K-theory and the Hopf invariant

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

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

then

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

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:

${\iota_n}$ restricts to a generator of ${\widetilde{K}(S^n)}$ (so, in particular, it is not canonically determined).
${\iota_{2n}}$ 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 ${n = 4, 8}$ .

The Quillen-Suslin theorem « Climbing Mount Bourbaki Says:

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). […]

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