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?

One can show that whenever {n = 2, 4, 8} (and {1} seems to be included as a degenerate case), there is a map of Hopf invariant one. I will describe a construction next time.

Here is an example to show that there are strong restrictions on {n}.

Example 2:  If {n} is not a power of {2}, there can be no map {f: S^{2n-1} \rightarrow S^n} of odd Hopf invariant. In fact, in that case {M_f} would be a complex whose {\mathbb{Z}/2} cohomology would be {\mathbb{Z}/2[\iota_n]/\iota_n^3} where {\iota_n} is in degree {n}.

This, however, is impossible. In fact, if {n} is not a power of two, the aforementioned {\mathbb{Z}/2}-cohomology ring does not arise.

Proposition 2If {n} is not a power of two and {X} is a complex with no {\mathbb{Z}/2}-cohomology between {n+1} and {2n-1}, then the square of any element in {H^n(X; \mathbb{Z}/2)} is zero.

To prove this, we use the fact that for {x \in H^n(X; \mathbb{Z}/2)}, we can write the cup square in terms of the Steenrod operations:

\displaystyle x^2 = \mathrm{Sq}^n x.

Now, by the Adem relations, the element {\mathrm{Sq}^n} in the Steenrod algebra is decomposable; that is, it can be written as a sum of products of smaller operations. But these smaller operations necessarily pass through {H^k} for {n < k < 2k}, so they are all zero.

In fact, using the Steenrod reduced powers in mod {p} cohomology and the corresponding Adem relations, one can show that there are very strong restrictions on when a cohomology ring is a (possibly truncated) polynomial ring on one generator.

It is a theorem of Adams that in fact, {n = 1, 2, 4, 8} are the only cases in which a map of Hopf invariant one exists. The original proof seems to have used some highly technical computations in the Adams spectral sequence. However, there is a much simpler and more transparent argument, due to Adams and Atiyah in their very enjoyable paper “K-theory and the Hopf invariant,” using a bit of K-theory.

The Adams operations

The relevant operations one needs for this argument are not the Steenrod operations in cohomology, but the Adams operations in K-theory. Let {E} be a vector bundle. Then for each {k}, there is supposed to be an operation {\psi^k} such that if

\displaystyle E = L_1 \oplus \dots \oplus L_m,


\displaystyle \psi^k E = L_1^k \oplus \dots \oplus L_m^k.

In other words, {\psi^k} is supposed to raise each line bundle to the {k}th power, and {\psi^k} is supposed to be additive. This determines {\psi^k}. By the splitting principle, one can basically pretend that any vector bundle is a sum of line bundles, as long as one restricts oneself to operations symmetric in these individual line bundles. One can then give an explicit formula for the {\psi^k} in terms of the exterior power operations

\displaystyle \wedge: K(X) \rightarrow 1 + tK(X)[t] , \quad \wedge E = \sum [\wedge^i E]t^i.

Anyway, in view of all this, one gets:

Proposition 3 {\psi^k} is a natural ring homomorphism {K(X) \rightarrow K(X)} for any compact space {X}.

For the proof, we need two more facts. The first is that for any element in the reduced K-theory {\widetilde{K}(S^{2n})}, {\psi^k} acts by multiplication by {k^{n}}. One can see this by using the Bott periodicity theorem to identify the generator of {\widetilde{K}(S^{2n})} as a power of the Hopf bundle {\mathcal{O}(-1)} minus one on {S^2}. Then, one can reduce to showing the claim for the Hopf bundle on {S^2}. For this, we have:

\displaystyle \psi^k ( \mathcal{O}(-1) - 1 ) = \mathcal{O}(-k) - 1.

But the Hopf bundle {H = \mathcal{O}(-1)} satisfies the relation for {k > 1},

\displaystyle (H - 1)^2 = 0, \quad \text{so} \quad \psi^2(H) = H^2 - 1 = 2(H - 1).

The higher operations can be handled similarly.

Another easy observation we’ll need is that the {\psi^k} commute with each other. This is immediate from the description on sums of line bundles and the splitting principle.

The real reason, though, that we’re interested in these operations {\psi^k} is that, for {k} prime, they’re not far from raising to a power. Namely:

Proposition 4 For {p} prime, {\psi^p(x) \equiv x^p \mod p}, for {x \in K(X)}.

This is exactly true if {x} comes a line bundle, and it is preserved under sums because we are working mod {p}. By the splitting principle, this proves the identity in general.

 The Adams-Atiyah argument

Suppose {f: S^{2n-1} \rightarrow S^n} is a map, with {n} even. The claim is that we can define the Hopf invariant solely in the setting of K-theory, which is why it is relevant at all. As usual, let’s say {n > 1}. Then {M_f} has, as before, a CW structure with three cells, in dimensions zero, {n}, and {2n}.

The claim is that {\widetilde{K}(M_f)} is free abelian on two generators {\iota_n} and {\iota_{2n}}. In fact, the cohomology ring is free, so by the Atiyah-Hirzebruch spectral sequence (which has torsion differentials and thus degenerates) we find that there are two generators {\iota_n, \iota_{2n}} of {\widetilde{K}(M_f)}. We can choose the generators such that:

  1. {\iota_n} restricts to a generator of {\widetilde{K}(S^n)} (so, in particular, it is not canonically determined).
  2. {\iota_{2n}} is the pull-back of a generator of {\widetilde{K}(S^{2n})} under the collapse map {M_f \rightarrow S^{2n}}.

Under the Chern character

\displaystyle \mathrm{ch}: \widetilde{K}(M_f) \rightarrow H^*(M_f; \mathbb{Q})

we find that {\iota_n} goes to a generator of {H^n} (plus possibly some extra in {H^{2n})}, and that {\iota_{2n}} is a generator of {H^{2n}}, even in {\mathbb{Z}}-cohomology. This is because the Chern character is an isomorphism for even-dimensional spheres of reduced K-theory and reduced integral cohomology (this encapsulates the Bott integrality theorem earlier). Consequently, we can use naturality to argue that {\mathrm{ch}(\iota_n)} restricted to the {n}-skeleton is a generator, etc.

Now, let’s apply {\psi^2} to {\iota_n}. We have that:

\displaystyle \psi^2(\iota_n) = \mu \iota_{2n} + 2^{n/2} \iota_n, \quad \mu.

This is for the following reason: {\psi^2 (\iota_n)} restricted to {S^n} is {2^{n/2}} times {\iota_n} restricted to {S^n}, by the earlier comments. For similar reasons,

\displaystyle \psi^3(\iota_n) = \nu \iota_{2n} + 3^{n/2} \iota_n, \quad \mu.

Here {\mu, \nu} are integers which we do not know. The main point is:

Observation: {\mu} mod 2 is the Hopf invariant.

This is because {\psi^2} is congruent to squaring mod 2. Consequently, we have {\iota_n^2 \equiv \mu \iota_{2n} \mod 2} in {\widetilde{K}(M_f)}. In fact, {\iota_n^2} restricted to the {n}-skeleton is zero, so we have

\displaystyle \iota_n^2 = \mu \iota_{2n} + 2a \iota_{2n} \in \widetilde{K}(M_f), \quad a \in \mathbb{Z}.

If we apply the Chern character (which is integral on {\iota_{2n}}), we find the analog in cohomology, that {\mu} is congruent to the Hopf invariant mod 2.

So, our goal will be to show that {\mu} is even. Now, we note that {\psi^2, \psi^3} are very easy to compute on {\iota_{2n}}; since this is obtained by pull-back from the sphere {S^{2n}}, it acts by multiplication by {2^{n}} or {3^n}.

With these preliminaries out of the way, we will simply play with the equation

\displaystyle \psi^2 \psi^3 \iota_n = \psi^3 \psi^2 \iota_n\ \ \ \ \ (1)

and a little number theory will force {\mu} to be even. Namely, from the definitions, the first side becomes

\displaystyle \psi^2 \left( \nu \iota_{2n} + 3^{n/2} \iota_n \right) = 2^n \nu \iota_{2n} + 3^{n/2} ( \mu \iota_{2n} + 2^{n/2} \iota_n).

Similarly, the second side becomes

\displaystyle \psi^3( \mu \iota_{2n} + 2^{n/2} \iota_n ) = 3^n \mu \iota_{2n} + 2^{n/2} \left( \nu \iota_{2n}+ 3^{n/2} \iota_n \right).

We can now equate coefficients of {\iota_{2n}} in both quantities. We get:

\displaystyle 2^{n }\nu + 3^{n/2}\mu = 3^n \mu + 2^{n/2} \nu ,

which gives

\displaystyle 2^{n/2}(2^{n/2}-1)\nu = 3^{n/2}(3^{n/2}-1) \mu.

The claim is that this is going to force {\mu} to be even, except in a couple of exceptional cases. Indeed, were {\mu} odd, we would have that

\displaystyle 2^{n/2} \mid 3^{n/2}-1.

It is a lemma now in number theory that the only possibilities for {n} are {2, 4, 8}. Granting this, the theorem is established:

Theorem 5 Suppose {n \neq 1, 2, 4, 8}. Then if {f: S^{2n-1} \rightarrow S^n} is a map, then all cup products in {H^*(M_f) } are zero.

The lemma from number theory

Actually, we don’t need the full strength of the lemma in number theory. Instead, we can modify the topological argument given above with {\psi^k} (for {k} odd) replacing {\psi^3}; everything goes through as before, except that we have

\displaystyle 2^{n/2} \mid k^{n/2} - 1, \quad k \ \mathrm{odd}.

This must be true for all {k}.

Suppose now {n > 4}. If we let {m = n/2}, then

\displaystyle k^m \equiv 1 \mod 2^m, \quad \forall k \ \text{odd}.

We see now that {m} is even. In fact, {m} must be a multiple of the order of {k} in the group {(\mathbb{Z}/2^m \mathbb{Z})^*}, which has order a power of two. So we can take {k = 1 + 2^{m/2}} and get

\displaystyle 1 + m 2^{m/2} \equiv 1 \mod 2^m.

This implies that {m 2^{m/2}} is divisible by {2^m}. This implies {2^{m/2} \mid m}, and the only possibilities for {m} are {m = 2, 4}. These correspond to {n = 4, 8}.