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

is called admissible if

\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


\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).