Let $X$ be a simply connected space of dimension $4k, k \geq 2$ for which the Poincaré duality theorem holds. In the previous post, I stated a theorem of Browder and Novikov which provides necessary and sufficient conditions for $X$ to be homotopy equivalent to a smooth manifold.

• $X$ must admit a candidate $\xi$ for a stable normal bundle (a lift of the Spivak normal fibration).
• Hirzebruch’s signature theorem, with this proto-normal bundle fed in, should accurately compute the signature of $X$.

The goal of this post is to sketch a proof of this theorem. The proof proceeds in two steps. The first step is to produce a degree one normal map $f: M \to X$ from a smooth manifold, i.e. a map which pulls $\xi$ back to the stable normal bundle of $M$ and which preserves the fundamental class. The second (and harder) step is to do surgery on $f$, to make it a homotopy equivalence. This surgery can be done “formally” before the middle dimension, but at the middle dimension a more careful bookkeeping process is required. It is here that the signature theorem, together with facts about quadratic forms over the integer, become necessary.

These are essentially the second half of the notes I prepared for the Kan seminar, for my second talk. My second talk was officially on Browder’s paper “Homotopy type of differentiable manifolds,” which announces (without proof) this result. In practice, I found the Kervaire-Milnor paper “Groups of homotopy spheres I” and the Milnor paper “A procedure for killing the homotopy groups of differentiable manifolds” very helpful in learning about this material.

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…)