This is the third in a series of posts on oriented cobordism. In the first post, we analyzed the spectrum $MSO$ at odd primes; in this post, we will analyze the prime $2$. 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 ${H_*(MSO; \mathbb{Z}/2)}$ as a comodule over ${\mathcal{A}_2^{\vee}}$. Alternatively, we can determine the structure of the cohomology ${H^*(MSO; \mathbb{Z}/2)}$ over the Steenrod algebra ${\mathcal{A}_2}$: this is a coalgebra and a module.

Theorem 8 (Wall) As a graded ${\mathcal{A}_2}$-module, ${H^*(MSO; \mathbb{Z}/2)}$ is a direct sum of shifts of copies of ${\mathcal{A}_2}$ and ${\mathcal{A}_2/\mathcal{A}_2\mathrm{Sq}^1}$.

This corresponds, in fact, to a splitting at the prime 2 of $MSO$ 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 ${t \in H^0(MSO; \mathbb{Z}/2)}$ is the Thom class, then the action of ${\mathcal{A}_2}$ on ${t}$ has kernel ${J = \mathcal{A}_2 \mathrm{Sq}^1}$: that is, the only way a cohomology operation can annihilate ${t}$ if it is a product of something with ${\mathrm{Sq}^1}$. Alternatively, we have to show that the complementary Serre-Cartan monomials in ${\mathcal{A}_2}$ applied to ${t}$,

$\displaystyle \mathrm{Sq}^{i_1} \mathrm{Sq}^{i_2} \dots \mathrm{Sq}^{i_n} t, \quad i_k \geq 2i_{k-1}, \quad i_n \neq 1,$

are linearly independent in ${H^*(MSO; \mathbb{Z}/2)}$.

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 ${H^*(BSO; \mathbb{Z}/2)}$ is the polynomial ring in the Stiefel-Whitney classes

$\displaystyle H^*(BSO ; \mathbb{Z}/2) \simeq \mathbb{Z}/2[w_2, w_3, \dots, ],$

where the first Stiefel-Whitney class ${w_1}$ is zero by orientability. The action of the Steenrod squares on the ${w_i}$ is given by the Wu formula

$\displaystyle \mathrm{Sq}^i w_j = \sum_{t=0}^i \binom{j+t -i - 1}{t} w_{i-t}w_{j+t};$

see for instance May’s Concise course, p. 197. Since ${\mathrm{Sq}^i w_j= 0}$ if ${i > j}$ in ${H^*(BSO; \mathbb{Z}/2)}$, we find that the extreme cases of monomials that can occur are ${w_{2j}}$ and ${w_i w_j}$. In fact, ${w_{2j}}$ does not occur (as ${\mathrm{Sq}^j w_j = w_j^2}$).

So the highest subscript of a ${w_k}$ that occurs in ${\mathrm{Sq}^i w_j}$ is ${2j-1}$. 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 ${(i_1, i_2, \dots, i_n)}$ ranges over sequences in ${\mathbb{Z}_{>0}}$ with ${i_n > 1}$ and ${i_k \geq 2i_{k-1}}$, the elements

$\displaystyle \mathrm{Sq}^{i_1} \dots \mathrm{Sq}^{i_n} t \in H^*(MSO; \mathbb{Z}/2)$

are linearly independent.

By the Thom isomorphism, we find that ${H^*(MSO; \mathbb{Z}/2)}$ is a free module of rank one over ${H^*(BSO; \mathbb{Z}/2)}$ (generated by the Thom class ${t}$). To prove this result, we will give an approximate expression for ${\mathrm{Sq}^{i_1} \dots \mathrm{Sq}^{i_n} t}$. Recall that ${\mathrm{Sq}^i t = w_i t}$ (by Thom’s construction of the Stiefel-Whitney classes). Applying the Steenrod squares repeatedly makes a mess, though, so we define an order on ${H^*(MSO;\mathbb{Z}/2)}$ to sort through it.

Definition 10 An order on monomials ${w_{i_1} \dots w_{i_k}t \in H^*(MSO; \mathbb{Z}/2)}$ is defined as follows: first order the ${i_1, \dots, i_k}$ in decreasing order, and then use the lexicographic order on sequences.

Then, the main tool is that, for a sequence ${(i_1, \dots, i_n)}$ with ${i_n > 1}$ and ${i_k \geq 2i_{k-1}}$, we have in ${H^*(MSO; \mathbb{Z}/2)}$,

$\displaystyle \mathrm{Sq}^{i_1} \dots \mathrm{Sq}^{i_n} t = w_{i_1} \dots w_{i_n} t + \text{lower order terms}.\ \ \ \ \ (8)$

It now follows that these terms are linearly independent in cohomology as ${(i_1, \dots, i_n)}$ 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 ${\mathrm{Sq}^{i_2} \dots \mathrm{Sq}^{i_n} t = w_{i_2} \dots w_{i_n} t + \text{lower terms}}$. Then, we find that

$\displaystyle \mathrm{Sq}^{i_1}\mathrm{Sq}^{i_2} \dots \mathrm{Sq}^{i_n} = \mathrm{Sq}^{i_1}(w_{i_2} \dots w_{i_n} t) + \mathrm{Sq}^{i_1}( \text{lower terms}).$

It thus follows that in order to make the inductive step, we might as well prove:

Lemma 11 Let ${w_{j_2} \dots w_{j_n}t \in H^*(MSO; \mathbb{Z}/2)}$ be an element with ${j_2 \geq \dots j_n}$, but we do not assume the strong increasing condition as for the ${i_k}$. Suppose ${i \geq 2j_2}$. Then

$\displaystyle \mathrm{Sq}^i ( w_{j_2} \dots w_{j_n}t) = w_i w_{j_2} \dots w_{j_n}t + \text{lower order terms}.$

Proof:

In fact, we use the Cartan formula to observe that

$\displaystyle \mathrm{Sq}^i ( w_{j_2} \dots w_{j_n}t) = \mathrm{Sq}^i t w_{j_2} \dots w_{j_n} + \sum_{0 < i' \leq i} \mathrm{Sq}^{i'}( w_{j_2} \dots w_{j_n} ) \mathrm{Sq}^{i-i'}t.$

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 ${w}$‘s is to a ${2j_2 - 1}$ or add a ${w_{i-i'}}$ from the ${\mathrm{Sq}^{i-i'}t}$, both of which are less than ${i}$. So the first term, which has a ${w_i}$, dominates all the others in the lexicographic ordering. $\Box$

Anyway, from all this analyze we can conclude the following: the terms ${\mathrm{Sq}^{i_1} \dots \mathrm{Sq}^{i_n}t}$ for the ${\left\{i_k\right\}}$ ranging through “admissible sequences” ${i_1 \geq 2i_2 \geq \dots}$ and with ${i_n \neq 1}$, are linearly independent in ${H^*(MSO; \mathbb{Z}/2)}$. It follows that the annihilator of the Thom class is precisely ${\mathcal{A}_2 \mathrm{Sq}^1}$, 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 $H^*(MO; \mathbb{Z}/2)$: as a result, we can conclude from the structure theory that $H^*(MO; \mathbb{Z}/2)$ is free over the Steenrod algebra. We proved that earlier using different means.