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Surgery theory aims to measure the difference between simple homotopy types and diffeomorphism types. In 3 dimensions, geometrization achieves something much more nuanced than that. Still, I wonder whether the surgeons' key problem has been solved. Is every simple homotopy equivalence between smooth, closed 3-manifolds homotopic to a diffeomorphism?

In related vein, it follows from J.H.C. Whitehead's theorem that a map of closed, connected smooth 3-manifolds is a homotopy equivalence if it has degree $\pm 1$ and induces an isomorphism on $\pi_1$. Is there a reasonable criterion for such a homotopy equivalence to be simple? One could, for instance, ask about maps that preserve abelian torsion invariants (e.g. Turaev's).

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Turaev defined a simple-homotopy invariant which is a complete invariant of homeomorphism type (originally assuming geometrization).

Here is the Springer link if you have a subscription: Towards the topological classification of geometric 3-manifolds

He claims in the paper that a map between closed 3-manifolds is a homotopy equivalence if and only if it is a simple homotopy equivalence, but he says that the proof of this result will appear in a later paper. I'm not sure if this has appeared though (I haven't searched through his later papers on torsion, and there's no MathScinet link).

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  • $\begingroup$ The main issue is the connect-sum decomposition, no? Ie: oriented simple homotopy equivalent 3-manifolds have oriented simple-homotopy equivalent prime summands. Once you're past that JSJ + geometrization tools take over. $\endgroup$
    – Ryan Budney
    Commented Dec 15, 2009 at 21:34
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    $\begingroup$ Agol gets the green box for pointing Turaev's paper which, alongside his 1988 paper "Homeomorphisms of geometric three-dimensional manifolds", MR0940851, apparently answers both questions affirmatively. However, Turaev's argument draws together threads that Paul, Daniel, John, algori, Henry and Ryan mentioned - in particular, Waldhausen's work on the Haken case. Thanks to all. $\endgroup$
    – Tim Perutz
    Commented Dec 16, 2009 at 2:58
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    $\begingroup$ Agol, Tim -- I can't access Turaev's paper right now, but if he claims that a homotopy equivalence is a simple homotopy equivalence for 3-manifolds, this would seem a little strange, since for lens spaces this is false: L(7,1) and L(7,2) are homotopy equivalent, but have different simple homotopy types. $\endgroup$
    – algori
    Commented Dec 17, 2009 at 16:21
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    $\begingroup$ @ algori: he's claiming that simple homotopy equivalent manifolds are homeomorphic. $\endgroup$
    – Ian Agol
    Commented Dec 18, 2009 at 17:56
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Waldhausen proved that homotopy equivalence is homotopic to homeo (and hence diffeo) for Haken 3-manifolds. Perelman extends that to irreducible/infinite pi_1.

It's an old conjecture that the Whitehead group of any torsion free group is trivial.

Irreducible 3-manifolds either have finite or torsion free pi_1, so given Perelman again only S^3/G have potentially non-simple homotopy equivalences.

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  • $\begingroup$ Thanks! Didn't know any of that. When you say "Perelman extends that", do you mean that, in light of Perelman, Waldhausen's arguments apply to all irreducible 3-manifolds with infinite $\pi_1$? $\endgroup$
    – Tim Perutz
    Commented Dec 15, 2009 at 5:08
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    $\begingroup$ yes, if I'm recalling correctly: if there exist incompressible tori (or surfaces) then Waldhausen applies. Perelman says that atoroidal manifolds are either hyperbolic, so Mostow applies, or (simple) Seifert-fibered (and hence classified), I assume it is known precisely what non-Haken SF manifolds admit h.e. that aren't homotopic to diffeos, presumably just lens spaces. $\endgroup$
    – Paul Kirk
    Commented Dec 15, 2009 at 23:46
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This doesn't actually answer the question, but concerns Tim's comment:-

"The core of the question - I think! - is whether group theory, plus a bit of extra topological input, recognizes the geometric pieces of a 3-manifold."

It's certainly true that the fundamental group sees a lot of the geometry. Scott and Swarup proved that you can reconstruct the JSJ (torus) decomposition of an irreducible 3-manifold from the fundamental group. If your manifold isn't irreducible then the Kneser--Milnor decomposition corresponds exactly to the Grushko decomposition of π1. And π1 also determines the geometry of geometric pieces---Seifert-fibered pieces have normal cyclic subgroups etc.

(Of course, you need the Poincare Conjecture to know that you didn't connect sum with a fake 3-sphere!)

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About the elliptic case $S^3/G$: elliptic 3-manifolds are classified up to homeomorphism by their $\pi_1$'s, except for the lens spaces. For the lens spaces the simple homotopy type classification is equivalent to the homeomorphism classification (see e.g Milnor, Whitehead torsion, Bulletin AMS, 72) but differs from the homotopy classification.

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Regarding your first question, in 1953 Moise proved the (manifold) Hauptvermutung for 3-manifolds (Ann. of Math. 58, pp. 458-480). One way to state his result is that every homeomorphism (diffeomorphism) between compact 3-manifolds is homotopic to a PL homeomorphism.

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I don't know in general, so I'll just the more obvious cases. For hyperbolic 3-manifolds, this is implied by Mostow's rigidity theorem, which states that a homotopy equivalence of hyperbolic manifolds $n$-manifolds is homotopic to an isometry. It's also true for $S^3$, since both $Diff(S^3)$ and $Aut^h(S^3)$ have two components.

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  • $\begingroup$ Right - indeed, MR is stronger still, in that it only requires an abstract isomorphism of $\pi_1$. The core of the question - I think! - is whether group theory, plus a bit of extra topological input, recognizes the geometric pieces of a 3-manifold. $\endgroup$
    – Tim Perutz
    Commented Dec 15, 2009 at 4:47
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    $\begingroup$ The $\pi_1$ issue isn't really part of Mostow rigidity proper. Hyperbolic manifolds are $K(\pi,1)$'s so an isomorphism of the fundamental group is the same thing as a homotopy-equivalence of the spaces. The technique falls under basic obstruction theory, which in this case is pretty much "follow your nose". $\endgroup$
    – Ryan Budney
    Commented Dec 15, 2009 at 21:59

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