Question about Milnor's talk at the Abel Prize

Solution 1:

The method in his slides differs from his original paper. First I will explain Milnor's original construction of exotic $7$-spheres, from his article

• On manifolds homeomorphic to the $7$-sphere, Annals of Mathematics, Vol. 64, No. 2, September 1956.

I'll answer your question about his construction mentioned in the slides after that.

First, he defines a smooth invariant $\lambda(M^7)$ for closed, oriented $7$-manifolds $M^7$. To do this, we first note that every $7$-manifold bounds an $8$-manifold, so pick an $8$-manifold $B^8$ bounded by $M^7$. Let $\mu \in H_7(M^7)$ be the orientation class for $M^7$ and pick an "orientation" $\nu \in H_8(B^8,M^7)$, i.e. a class satisfying $$\partial \nu = \mu.$$ Define a quadratic form $$H^4(B^8,M^7)/\mathrm{Tors} \longrightarrow \mathbb{Z},$$ $$\alpha \mapsto \langle \alpha \smile \alpha, \nu \rangle,$$ and let $\sigma(B^8)$ be the signature of this form. Milnor assumes that $M^7$ has $$H^3(M^7) \cong H^4(M^7) \cong 0,$$ so that $$i: H^4(B^8,M^7) \longrightarrow H^4(B^8)$$ is an isomorphism. Hence the number $$q(B^8) = \langle i^\ast p_1(B^8) \cup i^\ast p_1(B^8), \nu \rangle$$ is well-defined. Then Milnor's $7$-manifold invariant is $$\lambda(M^7) \equiv 2q(B^8) - \sigma(B^8) \pmod 7.$$ Milnor shows that $\lambda(M^7)$ does not depend on the choice of $8$-manifold $B^8$ bounded by $M^7$.

Now let us turn to the specific $7$- and $8$-manifolds that Milnor considers. As you noted, he looks at the total spaces of $S^3$-bundles over $S^4$. The total space of such a bundle is a $7$-manifold bounding the total space of the associated disk bundle. $S^3$-bundles over $S^4$ (with structure group $\mathrm{SO}(4)$) are classified by elements of $$\pi_3(\mathrm{SO}(4)) \cong \mathbb{Z} \oplus \mathbb{Z}.$$ An explicit isomorphism identifies the pair $(h,j) \in \mathbb{Z} \oplus \mathbb{Z}$ with the $S^3$-bundle over $S^4$ with transition function $$f_{hj}: S^3 \longrightarrow \mathrm{SO}(4),$$ $$f_{hj}(u) \cdot v = u^h v u^j$$ on the equatorial $S^3$, where here we consider $u \in S^3$ and $v \in \mathbb{R}^4$ as quaternions, i.e. the expression $u^h v u^j$ is understood as quaternion multiplication.

Let $\xi_{hj}$ be the $S^3$ bundle on $S^4$ corresponding to $(h,j) \in \mathbb{Z} \oplus \mathbb{Z}$. For each odd integer $k$, let $M^7_k$ be the total space of the bundle $\xi_{hj}$, where \begin{align*} h + j & = 1, \\ h - j & = k. \end{align*} Milnor shows that $$\lambda(M^7_k) \equiv k^2 - 1 \pmod 7.$$ Furthermore, he shows that $M^7_k$ admits a Morse function with exactly $2$ critical points, and hence is homeomorphic to $S^7$. Clearly we have $$\lambda(S^7) \equiv 0,$$ so if $$k \not\equiv \pm 1 \pmod 7,$$ then $M^7_k$ is homeomorphic but not diffeomorphic to $S^7$, and hence is an exotic sphere. In particular, $S^7$, $M^7_3$, $M^7_5$, and $M^7_7$ are all homeomorphic to one another but all pairwise non-diffeomorphic.

Now, in the slides, the space $E$ should be the total space of the disk bundle associated to an $S^3$ bundle $\xi_{hj}$ over $S^4$ with \begin{align*} h + j & = 1, \\ h - j & = k \end{align*} for some odd integer $k$, as described above. Then $E$ is an $8$-manifold with boundary $\partial E$ homeomorphic to $S^7$. Now, if $\partial E$ is diffeomorphic to $S^7$, then we can glue $D^8$ to $E$ along their common boundary via a diffeomorphism $$f: \partial E \longrightarrow S^7$$ in order to get a smooth manifold $$E' = E \cup_f D^8.$$ If $f$ is not a diffeomorphism, then $E'$ is not necessarily smooth. So in showing that $$p_2(E') \notin \mathbb{Z},$$ Milnor proves by contradiction that no such diffeomorphism $f$ can exist, since Pontrjagin numbers of a manifold are integers. So in that case $\partial E$ would be homeomorphic to $S^7$ but not diffeomorphic, and hence an exotic $7$-sphere.