I’d like to use the next couple of posts to compute the first three stable stems, using the Adams spectral sequence. Recall from the linked post that, for a connective spectrum ${X}$ with appropriate finiteness hypotheses, we have a first quadrant spectral sequence

$\displaystyle \mathrm{Ext}^{s,t}_{\mathcal{A}_2^{\vee}}(\mathbb{Z}/2, H_*( X; \mathbb{Z}/2)) \implies \widehat{\pi_{t-s} X} ,$

where the ${\mathrm{Ext}}$ groups are computed in the category of comodules over ${\mathcal{A}_2^{\vee}}$ (the dual of the Steenrod algebra), and the convergence is to the ${2}$-adic completion of the homotopy groups of ${X}$. In the case of ${X}$ the sphere spectrum, we thus get a spectral sequence

$\displaystyle \mathrm{Ext}^{s,t}_{\mathcal{A}_2^{\vee}}(\mathbb{Z}/2, \mathbb{Z}/2) \implies \widehat{\pi_{t-s} S^0},$

converging to the 2-torsion in the stable stems. In this post and the next, we’ll compute the first couple of ${\mathrm{Ext}}$ groups of ${\mathcal{A}_2^{\vee}}$, or equivalently of ${\mathcal{A}_2}$ (this is usually called the cohomology of the Steenrod algebra), and thus show:

1. ${\pi_1 S^0 = \mathbb{Z}/2}$, generated by the Hopf map ${\eta}$ (coming from the Hopf fibration ${S^3 \rightarrow S^2}$).
2. ${\pi_2 S^0 = \mathbb{Z}/2}$, generated by the square ${\eta^2}$ of the Hopf map.
3. ${\pi_3 S^0 = \mathbb{Z}/8}$, generated by the Hopf map ${\nu}$ (coming from the Hopf fibration ${S^7 \rightarrow S^4}$). We have ${\eta^3 = 4 \nu}$. (This is actually true only mod odd torsion; there is also a ${\mathbb{Z}/3}$, so the full thing is a ${\mathbb{Z}/24}$.)

In fact, we’ll be able to write down the first four columns of the Adams spectral sequence by direct computation. There are numerous fancier tools which let one go further.

1. The cobar complex

Let ${M}$ be a comodule over a coalgebra ${A}$ over some field (for instance, ${\mathcal{A}_2^{\vee}}$). There is a useful cofree resolution which one can use for computing ${\mathrm{Ext}}$ groups. Namely, consider the cosimplicial object

$\displaystyle A \otimes M \rightrightarrows A \otimes A \otimes M \dots.$

The various cosimplicial arrows come from the comultiplications on various factors and the counit maps. We can extract from this a normalized chain complex ${C_u^*(M; A)}$, which we can describe as follows:

1. In degree ${s \geq 0}$, ${C_u^s(M, A) = A^{\otimes (s+1)} \otimes M}$. Elements are written as ${a[a_1 | \dots | a_s ] m}$.
2. The coboundary ${\delta: C_u^s(M, A) \rightarrow C_u^{s+1}(M, A)}$ is described via
3. Here the notation is such that the comultiplication ${\Delta: A \rightarrow A \otimes A}$ satisfies ${\Delta(a) = a' \otimes a'}$, ${\Delta(a_i) = a_i ' \otimes a_i''}$, and the map ${ M \rightarrow A \otimes M}$ sends ${m \mapsto a_m' \otimes m'}$. This is an abuse of notation—really ${\Delta(a_i)}$ may be a sum of pure tensors, etc. Note that this is a chain complex of cofree ${A}$-modules (with ${A}$ coacting on the first factor), and (by a formal argument which works in a fair bit of generality), it is a resolution of ${M}$.
4. If ${A}$ is an augmented coalgebra and ${\overline{A}}$ is the cokernel of the augmentation, then we can use a smaller version of the cobar complex: instead take

$\displaystyle C^s(M, A) = A \otimes \overline{A}^{\otimes s} \otimes M.$

This will still be a resolution of ${M}$ by cofree ${A}$-modules, and we will denote this by the ${C^s(M, A)}$. This will be the cobar complex that we use below. The formula for the differential is the same.

As an example (the purpose of this post), let’s write down what the complex to compute ${\mathrm{Ext}^{s,t}_{\mathcal{A}_2^{\vee}}(\mathbb{Z}/2, \mathbb{Z}/2)}$ (the ${E_2}$ page of the ASS for the sphere) looks like. Namely, we have to take the complex

$\displaystyle \mathcal{A}_2^{\vee} \rightarrow \mathcal{A}_2^{\vee} \otimes \overline{\mathcal{A}_2^{\vee}} \rightarrow \dots,$

which is a cofree resolution of ${\mathbb{Z}/2}$, and then take comodule maps of ${\mathbb{Z}/2}$ into this. Taking comodule maps of ${\mathbb{Z}/2}$ into this peels of the first (cofree) factor in each case, so we are left with a complex

$\displaystyle \mathbb{Z}/2 \rightarrow \overline{\mathcal{A}_2^{\vee}} \rightarrow \overline{\mathcal{A}_2^{\vee}} \otimes \overline{\mathcal{A}_2^{\vee}} \rightarrow \dots.$

This cobar complex ${\hom(\mathbb{Z}/2, C^*(\mathbb{Z}/2, \mathcal{A}_2^{\vee})}$ has cohomology which is the ${E_2}$ page of the ASS. So, in degree ${s}$, we have the free vector space on elements

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

and the coboundary is given by

$\displaystyle \delta( [x_1 | \dots | x_s]) = \sum_{i=1}^s [x_1 | \dots | x_{i-1} | x_i' | x_i''| \dots | x_s ],$

where ${\Delta(x_i) = x_i' \otimes x_i''}$ after quotienting out by the image of ${1}$ (so as to get into the reduced thing ${\overline{\mathcal{A}_2^{\vee}}}$). The nice thing about working mod 2 is that we don’t have to worry about signs.

Note that this is a bigraded complex, which means it lets us see the bigrading of ${\mathrm{Ext}}$ (which we need for the ASS). Note also that the cobar complex has a multiplicative structure given by juxtaposing two “bars”: that is, one multiplies ${[ x_1 | \dots |x_s]}$ by ${[y_1 | \dots | y_t]}$ to get ${[x_1 |\dots | x_s | y_1 | \dots | y_t]}$. It turns out that this coincides with the Yoneda product in ${\mathrm{Ext}}$, which corresponds to the ring structure in the stable stems.

2. First steps

We now have a recipe for working out the ${E_2}$ page of the ASS for the sphere. This is a fairly large complex of which we have to compute the cohomology, though, and so we do it only in small dimensions.

For ${s = 0}$, there is only one element ${[]}$, whose differential is trivial: this gives a cycle in ${\mathrm{Ext}^{0, 0}}$.

Let’s start filling in the ${E_2}$ page of the ASS. The topologists like to typeset it with the ${s}$ vertically and the ${t-s}$ horizontally, so right now what we have computed looks like:

The dot in the position ${(0, 0)}$ indicates that there’s a ${\mathbb{Z}/2}$ there. The empty space here means that we don’t know what is there, not that it is zero.

Anyway, the point of drawing the spectral sequence this way is that the vertical lines are parametrized by ${t-s}$, so that any given homotopy group is obtained by going up one of them. The horizontal level is the “Adams filtration.”

For the rest of the post, I’ll be using the notation for the dual Steenrod algebra as in this post: that is, $\mathcal{A}_2^\vee$ is a polynomial algebra on generators $\zeta_i, i = 1, 2, \dots$ of degrees $2^i - 1$. The formula for the coproduct is described there.

Admittedly that wasn’t very interesting. Let’s now get the remaining elements on the zeroth vertical line. These are cocycles in the cobar complex of degree ${t - s =0}$. Any such cocycle is of the form ${[a_1 | \dots |a_s]}$ where the ${a_i}$ are in the augmentation ideal of ${\mathcal{A}_2^{\vee}}$, i.e. have positive degree. So the only such cocycles are

$\displaystyle [ \zeta_1 | \dots | \zeta_1 ] \quad \text{(s times)}$

These can’t be coboundaries, because there are no terms in the cobar complex with ${t-s < 0}$ to annihilate them. So, if ${h_0 \in \mathrm{Ext}^{1,1}_{\mathcal{A}_2^{\vee}}(\mathbb{Z}/2, \mathbb{Z}/2)}$ is represented by ${[\zeta_1]}$, we get a chain of dots:

The zero stem is ${\mathbb{Z}}$ and its completion is the 2-adic integers ${\mathbb{Z}_2}$, so ${h_0}$ must represent multiplication by ${2}$. It isn’t surprising at all that we’ve gotten the above associated graded for the 0-stem. It also isn’t surprising that the multiples of 2 should live in Adams filtration higher than one.

3. The 1-stem

Now let’s do the 1-stem. We need to look for cocycles in the cobar complex with ${t - s =1}$. That is, we need to look for elements of the form

$\displaystyle [x_1 | \dots | x_s]$

where the total degree of the ${x_i}$ is ${s + 1}$. This means that all the ${x_i}$ but one must have degree ${1}$ (i.e., all but one of them has to be ${\zeta_1}$) and the other must have degree two: that is, it is ${\zeta_1^2}$. So the only possible candidates for cycles in the cobar complex with ${t-s = 1}$ are the elements (and their permutations)

$\displaystyle [\zeta_1^2 | \zeta_1 | \zeta_1 |\dots | \zeta_1].$

Now, if we let ${h_1 = [\zeta_1^2] \in \mathrm{Ext}^{1, 2}(\mathbb{Z}/2, \mathbb{Z}/2)}$, then we easily check that ${h_1}$ is a cycle: this corresponds to the fact that ${\zeta_1^2}$ is a primitive element of the Hopf algebra ${\mathcal{A}_2^{\vee}}$. (The dual indecomposable element of the Steenrod algebra is $\mathrm{Sq}^2$.)

So, the elements ${[\zeta_1^2 | \zeta_1 | \zeta_1 |\dots | \zeta_1]}$ are cycles and they are the only cycles with ${t-s = 1}$ in the cobar complex. These represent the classes ${h_0^i h_1}$ in the Adams spectral sequence.

Claim: ${h_1}$ is not zero (i.e., ${[\zeta_1^2]}$ is not a coboundary) but ${h_0 h_1 = 0}$.

In fact, ${[\zeta_1^2]}$ cannot be a coboundary because there is nothing with ${s =0 }$ to cobound it. So ${h_1 \neq 0}$. However,

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

as one sees by going back to the formulas in ${\mathcal{A}_2^{\vee}}$. This exhibits ${h_0 h_1}$ as a coboundary (and corresponds to the fact that twice the Hopf map is stably zero).

So we can extend the spectral sequence:

This means that we’ve computed the first two columns of the ASS, and we find that ${\pi_1 S^0 = \mathbb{Z}/2}$. Using the Freudenthal suspension theorem, we find that the image of the Hopf map ${S^3 \rightarrow S^2}$ must be a generator.

4. The second stem

Now let’s move on to the second stem. We need to look for cocycles in the cobar complex with ${t-s = 2}$. This means that we have terms of the form

$\displaystyle [x_1 | \dots | x_s]$

where the total degree of the ${x_i}$ amounts to ${s + 2}$. This means that any such term is a permutation of

$\displaystyle [\zeta_1^2 | \zeta_1^2 | \zeta_1 | \zeta_1 | \dots | \zeta_1 ] , \quad [ \zeta_1^3 | \zeta_1 | \zeta_1 | \dots | \zeta_1].$

The second term is not a cocycle. The first term is, and represents a power of ${h_0}$ times ${h_1^2}$. Since we saw ${h_0 h_1 =0 }$ earlier, the only possibility for a nontrivial cohomology class is ${h_1^2 = [\zeta_1^2 | \zeta_1^2]}$.

Claim: ${h_1^2 \neq 0}$.

This corresponds to the fact that the square of the Hopf map is stably essential. We can prove it in the cobar complex, though: we have to show that ${[\zeta_1^2 | \zeta_1^2]}$ is not a coboundary. In fact, a cobounding element would have to be something like ${[x]}$ where ${x}$ has degree four in ${\mathcal{A}_2^{\vee}}$: that is, either ${x = \zeta_1^4}$ or ${x = \zeta_2 \zeta_1}$. It’s easy to check that no combination of those possibilities works. So the spectral sequence now looks like:

Now this takes care of the first two stems (it’s easy to see that the spectral sequence has to degenerate here). I think we can treat the third stem in the same way, but it’ll require a little more effort, so it’ll be the subject of the next post.