Question

Can someone help explain the following result:

If the dimension is at least five, there are at most finitely many different smooth manifolds with given homotopy type and Pontryagin classes.

Thank you so much!

Answer 1

In the $1$-connected case, one may argue as follows:

Let $X$ be a closed $1$-connected smooth $n$-manifold, $n \ge 5$. The theory of the Spivak fibration shows that any homotopy equivalence $f: M^n \to X$ with $M$ smooth is covered by a stable fiber homotopy equivalence of underlying stable tangent spherical fibrations of $M$ and $X$. Call $f$ stably tangential if this equivalence of stable spherical fibrations lifts to an isomorphism of stable tangent vector bundles.

Then the surgery exact sequence shows that any stable tangential homotopy equivalence $f: M \to X$ is homotopic to a diffeomorphism $f': M \sharp \Sigma \to X$, where $\Sigma$ is a homotopy sphere, and $\sharp$ means connected sum.

(You can either quote here Corollary II.3.8 of Browder's book, or you can deduce it directly from the surgery exact sequence. The point is that connected sum gives an action of the homotopy $n$-spheres on the the structure set of $X$, and one can compare the surgery exact sequence for $M$ and the sphere to deduce the above statement.)

To finish the proof of what you want, notice:

1. Kervaire and Milnor showed that there are only finitely many homotopy spheres in each dimension $\ge 5$.

2. The obstruction to the homotopy equivalence being stably tangential is given by its normal invariant, which is an element of $[X,\text{G/O}]$ (it's given by the "difference" between the stable tangent bundles of $M$ and $X$, i.e., $(f^{-1})^*\tau_M - \tau_X$, appropriately interpreted).

3. The map $\text{G/O} \to B\text{O}$ is a rational homotopy equivalence (by Serre). The image of the normal invariant in $[X,B\text{O}]$ is rationally detected by the difference of the Pontryagin classes of $M$ and $X$ using the fact that $H^*(B\text{O}; \Bbb Q)$ is a polynomial algebra on the $p_i$.