# Hopf invariant one, part I

Computing $\pi_k(S^n)$ for $k 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. 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 $k$th 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.