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My question concerns whether there is a contradiction between two particular papers on exotic smoothness, arxiv:0807.4248v1 and arxiv:gr-qc/9404003v1. The former asserts:

"Let $M$ be a smooth closed simply connected $4$-manifold, and $M'$ be an exotic copy of $M$ (a smooth manifold homeomorphic but not diffeomorphic to $M$). Then we can find a compact contractible codimension zero submanifold $W\subset M$ with complement $N$, and an involution $f:\partial W\to \partial W$ giving a decomposition: $M=N\cup_{id}W$, $M'=N\cup_{f}W$."

The latter states:

"Gompf's end-sum techniques are used to establish the existence of an infinity of non-diffeomorphic manifolds, all having the same trivial ${\bf R^4}$ topology, but for which the exotic differentiable structure is confined to a region which is spatially limited. Thus, the smoothness is standard outside of a region which is topologically (but not smoothly) ${\bf B^3}\times {\bf R^1}$, where ${\bf B^3}$ is the compact three ball. The exterior of this region is diffeomorphic to standard ${\bf R^1}\times {\bf S^2}\times{\bf R^1}$. In a space-time diagram, the confined exoticness sweeps out a world tube..."

and further:

"The smoothness properties of the ${\bf R^4_\Theta}$... can be summarized by saying the global $C^0$ coordinates, $(t,x,y,z)$, are smooth in the exterior region $[a,\infty){\bf\times S^2\times R^1}$ given by $x^2+y^2+z^2>a^2$ for some positive constant $a$, while the closure of the complement of this is clearly an exotic ${\bf B^3\times_\Theta R^1}$. (Here the 'exotic' can be understood as referring to the product which is continuous but cannot be smooth...)"

The theorem from the first paper applies to closed manifolds. Is it generalizable to open manifolds (such as $\mathbb{R}^4$)? If so, then its confinement of "exoticness" to a compact submanifold seems inconsistent with the world tube construction implied in the statements from the second paper.

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No, it does not generalize to non-compact manifolds. If it did, there would be a continuum of differentiable structures on the $4$-ball, but since the 4-ball is compact, there can only be countably-many smooth structures (there are only countably-many compact smooth 4-manifolds). –  Ryan Budney Aug 14 '10 at 21:52
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1 Answer

The compactness of the 4-manifold is really necessary for the existence of the Akbulut cork. The non-compact case cannot be done with this method. If there is a compact subset in the exotic $\mathbb{R}^4$ determing the exoticness, then one can make a one-point compactifications of the $\mathbb{R}^4$ getting an exotic 4-sphere $S^4$ (constructing a counterexample to the smooth Poince conjecture in dimension 4 (SPC4)).

More exact: Let $X$ be an exotic $\mathbb{R}^4$ and $R$ the standard $\mathbb{R}^4$. Let $C\subset X, C'\subset R$ be compact, contractable subsets. Now we assume that the subsets $C,C'$ are like Akbulut corks, i.e. $C$ and $C'$ are homeomorphic but non-diffeomorphic and $X-C$ is diffeomorphic to $R-C'$. The compactification of $X$ and $R$ result in a homotopy $S^4$ homeomorphic to the $S^4$. Then $X-C$ and $R-C'$ are changed to $\hat{X}-C$ and $\hat{R}-C'$ where $\hat{X},\hat{R}$ are homeomorphic to the $S^4$. Then $\hat{X}-C$ and $\hat{R}-C'$ are diffeomorphic (the assumption above) but $\hat{X},\hat{R}$ are not diffeomorphic. Thus we produce a counterexample to SPC4. But there is not such a compact subset (someday I heart the formulation: "the exoticness is located at infinity").

Therefore the approach in the 94'paper is correct, the world tube is a non-compact area. No contradiction.

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"If there is a compact subset in the exotic 4 determining the exoticness". Does this mean something like "$X$ is homeomorphic to $\mathbb{R}^4$ and there exist compact domains $C$ in $X$ and $C'$ in $\mathbb{R}^4$, with smooth boundary, and diffeomorphisms $C\to C'$ and $X−int(C)\to \mathbb{R}^4−int(C')$? If this were the case, one could compactify $X$ to a smooth homotopy 4-sphere, but I don't know how you would prove that is not diffeomorphic to $S^4$. –  Tim Perutz Aug 12 '10 at 15:26
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Besides, saying to oneself that "this can't work because if it did we could solve a famous problem" may be a useful guiding principle, but it's not a mathematical argument... –  Tim Perutz Aug 12 '10 at 15:31
    
Thanks for the comments. Of course you are right and I changed the answer. –  Torsten Asselmeyer-Maluga Aug 14 '10 at 20:32
    
Ah, now I see what you mean about SPC4: if there were a diffeomorphism $\hat{X}\to S^4$ then there would also be one preserving the points at infinity, which then restricts to a diffeo $X\to \mathbb{R}^4$. (I forget - is it in fact known whether a potentially-exotic $\mathbb{R}^4$ that's standard at infinity must be globally standard?) –  Tim Perutz Aug 14 '10 at 22:52
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Removing a standard $D^4$ from a potentially-exotic smooth $S^4$ yields a potentially-exotic $\mathbb{R}^4$ that's standard at infinity. So if it were known that the latter must be globally standard, then replacing the $D^4$ would imply the original $S^4$ is standard, and hence the 4-dimensional smooth Poincaré conjecture. And there's essentially only one way to replace the $D^4$, since Gamma_4 = 0 (Cerf) implies oriented diffeos of $S^3$ are smoothly isotopic. –  Daniel Asimov Aug 16 '10 at 6:37
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