This is the third in a series of posts on oriented cobordism. In the first post, we analyzed the spectrum at odd primes; in this post, we will analyze the prime
. After this, we’ll be able to deduce various classical geometric facts about manifolds.
The next goal is to determine the structure of the homology as a comodule over
. Alternatively, we can determine the structure of the cohomology
over the Steenrod algebra
: this is a coalgebra and a module.
Theorem 8 (Wall) As a graded
-module,
is a direct sum of shifts of copies of
and
.
This corresponds, in fact, to a splitting at the prime 2 of into a wedge of Eilenberg-MacLane spectra.
In fact, this will follow from the comodule structure theorem of the previous post once we can show that if is the Thom class, then the action of
on
has kernel
: that is, the only way a cohomology operation can annihilate
if it is a product of something with
. Alternatively, we have to show that the complementary Serre-Cartan monomials in
applied to
,
are linearly independent in .
This is going to require a bit of bookkeeping. (I would be curious if any readers of this blog have a better argument.) Note that is the polynomial ring in the Stiefel-Whitney classes
where the first Stiefel-Whitney class is zero by orientability. The action of the Steenrod squares on the
is given by the Wu formula
see for instance May’s Concise course, p. 197. Since if
in
, we find that the extreme cases of monomials that can occur are
and
. In fact,
does not occur (as
).
So the highest subscript of a that occurs in
is
. This is a useful fact which will be used below (and which can be seen directly from degree considerations, without use of the Wu formulas).
Proposition 9 As
ranges over sequences in
with
and
, the elements
are linearly independent.
By the Thom isomorphism, we find that is a free module of rank one over
(generated by the Thom class
). To prove this result, we will give an approximate expression for
. Recall that
(by Thom’s construction of the Stiefel-Whitney classes). Applying the Steenrod squares repeatedly makes a mess, though, so we define an order on
to sort through it.
Definition 10 An order on monomials
is defined as follows: first order the
in decreasing order, and then use the lexicographic order on sequences.
Then, the main tool is that, for a sequence with
and
, we have in
,
It now follows that these terms are linearly independent in cohomology as ranges through sequences satisfying the relevant conditions. So, if we can prove (8), then we will be done with the proposition (and thus with Wall’s theorem).
In fact, we can prove this by induction. Suppose . Then, we find that
It thus follows that in order to make the inductive step, we might as well prove:
Lemma 11 Let
be an element with
, but we do not assume the strong increasing condition as for the
. Suppose
. Then
Proof:
In fact, we use the Cartan formula to observe that
The terms in the sum, which can be expanded using the Cartan formula iteratively, are all lower order. We see this by noting that the worst the Steenrod squares can do (see the remarks about the Wu formula above) to increase the subscripts of the ‘s is to a
or add a
from the
, both of which are less than
. So the first term, which has a
, dominates all the others in the lexicographic ordering.
Anyway, from all this analyze we can conclude the following: the terms for the
ranging through “admissible sequences”
and with
, are linearly independent in
. It follows that the annihilator of the Thom class is precisely
, and we can now conclude Wall’s theorem by the comodule structure theorems.
This argument was a little tedious (at least, it took me a while to figure out), though the literature treats it as obvious, so I’m curious if there’s something simpler than this. One thing to note is that the exact same argument works to show that the Steenrod algebra acts freely on the Thom class in : as a result, we can conclude from the structure theory that
is free over the Steenrod algebra. We proved that earlier using different means.
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