Hopf invariant one, part I

Computing \pi_k(S^n) for k<n is easy. By definition, \pi_k(S^n)=[S^k,S^n], and by cellular approximation, we note that every map S^k\to S^n can be homotoped to a cellular map. It is obvious that all such cellular maps are trivial, so that \pi_k(S^n)=0 if k<n. The homotopy groups \pi_n(S^n) are more interesting, and can be computed via the Hurewicz theorem:

Theorem (Hurewicz): There is a homomorphism \pi_k(X)\to  H_k(X). This map is an isomorphism when k\leq n if X is (n-1)-connected and n\geq 2. If n=1, this map is abelianization. When k=n+1, this map is an epimorphism.

In particular, when X=S^n, \pi_n(S^n)\cong  H_n(S^n)\cong\mathbf{Z}. Given this computation and the Hurewicz theorem, one might hope that \pi_k(S^n) is trivial for k>n. But the Hurewicz theorem does not give any information about higher values of k, and one is forced to try to find other methods. Recall the Freudenthal suspension theorem.

Theorem (Freudenthal): Let X be an n-connected space, i.e., such that \pi_k(X)=0 for k\leq n. Then \pi_k(X)\to\pi_{k+1}(\Sigma X) is an isomorphism for k\leq 2n and is an epimorphism for k=2n+1.

The most important example of this is the n-sphere S^n. This is an (n-1)-connected space, and thus \pi_k(S^n)\to\pi_{k+1}(\Sigma S^n)\simeq\pi_{k+1}(S^{n+1}) is an isomorphism for k\leq 2n-2. In particular, \pi_{n+k}(S^n)\to\pi_{n+k+1}(S^{n+1}) is an isomorphism for n+k\leq 2n-2, i.e., k+2\leq n. Thus \pi_{n+k}(S^n)=\pi^S_k are called the stable homotopy groups of spheres. For those familiar with spectra, this is the kth stable homotopy group of the sphere spectrum. In this notation, \pi_n(S^n)=\pi^S_0=\mathbf{Z}. It is extremely hard (and probably impossible) to compute \pi^S_k in general, but there has been a lot of deep mathematics motivated by \pi^S_k. To provide an example of a nontrivial \pi^S_k, I will now compute \pi^S_1=\pi_3(S^2).

We will define a non-nullhomotopic map \eta:S^3\to S^2, called the Hopf fibration. To explain the construction of this map, think of S^3 as sitting inside \mathbf{R}^4=\mathbf{C}^2. Also recall that S^2=\mathbf{CP}^1. Given a complex line in \mathbf{C}^2, there is a unit circle on it. Thus one is supposed to think of S^3 as a collection of circles parametrized by \mathbf{CP}^1. Let us make this precise.

Definition: The Hopf fibration \eta:S^3\to S^2 is defined as (z_1,z_2)\mapsto [z_1:z_2].

Lemma: \eta:S^3\to S^2 is a fiber bundle with fibers S^1.

Proof. Suppose (z_1,z_2),(z_1^\prime,z_2^\prime)\mapsto[z_1:z_2]. Then there exists \lambda\in\mathbf{C} such that z_1^\prime=\lambda z_1 and z_2^\prime=\lambda z_2. Consider z_1^\prime\overline{z_1^\prime}+z_2^\prime\overline{z_2^\prime}=1. The left hand side can be expanded to give \lambda\overline{\lambda}(z_1\overline{z_1}+z_2\overline{z_2})=\lambda\overline{\lambda}. In particular, |\lambda|=1. This means that \lambda=e^{i\theta} for some 0<\theta\leq 2\pi. This analysis shows that \eta^{-1}[z_1:z_2] is homeomorphic to S^1.

Let x\in S^2. For i=1,2, let U_i denote the open set of equivalence classes [z_1:z_2] such that z_i\neq 0. Define a local trivialization h:\eta^{-1}(U_i)\to U_i\times S^1 via (z_1,z_2)\mapsto ([z_1:z_2],z_i/||z_i||). This is a homeomorphism with inverse map ([z_1:z_2],e^{i\theta})\mapsto e^{i\theta}\left(z_1\frac{z_i}{||z_i||},z_2\frac{z_i}{||z_i||}\right). QED.

The proof works equally well for the maps S^{2n+1}\to \mathbf{CP}^n (which has fiber S^1), and S^{4n+3}\to\mathbb{H}\mathbf{P}^n (which has fiber S^3). There is also a fiber bundle S^{15}\to S^8 defined analogously that uses the octonions. The proof that this is a fiber bundle is a little trickier. We will not prove this here.

Anyway, one now has a fiber bundle S^1\to S^3\to S^2. Every fiber bundle is a Serre fibration We can now apply the long exact sequence in homotopy groups to obtain: \cdots\to\pi_{k+1}(S^2)\to\pi_k(S^1)\to\pi_k(S^3)\to\pi_k(S^2)\to\pi_{k-1}(S^1)\to\cdots.

In particular, because S^1 is a K(\mathbf{Z},1), it follows that \pi_k(S^2)\cong\pi_k(S^3) for k\geq 3. We have computed that \pi_3(S^3)\cong\mathbf{Z}, and hence \pi_3(S^2)\cong\mathbf{Z}. In particular, this group is infinite cyclic, and we already have a nontrivial map \eta:S^3\to S^2. Thus the Hopf fibration is a generator of this group.

The Hopf fibration is rather special. We can consider the mapping cylinder of \eta. This is defined to be M_\eta:=S^2\cup_\eta D^4, where we are attaching a 4-cell via \eta. Thus M_\eta=\mathbf{CP}^2. Recall that  H^\ast(\mathbf{CP}^2;\mathbf{Z})=\mathbf{Z}[\alpha]/(\alpha^3) where |\alpha|=2. In particular, \alpha^2 generates  H^4(\mathbf{CP}^2;\mathbf{Z}), so we say that the Hopf invariant of \eta is 1. We can define the Hopf invariant in more generality.

Definition: Suppose we are given a map f:S^{2n-1}\to S^n. Consider the mapping cylinder M_f:=S^n\cup_f D^{2n}. This has cohomology concentrated in dimensions 0,n and 2n, and these cohomology groups are \mathbf{Z}. Let \alpha be the generator of  H^n(M_f) and let \beta be the generator of  H^{2n}(M_f). Then \alpha^2 is some integer multiple of \beta. This integer is called the Hopf invariant of f.

In light of the above observation that \eta has Hopf invariant 1, it is natural to ask what maps f:S^{2n-1}\to S^n have Hopf invariant 1. As remarked above, there are fiber bundles S^{4n+3}\to\mathbb{H}\mathbf{P}^n, which has fiber S^3, and S^{15}\to S^8, which has fiber S^7. In particular, the case that n=1 above gives a fiber bundle S^3\to S^7\to S^4. There is also the trivial fiber bundle S^0\to S^1\to S^1.

Claim: The fiber bundles S^0\to S^1\to S^1, S^1\to S^3\to S^2, S^3\to S^7\to S^4, and S^7\to S^{15}\to S^8 are the only fiber bundles with fiber, total space, and base space as spheres.

Let us try to prove this claim. Suppose S^k\to S^m\xrightarrow{p} S^n is a fiber bundle. Consider the Gysin sequence \cdots\to  H^{i-k-1}(S^n)\to  H^i(S^n)\to  H^i(S^m)\to  H^{i-k}(S^n)\to\cdots. From local trivialization, it follows that m=n+k. Suppose i=n. Then there is an isomorphism  H^{n-k-1}(S^n)\cong  H^n(S^n)\cong\mathbf{Z}, so that k=n-1, and hence m=2n-1.

Consider the disk bundle D^{n}\to M_p\to S^n which has S^{n-1}\to S^{2n-1}\to S^n as its boundary sphere bundle. Let \tau be a Thom class for (M_p,S^{2n-1})\to S^n. This means that the restriction of \tau to each fiber (D^n,S^{n-1}) generates  H^n(D^n,S^{n-1})\cong\mathbf{Z}. The Leray-Hirsch theorem says that \tau generates  H^\ast(M_p,S^{2n-1}).

There is an isomorphism  H^{i+n}(M_p,S^{2n-1})\cong\widetilde{ H}^{i+n}(M_p/S^{2n-1})=\widetilde{ H}^{i+n}(C_p) where C_p is the mapping cone. Consider a generator \alpha of  H^n(S^n). Then p^\ast(\alpha) is also a generator of \widetilde{ H}^n(C_p)\cong\mathbf{Z}. This means that \tau=\pm p^\ast(\alpha).

The Thom isomorphism theorem tells us that the map  H^i(S^n)\to  H^{i+n}(M_p,S^{2n-1}) sending x\mapsto p^\ast(x)\smile \tau is an isomorphism. Consequently, p^\ast(\alpha)\smile\tau=\pm (p^\ast(\alpha))^2 generates \widetilde{ H}^{2n}(C_p). This means that the Hopf invariant of p is (up to sign) 1. We have therefore reduced the problem to proving the following result.

Theorem (Adams-Atiyah): If f:S^{2n-1}\to S^n has Hopf invariant 1, then n=1,2,4, or 8.

This is a nontrivial result. Our goal in the next few blog posts is to present Adams’ and Atiyah’s proof of this result, through a beautiful argument using K-theory.


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