Halloween 4: The Return of Adams Spectral Sequence 🎃

Happy Halloween! We're first going to define the Steenrod algebra $\mathcal{A}$ which by the end of this post will be the best friend we never knew we always wanted.

The Steenrod algebra, denoted $\mathcal{A},$ is the graded, associative, non-commutative $\mathbb{Z}_2$-algebra generated by the Steenrod squares $\{Sq^i\}_{i\geq 0}$ subject to the Adem relations:

where $a,b$ are integers such that $0 < a < 2b,$ and where $Sq^0 = 1$ is the identity [1]. The grading of $\mathcal{A}$ is given by $\deg(Sq^i) = i,$ and $\deg(Sq^a \circ Sq^b) = a+b.$ While $\{Sq^i\}_{i\geq 0}$ generates $\mathcal{A},$ it does not form a basis. For that, we'll need the concept of admissible monomials.

A monomial $Sq^{i_n} \circ \cdots \circ Sq^{i_1}$ is called admissible iff $i_k \geq 2 i_{k-1}$ for all $k.$ For example, $Sq^2 \circ Sq^2$ is not an admissible monomial since $2 \ngeq 4.$ However, we can use the Adem relations to get that

where $Sq^3 \circ Sq^1$ is admissible since $3 \geq 2\cdot 1.$

Theorem: The set of all admissible monomials forms a vector space basis for the $\mathcal{A},$ i.e. the Steenrod algebra [2]. This means that any composition of Steenrod squares in $\mathcal{A}$ can be written uniquely as a linear combination of admissible monomials.

The main goal this post is to introduce the Adams spectral sequence and construct the $E_2$ page for the sphere spectrum $\mathbb{S},$ which is the bigraded algebra:

While the Serre spectral sequence was concerned about the (co)homology of fibrations, the Adams spectral sequence is a machine for computing stable homotopy groups of spheres, i.e. $\pi_*^S.$ These guys are way spookier than homology or even cohomology 👻. Note that the $E_2$ page for $\mathbb{S}$ is:

and while we won't prove this here, it turns out that $H^*(\mathbb{S};\mathbb{Z}_2) \cong \mathbb{Z}_2,$ hence why we'll be doing a deep-dive into $\text{Ext}_{\mathcal{A}}^{s, t}(\mathbb{Z}_2, \mathbb{Z}_2).$

$\text{Ext}$: What is $\text{Ext}$? $\text{Ext}$ is short for extension and is a regular in the stage of homological algebra. Consider the following short exact sequence (SES) of modules:

Let's apply the (contravariant) functor $\text{Hom}_{\mathcal{A}}(-, \mathbb{Z}_2)$ to get:

Suppose our modules are left $\mathcal{A}$-modules, then this new sequence will be left-exact which just means that the induced map $\text{Hom}_{\mathcal{A}}(C, \mathbb{Z}_2) \xrightarrow{\,\, \,\,} \text{Hom}_{\mathcal{A}}(B, \mathbb{Z}_2)$ is injective. Now, the right map $\text{Hom}_{\mathcal{A}}(B, \mathbb{Z}_2) \xrightarrow{\,\, \,\,} \text{Hom}_{\mathcal{A}}(A, \mathbb{Z}_2)$ is not generally surjective, so the sequence as a whole is not (in general) exact. The $\text{Ext}$ groups are the algebraic objects that measure this "failure of exactness."

$\text{Ext}_{\mathcal{A}}^{s}(M, N)$: Let's see how we construct these extensions. We'll start by looking at a general $\text{Ext}_{\mathcal{A}}^s(M,N)$ case which is equivalent to our case study, just with $M=\mathbb{Z}_2$ and $N=\mathbb{Z}_2.$ We construct $\text{Ext}_{\mathcal{A}}^s(M,N)$ in three simple steps:

1. Resolve: We first look at the projective resolution of $M.$ This is an exact sequence of projective modules (i.e. projective resolution). These are modules that are "nice" and behave like free modules. This LES "approximates" $M$:

   $$\cdots \xrightarrow{\,\, \partial_2 \,\,} P_1 \xrightarrow{\,\, \partial_1 \,\,} P_0 \xrightarrow{\,\, \epsilon \,\,} M \xrightarrow{\,\, \,\,} 0,$$

   where each $P_s$ is a projective $\mathcal{A}$-module and $\epsilon$ is a surjective map.

2. Apply Hom: We apply the $\text{Hom}_{\mathcal{A}}(-,N)$ functor to the sequence after first removing the $M$ object from the sequence to get:

   $$0 \xrightarrow{\,\, \,\,} \text{Hom}_\mathcal{A}(P_0, N) \xrightarrow{\,\, \partial_1^* \,\,} \text{Hom}_\mathcal{A}(P_1, N) \xrightarrow{\,\, \partial_2^* \,\,} \cdots$$

   This new sequence is not guaranteed to be exact, but we do have the property $\partial_{s+1}^* \circ \partial_{s}^* = 0,$ like that of cochain complexes in singular cohomology.

3. Take the Cohomology: There's not much else to it. The $\text{Ext}$ groups are defined as the cohomology of this complex:

   $$\text{Ext}_\mathcal{A}^s(M,N) := \frac{\ker(\partial_{s+1}^*)}{\text{im}(\partial_{s}^*)},$$

   where $\ker(\partial_{s+1}^*) / \text{im}(\partial_{s}^*)$ is the $s$-th cohomology group $H^s(\text{Hom}_\mathcal{A}(P_*, N)).$

Unfortunately, for the Steenrod algebra $\mathcal{A},$ finding a simple projective resolution of $\mathbb{Z}_2$ is extremely difficult. Instead, we'll be using a different resolution that works for any algebra: the bar resolution. For this, we'll be using the normalized bar resolution which is defined using the augmentation ideal of $\mathcal{A}.$

Let $\bar{\mathcal{A}}:= \ker(\epsilon: \mathcal{A}\to\mathbb{Z}_2)$ denote the augmentation ideal. This is the vector space spanned by all the Steenrod squares of positive degree, i.e. $\{Sq^i\}_{i>0}.$ Importantly, note that $Sq^0 = 1 \notin \bar{\mathcal{A}}.$ The bar resolution for $\mathbb{Z}_2$ is given by

where the $s$-th term is:

For example, we have: $P_0 = \mathcal{A},$ $P_1 = \mathcal{A} \otimes \bar{\mathcal{A}},$ $P_2 = \mathcal{A} \otimes \bar{\mathcal{A}} \otimes \bar{\mathcal{A}},$ and so on [4]. The differential

makes use of the following notation $[a_1 \mid a_2 \mid \cdots \mid a_s]$ for elements in $(\bar{\mathcal{A}})^{\otimes s}.$ Hence, we denote elements in $P_s = \mathcal{A} \otimes (\bar{\mathcal{A}})^{\otimes s}$ as $a \otimes [a_1 \mid \cdots \mid a_s].$ The differential maps these elements to $\partial_{s-1}$ the following way:

Note that the $(-1)^i$ term is just $1$ in our case since $\mathcal{A}$ is over $\mathbb{Z}_2.$ Additionally, note that the first term has the $\mathcal{A}$-action with $a \cdot a_1.$ In the sum, we just multiply all adjacent $a_{i}a_{i+1}$ terms inside the $(\bar{\mathcal{A}})^{\otimes}$ bracket. Example:

Note that here we use + to show that $-1 \equiv 1$ (mod 2). Side note: the summation for $\partial_s$ usually includes a term

but since $a_s \in \bar{\mathcal{A}}$ then $\epsilon(a_s) = 0$, so this term is just zero in our case.

Before we proceed, let's briefly look at the following map called the augmentation map:

$\epsilon$ is a homomorphism that strips away all the complexity in $\mathcal{A}.$ Note that for any element $a' \in \bar{\mathcal{A}} \subset \mathcal{A},$ we get $\epsilon(a') = 0$ (since we defined $\bar{\mathcal{A}}$ to be the kernel of this $\epsilon$ map). Additionally, note that $\epsilon(1) = 1$ since it's a homomorphism. We define the trivial action of $\mathcal{A}$ on $\mathbb{Z}_2$ using this $\epsilon$ map , where we have the trivial action $a \cdot k$ for any $a \in \mathcal{A},$ $k \in \mathbb{Z}_2$ such that:

Example 1: $Sq^1 \cdot 1 = \epsilon(Sq^1)1 = 0,$ since $\epsilon(Sq^1) = 0.$

Example 2:

An $\mathcal{A}$-module map $\phi: P_s \to \mathbb{Z}_2$ is a map

that satisfies the trivial action we defined above, i.e.

since $\phi(x) \in \mathbb{Z}_2,$ hence why $a \in \mathcal{A}$ acts trivially on it via the augmentation map $\epsilon.$ Note that $\epsilon(a) = 0,$ as well as $\phi(a \cdot x) = 0,$ for anything not starting with 1. This means that $\phi$ is completely determined by what it does to elements of the form

The next thing we do is apply $\text{Hom}_{\mathcal{A}}(-, \mathbb{Z}_2)$ to our bar complex, so we'll have the groups $\text{Hom}_\mathcal{A}(P_s, \mathbb{Z}_2).$ However, note that these map homomorphisms are the same as our $\phi,$ so we get the following isomorphism of vector spaces:

where we get this $\text{Hom}_{\mathbb{Z}_2}(-)$ simplification since we're now dealing with just $\bar{\mathcal{A}}$ and we don't have to worry about the $\mathcal{A}$-module structure from $\mathcal{A}$ since we've basically factored it out. Furthermore, since $\bar{\mathcal{A}}$ is just a $\mathbb{Z}_2$ vector space, then note that this $\text{Hom}_{\mathbb{Z}_2}( (\bar{\mathcal{A}})^{\otimes s}, \mathbb{Z}_2 )$ is precisely the definition of the dual of $(\bar{\mathcal{A}})^{\otimes s}.$ We'll denote this dual as $(\bar{\mathcal{A}}^{\otimes s})^*$ and will denote most other duals with $V^*$-notation, however, we'll be making a special exception for the Steenrod algebra itself which is denoted $\mathcal{A}_*$ and we'll explain why later (spoiler alert: it's because $\mathcal{A}$ is a Hopf algebra). Altogether, we have the following isomorphism:

We also get the following induced differentials $\partial^*_s$ which map from $\text{Hom}_\mathcal{A}(P_{s-1}, \mathbb{Z}_2)$ to $\text{Hom}_\mathcal{A}(P_{s}, \mathbb{Z}_2)$ (reversed due to contravariant functor), i.e.

We denote this entire dual bar complex object as $\left( (\bar{\mathcal{A}}^{\otimes s})^*, \partial_s^* \right) .$ Before in our general $\text{Hom}_\mathcal{A}(-,N)$ case we took the cohomology of the complex $\left( \text{Hom}_\mathcal{A}(P_*, N), \partial_s^* \right),$ but now we'll be taking the cohomology of $\left( (\bar{\mathcal{A}}^{\otimes s})^*, \partial_s^* \right)$ to get $\text{Ext}_\mathcal{A}(\mathbb{Z}_2, \mathbb{Z}_2)$ [5].

Let's compute $\text{Ext}_\mathcal{A}^1(\mathbb{Z}_2, \mathbb{Z}_2).$ Looking at the cohomology of $\left( (\bar{\mathcal{A}}^{\otimes s})^*, \partial_s^* \right),$ we get:

Let's first look at $\partial_1^*: (\bar{\mathcal{A}}^{\otimes 0})^* \to (\bar{\mathcal{A}}^{\otimes 1})^*.$ Note that this dual $(\bar{\mathcal{A}}^{\otimes 0})^* \cong \text{Hom}_{\mathbb{Z}_2}( \bar{\mathcal{A}}^{\otimes 0}, \mathbb{Z}_2 ),$ and since $\bar{\mathcal{A}}$ is a $\mathbb{Z}_2$ vector space then $\bar{\mathcal{A}}^{\otimes 0} \cong \mathbb{Z}_2,$ so we get:

so we just get the following differential:

This is dual to $\partial_1: \mathcal{A} \otimes \bar{\mathcal{A}} \to \mathcal{A}$ such that $\partial_1(a \otimes [a_1]) = a\cdot a_1.$ Let $\phi \in \text{Hom}_\mathcal{A}(P_0, \mathbb{Z}_2) \cong \mathbb{Z}_2,$ where this space is spanned by the augmentation map $\epsilon: \mathcal{A} \to \mathbb{Z}_2.$ Consider the pull-back $\partial_1^*(\phi) = \phi \circ \partial_1$ where we use inputs from $\mathcal{A} \otimes \bar{\mathcal{A}}.$ Let $a \otimes [a_1] \in \mathcal{A} \otimes \bar{\mathcal{A}},$ then consider:

Note that since $\phi \in \mathbb{Z}_2$ then $\phi(a_1) = 0$ if $\phi$ is the zero map, so suppose $\phi = \epsilon.$ Note that $\phi(a_1) = \epsilon(a_1) = 0$ for all $a_1$ in $\bar{\mathcal{A}}.$ This means that $\epsilon(a)\epsilon(a_1) = 0,$ hence

for any $a \otimes [a_1] \in \mathcal{A} \otimes \bar{\mathcal{A}}.$ This means that $\partial_1^*$ is the zero map, and hence:

Therefore, we get a major simplification for $\text{Ext}^1$ where we have:

Let's look at this differential $\partial_2^*: \bar{\mathcal{A}}^* \to (\bar{\mathcal{A}} \otimes \bar{\mathcal{A}})^*.$ Note that $\bar{\mathcal{A}}^* \cong \text{Hom}_{\mathbb{Z}_2}(\bar{\mathcal{A}}, \mathbb{Z}_2)$ and $\ker(\partial_2^*) \subset \bar{\mathcal{A}}^*.$ This means that for $\ker(\partial_2^*),$ we're looking for all the $\bar{\mathcal{A}} \to \mathbb{Z}_2$ map homomorphisms, i.e. let $f \in \text{Hom}_{\mathbb{Z}_2}(\bar{\mathcal{A}}, \mathbb{Z}_2),$ such that $\partial_2^*(f) = 0$ in $(\bar{\mathcal{A}} \otimes \bar{\mathcal{A}})^*.$ Consider the pull-back:

where $\partial_2^*$ is the dual of $\partial_2$ such that $\partial_2: \mathcal{A} \otimes \bar{\mathcal{A}} \otimes \bar{\mathcal{A}} \to \mathcal{A} \otimes \bar{\mathcal{A}}.$

Let $a \otimes [a_1 \mid a_2] \in \mathcal{A} \otimes \bar{\mathcal{A}} \otimes \bar{\mathcal{A}}$, and note that

Then when we apply $f$ we get

Note that this term $f(a\cdot a_1 \otimes [a_2]) = 0$ since we have the term $\epsilon(a\cdot a_1) = \epsilon(a)\epsilon(a_1) = 0.$ For the second term we get:

Which means that $f \circ \partial_2 = 0$, hence we want $f(1 \otimes [a_1 a_2]) = 0.$ This is equivalent to looking for maps $f: \bar{\mathcal{A}} \to \mathbb{Z}_2$ such that $f([a_1a_2]) = 0$ for all $a_1,a_2 \in \bar{\mathcal{A}}$ since we can just factor out the $(1 \otimes -)$ term. Additionally, another way to see this is that we're looking for maps $f$ such that for any elements $[a] \in \bar{\mathcal{A}}$ that can be written as $a = a_1a_2$ for $a_1,a_2 \in \bar{\mathcal{A}}$ we get $f([a]) = 0.$

Let's take a tiny step back and then tie this back to $f([a]) = 0$ for $a = a_1a_2.$ Elements in the Steenrod algebra $\mathcal{A}$ are either decomposable or indecomposable. A simple example is $Sq^3$ which can be rewritten as $Sq^3 = Sq^1 \circ Sq^2$ via the Adem relations. Since we can write $Sq^3 = Sq^1 \circ Sq^2,$ this means that $Sq^3$ is decomposable [6]. Examples of indecomposable elements are $Sq^1$, $Sq^2$, and $Sq^4.$ Spoiler alert: it turns out that indecomposable elements are of the form $Sq^{2^i}$ which becomes relevant very soon. Recall that $\bar{\mathcal{A}}$ is spanned by $\{ Sq^i \}_{i > 0},$ so $Sq^3 \in \bar{\mathcal{A}}$ such that we can write $Sq^3 = a_1a_2$ where $a_1 = Sq^1$ and $a_2 = Sq^2.$ Note that this means that $Sq^3$ is such that $f([Sq^3]) = 0$ for $f \in \ker(\partial_2^*).$

So we want our maps $f$ to vanish for any decomposable element in $\bar{\mathcal{A}}.$ Note that when we write these decomposable elements like $a_1a_2,$ these are elements in $\bar{\mathcal{A}}\cdot\bar{\mathcal{A}} \subset \bar{\mathcal{A}}.$ This means that $\ker(\partial_2^*)$ is the set of all $\bar{\mathcal{A}} \to \mathbb{Z}_2$ maps that are zero at $\bar{\mathcal{A}}\cdot\bar{\mathcal{A}}.$ Note the following equivalence:

And then note that the set of maps from $\bar{\mathcal{A}} / (\bar{\mathcal{A}} \cdot \bar{\mathcal{A}}) \to \mathbb{Z}_2$ is just the dual:

Putting everything together, we get that

Importantly, note that since we're modding out by all the decomposable elements in $\bar{\mathcal{A}}$ then we're left with the indecomposable elements.

Theorem: The vector space of indecomposable elements of $\mathcal{A},$ i.e. $\bar{\mathcal{A}} / (\bar{\mathcal{A}} \cdot \bar{\mathcal{A}}) \subset \mathcal{A},$ denoted as $Q(\mathcal{A}),$ has a $\mathbb{Z}_2$-basis given by the elements $\{ Sq^{2^i}\}_{i\geq 0}.$ Note that $\text{Ext}^1_{\mathcal{A}}(\mathbb{Z}_2, \mathbb{Z}_2)$ is the dual of $Q(\mathcal{A})$ which means that $\text{Ext}^1$ must have a dual basis, and this is dual to the indecomposable elements in $\bar{\mathcal{A}} / (\bar{\mathcal{A}} \cdot \bar{\mathcal{A}}).$

The dual basis of $\text{Ext}^1$ is the set of generators we call $h_i$:

• The class dual to the first indecomposable element, which is $Sq^1 = Sq^{2^0},$ is $h_0.$ Its bidegree $(s,t)$ is $(1, \deg(Sq^1)) = (1, 1).$
• The class dual to $Sq^2 = Sq^{2^1}$ is $h_1$ with bidegree $(1, \deg(Sq^2)) = (1, 2).$
• In general, the class dual to $Sq^{2^i}$ is $h_i$ with bidegree $(1, 2^i).$

With this, we've constructed the ${s=1}$ column of the Adams $E_2$ page. It's just an infinite column of generators $h_i,$ each corresponding to the indecomposable elements $Sq^{2^i}.$

Next, we'll be building the ${s=2}$ column of $\text{Ext}_{\mathcal{A}}^2(\mathbb{Z}_2, \mathbb{Z}_2)$ using the $h_i$ generators from the ${s=1}$ column. These are of the form $h_{i}h_{j}$ such that:

• $h_0^2 = h_0h_0$ has bidegree $(s, t)$ such that $(1, 1) + (1, 1) = (2, 2).$
• $h_0h_1$ has bidegree $(1,1) + (1, 2) = (2, 3).$
• $h_1^2 = h_1h_1$ has bidgree $(1, 2) + (1, 2) = (2, 4).$

Warning: For $h_0^2 = h_0h_0,$ one may be tempted to look at its corresponding (dual) $Sq^1 \circ Sq^1$ which is just zero due to the Adem relations in $\mathcal{A},$ and then incorrectly conclude that $Sq^1 \circ Sq^1 = 0$ implies that $h_0^2 = 0$ in $\text{Ext}^2.$ This turns out to not be the case. Since we're looking at these $h_ih_j$ elements in

then we'll have to deal with lots of messy Adem relations. This would be a nightmare. Is there something else we can do? We'll be switching to a different machine that will at first look more spooky 👻, but turns out to make life in $\text{Ext}^2$ a lot easier. Since $\mathcal{A}$ is a Hopf algbra, then we can use the following theorem:

where the cohomology of the bar complex of the Steenrod algebra $\mathcal{A}$ is isomorphic to the cohomology of the cobar complex of the dual for $\mathcal{A}$ which we denote as $\mathcal{A}_*$ [7, 8]. This is the bar-cobar isomorphism. So far we've been using the dual of the reduced bar complex:

where we have the differentials $\partial_s^*: (\bar{\mathcal{A}}^{\otimes s-1})^* \to (\bar{\mathcal{A}}^{\otimes s})^*.$ When we switch to this new cobar complex, we get the following:

where we have these new differentials $d^s: (\bar{\mathcal{A}}^*)^{\otimes s} \to (\bar{\mathcal{A}}^*)^{\otimes s+1}.$ Importantly, note that these are not isomorphic when $s > 1$:

So we build this new cobar complex from the dual algebra $\mathcal{A}_*$ which is a graded-commutative polynomial algebra which we denote:

where these generators $\xi_k$ have internal degree $\deg(\xi_k) = 2^k-1.$ Similar to the $\mathcal{A}$ case, we look at the augmentation ideal $\bar{\mathcal{A}}^*$ of $\mathcal{A}_*$ to get $C^s = (\bar{\mathcal{A}}^*)^{\otimes s}$ where our differentials are maps $d^s: C^{s} \to C^{s+1}.$ Furthermore, note that $s$ in this case is the same $s$-degree in $\text{Ext}^{s,*},$ and we also denote the degree $t$ with a subscript $C^s_t.$ Let's use this new cobar framework to view our previous results for $\text{Ext}^1$ again:

• $h_0 \in \text{Ext}^{1, 1}$ is represented by the cocycle $\xi_1 \in C_1^1$ since $\deg(\xi_1) = 2^1-1 = 1.$
• $h_1 \in \text{Ext}^{1, 2}$ is represented by the cocycle $\xi_1^2 \in C_2^1$ since $\deg(\xi_1^2) = \deg(\xi_1) + \deg(\xi_1) = 2.$
• $h_2 \in \text{Ext}^{1, 4}$ is represented by the cocycle $\xi_1^4 \in C_4^1$ since $\deg(\xi_1^4) = 4.$

In general, $h_i \in \text{Ext}^{1, 2^i}$ is represented by the cocycle $\xi_1^{2^i} \in C_{2^i}^1$ with such that $\deg(\xi_1^{2^i}) = 2^i.$ Note that this degree is different from the degree of $\xi_k$ as a generator which is $2^k-1.$

We take the cohomology of this cobar complex $\left( (\bar{\mathcal{A}}^*)^{\otimes s} , d^s \right) = (C^s, d^s)$ to get $\text{Ext}:$

$h_0^2$: We're finally ready to compute $h_0^2.$ Recall that $h_0^2 = h_0h_0$ has bidegree ${(s, t) = (2, 2)},$ so we're computing $\text{Ext}_{\mathcal{A}}^{2, 2} \cong \ker(d^2: C_2^{2} \to C_2^{3} ) / \text{im}(d^1: C_2^{1} \to C_2^{2} ).$ Furthermore, $h_0$ is represented by the $1$-cocycle $\xi_1 \in C_1^1,$ so $h_0^2 = h_0h_0$ is represented by the $2$-cochain $\xi_1 \otimes \xi_1 \in C_2^2.$ Side note: we call general elements $x \in C^s$ cochains, and if $x \in \ker(d^s)$ then it's a cocycle, and if $x \in \text{im}(d^{s-1})$ then it's a coboundary. Additionally, as we do a deep dive into $\ker(d^s)$ and $\text{im}(d^{s-1}),$ note that the differential $d^s$ follows the Leibniz rule (see blog post #3) since it's induced by the cup product. Let's see this in action to see if $\xi_1 \otimes \xi_1 \in \ker(d^2),$ i.e. to check if $\xi_1 \otimes \xi_1$ is a cocycle:

Note that $(-1)^{\deg(\xi_k)} = 1$ for any $k$ since we're working in $\mathbb{Z}_2$ (mod 2). Note that our $d^1$ operator is the same as the reduced coproduct $\bar{\Delta}: \bar{\mathcal{A}}^* \to \bar{\mathcal{A}}^* \otimes \bar{\mathcal{A}}^*.$ This map is derived from the full coproduct $\Delta$ on the dual Steenrod algebra $\mathcal{A}_*$ by the relation:

where $\bar{\Delta}$ captures the non-trivial part of the coproduct by subtracting the trivial parts $x\otimes 1$ and $1 \otimes x,$ etc. With this, we get:

where $\bar{\Delta}(\xi_1) = 0$ since $\Delta(\xi_1) = \xi_1 \otimes 1 + 1\otimes \xi_1.$ Therefore, $\xi_1 \otimes \xi_1 \in \ker(d^2),$ so it's a cocycle. Next, let's see if it's a coboundary, i.e. if $\xi_1 \otimes \xi_1 \in \text{im}( d^1: C_2^1 \to C_2^2 ).$ Looking at the source space $C_2^1,$ note that this has degree 2, and we have the following available generators from $\bar{\mathcal{A}}^*$: $\xi_1, \xi_2, \ldots$ such that $\deg(\xi_1) = 1,$ $\deg(\xi_2) = 3,$ etc. There's no single degree 2 generator, so the only way to get a generator for $C_2^1$ is to consider $\xi_1^2$ with degree $\deg(\xi_1^2) = 2.$ We don't have anything else (since even considering $\xi_1\xi_2$ has degree 4), so our source space $C_2^1$ is spanned by the single element $\xi_1^2,$ i.e.

Let's see how $d^1 = \bar{\Delta}$ maps $\xi_1^2$ to $C_2^2.$ First, let's look at $\Delta(\xi_1^2):$

where we get $(\xi_1 \otimes 1 + 1 \otimes \xi_1)^2 = \xi_1^2 \otimes 1 + 1 \otimes \xi_1^2$ since we're working over $\mathbb{Z}_2$ (characteristic 2). Hence, when we look at $d^1,$ we get:

Since $C_2^1 = \text{span}\{\xi_1^2\}$ and $d^1(\xi_1^2) = 0,$ then this means that $\text{im}(d^1) = \{0\}.$ We conclude that $\xi_1 \otimes \xi_1$ is not a coboundary, and get:

Thus, $h_0^2$ is a non-zero generator of $\text{Ext}^{2,2}.$

$h_0h_1$: As an additional example, let's look at $h_0h_1.$ Similar to before, recall that $h_0h_1$ has bidegree $(s,t) = (2,3),$ so we're looking at

Let's look at $\ker(d^2).$ We have the source space $C_3^2$ with elements from $\bar{\mathcal{A}}^* \otimes \bar{\mathcal{A}}^*.$ Note that since $1 \notin \bar{\mathcal{A}}^*$ then we can't have elements like $1 \otimes \xi_2$ or $\xi_2 \otimes 1.$ Since any variation of $\xi_2\otimes \xi_1$ will have degree $\geq 4$ then we can only span $C_3^2$ using $\xi_1.$ We get:

where the degree of both $\xi_1^2 \otimes \xi_1$ and $\xi_1 \otimes \xi_1^2$ is 3. Let's see how $d^2$ maps these. Let's start with $\xi_1^2 \otimes \xi_1:$

where $d^1(\xi_1^2) = 0$ from our previous $h_0^2$ calculation, and $d^1(\xi_1) = 0$ since $\xi_1$ is primitive. Note that $d^2(\xi_1 \otimes \xi_1^2)$ is a similar story where we get $d^2(\xi_1 \otimes \xi_1^2) = 0.$ Since $d^2$ maps everything from the source space to zero, then we get:

Next, let's look at the coboundaries $\text{im}( d^1 ).$ In this case, since we're getting elements of $C_3^1$ from $\bar{\mathcal{A}}^*$ then we get the following:

Let's apply $d^1 = \bar{\Delta}$ to this basis. First, let's look at $\Delta$ for $\xi_1^3$ and then just remove the trivial terms $\xi_1^3 \otimes 1$ and $1 \otimes \xi_1^3.$ Note that we'll be using the (mod 2) expansion of $(a+b)^3 = a^3 + a^2b + ab^2 + b^3:$