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 ${X}$ homotopy equivalent to a (compact) ${n}$-dimensional smooth manifold?

A compact manifold is homotopy equivalent to a finite CW complex, so ${X}$ must be one itself. Equivalently (since ${X}$ is simply connected), the homology ${H_*(X; \mathbb{Z})}$ must be finitely generated.

More interestingly, we know that a hypothetical compact ${n}$-manifold ${M}$ homotopy equivalent to ${X}$ (which is necessarily orientable) satisfies Poincaré duality. That is, we have a fundamental class ${[M] \in H_n(M; \mathbb{Z})}$ with the property that cap product induces an isomorphism

$\displaystyle H^r(M; G) \stackrel{\cap [M]}{\simeq} H_{n-r}(M; G),$

for all groups ${G}$ and for all ${r}$. It follows that, if the answer to the above question is positive, an analogous condition must hold for ${X}$. This motivates the following definition:

Definition 1 A simply connected space ${X}$ is a Poincaré duality space (or Poincaré complex) of dimension ${n}$ if there is a class ${[X] \in H_n(X; \mathbb{Z})}$ such that for every group ${G}$, cap product induces an isomorphism

$\displaystyle H^r(X; G) \stackrel{\cap [M]}{\simeq} H_{n-r}(X; G).$

A consequence is that the cohomology (and homology) groups of ${X}$ vanish above dimension ${n}$, and ${H_n(X; \mathbb{Z})}$ is generated by ${[X]}$. In other words, ${[X]}$ behaves like the fundamental class of an ${n}$-dimensional manifold.

We can thus pose a refined version of the above question.

Question 2: Given a simply connected Poincaré duality space ${X}$ of dimension ${n}$, is there a homotopy equivalence ${f: M \rightarrow X}$ for ${M}$ a compact (necessarily ${n}$-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 “${E_8}$ manifold”). Such a manifold cannot be homotopy equivalent to any smooth manifold. In fact, if it were homotopy equivalent to a smooth 4-manifold ${M}$, then the evenness of the intersection form on ${H^2(M; \mathbb{Z})}$ shows that ${\mathrm{Sq}^2}$ (and in fact all ${\mathrm{Sq}^i, i \neq 0}$) act trivially on ${M}$. The Wu formulas imply that the Stiefel-Whitney classes of ${M}$ vanish. In particular, ${M}$ admits a spin structure, and a theorem of Rohlin asserts that the signature of a spin 4-manifold is divisible by ${16}$.

2. Atiyah duality and Spivak fibrations

Let us return to the question above. Consider a simply connected Poincaré duality space ${X}$ of dimension ${n}$. If ${X}$ has the homotopy type of a manifold, we should in particular obtain an ${n}$-dimensional vector bundle ${\xi}$ on ${X}$, 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, ${\xi}$ is not totally arbitrary either. It turns out the stable spherical fibration associated to ${\xi}$ is homotopy invariant, and can in fact be reconstructed simply from ${X}$.

To see this, let’s recall the homotopy properties of the tangent bundle. Let ${M}$ be a smooth, compact manifold. The tangent bundle ${TM }$ cannot be reconstructed homotopy-theoretically from ${M}$. However, one important homotopy-theoretic property of the normal bundle ${\nu}$ of an imbedding ${M \hookrightarrow S^N}$ is the Pontryagin-Thom collapse map

$\displaystyle S^N \rightarrow M^{\nu},$

mapping onto the Thom complex ${M^{\nu}}$ of ${\nu}$. It crushes everything outside a tubular neighborhood of ${M}$ to the “point at infinity” in the Thom complex. It has the property that ${\widetilde{H}_N(S^N, \mathbb{Z}) \rightarrow \widetilde{H}_N(M^{\nu}, \mathbb{Z})}$ has the property that ${[S^N]}$ is sent to the generator of ${\widetilde{H}_N(M^{\nu}; \mathbb{Z})}$ (corresponding to the fundamental class of ${M}$). This is called “${S}$-reducibility” and can be phrased by saying that the top cell of ${M^{\nu}}$ splits off.

We can use the Pontryagin-Thom map to produce a map

$\displaystyle S^N \rightarrow M^{\nu} \rightarrow M^{\nu} \wedge M_+$

and desuspending produces a map of spectra

$\displaystyle \mathrm{coev}: S^0 \rightarrow M^{-TM} \wedge \Sigma^\infty_+ M,$

where ${M^{-TM}}$ is the Thom spectrum ${M^{-TM} \stackrel{\mathrm{def}}{=} \Sigma^{-N} M^{\nu}}$. There is also a map in the opposite direction,

$\displaystyle M^{\nu} \wedge M_+ \rightarrow S^N,$

obtained as follows. First, ${M^{\nu} \wedge M_+ \simeq (M \times M)^{p_1^*\nu}}$. There is a (diagonal) submanifold ${M \stackrel{\Delta}{\rightarrow} M \times M}$ whose normal bundle is isomorphic to the tangent bundle of ${M}$. That gives a Pontryagin-Thom collapse map

$\displaystyle M \times M \rightarrow M^{TM}$

crushing the exterior of a tubular neighborhood of the diagonal, and therefore a map

$\displaystyle M^{\nu} \wedge M_+ \simeq (M \times M)^{p_1^*\nu} \rightarrow M^{\nu \oplus TM} \rightarrow S^N$

where the final map is ${M^{\nu \oplus TM} \simeq M^{\mathbb{R}^N} \rightarrow S^N}$ that crushes ${M}$ to a point. Desuspending produces a map of Thom spectra

$\displaystyle \mathrm{ev} : M^{-TM} \wedge \Sigma^\infty_+ M \rightarrow S^0.$

Theorem 2 (Atiyah ) Let ${M}$ be a compact smooth manifold. The Spanier-Whitehead dual to ${\Sigma^\infty_+ M}$ is the Thom spectrum ${M^{-TM}}$, under the above maps ${\mathrm{coev}: S^0\rightarrow M^{-TM} \wedge \Sigma^\infty_+ M}$ and ${\mathrm{ev}: M^{-TM} \wedge \Sigma^\infty_+ M \rightarrow S^0}$.

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 ${M^{-TM}}$ is determined by ${M}$ solely in terms of (stable) homotopy theory. However, the Thom spectrum associated to a vector bundle ${E}$ over a space ${X}$ does not determine the vector bundle. For one thing, it depends only on the class of the vector bundle in ${KO}$-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

$\displaystyle \mathrm{Sph}(E) \rightarrow X .$

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 ${M}$ be a manifold, and let ${E}$ be a vector bundle on ${M}$ such that ${M^{E}}$ is ${S}$-reducible. Then the stable spherical fibration associated to ${E}$ is homotopy equivalent to the one associated to the stable normal bundle. In particular, the stable spherical fibration associated to ${TM}$ is a homotopy invariant of ${M}$.

Proof: The Spanier-Whitehead dual of ${\Sigma^\infty M^{E}}$ is ${M^{-TE - E}}$: one can produce the appropriate maps as above by taking Thom spaces. By hypothesis, there is a map of spectra

$\displaystyle S^N \rightarrow \Sigma^\infty M^E, \quad N = \dim M + \dim E$

splitting off the top cell. Taking duals produces a stable map

$\displaystyle M^{-TE - E} \rightarrow S^{-N} .$

We can represent this by an actual map of spaces

$\displaystyle M^{n - TE - E} \rightarrow S^{n-N}, \quad n \gg N,$

where ${n - TE - E}$ is an honest (not virtual) vector bundle for ${n}$ large enough. By assumption, this map splits off the bottom cell.

Now, let ${V}$ be the vector bundle ${n - TE - E}$: for ${n \gg 0}$ it is actually well-defined. Let ${B(V)}$ be the ball bundle of ${V}$, and ${S(V)}$ be the sphere bundle; we have ${M^{V} = B(V)/S(V)}$. By assumption, there is a map of pairs

$\displaystyle (B(V), S(V)) \rightarrow (B^{\dim V}/S^{\dim V-1}, \ast)$

such that, when restricted to the fiber ${(B(V_x), S(V_x))}$ over any ${x \in M}$, it induces a degree one map ${B(V_x)/S(V_x) \rightarrow S^{\dim V}}$. When ${\dim V}$ is large, we can replace this by a map of pairs

$\displaystyle (B(V), S(V)) \rightarrow (B^{\dim V}, S^{\dim V - 1}).$

To see this, observe that the map ${M \simeq B(V) \rightarrow S^{\dim V}}$ is nullhomotopic (for ${\dim V \gg 0}$), so we can extend this to a map ${\Sigma S(V) \rightarrow S^{\dim V}}$. If ${\dim V \gg 0}$, the Freudenthal suspension theorem lets us desuspend this to a map ${S(V) \rightarrow S^{\dim V-1}}$.

We in particular get a map

$\displaystyle S(V) \rightarrow S^{\dim V - 1}$

which restricts to an equivalence over any ${x \in M}$. But this is precisely a trivialization of the spherical fibration ${S(V) \rightarrow M}$. Since ${V}$ is a trivial bundle plus ${-TE - E}$, we find that ${E}$ is stably fiber homotopy equivalent to the normal bundle of ${M}$. $\Box$

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 ${S}$-reducibility of the Thom space of the normal bundle as a distinguishing feature.

Definition 4 Let ${X}$ be a simply connected Poincaré complex. A Spivak fibration for ${X}$ is a spherical fibration ${T \rightarrow X}$ such that the Thom complex ${X^{T}}$ of ${T}$ (that is, the mapping cone of ${T \rightarrow X}$) admits a map$\displaystyle S^N \rightarrow X^{T}$

sending ${[S^N] \in H_N(S^N; \mathbb{Z})}$ to the image of the fundamental class ${[X]}$ 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 ${\xi}$ over the simply connected Poincaré complex ${X}$, with the property that there exists a stable map

$\displaystyle S^N \rightarrow X^{\xi}$

sending ${[S^N]}$ to the fundamental class ${[X]}$ (fed into the Thom isomorphism): in other words, ${\xi}$ is a vector bundle lifting the Spivak normal fibration. Is this enough to show that ${X}$ is homotopy equivalent to a manifold ${M}$ such that ${\xi}$ corresponds to the stable normal bundle? Suppose ${\dim X = 4k}$. Then there is an obstruction that comes from cobordism theory. One has:

Theorem 6 (Hirzebruch signature formula) If ${M}$ is an oriented ${4k}$-dimensional manifold, then there exists a polynomial ${\mathbf{L}(x_1, \dots, x_k) \in \mathbb{Q}[x_1, \dots, x_k]}$ such that$\displaystyle \sigma(M) = \int_M \mathbf{L}(p_1, \dots, p_k) .$

In other words, the signature ${\sigma(M)}$ of ${M}$ can be computed in terms of the Pontryagin classes of the tangent bundle of ${M}$.

In the above situation, the signature is a homotopy invariant: in particular, we should expect the signature formula to hold for ${(X, \xi)}$ as well if we are to realize ${(X,\xi)}$ from a manifold. The following result states that it is the only obstruction.

Theorem 7 (Browder-Novikov)  Let ${k \geq 2}$. Then a simply connected, ${4k}$-dimensional Poincaré complex ${X}$ is homotopy equivalent to a manifold if and only if there exists a stable bundle ${\xi}$ on ${M}$ such that:

1. The complex ${X^{\xi}}$ is ${S}$-reducible: there exists a map ${S^N \rightarrow X^{\xi}}$ inducing an isomorphism on top-dimensional homology.
2. Hirzebruch’s signature formula is valid for the pair ${(X, \xi)}$.

The proof of the Browder-Novikov theorem proceeds in two stages. The first stage uses cobordism theory to produce a degree one normal map

$\displaystyle f: M \rightarrow X$

from a ${4k}$-dimensional manifold: in other words, ${f( [M]) = [X]}$ and there is an isomorphism between ${f^*\xi}$ and the stable normal bundle of ${M}$. The second (more involved) stage involves doing surgery on ${M}$ (actually, on ${f}$) 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.