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There exist topological manifolds which don't admit a smooth structure in dimensions > 3, but I haven't seen much discussion on homotopy type. It seems much more reasonable that we can find a smooth manifold (of the same dimension) homotopy equivalent to a given topological manifold. Is this true, or is there a counterexample?

flag	asked Aug 7 2010 at 16:34 Jake 355●1●6

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Your question is answered here: en.wikipedia.org/wiki/Hauptvermutung – Ryan Budney Aug 7 2010 at 17:14

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@Ryan's Comment: It seems to me that the wiki article doesn't explicitly address (at least the stronger form of) this question. The triangulation obstruction is for lifting a given $BTop$ structure, and this structure (and thus the obstruction) is preserved by homeomorphism of topological manifolds but not homotopy equivalence. (There are non-triangulable topological manifolds homotopy equivalent to triangulable ones.) [It indirectly addresses it, by mentioning $E8$. But it doesn't mention the fact that it's ruled out by its signature form.] – Anatoly Preygel Aug 7 2010 at 18:44

This is an old question but I'd still like to see a proper answer to it in dimensions above 4. The counterexample presented only works in dimension 4 which is very special. And as Anatoly Preygel points out the wiki link that Ryan gave does not address the issue in dimensions above 4. It only discusses the obstructions to existence of PL structures on a fixed topological manifold but says nothing about what can happen on manifolds homotopy equivalent to a given one which is what the question is really asking. – Vitali Kapovitch Jan 3 2012 at 18:42

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The first example of (12-dimensional) manifold which is not homotopy-equivalent to a smooth manifold is in Smale's "Generalized Poincaré's conjecture in dimensions greater than four", see Hirsch's review of this paper in Math. Reviews (since Smale does not state the homotopy-equivalence part). Examples of closed aspherical manifolds (in all dimensions $\ge 13$) which are not homotopy-equivalent to smooth ones were constructed by Davis and Hausmann in "Aspherical manifolds without smooth or PL structure" (1986). Remark: Kirby-Siebenmann invariant is not (always) homotopy-invariant. – Misha Mar 29 2012 at 16:03

Richard Borcherds · Accepted Answer · 2010-08-07 18:26:54Z UTC

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It is false for compact manifolds in 4 dimensions. Freedman showed that there is a compact simply connected topological 4-manifold with intersection form E8, but Donaldson showed that there is no such smooth manifold.

answered Aug 7 2010 at 18:26

Richard Borcherds
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It should be noted that the mention of the intersection form is particularly germane to your question. Among such manifolds, the intersection form (of 2-homology classes) completely determines the manifold up to homotopy (=h-cobordism=homeomorphism). – Tom Boardman Aug 7 2010 at 18:37

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If you pose a relative form of the same question -- what's a topological manifold with smooth boundary that is not homotopy equivalent to a smooth manifold with the same boundary (by a map that's the identity on the boundary)? -- then the original exotic spheres are examples: any smooth $7$-manifold homeomorphic but not diffeomorphic to $S^7$ bounds a topological $8$-disk but cannot possibly bound a smooth contractible manifold, by the $h$-cobordism theorem. – Tom Goodwillie Aug 7 2010 at 20:16

Misha · Answer 2 · 2012-04-02 11:19:18Z UTC

For every $n\ge 4$ there exists a closed aspherical topological $n$-manifold $N$ which is not homotopy-equivalent to a PL manifold. Furthermore, $\pi_1(N)$ is a CAT(0) group. This is a theorem of Davis and Januszkiewicz, see theorems 5a1, 5b1 in their "Hyperbolziation of polyhedra" paper http://intlpress.com/JDG/archive/1991/34-2-347.pdf The construction relies heavily on Freedman's result about E8-manifold. As Vitali said, this is an old post, but it is good to have an answer that works in all dimensions since dimension 4 is exceptional in many ways.

Are topological manifolds homotopy equivalent to smooth manifolds?

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