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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.

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If you don't get any answers here in a day or so, this question could be suitable for Math Overflow. –  Ragib Zaman Jun 24 '12 at 18:41

2 Answers 2

up vote 24 down vote accepted

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.

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Thank you for this illuminating answer. For the last part, I think there is a typo when you say that $E$ should be the total space of the $S^3$ bundles $\xi_{hj}$ over $S^4$, do you mean the disk bundle? –  Olivier Bégassat Jun 25 '12 at 23:38
    
@OlivierBégassat: That was indeed a typo, and I did indeed mean the associated disk bundle. The error has been fixed. –  Henry T. Horton Jun 26 '12 at 0:05
    
Dear Henry, In your fifth last line, should "not nec. a homeo." read "not nec. a diffeo"? Regards, –  Matt E Jul 24 '12 at 6:08
    
@MattE That is correct. I've edited that part of the post now. Thanks. –  Henry T. Horton Jul 24 '12 at 15:11

There is a short article on Climbing Mountain Bourbaki which I think should be helpful. Alternatively you may access Jstor to read his original paper, which may be hard to read but coupled with the slides, etc you should be able to get the answers.

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