11
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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 $\mathbb{R}^3$. $\endgroup$ Jul 29, 2010 at 20:02
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 29, 2010 at 20:05
  • 10
    $\begingroup$ 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. $\endgroup$ Jul 29, 2010 at 20:18

2 Answers 2

2
$\begingroup$

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).

$\endgroup$
0
$\begingroup$

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).

$\endgroup$
2
  • 9
    $\begingroup$ 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? $\endgroup$ Aug 13, 2010 at 15:39
  • $\begingroup$ Yes, you are right but I was not sure. I edited the answer. Thanks for the comment. $\endgroup$ Aug 14, 2010 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.