The goal of this post is to describe a small portion of the answer to the following question:
Question 1: When is a simply connected space homotopy equivalent to a (compact) -dimensional smooth manifold?
A compact manifold is homotopy equivalent to a finite CW complex, so must be one itself. Equivalently (since is simply connected), the homology must be finitely generated.
More interestingly, we know that a hypothetical compact -manifold homotopy equivalent to (which is necessarily orientable) satisfies Poincaré duality. That is, we have a fundamental class with the property that cap product induces an isomorphism
for all groups and for all . It follows that, if the answer to the above question is positive, an analogous condition must hold for . This motivates the following definition:
Definition 1 A simply connected space is a Poincaré duality space (or Poincaré complex) of dimension if there is a class such that for every group , cap product induces an isomorphism
A consequence is that the cohomology (and homology) groups of vanish above dimension , and is generated by . In other words, behaves like the fundamental class of an -dimensional manifold.
We can thus pose a refined version of the above question.
Question 2: Given a simply connected Poincaré duality space of dimension , is there a homotopy equivalence for a compact (necessarily -dimensional) manifold?
The answer to Question 2 is not always positive.
Example 1 The Kervaire manifold is a topological 4-connected 10-manifold of (suitably defined) Kervaire invariant one. Since, as Kervaire showed, all smooth framed 10-manfolds have Kervaire invariant zero, he concluded that there was no smooth manifold in its homotopy type.
Example 2 It is known that there exists a simply connected compact topological 4-manifold whose intersection form is even and has signature eight (the “ manifold”). Such a manifold cannot be homotopy equivalent to any smooth manifold. In fact, if it were homotopy equivalent to a smooth 4-manifold , then the evenness of the intersection form on shows that (and in fact all ) act trivially on . The Wu formulas imply that the Stiefel-Whitney classes of vanish. In particular, admits a spin structure, and a theorem of Rohlin asserts that the signature of a spin 4-manifold is divisible by .
2. Atiyah duality and Spivak fibrations
Let us return to the question above. Consider a simply connected Poincaré duality space of dimension . If has the homotopy type of a manifold, we should in particular obtain an -dimensional vector bundle on , coming from the tangent bundle of that manifold. We cannot expect this bundle to be uniquely determined, because the tangent bundle of a smooth manifold is not a homotopy invariant. Nonetheless, is not totally arbitrary either. It turns out the stable spherical fibration associated to is homotopy invariant, and can in fact be reconstructed simply from .
To see this, let’s recall the homotopy properties of the tangent bundle. Let be a smooth, compact manifold. The tangent bundle cannot be reconstructed homotopy-theoretically from . However, one important homotopy-theoretic property of the normal bundle of an imbedding is the Pontryagin-Thom collapse map
mapping onto the Thom complex of . It crushes everything outside a tubular neighborhood of to the “point at infinity” in the Thom complex. It has the property that has the property that is sent to the generator of (corresponding to the fundamental class of ). This is called “-reducibility” and can be phrased by saying that the top cell of splits off.
We can use the Pontryagin-Thom map to produce a map
and desuspending produces a map of spectra
where is the Thom spectrum . There is also a map in the opposite direction,
obtained as follows. First, . There is a (diagonal) submanifold whose normal bundle is isomorphic to the tangent bundle of . That gives a Pontryagin-Thom collapse map
crushing the exterior of a tubular neighborhood of the diagonal, and therefore a map
where the final map is that crushes to a point. Desuspending produces a map of Thom spectra
Theorem 2 (Atiyah ) Let be a compact smooth manifold. The Spanier-Whitehead dual to is the Thom spectrum , under the above maps and .
This result can be viewed as a refinement of Poincaré duality: it implies, for instance, the Poincaré duality theorem in generalized cohomology. The Thom spectrum is determined by solely in terms of (stable) homotopy theory. However, the Thom spectrum associated to a vector bundle over a space does not determine the vector bundle. For one thing, it depends only on the class of the vector bundle in -theory. It even depends only on the stable spherical fibration associated to the vector bundle, because the Thom complex can be described (up to homotopy) as the mapping cone of
The next proposition shows that, on a manifold, this is the only indeterminacy: that is, the stable spherical fibration associated to the normal (or tangent) bundle is determined by homotopy theory.
Proposition 3 Let be a manifold, and let be a vector bundle on such that is -reducible. Then the stable spherical fibration associated to is homotopy equivalent to the one associated to the stable normal bundle. In particular, the stable spherical fibration associated to is a homotopy invariant of .
Proof: The Spanier-Whitehead dual of is : one can produce the appropriate maps as above by taking Thom spaces. By hypothesis, there is a map of spectra
splitting off the top cell. Taking duals produces a stable map
We can represent this by an actual map of spaces
where is an honest (not virtual) vector bundle for large enough. By assumption, this map splits off the bottom cell.
Now, let be the vector bundle : for it is actually well-defined. Let be the ball bundle of , and be the sphere bundle; we have . By assumption, there is a map of pairs
such that, when restricted to the fiber over any , it induces a degree one map . When is large, we can replace this by a map of pairs
To see this, observe that the map is nullhomotopic (for ), so we can extend this to a map . If , the Freudenthal suspension theorem lets us desuspend this to a map .
We in particular get a map
which restricts to an equivalence over any . But this is precisely a trivialization of the spherical fibration . Since is a trivial bundle plus , we find that is stably fiber homotopy equivalent to the normal bundle of .
The definition suggests that there might be an analog of the spherical fibration to a general Poincaré duality, and that turns out to be the case. For our purposes, we will take the -reducibility of the Thom space of the normal bundle as a distinguishing feature.
Definition 4 Let be a simply connected Poincaré complex. A Spivak fibration for is a spherical fibration such that the Thom complex of (that is, the mapping cone of ) admits a map
sending to the image of the fundamental class under the Thom isomorphism.
It can be shown that any Poincaré complex admits a Spivak fibration, and that it is (stably) unique. For our purposes, we will take this mostly as motivation.
3. Statement of the main result
Let us suppose now that there exists a vector bundle over the simply connected Poincaré complex , with the property that there exists a stable map
sending to the fundamental class (fed into the Thom isomorphism): in other words, is a vector bundle lifting the Spivak normal fibration. Is this enough to show that is homotopy equivalent to a manifold such that corresponds to the stable normal bundle? Suppose . Then there is an obstruction that comes from cobordism theory. One has:
Theorem 6 (Hirzebruch signature formula) If is an oriented -dimensional manifold, then there exists a polynomial such that
In other words, the signature of can be computed in terms of the Pontryagin classes of the tangent bundle of .
In the above situation, the signature is a homotopy invariant: in particular, we should expect the signature formula to hold for as well if we are to realize from a manifold. The following result states that it is the only obstruction.
Theorem 7 (Browder-Novikov) Let . Then a simply connected, -dimensional Poincaré complex is homotopy equivalent to a manifold if and only if there exists a stable bundle on such that:
- The complex is -reducible: there exists a map inducing an isomorphism on top-dimensional homology.
- Hirzebruch’s signature formula is valid for the pair .
The proof of the Browder-Novikov theorem proceeds in two stages. The first stage uses cobordism theory to produce a degree one normal map
from a -dimensional manifold: in other words, and there is an isomorphism between and the stable normal bundle of . The second (more involved) stage involves doing surgery on (actually, on ) to make the map a homotopy equivalence, while preserving these conditions. The condition on the signature is necessary to make the last step of the surgery work.