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Since $\mathbb{R}$ and any 3-manifold $N$ must be non-exotic, their product $\mathbb{R}\times N$ cannot possibly be diffeomorphic to exotic $\mathbb{R}^4$, correct?

Update: Andy Putman already answered this question in a different thread, as pointed out by Steven Sivek below. The answer is yes, but not for the reasoning I implied above, because, I gather, the product could in principle be taken in a nontrivial way that alters the differentiable structure.

The proof outlined by Andy relies on $\mathbb{R}\times N$ being piecewise linearly isomorphic to $\mathbb{R}^4$, which is said to be proved in "Cartesian products of contractible open manifolds" by McMillan, which happens to be freely available here: http://www.ams.org/journals/bull/1961-67-05/S0002-9904-1961-10662-9/S0002-9904-1961-10662-9.pdf . The relevant part of that paper is as follows:

"A recent result of M. Brown asserts that a space is topologically $E^n$ if it is the sum of an ascending sequence of open subsets each homeomorphic to $E^n$.

THEOREM 2. Let $U$ be a $W$-space. Then $U\times E^1$ is topologically $E^4$

Proof. Let $U=\sum_{i=1}^{\infty}H_i$ where $H_i$ is a cube with handles and $H_i\subseteq \text{Int } H_{i+1}$. By the above result of Brown, it suffices to show that if $i$ is a positive integer and $[a,b]$ an interval of real numbers ($a\lt b$), then there is a 4-cell $C$ such that

$H_i\times[a,b]\subseteq\text{Int }C\subseteq C\subseteq U\times E^1$.

There is a finite graph $G$ in $(\text{Int }H_i)\times\{(a+b)/2\}$ such that if $V$ is an open set in $U\times E^1$ containing $G$ then there is a homeomorphism $h$ of $U\times E^1$ onto itself such that $h(H_i\times[a,b])\subseteq V$. But $G$ is contractible to a point in $U\times E^1$. Hence, by Lemma 8 of [Bull. Amer. Math. Soc. 66, 485 (1960)], a 4-cell in $U\times E^1$ contains $G$, and the result follows."

A $W$-space was earlier defined as a contractible open 3-manifold, each compact subset of which is embeddable in a 3-sphere.

I'm not sure what it means for a simply connected manifold such as $\mathbb{R}^3$ to be equal to an infinite sum of cubes with handles, but given that, can we say that the above machination qualifies as a piecewise linear isomorphism because each $H_i\times[a,b]$ can be covered with a chart, and each $C$ can be covered with a chart, such that there is a linear mapping between the two?

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I don't know the answer to this question, but I'm pretty sure it doesn't follow from this argument. Factoring $\mathbb{R}^4$ as $\mathbb{R}\times N$ does not imply that $N$ is homeomorphic to $\mathbb{R}^3$, and thus the structure on $\mathbb{R}^4$ need not induce an exotic structure on $\mathbb{R}^3$. –  Daniel Litt Jul 29 '10 at 20:02
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That's not a valid line of reasoning, if that's what you mean. It's a theorem of Morton Brown's that if you take a contractible open 3-manifold cross $\mathbb R$, you get a manifold homeomorphic to $\mathbb R^4$. I believe it has since been proven that you get the standard smooth $\mathbb R^4$ regardless of which contractible open $3$-manifold you use as input but it's a theorem -- due to whom, at present I forget. –  Ryan Budney Jul 29 '10 at 20:05
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See Andy Putman's first comment on mathoverflow.net/questions/24970/… -- he cites McMillan for the fact that it has the standard PL structure, and then Munkres to show that this implies its smooth structure is standard. –  Steven Sivek Jul 29 '10 at 20:18
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2 Answers

To say that a 3-manifold $W$ is an infinite sum of cubes with handles means that there is an exhausting sequence $W_0 \subset W_1 \subset W_2 \subset \dots$ such that $W = \cup W_i$ and each $W_i$ is a compact handlebody (i.e. homeomorphic to a closed regular neighborhood of a finite graph in $\mathbb{R}^3$). It is a theorem that any contractible 3-manifold has such an exhaustion. In fact, if $W$ is open, irreducible and contains no closed essential surface then $W$ has such an exhaustion. The proof is not too difficult and can be found in several places including Theorem 2 of Freedman and Freedman's article "Kneser-Haken finiteness for bounded 3-manifolds, locally free groups, and cyclic covers" (Topology Vol 37 No 1).

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Steven Sievik comment is very important. Following the approach of Munkres and McMillan, then every 4-manifold $N\times\mathbb{R}$ with $N$ a contractable 3-manifold is diffeomorphic to the standard $\mathbb{R}^4$. Therefore the exotic $\mathbb{R}^4$ cannot be splitted like $N \times\mathbb{R}$ and esspecially not like $\mathbb{R}^3 \times\mathbb{R}$. Or, there is no diffeomorphism between the exotic $\mathbb{R}^4$ and $\mathbb{R}^3 \times\mathbb{R}$. But by definition there is a homeomorphism between the exotic $\mathbb{R}^4$ and the standard $\mathbb{R}^4$. Then we have a homeomorphism between the exotic $\mathbb{R}^4$ and $\mathbb{R}^3 \times\mathbb{R}$.

In the topological category we have a splitting $\mathbb{R}^3 \times\mathbb{R}$ but not in the smooth category.

Addendum: In contrast, every exotic $\mathbb{R}^4$ admits a $C^\infty$ codimension-1 foliation because any non-compact manifold admits one (see the BAMS article of Lawson "Foliations", Corollary 1.2).

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I don't understand this answer. Why do you need a fancy theorem to conclude that an exotic $\mathbb{R}^4$ is homeomorphic to a standard $\mathbb{R}^4$? Isn't that the definition? –  Andy Putman Aug 13 '10 at 15:39
    
Yes, you are right but I was not sure. I edited the answer. Thanks for the comment. –  Torsten Asselmeyer-Maluga Aug 14 '10 at 20:19
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