Let be a map with . Associated to this, one can form a CW complex ; that is, we attach a -cell to via the map . This CW complex has one cell in dimension and one cell in dimension (and one cell in dimension ). The map determines a generator of and the map determines a generator of ; there are no other elements in cohomology other than the unit. Consequently, we have

Definition 1The number as above such that is theHopf invariantof .

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

**Example 1** The Hopf fibration is, by definition, the map such that the mapping cone is ; it follows that the Hopf fibration has Hopf invariant one.

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

which for even has image containing the even integers. (This is where the exceptional summand in the homotopy groups of spheres comes from.)

A classical problem in topology was the following:

Question:For which does there exist a map of Hopf invariant one?

One can show that whenever (and 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 .

**Example 2: **If is not a power of , there can be no map of odd Hopf invariant. In fact, in that case would be a complex whose cohomology would be where is in degree .

This, however, is impossible. In fact, if is not a power of two, the aforementioned -cohomology ring does not arise.

Proposition 2If is not a power of two and is a complex with no -cohomology between and , then the square of any element in is zero.

To prove this, we use the fact that for , we can write the cup square in terms of the Steenrod operations:

Now, by the Adem relations, the element 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 for , so they are all zero.

In fact, using the Steenrod reduced powers in mod 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, 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 be a vector bundle. Then for each , there is supposed to be an operation such that if

then

In other words, is supposed to raise each line bundle to the th power, and is supposed to be additive. This determines . 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 in terms of the *exterior power operations*

Anyway, in view of all this, one gets:

Proposition 3is a natural ring homomorphism for any compact space .

For the proof, we need two more facts. The first is that for any element in the reduced K-theory , acts by multiplication by . One can see this by using the Bott periodicity theorem to identify the generator of as a power of the Hopf bundle minus one on . Then, one can reduce to showing the claim for the Hopf bundle on . For this, we have:

But the Hopf bundle satisfies the relation for ,

The higher operations can be handled similarly.

Another easy observation we’ll need is that the 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 is that, for prime, they’re not far from raising to a power. Namely:

Proposition 4For prime, , for .

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

** The Adams-Atiyah argument**

Suppose is a map, with 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 . Then has, as before, a CW structure with three cells, in dimensions zero, , and .

The claim is that is free abelian on two generators and . 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 of . We can choose the generators such that:

- restricts to a generator of (so, in particular, it is not canonically determined).
- is the pull-back of a generator of under the collapse map .

Under the Chern character

we find that goes to a generator of (plus possibly some extra in , and that is a generator of , even in -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 restricted to the -skeleton is a generator, etc.

Now, let’s apply to . We have that:

This is for the following reason: restricted to is times restricted to , by the earlier comments. For similar reasons,

Here are integers which we do not know. The main point is:

**Observation**: mod 2 is the Hopf invariant.

This is because is congruent to squaring mod 2. Consequently, we have in . In fact, restricted to the -skeleton is zero, so we have

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

So, our goal will be to show that is even. Now, we note that are very easy to compute on ; since this is obtained by pull-back from the sphere , it acts by multiplication by or .

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

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

Similarly, the second side becomes

We can now equate coefficients of in both quantities. We get:

which gives

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

It is a lemma now in number theory that the only possibilities for are . Granting this, the theorem is established:

Theorem 5Suppose . Then if is a map, then all cup products in 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 (for odd) replacing ; everything goes through as before, except that we have

This must be true for all .

Suppose now . If we let , then

We see now that is even. In fact, must be a multiple of the order of in the group , which has order a power of two. So we can take and get

This implies that is divisible by . This implies , and the only possibilities for are . These correspond to .

January 16, 2012 at 6:03 pm

[…] This observation allows one to get a simple example of a stably free module which is not free. The tangent bundle to is stably trivial (in fact, its one-dimensional normal bundle is trivial), but it is not trivial unless by a famous theorem (which is in fact a consequence of the Hopf invariant one theorem). […]