This is a continuation of yesterday’s post, which used the Adams spectral sequence to compute the first two stable homotopy groups of spheres (only as a toy example for myself: one can use more elementary tools). In this post, I’d like to describe the third stable stem.The claim is that the first four columns of the ${E_2}$-page of the Adams spectral sequence for the sphere look like:

Furthermore, we have the relation ${h_0^2 h_2 = h_1^3}$. This is the complete picture for the first four columns.

Note that there can be no nontrivial differentials in this range for dimensional reasons. Since ${h_0}$ corresponds to multiplication by 2 in the stable stems, this corresponds to the fact that ${\pi_3(S^0) = \mathbb{Z}/8}$: in fact, we find that ${\pi_3(S^0)}$ has a three-term filtration with successive quotients ${\mathbb{Z}/2}$, and that passage down each step of the filtration is given by multiplication by ${2}$. The relation ${h_0^2 h_2 = h_1^3}$ corresponds to the fact that the Hopf map ${\nu \in \pi_3(S^0)}$ (which corresponds to the element of Hopf invariant one in ${\pi_3(S^0)}$) satisfies

$\displaystyle 4 \nu = \eta^3,$

for ${\eta}$ the element of Hopf invariant one in ${\pi_1(S^0)}$ represented by ${h_1}$.

1. First steps

Proving this will take a little more work than the first two stems. Let’s first show:

Lemma 1 In the ${E_2}$ page of the ASS, we have ${h_0^2 h_2 = h_1^3}$.

This amounts to saying that the two cycles in the cobar complex

$\displaystyle [\zeta_1^4 | \zeta_1 | \zeta_1 ], \quad [\zeta_1^2 | \zeta_1^2 | \zeta_1^2]$

are cohomologous (recall that multiplication corresponds to the juxtaposition product in the cobar complex). So, we’ll need elements of total ${t}$-degree six of the form ${[x|y]}$ to provide a cobounding element. There aren’t too many such elements, and writing out the cobar formula gives

$\displaystyle \delta( [ \zeta_1^2 | \zeta_2 \zeta_1 ]) = [ \zeta_1^2 | \zeta_2 | \zeta_1 ] + [ \zeta_1^2 | \zeta_1 | \zeta_2 ] + [ \zeta_1^2 | \zeta_1^2 | \zeta_1^2] + [ \zeta_1^2 | \zeta_1^3 | \zeta_1 ] .$

Next, we get

$\displaystyle \delta( [ \zeta_2 | \zeta_2 ] ) = [ \zeta_1^2 | \zeta_1 | \zeta_2 ] + [ \zeta_2 | \zeta_1^2 | \zeta_1 ] .$

Finally, we have

$\displaystyle \delta( [ \zeta_2 \zeta_1^2 | \zeta_1 ] ) = [ \zeta_2 | \zeta_1^2 | \zeta_1 ] + [ \zeta_1^2 | \zeta_2 | \zeta_1 ] + [ \zeta_1^4 | \zeta_1 | \zeta_1 ] + [ \zeta_1^2| \zeta_1^3 | \zeta_1 ] .$

Adding these three relations shows that almost all the terms cancel, and we are left with the fact that ${[ \zeta_1^2 | \zeta_1^2 | \zeta_1^2 ] + [ \zeta_1^4 | \zeta_1 | \zeta_1]}$ is a coboundary in the cobar complex. This is precisely the meaning of the assertion claimed in the lemma.

We haven’t yet shown yet that these elements are nonzero; that will come last. Now, let’s show that there can’t be any other elements in this column.

Lemma 2 The only elements in the ${E_2}$ page of the ASS with ${t - s = 3}$ are the multiples of ${[ \zeta_1^4] = h_2}$.

To see this, let’s look for cocycles in the cobar complex with ${t -s = 3}$. These will be, necessarily, of the form

$\displaystyle [x_1 | \dots | x_s ], \quad x_i \in \overline{\mathcal{A}_2^{\vee}} ,$

where the total degree of the ${x_i}$ is ${s + 3}$. This means that all but three of the ${x_i}$‘s have to be degree one, and so must be ${\zeta_1}$—these just correspond to multiplying by ${h_0}$. The other three will have to have degrees summing to ${6}$. We have a few possibilities for the remaining three of the ${x_i}$ corresponding to the various partitions of six into three natural numbers:

1. ${\zeta_2 \zeta_1, \zeta_1, \zeta_1}$.
2. ${\zeta_1^4, \zeta_1, \zeta_1}$.
3. ${\zeta_2, \zeta_1^2, \zeta_1}$.
4. ${\zeta_1^3, \zeta_1^2, \zeta_1}$.
5. ${\zeta_1^2, \zeta_1^2, \zeta_1^2}$.

This is getting somewhat tedious to analyze, so let’s abandon this and introduce a new tool.

2. The change-of-rings spectral sequence

Consider the subHopf algebra ${A}$ of ${\mathcal{A}_2^{\vee}}$ given by ${\mathbb{Z}/2[\zeta_1, \zeta_2]}$: the coproduct formulas show easily that it is closed under the coproduct. This is a normal subHopf algebra, and a useful observation here is that:

Proposition 3 The map ${\mathrm{Ext}^{s,t}_{A}(\mathbb{Z}/2, \mathbb{Z}/2) \rightarrow \mathrm{Ext}^{s, t}_{\mathcal{A}_2^{\vee}}( \mathbb{Z}/2, \mathbb{Z}/2)}$ is an isomorphism for ${t - s \leq 4}$ and a surjection for ${t -s = 5}$.

The map in question comes from the inclusion of cobar complexes: the cobar complex of ${A}$ maps to that of ${\mathcal{A}_2^{\vee}}$. The elements with ${t-s \leq 5}$ in the cobar complex of ${A}$ are the same as the elements with ${t - s \leq 5}$ in the cobar complex of ${\mathcal{A}_2^{\vee}}$ (because the ${\zeta_i, i \geq 3}$ have ${t -s \geq 6}$), which is the argument.

So our goal will be to compute the cohomology of the subalgebra ${A}$. Notice that ${A}$ has a subHopf algebra

$\displaystyle \mathbb{Z}/2[\zeta_1] \subset A,$

and ${A // \mathbb{Z}/2[\zeta_1] = \mathbb{Z}/2[\overline{\zeta_2}]}$ where the quotient element ${\overline{\zeta_2}}$ is primitive. In other words, we have a sequence

$\displaystyle 0 \rightarrow \mathbb{Z}/2[\zeta_1] \subset A \twoheadrightarrow \mathbb{Z}/2[\overline{\zeta_2}] \rightarrow 0,$

which is not exact, but which is an extension of Hopf algebras in the appropriate sense. The coalgebras ${\mathbb{Z}/2[\zeta_1], \mathbb{Z}/2[\overline{\zeta_2}]}$ are simple and their cohomology is easy to compute (as they are tensor products of exterior coalgebras, the cohomology is a big polynomial algebra on a countable number of generators).

Proposition 4 (Cartan-Eilenberg spectral sequence) There is a spectral sequence of algebras converging to ${\mathrm{Ext}^{s}_A(\mathbb{Z}/2, \mathbb{Z}/2)}$. The ${E_2}$ page is given by

$\displaystyle E_2^{p,q} \simeq H^q(\mathbb{Z}/2[\zeta_1]) \otimes H^p(\mathbb{Z}/2[\overline{\zeta_2}])$

(where ${H^p}$ means the ${\mathrm{Ext}}$ groups). The ${r}$th differential runs upward: ${d_r: E_r^{s,t} \rightarrow E_r^{s-r + 1, t+ r} }$.

This is proved in Adams’s paper on the Hopf invariant one problem. I don’t really understand the details, but for now I’d like to take it as a black box. This is a somewhat more cavalier attitude than I like to take, but I think for now I’d rather see how the machinery is used before getting into the details. Let’s just try to use it and see what we can get.

3. Computations

Now, the cohomology of the coalgebra ${\mathbb{Z}/2[\overline{\zeta_2}]}$ is a polynomial ring on the generators ${h_i' = [\overline{\zeta_2}^{2^i}], i = 0, 1, 2, \dots}$. This is because the coalgebra is a tensor product of (primitive) exterior algebras on the generators ${\overline{\zeta_2}^{2^i}}$, and the cohomology of an exterior algebra is a polynomial algebra. Similarly, the cohomology of the coalgebra ${\mathbb{Z}/2[\zeta_1]}$ is a polynomial ring on the ${h_i = [\zeta_1^{2^i}]}$. It follows that we can draw the ${E_2}$ page of this spectral sequence:

This spectral sequence comes from a filtration of the cobar complex on ${A}$. Namely, an element ${[x_1 | \dots | x_s]}$ in the cobar complex of ${A}$ belongs to the ${p}$th stage of the filtration if ${p}$ of the ${x_i}$ belong to ${\mathbb{Z}/2[\zeta_1]}$. The exact identification of the ${E_2}$ page is not really necessary to our purposes, but the point that the spectral sequence comes from such a filtration is crucial.

For instance, we can work out with the second differential ${d_2}$ does to ${\left\{h_i'\right\}}$. This element is represented by ${[\overline{\zeta_2}^{2^i}]}$ in the cobar complex of the quotient, so we lift that to ${[\zeta_2^{2^i}]}$ in the cobar complex of ${A}$. Then compute the coboundary to get ${[\zeta_1^{2^{i+1}} | \zeta_1^{2^i}]}$. This means

$\displaystyle d_2( h_i') = h_ih_{i+1} .$

So these elements must be killed in the ${E_3}$ page and hence in the cohomology of ${A}$: this corresponds, for instance, to the fact that ${h_0 h_1 = 0}$ in the Adams spectral sequence as earlier (i.e., the Hopf fibration is stably of order two).

Anyway, it follows from this that in the ${E_3}$ page, none of the ${\left\{h_i'\right\}}$ terms survive: any linear combination of them will have nonzero ${d_2}$-value. Also, the ${h_i h_{i+1}}$ terms in ${(0, 2)}$ in ${E_2}$ will be killed in ${E_3}$ because they are hit by boundaries, and by a similar logic, the terms ${h_i' h_j'}$ for ${i \neq j}$ will have nonzero ${d_2}$:

$\displaystyle d_2(h_i' h_j') = h_i' h_{j} h_{j+1} + h_j' h_{i}h_{i+1},$

and these are linearly independent.

Finally, we have to figure out what happens to the ${\left\{h_i h'_j\right\}}$. We have

$\displaystyle d_2(h_i h'_j) = h_i h_{j}h_{j+1},$

so no new elements are killed in ${E_2^{0, 3}}$ (that is, only multiples of the old relation ${h_j h_{j+1} = 0}$ are introduced). However, we observe that

$\displaystyle g_j = h_{j-1} h'_j + h'_{j-1} h_{j+2} \in E_2^{1, 1}, \quad j \geq 1.$

is a cycle under ${d_2}$, and so survives to the ${E_3}$ page. These are the linear combinations that do survive.

So ${E_3}$ looks like:

The dots indicate terms which are not determined. The term in ${(3, 0)}$ is already quotiented by the relations ${h_i h_{i+1} = 0}$ and those deriving from it. In order to proceed, we will determine the transgression of the ${h_i'^2}$.

Morally, the idea is that ${h_i'^2 =\mathrm{Sq}^1 h_i'}$ and the transgression should commute with the Steenrod squares, as it does in the Serre spectral sequence. Consequently we can determine the transgression of ${h_i'^2}$ from that of ${h_i'}$, which we found to be ${h_i h_{i+1}}$ earlier. This can be done and made precise, but let’s work with the cobar complex. Here ${h_i'^2}$ in ${E_2}$ is represented by

$\displaystyle [ \zeta_2^{2^i} | \zeta_2^{2^i} ].$

Unfortunately this is not a representative in ${E_3}$, as the differential of this is

$\displaystyle [ \zeta_1^{2^{i+1}} | \zeta_1^{2^i}| \zeta_2^{2^i} ] + [\zeta_2^{2^i} | \zeta_1^{2^{i+1}} | \zeta_1^{2^i}],$

which is not sufficiently good enough for the filtration degree: it’s in the second, but not the third, level of the filtration. So to get a representative of this class in ${E_3}$, we need to modify this class in the cobar complex to get something whose boundary has three terms involving ${\zeta_1}$.

For simplicity, let’s take ${i = 0}$, so we need to modify ${[\zeta_2| \zeta_2]}$ to get a representative of ${h_0'^2}$. The strategy is to take the modification

$\displaystyle [\zeta_2 | \zeta_2 ] + [ \zeta_1^2 | \zeta_1 \zeta_2] + [ \zeta_2 \zeta_1^2 | \zeta_1]$

which is a modification by terms of lower filtration, but which has the advantage that the cobar differential takes this to something in filtration 3. This is a computation we have done before:

$\displaystyle d_3( h_0'^2) = d( [\zeta_2 | \zeta_2 ] + [ \zeta_1^2 | \zeta_1 \zeta_2] + [ \zeta_2 \zeta_1^2 | \zeta_1] ) = [ \zeta_1^2 | \zeta_1^2 | \zeta_1^2 ] + [ \zeta_1^4 | \zeta_1 | \zeta_1].$

This gives

$\displaystyle d_3(h_0'^2) = h_1^3 + h_0^2 h_2.$

More generally,

$\displaystyle d_3(h_i'^2) = h_{i+1}^3 + h_i^2 h_{i+2}.$

It follows that in the ${E_4}$ term, the space ${(2, 0)}$ becomes empty, and we have something looking like

The relation ${\sim}$ means that the ${(0, 3)}$ term is quotiented out by the relations arising from ${h_i h_{i+1} = 0}$ and ${h_{i+1}^3 + h_i^2 h_{i+2} = 0}$. These relations are exactly those that we want to claim in ${H^\bullet(\mathcal{A}_2^{\vee})}$.

Now, with this ${E_4}$ term, we can determine the cohomology of the Steenrod algebra in the relevant range of ${t-s}$. We note that ${h_i'}$ has ${t-s}$ degree equal to ${3( 2^i) - 1 \geq 2}$ and ${h_i}$ has ${t-s}$ degree equal to ${2^{i}-1}$. Consequently, ${g_j}$ has ${t-s}$ degree equal to ${2^{j-1} + 3 ( 2^j ) - 2 \geq 6}$ for ${j \geq 1}$. Any higher terms in the middle column that survive can also be seen to have degree at least ${3}$ (something like ${h_0^n h_0'}$ is the only possible counterexample, and that doesn’t even live to ${E_3}$).

It follows that all the terms not shown have ${t-s}$ degree at least ${6}$ and so do not contribute to the cohomology of ${\mathcal{A}_2^{\vee}}$ in the range ${t - s \leq 3}$, except possibly ${g_1}$.

Proposition 5 The cohomology of ${A}$ in degrees ${t-s = 3 5}$ is: free on ${h_2, h_0 h_2, h_0^2 h_2 = h_1^3}$.

As a corollary, we find that the cohomology of ${\mathcal{A}_2^{\vee}}$ is the same in ${t - s = 3}$. It follows that the picture of the Adams spectral sequence drawn at the beginning of this post is correct, and we get the third stable stem.

This technique is used more generally in Adams’s paper on the Hopf invariant one problem to get a picture of the first three rows of the Adams spectral sequence. Hopefully I’ll be able to return to this in a later post.