Let’s try to do some (baby) examples of the Adams spectral sequence. The notation used will be that of yesterday’s post.

1. ${H\mathbb{Z}}$

So let’s start with a silly example, where the answer is tautological: ${H \mathbb{Z}}$. We could try to compute the homotopy groups of this using the Adams spectral sequence. At a prime ${p}$, this means that we should get the ${p}$-adic completion ${\mathbb{Z}_p}$ in degree zero, and nothing elsewhere.

It turns out that we can write down a very explicit Adams resolution for ${H \mathbb{Z}}$. To start with, we need a map ${f: H \mathbb{Z} \rightarrow X}$ where ${X}$ is a wedge of ${H \mathbb{Z}/p}$ and shifts, and such that ${f}$ is a monomorphism on ${\mathbb{Z}/p}$-homology. We can take the map

$\displaystyle f: H \mathbb{Z} \rightarrow H \mathbb{Z}/p;$

the fact that ${f}$ is a monomorphism on ${\mathbb{Z}/p}$-homology follows because ${f}$ is an epimorphism on ${\mathbb{Z}/p}$-cohomology, by Serre’s computation of the cohomology of Eilenberg-MacLane spaces. Serre’s computation tells us, in fact, that the cohomology of ${H \mathbb{Z}}$ is the Steenrod algebra mod the ideal generated by the Bockstein. (more…)

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…)