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 29 The
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 30 The excess
of an admissible sequence
is defined to be
If is an admissible sequence of excess
, then the instability condition forces
. However, we have:
Proposition 31 The
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 3
is
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 32 We 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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