[Math] How well can we localize the “exoticness” in exotic R^4

differential-topologygt.geometric-topologysmooth-manifolds

My question concerns whether there is a contradiction between two particular papers on exotic smoothness, Exotic Structures on smooth 4-manifolds by Selman Akbulut and Localized Exotic Smoothness by Carl H. Brans. 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.

Best 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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