The purpose of this post (like the previous one) is to go through some of the basic properties of the category ${\mathcal{U}}$ of unstable modules over the (mod ${2}$) Steenrod algebra. An analysis of ${\mathcal{U}}$ will ultimately lead to the proof of the Sullivan conjecture. Most of this material, again, is from Schwartz’s Unstable modules over the Steenrod algebra and Sullivan’s fixed point set conjecture; another useful source is Lurie’s notes.

1. The modules ${F(n)}$

In the previous post, we showed that the category ${\mathcal{U}}$ had enough projectives. More specifically, we constructed — using the adjoint functor theorem — an object ${F(n)}$, for each ${n}$, which we called the free unstable module on a class of degree ${n}$.The object ${F(n)}$ had the universal property

$\displaystyle \hom_{\mathcal{U}}(F(n), M) \simeq M_n,\quad M \in \mathcal{U}.$

To start with, we’d like to have a more explicit description of the module ${F(n)}$.

To do this, we need a little terminology. A sequence of positive integers

$\displaystyle i_k, i_{k-1}, \dots, i_1$

$\displaystyle i_j \geq 2 i_{j-1}$

for each ${j}$. It is a basic fact, which can be proved by manipulating the Adem relations, that the squares

$\displaystyle \mathrm{Sq}^I \stackrel{\mathrm{def}}{=} \mathrm{Sq}^{i_k} \mathrm{Sq}^{i_{k-1}} \dots \mathrm{Sq}^{i_1}, \quad I = (i_k, \dots, i_1) \ \text{admissible}$

form a spanning set for ${\mathcal{A}}$ as ${I}$ ranges over the admissible sequences. In fact, by looking at the representation on various cohomology rings, one can prove:

Proposition 29 The ${\mathrm{Sq}^I}$ for ${I }$ admissible form a basis for the Steenrod algebra ${\mathcal{A}}$.

The free unstable module ${F(n)}$ has a canonical class ${\iota_n \in F(n)_n}$ (and no classes in lower dimensions). However, it is not true that ${F(n)}$ is free as an ${\mathcal{A}}$-module. For instance, ${\mathrm{Sq}^{n+1} \iota_n = 0}$ by the instability condition. We need a new definition:

Definition 30 The excess ${e(I)}$ of an admissible sequence ${I = (i_k, \dots, i_1)}$ is defined to be

$\displaystyle (i_k - 2i_{k-1}) + (i_{k-1} - 2i_{k-1}) + \dots + i_1 = i_k - (i_{k-1} + \dots + i_1).$

If ${I}$ is an admissible sequence of excess ${e(I) > n}$, then the instability condition forces ${\mathrm{Sq}^I \iota_n = 0}$. However, we have:

Proposition 31 The ${\mathrm{Sq}^I \iota_n}$ for ${I}$ admissible and ${e(I) \leq n}$ form a basis for ${F(n)}$. In particular, a ${\mathbb{F}_2}$ basis for the kernel of

$\displaystyle \mathcal{A} \rightarrow F(n), \quad 1 \mapsto \iota_n$

is given by ${\left\{\mathrm{Sq}^I, e(I) > n\right\}}$.

Proof: We already know that ${\mathrm{Sq}^I \iota_n, e(I) > n}$ all vanish. Therefore, to prove the result, we need to show that the ${\mathbb{F}_2}$-vector space spanned by the ${\mathrm{Sq}^I, e(I) > n}$ is in fact an ${\mathcal{A}}$-submodule ${J}$ of ${\mathcal{A}}$ (that is, a left ideal). In fact, then we will get a map of ${\mathcal{A}}$-modules

$\displaystyle \Sigma^n\mathcal{A}/J \rightarrow F(n), \quad 1 \mapsto \iota_n,$

where ${\Sigma^n}$ denotes an ${n}$-fold shift in the grading. Using the basis ${\left\{\mathrm{Sq}^I, e(I) \leq n\right\}}$ for ${\Sigma^n\mathcal{A}/J}$, it follows that ${\Sigma^n\mathcal{A}/J}$ is actually an unstable ${\mathcal{A}}$-module. The map ${\Sigma^n \mathcal{A}/J \rightarrow F(n)}$ thus has a section (sending ${\iota_n \mapsto 1}$), but since ${\Sigma^n \mathcal{A}/J}$ is indecomposable, we see that our maps are isomorphisms.

So let’s prove the claim: given an admissible sequence ${I = (i_k, \dots, i_1)}$ of excess ${e(I) >n}$ and ${j \geq 0}$, we’d like to show that

$\displaystyle \mathrm{Sq}^j \mathrm{Sq}^I$

lies in the ${\mathbb{F}_2}$-vector space spanned by the ${\mathrm{Sq}^{I'}}$ for ${I'}$ admissible and ${e(I')>n}$.

Now, if ${j \geq 2i_k}$, the sequence ${(j, i_k, \dots, i_1)}$ is already admissible of excess ${>n}$ and there is nothing to prove. If ${j < 2i_k}$, we use the Adem relations to write

$\displaystyle \mathrm{Sq}^j \mathrm{Sq}^{I} = \sum_{l \leq j/2} c_{l,j,i_k} \mathrm{Sq}^{j+i_k - l} \mathrm{Sq}^l \mathrm{Sq}^{i_{k-1}} \dots \mathrm{Sq}^{i_1},$

for the ${c_{l, j, i_k}}$ appropriate constants. The terms ${\mathrm{Sq}^l \mathrm{Sq}^{i_{k-1}} \dots \mathrm{Sq}^{i_1}}$ can be expressed as a sum of terms ${\mathrm{Sq}^J}$ with ${J}$ admissible of total degree ${l + i_{k-1} + \dots + i_1}$. For ${l \leq j/2}$, we have

$\displaystyle j + i_k - l \geq l + i_{k-1} + \dots + i_1$

because the original sequence ${I}$ was admissible. In particular, for these terms ${\mathrm{Sq}^J}$ (for fixed ${l}$), ${\mathrm{Sq}^{j+i_k-l} \mathrm{Sq}^J}$ is admissible, and in fact is of excess at least ${e(I) > n}$. This proves the claim and the proposition. $\Box$

Example 3 ${F(0)}$ is ${\mathbb{F}_2}$ concentrated in degree zero.

In the next section, we’ll see consider the (more interesting) example of ${F(1)}$.

2. ${F(1)}$

Let’s try to make this analysis even more concrete and connect it to topology. In order to do this, let’s consider a simple example of an unstable ${\mathcal{A}}$-module: the reduced cohomology ${\widetilde{H}^*(\mathbb{RP}^\infty; \mathbb{F}_2)}$. We will show below that this contains ${F(1)}$ and use tensor powers of it to get another description of ${F(n)}$.

Now we can write:

$\displaystyle \widetilde{H}^*(\mathbb{RP}^\infty; \mathbb{F}_2) = \mathbb{F}_2 \left\{t^n, n \geq 1\right\}.$

This contains the class ${t}$ of degree ${1}$, which determines a map

$\displaystyle F(1) \rightarrow \widetilde{H}^*(\mathbb{RP}^\infty; \mathbb{F}_2).$

We’ll show that this is injective and determine the image.

The sequences of excess ${\leq 1}$ are given by ${\mathrm{Sq}^1}$, ${\mathrm{Sq}^2 \mathrm{Sq}^1, \mathrm{Sq}^4 \mathrm{Sq}^2 \mathrm{Sq}^1}$, and so forth. Using the multiplicative properties of the Steenrod operations, these carry ${t}$ to the linearly independent classes

$\displaystyle t^2, t^4, t^8, \dots.$

We conclude that

$\displaystyle F(1) = \mathbb{F}_2\left\{t, t^2, t^4, \dots\right\} \subset \widetilde{H}^*( \mathbb{RP}^\infty; \mathbb{F}_2).$

Let’s use this to get another description of ${F(n)}$. Now consider the cohomology ${H^*( (\mathbb{RP}^\infty)^n; \mathbb{F}_2) = \mathbb{F}_2[t_1, \dots, t_n] = (F(1) \oplus \mathbb{F}_2)^{\otimes n}}$. We will look at the action of the Steenrod squares on the class ${t_1 \dots t_n}$ (which determines a map ${F(n) \rightarrow H^*( (\mathbb{RP}^\infty)^n; \mathbb{F}_2)}$). Another way of saying this is that ${F(1)^{\otimes n}}$ contains a class ${t_1 \dots t_n}$ in degree ${n}$ (in fact, this is the unique nonzero class in degree ${n}$), which determines a map

$\displaystyle F(n) \rightarrow F(1)^{\otimes n},$

although we have not yet spelled out the symmetric monoidal structure on ${\mathcal{U}}$.

Proposition 32 We have an equivalence

$\displaystyle F(n) \simeq F(1)^{\Sigma_n}.$

More explicitly, ${F(n)}$ imbeds inside ${H^*(( \mathbb{RP}^\infty)^n; \mathbb{F}_2) = \mathbb{F}_2[t_1, \dots, t_n]}$ as the subgroup of symmetric polynomials such that every monomial only contains powers of two in the exponents.

Indeed, the map

$\displaystyle F(n) \rightarrow H^*( (\mathbb{RP}^\infty)^n; \mathbb{F}_2) = \mathbb{F}_2[t_1, \dots, t_n] , \quad \iota_n \mapsto t_1 \dots t_n$

sends

$\displaystyle \mathrm{Sq}^I \iota_n \mapsto \mathrm{Sq}^I (t_1 \dots t_n).$

This is clearly going to be a symmetric polynomial in the ${t_i}$‘s. Using the Cartan formula repeatedly and the known action of the squares on ${t_i}$, we also see that each of the monomials in ${\mathrm{Sq}^I (t_1 \dots t_n)}$ have only powers of ${2}$ in the exponents.

One can now prove that the map is a bijection by explicitly playing with the combinatorics of the two sides. It’s also possible to give a more homological proof, which we’ll do later (following Schwartz’s book).