Some remarks on Homotopy Groups of Spheres

Computing homotopy groups of spheres is an extremely complex topological problem, where much progress has been made, but there is much more still to do. There are many different approaches to their computations: fiber bundles and fibrations, spectral sequences etc. Here I will give an overview of some results with elementary methods which have been found. To get you excited: here is a table of their homotopy groups (taken from Hatcher’s text):

Clearly, many interesting patterns appear: one of the most obvious to see is that

$\pi_n(S^n)=\mathbb{Z}$

Again, there are many zeros and it does follow that

$\pi_i(S^n)=0, \text{when } i<n$

Furthermore, we can see that

We can see that $S^1$ is the Eilenberg MacLane space $S^1=K(\mathbb{Z}, 1)$

There are also more subtle patterns that emerge from this table: for example

$\text{when } i\geq 3$ , it appears that $\pi_{i}(S^2)=\pi_{i}(S^3)$

Also the only infinite homotopy groups of spheres are

$\pi_n(S^n)$ and $\pi_{4n-1}(S^{2n})$

And it is fun to just look at the table and see if you can find some patterns. I will prove some of these theorems (all of the patterns spotted in this blog are true).

$\textbf{Theorem:}$

$\pi_n(S^n)\cong\mathbb{Z}$

$\textit{Proof:}$

To do this, we shall use that $\pi_1(S^1)\cong\mathbb{Z}$ and use the Freudenthal suspension theorem to prove it in general. Let’s quickly recap what it says:

$\textbf{Theorem:} (\text{Freudenthal suspension theorem})$

Let $X$ be an $(n-1)$ connected CW complex. Then the suspension homomorphism

$\Sigma:\pi_{q}(X)\to\pi_{q+1}(\Sigma X)$

is a bijection for $q<2n-1$ and a surjection for $q=2n-1$

where $\Sigma f =f\wedge \text{id}$

I will be assuming this theorem, and can now use an inductive step to prove the general result I was trying to prove to begin with. One must take care however, since we can only show by this theorem that:

$\Sigma:\pi_1(S^1)\to\pi_2(\Sigma S^1)\cong\pi_2(S^2)$

is a surjection. Therefore, all we need is to show that $\pi_2(S^2)=\mathbb{Z}$ and the result will follow easily. To do this consider the following Hopf fibration (which is a fiber bundle)

$S^1\hookrightarrow S^3\to S^2$

Fiber bundles are very nice in the sense that a fiber bundle

$F\to E\to B$

passes to a long exact sequence of homotopy groups

$\dots\to\pi_{i}(F)\to\pi_{i}(E)\to\pi_{i}(B)\to\pi_{i-1}(F)\to\dots\to\pi_{0}(E)\to\pi_{0}(B)$

The long exact sequence breaks down into split short exact sequences

$0\to\pi_{i}(S^3)\to\pi_{i}(S^2)\to\pi_{i-1}(S^1)\to 0$

which splits because of the suspension homomorphism

$\Sigma:\pi_{i-1}(S^1)\to\pi_{i}(S^2)$

The fact that this short exact sequence splits gives us:

$\pi_{i}(S^2)\cong\pi_{i}(S^3)\bigoplus\pi_{i-1}(S^1)$

Since $\pi_{2}(S^3)$ vanishes, we get that $\pi_2(S^2)\cong\pi_{1}(S^1)\cong\mathbb{Z}$

Now that we have $\pi_{2}(S^2)\cong\mathbb{Z}$ , the general result follows from the Freudenthal suspension theorem $\Box$

Note that the last part of this proof also showed that

$\pi_{i}(S^2)\cong\pi_{i}(S^3)$ for all $i\geq 3$ since that is when $\pi_{i-1}(S^1)$ vanishes.

The next thing we shall look at is called the Hopf invariant. Disclaimer: knowledge of cohomology is assumed.

Take a continuous map

$f:S^{2n-1}\to S^n$

Then, define the space $X_f$ by

$X_f=S^{n}\cup_{f} e^{2n}$

for a $2n$ cell $e^{2n}$

Then choose generators $\alpha\in H^n(X_f)$ and $\beta\in H^{2n}(X_f)$ . The multiplicative structure of the cohomology ring $H^{*}(X_f)$ is determined by the relation $\alpha^2=H(f)\beta$ . The number $H(f)$ is called the Hopf invariant of $f$ . The Hopf invariant one problem asks the natural question: When is $H(f)=1$ ? The answer, proven by Adams, is when $n=2,4,8$ . This has interesting applications such as the fact that the only fiber bundles

$S^p\to S^q\to S^r$

are when

$(p,q,r)=(0,1,1), (1,3,2), (3,7,4), (7,15,8)$ .

Furthermore, the Hopf invariant defines a homomorphism

$H:\pi_{2n-1}(S^n)\to\mathbb{Z}$

Consider the CW cell complex $J(S^n)$ . The CW cell structure consists of a single cell, a multiple of $n$ . For example, the CW cell structure on $J_2(S^n)$ consists of cells in dimension $0,n,2n$ .

The next computation I shall do is $\pi_{n+1}(S^n)$ . We have already shown that $\pi_{3}(S^2)\cong\mathbb{Z}$ . I claim that, for $n\geq 3, \pi_{n+1}(S^n)\cong\mathbb{Z}_2$ . The suspension theorem gives us the chain of isomorphisms:

$\pi_{4}(S^3)\cong\pi_{5}(S^4)\cong\pi_{6}(S^5)\cong\dots$

So the problem of what the homotopy groups $\pi_{n+1}(S^n)$ are for $n\geq 3$ boils down to computing $\pi_4(S^3)$

$\textbf{Theorem:}$

$\pi_4(S^3)\cong\mathbb{Z}_2$

$\textit{Proof:}$

(1) Use the exactness of

$\pi_{2n}(J(S^n),S^n)\overset{\partial}{\rightarrow} \pi_{2n-1}(S^n)\overset{\Sigma}{\rightarrow} \pi_{2n}(S^{n+1})$

(2) The kernel of $\Sigma$ is generated by the attaching map of the $2n$ cell of $J(S^n)$ . This attaching map is precisely the Whitehead product $[\text{id}, \text{id}]$ .

(3) When $n=2$ , we have that $\pi_{3}(S^2)$ is generated by the hopf map with Hopf invariant 1, therefore $\pm[\text{id}, \text{id}]=2\eta$ generating the kernel of the suspension, so $\pi_{4}(S^3)=\mathbb{Z}_2$ generated by the suspension of the Hopf map.

Corollary:

$\pi_{n+1}(S^n)=\mathbb{Z}_2$ ,generated by the iterated suspension of the Hopf map.

Corollary:

$\pi_{4}(S^2)=\mathbb{Z}_2$ by the fact that $\pi_{n}(S^3)=\pi_{n}(S^4)$ for all $n\geq 3$ .

Finally we take a brief look at rational homotopy theory. The idea of rational homotopy theory is to “ignore torsion” of the homotopy groups of a space $X$ . For the spheres, the rational homotopy groups

$\pi_{i}(S^n)\otimes\mathbb{Q}$