I don't quite follow the rough outline Milnor gives of the fact that the 7-sphere has different differentiable structures. The video is available here, and the slides he used can be found here.
Here's what I got from the talk. Unless otherwise stated, all manifolds are compact, orientable smooth (most of the time I state these hypotheses explicitely anyways), and $M$ is an ($n$-dimensional) manifold. Homology and cohomology are taken with coefficients in $\mathbb{Z}$.
- By a theorem of Whitney, $M$ embeds in $\mathbb{R}^{n+k}$ from which we get a Gauss map $g:M\rightarrow G_n(\mathbb{R}^{n+k})\subset G_n$, where $G_n(\mathbb{R}^{n+k})$ is the grassmannian manifold of $n$-planes in $\mathbb{R}^{n+k}$, and $G_n$ is the colimit of those grassmannians, i.e. the grassmannian of $n$-planes in $\mathbb{R}^{\infty}$. All Gauss maps $g$ (viewed as maps $M\rightarrow G_n$) are homotopic, and we obtain a well defined homology class $\langle M\rangle=g_*\mu\in H_n(G_n)$. The letter $\mu$ stands for the fundamental class of $M$.
- If $M$ is a compact orientable topological manifold of dimension divisible by $4$, say $\dim~M=4k$, it comes equipped with a symmetric bilinear form given by the cup product in dimensions $2k$: $$H^{2k}(M)\times H^{2k}(M)\rightarrow H^{4k}(M)\simeq \mathbb{Z},~(x,y)\mapsto x\cup y$$ This form must kill torsion, so we get a quadratic form on the finitely generated free abelian group $H^{2k}(M)/\mathrm{Torsion}\simeq \mathbb{Z}\oplus\cdots\oplus\mathbb{Z}$, and we can define its signature (as a real quadratic form). The signature of the manifold $M$ is then defined as $\sigma (M)=p-q$ where $p=\#$ of positive eigen values and $q=\#$ of negative eigenvalues.
- Going back to the differentiable case, we define the Pontrjagin numbers of $M$ by looking at the cohomology ring of $G_n$. This ring is concentrated in dimensions that are multiples of $4$ and has one generator $p_i\in H^{4i}(G_n)$ for each $i\geq 1$, so that all $1,p_1, p_1^2,p_2,p_1^3,p_1\cup p_2,p_3,\dots$ generate the cohomology (plus some torsion elements). We then define the Pontrjagin numbers of $M$ by evaluating these cohomology classes on the homology class $\langle M\rangle$. This gives potentially non zero numbers provided $M$ has dimension $4k$. In particular, if $M$ has dimension $8$ and is smooth, compact, orientable, there are two Pontrjagin numbers: $p_1^2(M)$ and $p_2(M)$.
- By a theorem of Hirzebruch, the signature of $M$ (with $\dim~M=4k$) ought to be a polynomial with rational coefficients in the Pontryagine numbers of $M$, and Milnor tells us that in case $\dim~M=8$, $$45\sigma(M)=7p_2(M)-p_1^2(M).$$ This is somehow related to oriented cobordism. Apparantly, two smooth oriented compact manifolds $M$ and $N$ of the same dimension are cobordant iff $\langle M\rangle=\langle N\rangle$. I understand that the signature is a cobordism invariant, so that it extends to a homomorphism $\sigma:\Omega(n)\rightarrow\mathbb{Z}$ where $\Omega(n)$ is the group of oriented cobordisms, and that since cobordism classes are determined by $\langle M\rangle\in H_n(G_n)$, there ought to be a linear relation between the non torsion bits of this homology class (which can be read off the Pontrjagin numbers) and the signature. The main idea of the proof will be to calculate $$\frac{1}{7}(45\sigma(M)+p_1^2(M))$$ for some eight dimensional manifold, to observe that it is not an integer thus showing that its boundary, while homeomorphic to the $7$-sphere, cannot be diffeomorphic to it.
Part of my confusion stems from here. Milnor considered $7$-dimensional smooth compact orientable manifolds $M$ that are the total space of a locally trivial fibre bundle with fibre $\mathbb{S}^3$ over $\mathbb{S}^4$. He was able to show explicitely that some of these $M$ were homeomorphic to the $7$-sphere by constructing explicit Morse functions with exactly two critical points. However, he found that some of these smooth manifolds that were topological $7$-spheres could not be diffeomorphic to the $7$-sphere. He then (as I understand it) calculated $\frac{1}{7}(45\sigma(E)+p_1^2(E))$ for some $8$ dimensional manifold (Which one?) and found that the result was not an integer.
Could you help me understand how he got to exotic $7$-spheres? What manifold $E$ would one consider? Why is the result a non integer? and how could it be a non integer? since in order to make sense of $p_1^2(E)$ in the first place we need it to be smooth, and then $\frac{1}{7}(45\sigma(M)+p_1^2(M))$ must be equal to $p_2(M)$ which is an integer.
