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I was reading a paper of Arnol'd ("Topological Properties of Eigenoscillations in Mathematical Physics") where he gives the following claim (hopefully I am stating it correctly).

One way to produce smooth 4-dimensional manifolds is to take some smooth, non-vanishing vector field $v$ on $\mathbb{R}^5$. The flow of this vector field defines a smooth $\mathbb{R}$-action on $\mathbb{R}^5$, and then we can just take the quotient of $\mathbb{R}^5$ by this action to produce some smooth 4-manifold $M$.

Arnol'd makes the interesting claim that, given any exotic $\mathbb{R}^4$ we know about, it can be produced by a careful choice of this vector field).

Could anyone shed some light on the details of this construction?

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$Exotic(\mathbb R^4) \times \mathbb R$ is diffeomorphic to $\mathbb R^5$ as we know that there exits unique smooth structure on $\mathbb R^5$ (proved by Stalling A reference for smooth structures on R^n). Now there exists a nice $\mathbb R$ action on $Exotic(\mathbb R^4)\times \mathbb R$, i.e, translation along the $\mathbb R$ axis. And push-forward of this action will generate a smooth action on $\mathbb R^5$ which is the one you are looking for.

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  • $\begingroup$ Great, thanks! This is probably what Arnol'd meant, but I was under the impression that he had some explicit construction method in mind. $\endgroup$ – Rohil Prasad Jun 23 '20 at 2:34
  • $\begingroup$ It would be nice to see if there exists any explicit construction. Most of the proofs that I know about exotic structures on $\mathbb R^4$ is in some-sense existential. $\endgroup$ – Anubhav Mukherjee Jun 23 '20 at 2:36
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    $\begingroup$ It would be nice if you could add references to your statements. $\endgroup$ – Piotr Hajlasz Jun 23 '20 at 2:44
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    $\begingroup$ @PiotrHajlasz I added a link of MO which contains nice explanations. $\endgroup$ – Anubhav Mukherjee Jun 23 '20 at 2:49

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