '18 at 17:07
  • 1
    $\begingroup$ 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. $\endgroup$ – Cheerful Parsnip Jun 25 '18 at 17:08
  • 1
    $\begingroup$ @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. $\endgroup$ – Jason DeVito Jun 25 '18 at 17:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.