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…)