It's a result of low-dimensional topology that in dimensions 3 and lower, two manifolds are homeomorphic if and only if they are diffeomorphic. Milnor's 7-spheres 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?

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    $\begingroup$ 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, 2010 at 2:20
  • $\begingroup$ Yes, thanks, realized that just after I wrote the post. $\endgroup$ Apr 22, 2010 at 2:38

2 Answers 2


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! $\endgroup$ Apr 22, 2010 at 4:37
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    $\begingroup$ 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. $\endgroup$ Apr 22, 2010 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 4-sphere is unresolved: I doubt it would be any simpler for any other closed 4-manifold, but perhaps I'm wrong? $\endgroup$ Apr 22, 2010 at 15:13
  • $\begingroup$ My understanding is that it is rather large industry. There're exotic smooth structures on compact simply connected 4-manifolds. 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. $\endgroup$ Apr 22, 2010 at 15:34
  • $\begingroup$ Andrey, 6 is correct; you could replace it by any $n\geq 2$ (cf. Fintushel-Stern'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, 2010 at 15:52

Any PL-manifold 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 1-1 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 1-1 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 Madsen-Milgram "Classifying spaces for surgery and cobordism of manifolds".

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    $\begingroup$ 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, 2010 at 4:34
  • $\begingroup$ Yes, that's right. $\endgroup$ Apr 22, 2010 at 10:43

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