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 ${16}$. (more…)