It's a result of lowdimensional topology that in dimensions 3 and lower, two manifolds are homeomorphic if and only if they are diffeomorphic. Milnor's 7spheres give nice counterexamples to this result in dimension 7, and exotic $\mathbb{R}^4$'s give nice counterexamples in dimension 4. But I don't know about dimensions 5 and 6. Is the result true or false in dimensions 5 and 6? And, if false, what are some classic counterexamples, and do stronger constraints  say compactness or closedness  happen to make it true?

2$\begingroup$ Well, the famous examples in dimension 4 are the exotic $\mathbb{R}^4$'s disovered by Friedman and Kirby. en.wikipedia.org/wiki/Exotic_R4 $\endgroup$ – HJRW Apr 22 '10 at 2:20

$\begingroup$ Yes, thanks, realized that just after I wrote the post. $\endgroup$ – symplectomorphic Apr 22 '10 at 2:38
It is false in dimension 5 and 6. Spheres happen to be standard, but some other (compact and closed) manifolds happen to admit different smooth (and PL) structures.
Simple example are tori. For example, $\mathbb T^5$ admits 3 different PL structures that give rise to 3 different differentiable structures. See, e.g., Hsiang, Shaneson "Fake tori" or Wall's book on surgery.

$\begingroup$ Wikipedia tells me that the result actually does hold for spheres in dimensions 5 and 6, contrary to what you say: en.wikipedia.org/wiki/… That's the main reason I asked the question. Or perhaps by calling the spheres "standard" you didn't mean they were standard counterexamples, but only that they carry unique differential structures. Otherwise, thanks! $\endgroup$ – symplectomorphic Apr 22 '10 at 4:37

1$\begingroup$ Yes, that's right, spheres in dimensions 5 and 6 are diffeomorphic to the standard $S^5$ ($S^6$). This is what I meant by calling them standard. I would like to point out that my and Igor's answer do not contradict each other. They complement each other very nicely. $\endgroup$ – Andrey Gogolev Apr 22 '10 at 14:37

$\begingroup$ While I'm at it, I might as well ask whether there's a simple compact or closed counterexample in dimension 4. As far as I know, this question for the 4sphere is unresolved: I doubt it would be any simpler for any other closed 4manifold, but perhaps I'm wrong? $\endgroup$ – symplectomorphic Apr 22 '10 at 15:13

$\begingroup$ My understanding is that it is rather large industry. There're exotic smooth structures on compact simply connected 4manifolds. For example $\Bbb{CP}^2\#6\overline{\Bbb{CP}^2}$ (not sure about 6) admits exotic structure. People are constantly making progress making the example "smaller" in second homology. Some names here are Park, Stipsicz, Szabo, Akhmedov. $\endgroup$ – Andrey Gogolev Apr 22 '10 at 15:34

$\begingroup$ Andrey, 6 is correct; you could replace it by any $n\geq 2$ (cf. FintushelStern's latest...). Moreover the number of different smooth structures is in each case countably infinite, while for simply connected manifolds of higher dimension it is always finite. $\endgroup$ – Tim Perutz Apr 22 '10 at 15:52
Any PLmanifold of dimension $\le 7$ is smoothable, and the smooth structure is unique in dimensions $5,6$. See e.g. remark 6.7 in Rudyak's paper for details.
EDIT: To explain the above, the smooth structures on a PL manifold $M$ of dimension $\ge 5$ are in 11 correspondence with $[M, PL/O]$, homotopy classes of maps from $M$ to the space $PL/O$, which is $6$connected. This implies the claim in the previous paragraph. Similarly, PL structures on a topological manifold $M$ of dimension $\ge 5$ are in 11 correspondnece with $[M,TOP/PL]$, and $TOP/PL$ is $K(\mathbb Z_2,3)$. Thus $[M,TOP/PL]$ is simply $H^3(M;\mathbb Z_2)$, the third cohomology group with $\mathbb Z_2$ coefficients, and if $H^3(M;\mathbb Z_2)$ is nonzero, then $M$ admits more than one PL structure. See MadsenMilgram "Classifying spaces for surgery and cobordism of manifolds".

2$\begingroup$ But a given topological manifold can host different PL structures, right? So if there are several nonisomorphic PL structures on a manifold $M$, then there are as many nonisomorphic differentiable structures, right? I was just figuring out why your answer is not in contradiction with Gogolev's answer. $\endgroup$ – Qfwfq Apr 22 '10 at 4:34
