Let {f: S^{2n-1} \rightarrow S^n} be a map with {n >1}. Associated to this, one can form a CW complex {M_f = D^{2n} \cup_f S^n}; that is, we attach a {2n}-cell to {S^n} via the map {f}. This CW complex has one cell in dimension {n} and one cell in dimension {2n} (and one cell in dimension {0}). The map {D^{2n} \rightarrow M_f} determines a generator {\iota_{2n}} of {H^{2n}(M_f; \mathbb{Z})} and the map {S^n \rightarrow M_f} determines a generator {\iota_n} of {H^{n}(M_f; \mathbb{Z})}; there are no other elements in cohomology other than the unit. Consequently, we have

\displaystyle \iota_n^2 = a \iota_{2n} , \quad a \in \mathbb{Z}.

Definition 1 The number {a} as above such that {\iota_n^2 = a \iota_{2n}} is the Hopf invariant of {f}.

The homotopy type of {M_f} determines only on the homotopy class of {f}, so the Hopf invariant is a homotopy invariant.

Example 1 The Hopf fibration {f: S^3 \rightarrow S^2} is, by definition, the map such that the mapping cone {M_f} is {\mathbb{CP}^2}; it follows that the Hopf fibration has Hopf invariant one.

The Hopf invariant is clearly identically zero for {n} odd, but when {n} is even the Hopf invariant is never identically zero; in fact, it defines a homomorphism

\displaystyle \pi_{2n-1}(S^n) \rightarrow \mathbb{Z},

which for {n} even has image containing the even integers. (This is where the exceptional {\mathbb{Z}} summand in the homotopy groups of spheres comes from.)

A classical problem in topology was the following:

Question: For which {n} does there exist a map of Hopf invariant one? (more…)