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.

The fiber of ${f}$ is ${H \mathbb{Z}}$ again, and we can repeat this process. So we have an Adams resolution for ${H \mathbb{Z}}$ which looks like

$\displaystyle \dots \rightarrow H \mathbb{Z} \stackrel{p}{\rightarrow} H \mathbb{Z} \stackrel{p}{\rightarrow} \dots \rightarrow H \mathbb{Z}.$

Each of the cofibers is ${H \mathbb{Z}/p}$, and the ${E_1}$-page of the spectral sequence ${E_1^{s,t}}$ is ${E_1^{s,t} = \mathbb{Z}/p}$ for ${s = t}$ and zero otherwise.

I didn’t go completely over the degree conventions for the ASS in the previous post, so let me discuss it here: if ${K_i}$ is the ${i}$th cofiber in the Adams resolution, then

$\displaystyle E_1^{s,t} = \pi_{t-s} K_s \implies \pi_{t-s} H \mathbb{Z}.$

Anyway, this spectral sequence completely degenerates at ${E_1}$, and all we have is a line of ${\mathbb{Z}/p}$‘s along the line ${s =t}$. This corresponds to the associated graded of the ${p}$-adic filtration of the group ${\mathbb{Z}_p}$, and we recover what we expected.

2. The sphere

Let’s now consider a much more interesting example of the ASS, the case of the sphere. The cohomology of the sphere is ${\mathbb{Z}/p}$ in degree zero, which means that the ASS for the sphere looks like

$\displaystyle E_2^{s,t} = \mathrm{Ext}^{s,t}_{\mathcal{A}_p}(\mathbb{Z}/p, \mathbb{Z}/p) \implies \widehat{\pi_{t-s} S^0},$

where ${\widehat{}}$ denotes ${p}$-adic completion. In practice, we know that the stable homotopy groups of spheres (other than the one in degree zero) are all finite, so ${p}$-adic completion just means taking the ${p}$-part.

Let me note a few features of this:

1. The differential in the ${r}$th stage goes ${d_r: E_r^{s,t} \rightarrow E_r^{s+t, t+ r - 1}}$.
2. The ${\mathrm{Ext}^{s,t}}$ means that ${s}$ is what affects the ${\mathrm{Ext}}$ (i.e. ${s = 1}$ corresponds to extensions). The ${t}$ is carried along from the grading of the Steenrod algebra, though.
3. We can compute all these things either in the category of modules over ${\mathcal{A}_p}$ or comodules over ${\mathcal{A}_p^{\vee}}$.
4. There is a multiplicative structure on this spectral sequence which corresponds to the smash (or composition) product on the stable homotopy groups of spheres. With respect to this, one actually has a spectral sequence of algebras converging to the graded ring ${\pi_* S^0}$.

Computing the ${E_2}$ page is not easy, but it can (theoretically at least) be done reasonably mechanically by writing down minimal resolutions. Unfortunately, the differentials are not so mechanical! Still, let’s try to see what we can do.

Let’s take ${p = 2}$. We can at least describe the ${\mathrm{Ext}}$ groups when ${s = 0, 1}$. When ${s =0 }$, it is sort of silly: there is a ${\mathbb{Z}/2}$ when ${t = 0}$ and zero elsewhere. When ${s =1}$, we start by writing down a very small piece of a resolution of ${\mathbb{Z}/2}$ over ${\mathcal{A}_2}$.

$\displaystyle \dots \rightarrow \bigoplus_k \mathcal{A}_2[2^k] \rightarrow \mathcal{A}_2 \rightarrow \mathbb{Z}/2 \rightarrow 0.$

The idea is that ${\mathcal{A}_2 \rightarrow \mathbb{Z}/2}$ is sending the generator to the nonzero element of ${\mathbb{Z}/2}$, of course. The idea of the second map is that the kernel of ${\mathcal{A}_2 \rightarrow \mathbb{Z}/2}$ is generated by the indecomposable elements in ${\mathcal{A}_2}$, which are given by the ${\mathrm{Sq}^{2^k}}$ for ${k \geq 0}$. So this defines the second map.

Now we’ve written down a very small piece of a minimal resolution for ${\mathbb{Z}/2}$ over ${\mathcal{A}_2}$. Since the resolution is minimal, we can compute the ${\mathrm{Ext}}$ groups by literally just taking ${\hom}$‘s into ${\mathbb{Z}/2}$—there’s no homology to compute. We find:

Proposition 1 The groups ${\mathrm{Ext}^{1, 2^k}_{\mathcal{A}_2}(\mathbb{Z}/2, \mathbb{Z}/2)}$ are ${\mathbb{Z}/2}$, generated by an element ${h_i}$. The groups ${\mathrm{Ext}^{1, t}_{\mathcal{A}_2}(\mathbb{Z}/2, \mathbb{Z}/2)}$ are otherwise zero.

3. The differentials and the Hopf invariant problem

Next, I’d like to describe what this has to do with the Hopf invariant one problem. One way of phrasing (a variant of) the Hopf invariant one problem is when there exists a map of spectra (i.e., a stable map)

$\displaystyle f: S^n \rightarrow S^0$

such that the cofiber ${C}$ is such that its cohomology ${H^*(C; \mathbb{Z}/2)}$ is a nontrivial extension of ${\mathbb{Z}/2}$ by ${\mathbb{Z}/2[n+1]}$ as ${\mathcal{A}_2}$-modules. In other words, if one starts with the generator in degree zero of the cofiber ${C}$, applying a suitable Steenrod square ${\mathrm{Sq}^{n+1}}$ gets you to a nonzero element in degree ${n+1}$.

If this happens, ${n+1}$ has to be a power of ${2}$. For otherwise ${\mathrm{Sq}^{n+1}}$ is decomposable and applying it cannot take the nonzero element in degree zero to the nonzero element in degree ${n+1}$. Alternatively, we’ve computed above that ${\mathrm{Ext}_{\mathcal{A}_2}^{1, t}(\mathbb{Z}/2, \mathbb{Z}/2)}$ is zero when ${t \neq 2^k}$ for some ${k}$.

With these preliminaries out of the way, let’s try to connect the existence of such a map with the question whether the ${h_i}$ are infinite cycles. Here is the answer:

Proposition 2 The element ${h_k \in \mathrm{Ext}_{\mathcal{A}_2}^{1, 2^k}(\mathbb{Z}/2, \mathbb{Z}/2)}$ is an infinite cycle in the ASS if and only if there exists a map ${S^{2^{k-1}} \rightarrow S^0}$ such that ${\mathrm{Sq}^{2^k}}$ is nontrivial on the cofiber.

In fact, suppose ${h_k}$ is an infinite cycle in the ASS. Then, in particular, ${h_k}$ corresponds to an actual map ${f: S^{2^{k}-1}\rightarrow S^0}$; the claim is that this has Hopf invariant one. To see this, let’s unwind the definitions. We can begin an Adams resolution for ${S^0}$

$\displaystyle \mathrm{fib}(S^0 \rightarrow H \mathbb{Z}/2) \rightarrow S^0$

and ${f}$ lifts to a map ${S^{2^{k}-1} \rightarrow \mathrm{fib}(S^0 \rightarrow H \mathbb{Z}/2)}$ because it is zero in homology. However, if ${f}$ is represented by something which comes from a nonzero element in row ${1}$, it can’t be lifted any higher in the Adams resolution: that is, ${f}$ induces a nonzero map on homology from

$\displaystyle S^{2^{k}-1} \rightarrow \mathrm{fib}(S^0 \rightarrow H \mathbb{Z}/2) .$

Anyway, we can now draw a commutative diagram of cofiber sequences

By assumption, ${\phi}$ is nonzero on homology, so using the diagram of long exact sequences, we find that the map ${\mathrm{cof}(f) \rightarrow H \mathbb{Z}/2}$ has trivial kernel in homology. In particular, the map

$\displaystyle H^* ( H \mathbb{Z}/2; \mathbb{Z}/2) \rightarrow H^*(\mathrm{cof}(f); \mathbb{Z}/2)$

is surjective, which is saying that ${f}$ has Hopf invariant one. This argument can be entirely reversed: if one has a map of Hopf invariant one, one gets an infinite cycle in the ASS in the right degree that it must be ${h_k}$.