Constructing exotic $\mathbb{R}^4$'s using vector fields on $\mathbb{R}^5$

Question

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?

Answer 1

$Exotic(\mathbb R^4) \times \mathbb R$ 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.