Welcome to MathOverflow

Question

11

4

Can one vary smooth structures on $\mathbb R^4$ smoothly/continuously?

This question popped out of Ben's answer here.

flag	asked Jan 16 2011 at 18:00 Mariano Suárez-Alvarez 30.6k●7●68●153

	I assume you are aware of Taubes "Gauge theory on asymptotically periodic 4-manifolds", proving the existence of a uncountable family of smooth structures on $\mathbb{R}^4$, and subsequent work by Gompf "An Exotic Menagerie" proving the existence of a two-parameter family. What exactly would you mean by "smoothy" or "continuously" in this case? – Kelly Davis Jan 16 2011 at 18:34
	@Kelly: I am aware of the first result you mention, bot not the second. You should add an answer! By smooth/continuous family I had in mind a smooth map $M\to N$ such that the fibers are non-diffeomorphic copies of $\mathbb R^4$; for continuous, I don't know what I mean, really :) – Mariano Suárez-Alvarez Jan 16 2011 at 18:38
	@Mariano: Probably you mean not just a smooth map $M\to N$ but a smooth submersion (as in Ryan's answer). That gives smooth structures to the fibers. – Tom Goodwillie Jan 16 2011 at 21:37
	@Mariano: I'm not sure how to construct the submersion. Maybe I am missing something obvious? – Kelly Davis Jan 17 2011 at 8:08
	Looks like this thread withered on the vine, but, in case anyone is interested: The $L_t$ of Theorem 9.4.10 in Gompf and Stipsicz could be used as a one-parameter family of exotic $\mathbb{R}^4$ and the construction in Theorem 9.4.16(a) in the same book could be able to be used for a two parameter family. – Kelly Davis Jan 18 2011 at 20:00

show 1 more comment

Ryan Budney · Answer 1 · 2011-01-16 18:17:35Z UTC

7

Yeah, sure. A $1$-parameter family of smooth structures on a topological manifold $M$ can be taken to be a smooth structure on $M \times I$ such that the projection map $M \times I \to I$ is a submersion. Similarly for higher families.

To relate it to your comments on Ben's thread, you can (apparently) find a $1$-parameter family of smooth structures on $\mathbb R^4$ such that all pairs of fibres $\mathbb R^4 \times \{a\}$ and $\mathbb R^4 \times \{b\}$ for all $a \neq b$ are not diffeomorphic.

That 2nd paragraph is really Larry Siebenmann talking. I don't believe I've ever seen such a construction.

edited Jan 16 2011 at 18:17

answered Jan 16 2011 at 18:11

Ryan Budney
21.6k●1●40●101

1

How does that play along the fact that on $\mathbb R^5$ there is only one smooth structure? – Mariano Suárez-Alvarez Jan 16 2011 at 18:15

(By the way: my intuition regarding smooth structures is simply non-existent... I may be asking something silly!) – Mariano Suárez-Alvarez Jan 16 2011 at 18:16

2

@Mariano: I guess in that case the "strange" thing would be the projection itself, not the smooth structure on $\mathbb{R}^5$. (my intuition about smooth structures is nonexistent as well, though..) – Qfwfq Jan 16 2011 at 18:31

@Ryan: For the required result in the second paragraph, couldn't you use the $L_t$ of Theorem 9.4.10 of Gompf and Stipsicz? – Kelly Davis Jan 17 2011 at 8:04

smoothly varying smooth structures

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.