The purpose of this post (like the previous one) is to go through some of the basic properties of the category of unstable modules over the (mod ) Steenrod algebra. An analysis of 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 **

In the previous post, we showed that the category had enough projectives. More specifically, we constructed — using the adjoint functor theorem — an object , for each , which we called the **free unstable module on a class of degree .**The object had the universal property

To start with, we’d like to have a more explicit description of the module .

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

is called **admissible** if

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

form a spanning set for as ranges over the admissible sequences. In fact, by looking at the representation on various cohomology rings, one can prove:

Proposition 29The for admissible form a basis for the Steenrod algebra .

The free unstable module has a canonical class (and no classes in lower dimensions). However, it is not true that is free as an -module. For instance, by the instability condition. We need a new definition:

Definition 30Theexcessof an admissible sequence is defined to be

If is an admissible sequence of excess , then the instability condition forces . However, we have:

Proposition 31The for admissible and form a basis for . In particular, a basis for the kernel of

is given by .

*Proof:* We already know that all vanish. Therefore, to prove the result, we need to show that the -vector space spanned by the is in fact an -submodule of (that is, a left ideal). In fact, then we will get a map of -modules

where denotes an -fold shift in the grading. Using the basis for , it follows that is actually an unstable -module. The map thus has a *section* (sending ), but since is indecomposable, we see that our maps are isomorphisms.

So let’s prove the claim: given an admissible sequence of excess and , we’d like to show that

lies in the -vector space spanned by the for admissible and .

Now, if , the sequence is already admissible of excess and there is nothing to prove. If , we use the Adem relations to write

for the appropriate constants. The terms can be expressed as a sum of terms with admissible of total degree . For , we have

because the original sequence was admissible. In particular, for these terms (for fixed ), is admissible, and in fact is of excess at least . This proves the claim and the proposition.

Example 3is concentrated in degree zero.

In the next section, we’ll see consider the (more interesting) example of .

**2. **

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 -module: the reduced cohomology . We will show below that this contains and use tensor powers of it to get another description of .

Now we can write:

This contains the class of degree , which determines a map

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

The sequences of excess are given by , , and so forth. Using the multiplicative properties of the Steenrod operations, these carry to the linearly independent classes

We conclude that

Let’s use this to get another description of . Now consider the cohomology . We will look at the action of the Steenrod squares on the class (which determines a map ). Another way of saying this is that contains a class in degree (in fact, this is the unique nonzero class in degree ), which determines a map

although we have not yet spelled out the symmetric monoidal structure on .

Proposition 32We have an equivalence

More explicitly, imbeds inside as the subgroup of symmetric polynomials such that every monomial only contains powers of two in the exponents.

Indeed, the map

sends

This is clearly going to be a symmetric polynomial in the ‘s. Using the Cartan formula repeatedly and the known action of the squares on , we also see that each of the monomials in have only powers of 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).

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