Is there some criterion for whether a space has the homotopy type of a closed manifold (smooth or topological)? Poincare duality is an obvious necessary condition, but it's almost certainly not sufficient. Are there any other special homotopical properties of manifolds?
5 Answers
In surgery theory (which is basically a whole field of mathematics which tries to answer questions as the above), the next obstruction to the existence of a manifold in the homotopy type is that every finite complex with Poincaré duality is the base space of a certain distinguished fibration (Spivak normal fibration) whose fibre is homotopy equivalent to a sphere. (In order to get a unique such fibration, identify two fibrations if they are fiber homotopy equivalent or if one is obtained from the other by fiberwise suspension.)
For manifolds, this fibration is the spherization of the normal bundle, so the Spivak normal fibration comes from a vector bundle. This is invariant under homotopy equivalence. Thus the next obstruction is: the Spivak normal fibration must come from a vector bundle.
If I remember right, then it was Novikov who first proved that for simplyconnected spaces of odd dimension at least 5, this is the only further obstruction.
In general, there is a further obstruction with values in a group $L_n(\pi_1,w)$ which depends on the fundamental group, first StiefelWhitney class and the dimension. See Lück's notes on surgery theory at https://www.him.unibonn.de/lueck/data/ictp.pdf

1$\begingroup$ There is also the wikipedia article on surgery theory, which was written at least partly by Andrew Ranicki. $\endgroup$– Martin OOct 6, 2009 at 23:11

1$\begingroup$ "For manifolds, this fibration is the sphere bundles of the normal bundle...."  I'm sorry, but the normal bundle of M inside of what? $\endgroup$ Dec 14, 2009 at 5:26

10$\begingroup$ Inside a highdimensional Euclidean space. Every manifold can be embedded in such a space, by Whitney. $\endgroup$ Mar 24, 2010 at 3:45
The main result of the BrowderNovikovSullivanWall surgery theory (19621969) is that for $n>4$ a space $X$ is homotopy equivalent to a compact ndimensional topological (resp. differentiable) manifold if and only if $X$ is homotopy equivalent to a finite $CW$ complex $M$ with $n$dimensional Poincaré duality, and there is a topological (resp. vector) bundle over $M$ for which the corresponding normal map $(f,b):N\to M$ from an $n$dimensional manifold $N$ has zero surgery obstruction in the Wall $L$group of quadratic forms over $Z[\pi_1(X)]$. Thus there are two obstructions, a primary topological $K$theory obstruction to the existence of a bundle, and (depending on the vanishing of the primary one, and a choice of reason) a secondary obstruction in algebraic $L$theory. The original theory was for differentiable manifolds: the extension to topological manifolds due to Kirby and Siebenmann (1970) remains a major success of surgery theory. All this is explained (at some length) in Wall's own book Surgery on compact manifolds (1970/1998) and also in my own books Algebraic Ltheory and topological manifolds (1992) and Algebraic and geometric surgery (2002), as well as many other references (such as Wolfgang Lück's notes listed in a previous post). I have made available a large number of surgeryrelated resources on my website.
Sean: this gives a Poincare space which is not homotopy equivalent to a closed manifold. the idea is that the Spivak fibration of the $5$ dimensional Poincare space doesn't lift to a stable vector bundle. One can prove this as follows: let $X^5$ be as in Madsen and Milgram. Then $X$ fibers over $S^3$ with fiber $S^2$. Call this fibration $\xi$, and let the projection $X \to S^3$ be $p$. Note that $p$ has a section, call it $s$.
It's not hard to prove that the Spivak fibration of $X$ in this case is just $p^*\xi$. One then needs to verify that $p^*\xi$ doesn't lift to a stable vector bundle. But if it did, then so would $$\xi = (p \circ s)^*\xi = s^*p^* \xi\ .$$ But it's very easy to see that $\xi$ doesn't lift ($\xi$ is given by the nontrivial element of $\pi_2(F) = \mathbb{Z}_2$, where $F$ classifies spherical fibrations with section, whereas $\pi_2(O)$ is trivial).
I'm relying on memory here. A good example, which is discussed in Madsen and Milgram's book on surgery and classifying spaces for topological, $PL$ and smooth manifolds is the set of $1$connected Poincare duality spaces of dimension 5 with 4skeleton h. e. to the $4$skeleton of $S^2\times S^3$, which is $S^2\vee S^3$.
a. $S^2\times S^3$, which is obviously a manifold
b. a manifold homeomorphic to $SU(3)/SO(3)$ (though this fact isn't mentioned in Madsen & Milgram).
c. a Poincaré duality space whose Spivak fibration cannot be reduced to a smooth vector bundle. This is proved using secondary cohomology operations based on Steenrod squares. This was first proven by Gitler and Spivak(?).

$\begingroup$ I am sorry, but what exactly are these examples of? $\endgroup$ Jul 29, 2010 at 20:37

$\begingroup$ They are examples of 1connected Poincare duality spaces of dimension 5. If you ignore cohomology operations they look homologically like a manifold. The space in c is a Poincare duality space which cannot be homotopy equivalent to a manifold, I believe. This is due to the "higher" homotopy structure which is implicit in (non)vanishing Steenrod operations. Sean, I'll try to find you a more explicit reference for this. For now, I would say, look at the first 24 chapters of the PUP book of BrumfielMadsenMilgram. $\endgroup$– MikeDec 8, 2010 at 2:27

$\begingroup$ @Mike: concerning your point c.: You don't need secondary Steenrod squares to get that: see my argument above. @MISC {MO33853, TITLE = {How can you tell if a space is homotopy equivalent to a manifold?}, AUTHOR = {John Klein (mathoverflow.net/users/8032)}, HOWPUBLISHED = {MathOverflow}, NOTE = {URL: mathoverflow.net/questions/33853 (version: 20100729)}, EPRINT = {mathoverflow.net/questions/33853}, URL = {mathoverflow.net/questions/33853}, } $\endgroup$ Jan 17, 2011 at 12:28

2$\begingroup$ The 4skeleton of $SU(3)/SO(3)$ is not $S^2\vee S^3$. One way to see this is to note that $H_2(SU(3)/SO(3); \mathbb{Z}) \cong \mathbb{Z}_2$ while $H_2(S^2\vee S^3; \mathbb{Z}) \cong \mathbb{Z}$. $\endgroup$ Jul 4, 2021 at 16:37
Jacob Lurie gave a talk last week at Peter May's birthday conference on noncommutative Poincaré duality. The idea is to take an $n$manifold $M$ and a $(n1)$connected space $X$. Then, he showed that the compact mapping space $\mbox{Map}_c(M,X)$ is isomorphic to a certain homotopy colimit over a certain category of open subsets of $M$. This is equivalent to the usual commutative Poincaré duality. However, it is not clear (to me) what the natural generalization of the statement is to nonmanifolds. So, I am not sure how to use it as a test. However, if you could use it as a test of being a manifold, it seems feasible that if the noncommutative statement held for your test space $M$ and all $(n1)$connected spaces $X$ for some $X$, then it would seem reasonable ask whether your test space is the homotopy type of an nmanifold.
The category of open sets over which Lurie takes the colimit is the category of disjoint balls (homeomorphic to $\mathbb{R}^n$) in $M$. Thus, a guess might be something like: if $M$ is a space, and if $U$ is a category of open sets of $M$ that cover $M$, and if $\mbox{Map}_c(M,X)$ is equivalent to the homotopy colimit of $\mbox{Map}_c(U_i,X)$ for all $U_i$ in $U$ for all $(n1)$connected $X$ for some $n$, then $M$ has the homotopy type of an $n$manifold.
I have no idea if this is true, and even if it is true, it is not clear if it would be useful.

$\begingroup$ maybe the U_i should be contractible or something like that. $\endgroup$ Jul 29, 2010 at 20:40