Computing for is easy. By definition, , and by cellular approximation, we note that every map can be homotoped to a cellular map. It is obvious that all such cellular maps are trivial, so that if . The homotopy groups are more interesting, and can be computed via the Hurewicz theorem:

Theorem (Hurewicz):There is a homomorphism . This map is an isomorphism when if is -connected and . If , this map is abelianization. When , this map is an epimorphism.

In particular, when , . Given this computation and the Hurewicz theorem, one might hope that is trivial for . But the Hurewicz theorem does not give any information about higher values of , and one is forced to try to find other methods. Recall the Freudenthal suspension theorem.

Theorem (Freudenthal):Let be an -connected space, i.e., such that for . Then is an isomorphism for and is an epimorphism for .

The most important example of this is the -sphere . This is an -connected space, and thus is an isomorphism for . In particular, is an isomorphism for , i.e., . Thus are called the stable homotopy groups of spheres. For those familiar with spectra, this is the th stable homotopy group of the sphere spectrum. In this notation, . It is extremely hard (and probably impossible) to compute in general, but there has been a lot of deep mathematics motivated by . To provide an example of a nontrivial , I will now compute .

We will define a non-nullhomotopic map , called the Hopf fibration. To explain the construction of this map, think of as sitting inside . Also recall that . Given a complex line in , there is a unit circle on it. Thus one is supposed to think of as a collection of circles parametrized by . Let us make this precise.

Definition:The Hopf fibration is defined as .

Lemma:is a fiber bundle with fibers .

*Proof. *Suppose . Then there exists such that and . Consider . The left hand side can be expanded to give . In particular, . This means that for some . This analysis shows that is homeomorphic to .

Let . For , let denote the open set of equivalence classes such that . Define a local trivialization via . This is a homeomorphism with inverse map . QED.

The proof works equally well for the maps (which has fiber ), and (which has fiber ). There is also a fiber bundle 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 . Every fiber bundle is a Serre fibration We can now apply the long exact sequence in homotopy groups to obtain: .

In particular, because is a , it follows that for . We have computed that , and hence . In particular, this group is infinite cyclic, and we already have a nontrivial map . Thus the Hopf fibration is a generator of this group.

The Hopf fibration is rather special. We can consider the mapping cylinder of . This is defined to be , where we are attaching a -cell via . Thus . Recall that where . In particular, generates , so we say that the Hopf invariant of is . We can define the Hopf invariant in more generality.

Definition:Suppose we are given a map . Consider the mapping cylinder . This has cohomology concentrated in dimensions and , and these cohomology groups are . Let be the generator of and let be the generator of . Then is some integer multiple of . This integer is called theHopf invariantof .

In light of the above observation that has Hopf invariant , it is natural to ask what maps have Hopf invariant . As remarked above, there are fiber bundles , which has fiber , and , which has fiber . In particular, the case that above gives a fiber bundle . There is also the trivial fiber bundle .

Claim:The fiber bundles , , , and are the only fiber bundles with fiber, total space, and base space as spheres.

Let us try to prove this claim. Suppose is a fiber bundle. Consider the Gysin sequence . From local trivialization, it follows that . Suppose . Then there is an isomorphism , so that , and hence .

Consider the disk bundle which has as its boundary sphere bundle. Let be a Thom class for . This means that the restriction of to each fiber generates . The Leray-Hirsch theorem says that generates .

There is an isomorphism where is the mapping cone. Consider a generator of . Then is also a generator of . This means that .

The Thom isomorphism theorem tells us that the map sending is an isomorphism. Consequently, generates . This means that the Hopf invariant of is (up to sign) . We have therefore reduced the problem to proving the following result.

Theorem (Adams-Atiyah):If has Hopf invariant , then , or .

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.