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What are known examples of two smooth, closed, oriented Manifolds $M,N$ of the same dimension that are simple homotopy equivalent, but not homeomorphic ?

It is well-known that the homotopy type of a given such $2$-manifold is the same as its homeomorphism type, so there won't be any such easy examples. Moreover (and this is where for me, the question becomes really interesting), I've heard at a recent conference that the simple homotopy type of a closed, oriented $3$-manifold is the same as its homeomorphism type, so any known examples must be in even higher dimension. Also, as the Borel conjecture is still open, any known such example must consist of non-aspherical manifolds.

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  • $\begingroup$ Note that the title is a little ambiguous and easy to misread, especially given your use of the word "simplest" in the first sentence: it could be parsed "(simple homotopy equivalent), non-homeomorphic manifolds" or "simple (homotopy equivalent, non-homeomorphic manifolds)." $\endgroup$ – Qiaochu Yuan Feb 25 '17 at 0:09
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    $\begingroup$ maths.ed.ac.uk/~aar/papers/connect.pdf theorem 1 gives counterexamples $\endgroup$ – Ian Agol Feb 25 '17 at 1:11
  • $\begingroup$ I've realized that the title is quite ambiguous. I've edited the text accordingly. Also, the paper is just what I've looked for. Thanks ! $\endgroup$ – Berni Waterman Feb 25 '17 at 6:33
  • $\begingroup$ There are quite possibly simpler or older examples, this was just the first thing I came across after a search. $\endgroup$ – Ian Agol Feb 25 '17 at 16:35
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The manifold $*\mathbb{C}P^2$ (or the Chern manifold) is homotopy equivalent to $\mathbb{C}P^2$, but it is not homeomorphic to $\mathbb{C}P^2$, since its Kirby-Siebenmann invariant is non-trivial.

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    $\begingroup$ Are they simple homotopy equivalent though? $\endgroup$ – Qiaochu Yuan Feb 25 '17 at 0:09
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    $\begingroup$ The fundamental group of both manifolds is trivial, so the Whitehead torsion of the homotopy equivalence vanishes. Thus, yes, they are simple homotopy equivalent. $\endgroup$ – Alex Suciu Feb 25 '17 at 1:39
  • $\begingroup$ Note however that the Chern manifold, unlike $\mathbb CP^2$, doesn't admit a PL-Structure (having non-trivial Kirby-Siebenmann class). From what I've read so far, Agol's reference seems to provide an example of two differentiable PL-manifolds that are simple homotopy equivalent and not homeomorphic. $\endgroup$ – Berni Waterman Feb 25 '17 at 10:32
  • $\begingroup$ I know this is from a long time ago; I'm wondering if you have a definition/construction or reference for the Chern manifold. I'm having trouble finding anything about it. $\endgroup$ – inkievoyd Jun 5 at 13:28
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It is easy to give infinite sets of smooth manifolds that are homotopy equivalent but not diffeomorphic. They are distinguished by their Pontrjagin classes and the argument requires little more than the theory of characteristic classes. It is much easier than Milnor’s construction of exotic spheres and I think it is easier than the analysis of lens spaces, but it is little known, I think because of the popularity historical recapitulation. Proving theorems about homeomorphism is much more difficult than diffeomorphism. It is a hard theorem of Novikov that the (rational) Pontrjagin classes are an invariant of topological manifolds, from which it follows that these actually yield infinite sets of manifolds which are homeomorphic but not diffeomorphic. In fact, there are only finitely many smooth manifolds homeomorphic to a given one.

Surgery theory approximately says manifolds homotopy equivalent to a given simply connected smooth manifold are determined by their tangent bundle, which can be anything, in particular arbitrary Pontrjagin classes. In particular, there are infinitely many smooth manifolds homotopy equivalent to $S^2\times S^4$ with differing first Pontrjagin class and thus clearly not diffeomorphic (nor even homeomorphic).

That’s a big hammer, but there is also an elementary construction, not for $S^2\times S^4$, but for $S^3\times S^4$. Consider $S^3$ bundles over $S^4$ with structure group $SO(4)\sim SO(3)\times SO(3)$. Such a bundle is trivial on the two hemispheres and classified by the gluing map on the equator: $\pi_4(BSO(4))=\pi_3(SO(4))=\pi_3(SO(3))\oplus\pi_3(SO(3))$. Two independent characteristic classes are the Euler class and the first Pontrjagin class. The Euler class controls the homology of the total space of the bundle. If it vanishes, the homology is that of the trivial bundle: $H^*(M)=H^*(S^3\times S^4)$; in particular $H^4(M)=\mathbb Z$ and there is room for infinitely many values of the Pontrjagin class. The Pontrjagin class (of the tangent space) of total space of the bundle is the Pontrjagin class of the (vector or sphere) bundle over $S^4$, so that gives infinitely many manifolds which are not pairwise diffeomorphic (nor homeomorphic).

However, these infinitely many manifolds aren’t all homotopy equivalent to $S^3\times S^4$. We must show that there are finitely many homotopy types. One way to see this is that we glued trivial bundles over the hemispheres by a map from the equator to the structure group $S^3\to SO(4)$. We could generalize to gluing by a family of homotopy equivalences: $S^3\to hAut(S^3)\subset Map(S^3,S^3)= S^3\times \Omega^3S^3$. A homotopy between maps $S^3\to Map(S^3,S^3)$ yields a homotopy equivalence between the total spaces of their bundles. Thus the set of homotopy classes of the total spaces of bundles is (at most) $\pi_3(S^3)\oplus \pi_6(S^3)$. The first factor is the Euler class and determines the homology of total space, while the second is a more subtle invariant, but is a finite group. Thus there are only finitely many homotopy types of these bundles, so there are infinitely many manifolds in one homotopy type with different Pontrjagin classes. The map of spaces $SO(4)\to Map(S^3,S^3)$ yields a homomorphism on homotopy groups $\pi_3$, so the element of $\pi_6(S^3)$ is additive in the Pontrjagin class, so every homotopy type realized is realized infinitely many times; in particular, there are infinitely many choices of Pontrjagin class with total space homotopy equivalent the product $S^3\times S^4$.

(Alternately, instead of restricting to linear bundles, one can show that there are only finitely many homotopy types of simply connected spaces with the homology of $S^3\times S^4$. Such a space receives a map from $S^3\vee S^4$ with homotopy cofiber a sphere. Thus its homotopy type is determined by how the top cell is attached. The attaching map lives in $\pi_6(S^3\vee S^4)$ which Hilton-Milnor computes as $\pi_6(S^3)\oplus \pi_6(S^4)\oplus \pi_6(S^6)$. This is the sum of an infinite cyclic group and a finite group. The infinite cycle group controls the cup product. Thus spaces with the desired homology are parameterized by a coset of a finite group, and thus finite.)

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  • $\begingroup$ "In fact, there are only finitely many smooth manifolds homeomorphic to a given one." This assumes that $M$ is compact, right ? $\endgroup$ – Berni Waterman Mar 3 '18 at 22:40
  • $\begingroup$ Right, for compact manifolds, possibly with boundary. . . For noncompact manifolds there is room to do infinitely many things. . . For example, infinite connected sums. If $X_n^1$ is a closed manifold homeomorphic to $X_n^2$ but not diffeomorphic, not even after removing a point, then the infinite connected sum of the $X_n^{i_n}$ has homeomorphism type independent of the $i_n$ but give an uncountable set of manifolds that are not diffeomorphic preserving the connected sum decomposition. With care in the choice of the $X_i$ they should be diffeomorphic at all. $\endgroup$ – Ben Wieland Mar 5 '18 at 0:43
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To expand a bit the answers given by others, the machinery of surgery theory outputs many examples of simple homotopy equivalent manifolds that are not homeomorphic, PL homeomorphic, or diffeomorphic (let's just say equivalent to avoid repeating all of these categories.) The only caveat is that what it produces are simple homotopy equivalences that are not homotopic to equivalences. The bible for this is Wall's book, Surgery on Compact Manifolds, although there are other treatments that are perhaps more approachable. Ben Wieland's answer has to do with one aspect of the theory, which has to do with the tangent bundle; the example of Cappell cited by Ian Agol has to do with the Wall surgery groups. I'll suggest some alternative examples along these same lines, although using easier to understand invariants.

If you want the manifolds to be non-equivalent, then a good collection of manifolds to look at are manifolds homotopy equivalent to higher dimensional lens spaces (quotients of $S^{2n-1}$ by a cyclic group $\mathbb{Z}_d$ acting linearly on $\mathbb{R}^{2n}$). In this setting, there are two fundamental sorts of invariants: Reidemeister torsion, and the multisignature (or Atiyah-Singer invariants). If you fix the simple homotopy type, that fixes the Reidemeister torsion, and the multisignature invariants can be varied more or less arbitrarily. This is detailed in Chapter 14E in Wall's book (which is not particularly easy reading). The executive summary of what is going on here is that you are using the action of the Wall surgery group $L^s_{2n}(\mathbb{Z}[\mathbb{Z}_d])$ on the structure set of a lens space $S^{2n-1}/\mathbb{Z}_d$, and the multisignature invariants are really invariants from that surgery group. They have the nice property that they are invariants of the manifold itself (together with an identification of the fundamental group with $\mathbb{Z}_d$) and so don't depend very strongly on a choice of homotopy equivalence.

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