The next goal of this series of posts (started here) is to analyze the oriented cobordism spectrum ${MSO}$ at the prime 2; the main result is that there is a splitting of ${MSO_{(2)}}$ into a direct sum of copies of ${H\mathbb{Z}_{(2)}}$ (the torsion-free part) and ${H \mathbb{Z}/2}$ (the torsion-part). In particular, it will follow that there is only torsion of order two in the cobordism ring — since we showed last time that there was no odd torsion. We will see this using the Adams spectral sequence at the prime ${2}$, once we’ve figured out what ${H_*(MSO; \mathbb{Z}/2)}$ looks like as a comodule over the dual Steenrod algebra. This, however, is apparently somewhat tricky to do directly.

In order to get there, we’ll need a bit of algebraic machinery (which we state in a dual context). Recall that a graded vector space ${V}$ is called connected if ${V_0}$ is one-dimensional and ${V_i = 0}$ for ${i < 0}$. The next result provides a sufficient criterion for a module over a graded, connected Hopf algebra to be free.

Theorem 5 (Milnor-Moore) Let ${A}$ be a connected, graded Hopf algebra over a field ${k}$, and let ${M}$ be a graded, connected ${{A}}$-module which is simultaneously a coalgebra (in such a way that ${M \rightarrow M \otimes_k M}$ is an ${A}$-homomorphism). Let ${1 \in M_0}$ be a generator, and suppose the map of ${A}$-modules

$\displaystyle A \rightarrow M, \quad a \mapsto a . 1$

is a monomorphism. Then ${M}$ is a free graded ${A}$-module.

This is a pretty surprising result, as a relatively minor hypothesis (coalgebra, and the action on ${1}$ is free) leads to freeness of the whole thing. The idea of the proof is going to be to produce generators of ${M}$ by lifting a vector space basis of ${\overline{M} = M \otimes_A k}$. The fact that these generators are forced to be linearly independent is an unexpected consequence of the coalgebra structure; the graded connectedness will be used to make certain inductive arguments.

So, let’s prove this. Let ${\overline{M} = M \otimes_A k}$ as before, and choose a vector space basis ${\overline{m_i}}$ of ${\overline{M}}$; we can assume that the ${\overline{m_i}}$ are homogeneous and that their degrees are nondecreasing, ${\deg \overline{m_i} \leq \deg \overline{m_{j}}}$ for ${i \leq j}$. If we lift the ${\overline{m_i}}$ to homogeneous elements ${m_i \in M}$, then we have a map

$\displaystyle \phi: \bigoplus_i A[-\deg m_i] \rightarrow M$

hitting the elements ${m_i}$. This map is an isomorphism when tensored with ${k}$ (that’s the definition of the ${m_i}$), so the graded version of Nakayama’s lemma implies that ${\phi}$ is surjective. If we can show that ${\phi}$ is injective as well, then it will follow that ${\phi}$ is an isomorphism and ${M}$ is free.

Suppose to the contrary that we had a nontrivial relation

$\displaystyle \sum_{i=1}^n a_i m_i = 0, \quad a_n \in A \setminus \left\{0\right\}.$

For convenience, we assume that the ${m_i}$ are indexed on ${\mathbb{N}}$. We can apply the comultiplication map ${\Delta: M \rightarrow M \otimes_k M}$ to this and find that

$\displaystyle \sum_{i=1}^n \Delta(a_im_i) = \sum_{i=1}^n \Delta(a_i)\Delta(m_i) = 0, \ \ \ \ \ (4)$

for ${\Delta: A \rightarrow A \otimes_k A}$ the comultiplication. We will now derive a contradiction by reducing modulo a certain subspace: we take the image of this identity in ${M \otimes \overline{M}/(m_1, \dots, m_{n-1})}$. In other words, we quotient out the second factor of the tensor product by the augmentation ideal as well as everything generated by ${m_1, \dots, m_{n-1}}$.

What does the above identity (4) become upon this reduction? We note that ${\Delta(m_i) = m_i \otimes 1 + 1 \otimes m_i + \sum m_{j}' \otimes m_{j}''}$. All but the second term have second factors in ${M_{, which is generated by the ${m_1, \dots, m_{n-1}}$ and these terms thus go to zero in ${M \otimes \overline{M}/(m_1, \dots, m_{n-1})}$. Also, the second term itself vanishes in ${M \otimes \overline{M}/(m_1, \dots, m_{n-1})}$ unless ${i = n}$. So the identity (4) becomes

$\displaystyle \Delta(a_n) (1 \otimes m_n) = 0 \in M \otimes \overline{M}/(m_1, \dots, m_{n-1}).$

But here ${\Delta(a_n) = a_n \otimes 1 + 1 \otimes a_n + \sum a_i' \otimes a_{i}''}$ where the degrees of the ${a_i', a_i'' }$ are positive; without loss of generality we can assume ${\deg a_n > 0}$. When this is multiplied by ${1 \otimes m_n}$ and one quotients by the augmentation ideal, the only thing that can contribute is the ${a_n \otimes 1}$. So we find

$\displaystyle (a_n \otimes 1) (1 \otimes m_n) = (a_n .1) \otimes m_n = 0 \in M \otimes \overline{M}/(m_1, \dots,m_{n-1}).$

But this is impossible, as ${a_n.1}$ is nonzero in ${M}$ by assumption and ${m_n}$ does not reduce to zero in ${\overline{M}/(m_1, \dots, m_{n-1})}$. This contradiction completes the proof.

A generalization

For the application to ${MSO}$, though, we’ll need a generalization of the previous Milnor-Moore result. The observation is that the cohomology of $MSO$ has a unit element (the Thom class) which is not acted upon freely by the Steenrod algebra: it is annihilated by the Bockstein $\mathrm{Sq}^1$ (this is orientability). So we need the following result, which I learned from Peterson’s lectures on cobordism:

Theorem 6 Let ${M}$ be a graded, connected ${\mathcal{A}_2}$-module and coalgebra. Let ${1 \in M_0}$ be a generator and suppose that the morphism of ${\mathcal{A}_2}$-modules ${\mathcal{A}_2 \rightarrow M, a \mapsto a.1}$ has kernel ${J = \mathcal{A}_2 \mathrm{Sq}^1 }$. Then ${M}$ is a direct sum of shifts of ${\mathcal{A}_2}$ and ${\mathcal{A}_2/J}$.

The proof will be a variant of the previous one, more complicated this time. Observe that ${J}$ is not a two-sided ideal; instead ${\mathcal{A}_2/J = \mathcal{A}_2 \otimes_{E(\mathrm{Sq}^1)} \mathbb{Z}/2}$. This is stated specifically for the Steenrod algebra, because the proof will use specific properties of it.

Proof: A useful observation is that ${\mathrm{Sq}^1}$ acts on ${M}$, or in fact any ${\mathcal{A}_2}$-module, as a differential, because it squares to zero. Consequently, we can define the homology ${H(M, \mathrm{Sq}^1)}$ and this will be a graded ${\mathbb{Z}/2}$-vector space. Note that

$\displaystyle H_*(\mathcal{A}_2, \mathrm{Sq}^1) = 0 \ \ \ \ \ (5)$

by a look at the Serre-Cartan basis for ${\mathcal{A}_2}$. However,

$\displaystyle H_*(\mathcal{A}_2/J, \mathrm{Sq}^1) = \mathbb{Z}/2, \ \ \ \ \ (6)$

generated by the image of ${1}$.

Our goal will now be to define a map from a sum of shifts of ${\mathcal{A}_2/J}$ to ${M}$ so as to induce an isomorphism on homology (${\mathrm{Sq}^1}$ homology, that is). We will then define a map from a sum of shifts of ${\mathcal{A}_2}$ to hit the remaining indecomposable elements; the goal is to show that this together defines a decomposition of ${M}$ as in the statement of the theorem.

Namely, choose classes ${\left\{x_\alpha\right\}_{\alpha \in I}}$ whose images are a basis for ${H_*(M; \mathrm{Sq}^1)}$, so the ${x_\alpha}$‘s are annihilated by ${\mathrm{Sq}^1}$, and define a map

$\displaystyle \phi_1: \bigoplus_I \mathcal{A}_2/J[-\deg x_\alpha] \rightarrow M$

hitting the ${x_\alpha}$‘s. We can define such a map, precisely because ${\mathrm{Sq}^1 x_\alpha = 0}$. Consider the image of ${\phi_1}$ in ${\overline{M} = M \otimes_{\mathcal{A}_2} \mathbb{Z}/2}$, and choose elements ${\left\{m_\beta\right\}_{\beta \in J}}$ which project to a basis of the complement. This gives a map ${\phi_2 : \bigoplus_J \mathcal{A}_2[-\deg m_\beta] \rightarrow M}$. Together, we have a map

$\displaystyle \phi = \phi_1 \oplus \phi_2: \bigoplus_I \mathcal{A}_2/J[-\deg x_\alpha] \oplus \bigoplus_J \mathcal{A}_2[-\deg m_\beta] \rightarrow M, \ \ \ \ \ (7)$

which is a surjection as it is a surjection on indecomposables.

Our goal is now to show that ${\phi}$ is an isomorphism. By construction and the calculations (5)(6), it is an isomorphism on ${\mathrm{Sq}^1}$ homology, which is a good start. Let ${N}$ be the module on the right side of (7), so that ${\phi: N \rightarrow M}$. We will show that ${\phi}$ is an isomorphism by an inductive process.

To do so, we will use an inductive argument. We know that ${N}$ is a direct sum of cyclic modules over ${\mathcal{A}_2}$. Arrange the generators in increasing order, and label them ${n_0, n_1, \dots}$ so that ${\deg n_i \leq \deg n_j}$ for ${i \leq j}$. By our choice, each of the ${n_i}$‘s is either freely acted upon by ${\mathcal{A}_2}$ or generates a submodule isomorphic to (a shift of) ${\mathcal{A}_2/J}$.

Let ${N^{(k)}}$ be submodule of ${N}$ generated by the ${n_i, i < k}$, and let ${M^{(k)}}$ be defined similarly (in terms of the images ${m_i}$ of the ${n_i}$). We have

$\displaystyle M = \varinjlim M^{(k)}, N = \varinjlim N^{(k)},$

so it suffices to show that each ${N^{(k)} \rightarrow M^{(k)}}$ is an isomorphism by induction on ${k}$. (It is clearly surjective.) Note that when ${k = 0}$, the map is an induction because of our assumption that the kernel of multiplication by ${1 \in M_0}$ is ${J \subset \mathcal{A}_2}$. So the induction can start.

Suppose we have shown inductively that ${N^{(k-1)} \rightarrow M^{(k-1)}}$ is injective (hence an isomorphism). If we show that ${\phi': N^{(k)}/N^{(k-1)} \rightarrow M^{(k)}/M^{(k-1)}}$ is injective, then the five-lemma will show that ${N^{(k)} \rightarrow M^{(k)}}$ is an isomorphism, and that will be the inductive step. Note that the map ${N/N^{(k-1)} \rightarrow M/M^{(k-1)}}$ (which we want to prove injective) induces an isomorphism on ${\mathrm{Sq}^1}$-homology, because this is true for ${N}$ and it is true for ${N^{(k-1)}}$.

Now, ${N^{(k)}/N^{(k-1)}}$ is generated by one element ${n_k}$, which is either freely acted upon by ${\mathcal{A}_2}$ or acted upon with kernel ${J}$. We need to show that the image ${\phi'(n_k) \in N^{(k)}/N^{(k-1)}}$ is acted upon similarly. There are two cases:

1. Suppose ${\mathrm{Sq}^1n_k = 0}$. Then any element of ${N^{(k)} /N^{(k-1)}}$ is of the form ${a n_k}$, ${a \in \mathcal{A}_2 \setminus J}$. Note that the image ${m_k}$ of ${n_k}$ is nonzero in ${M^{(k)}/M^{(k-1)}}$: it is nonzero after reducing mod the augmentation ideal, after which it generates ${(\overline{M}^{(k)}/\overline{M}^{(k-1)})_{\deg m_k}}$.If ${\phi'(an_k) = 0 \in M^{(k)}/M^{(k-1)}}$, then we find that ${\phi(an_k) \in M^{(k-1)}}$ and taking the coproduct, we find in ${M \otimes M }$,

$\displaystyle (a \otimes 1 + 1 \otimes a + \sum a_i' \otimes a_{i}'') (m_k \otimes 1 + 1 \otimes m_k + \sum_j m_j' \otimes m_{j}'') \in M \otimes M^{(k-1)}.$

This is because the terms with primes on them are intermediate terms: for instance, the ${m_{j}'}$ have smaller degree and are in the submodule generated by ${m_i, i < k}$. (We note, in fact, that ${M^{(k-1)}}$ is a coalgebra for this reason.)Now, when we take the image of this in ${M \otimes \overline{M/M^{(k-1)}}}$, we find that most of the terms disappear and that

$\displaystyle a.1 \otimes m_k$

must go to zero in ${M \otimes \overline{M/M^{(k-1)}}}$. This is a contradiction, since ${a .1}$ is not zero (as ${a \notin J}$), and ${m_k}$ does not go to zero in ${\overline{M/M^{(k-1)}}}$. This was the easier case, and it is completely analogous to the Milnor-Moore result above.

2. Suppose ${\mathrm{Sq}^1 n_k \neq 0}$, so that ${\mathcal{A}_2}$ acts freely on ${n_k}$. Then any element of ${N^{(k)}/N^{(k-1)}}$ is of the form ${a_1n_k + a_2 \mathrm{Sq}^1 n_k}$ for ${a_1, a_2 \notin J}$.The first claim is that ${m_k, \mathrm{Sq}^1 m_k}$ are linearly independent (over ${\mathbb{Z}/2}$) in ${M/M^{(k-1)}}$; since they live in different degrees, this equates to saying that ${m_k, \mathrm{Sq}^1 m_k \notin M^{(k-1)}}$. The first is evident because ${m_k}$ is nonzero in ${\overline{M/M^{(k-1)}}}$. For the second, if ${\mathrm{Sq}^1 m_k }$ went to zero in ${M/M^{(k-1)}}$, then ${m_k}$ would be a cycle in lowest degree in ${M/M^{(k-1)}}$ which is not hit by a cycle from ${N/N^{(k-1)}}$. This contradicts the fact that ${N/N^{(k-1)} \rightarrow M/M^{(k-1)}}$ is an isomorphism on ${\mathrm{Sq}^1}$-homology. The linear independence assertion follows.

Consider an element ${a_1n_k + a_2 \mathrm{Sq}^1 n_k}$ in ${N}$, where ${a_1, a_2}$ are not both zero. If such an element goes to zero in ${M^{(k)}/M^{(k-1)}}$, i.e.

$\displaystyle a_1 m_k + a_2 \mathrm{Sq}^1 m_k \in M^{(k-1)} ,$

then we take coproducts again to get that the element

$\displaystyle (a_1 \otimes 1 + 1 \otimes a_1 + \dots ) ( m_k \otimes 1 + 1 \otimes m_k + \dots) + (a_2 \otimes 1 + 1 \otimes a_2) (\mathrm{Sq}^1 \otimes 1 + 1 \otimes \mathrm{Sq}^1)( m_k \otimes 1 + 1\otimes m_k + \dots)$

belongs to ${M \otimes M^{(k-1)}}$.Now we let ${\overline{\overline{M/M^{(k-1)}}}}$ be the quotient of ${M/M^{(k-1)}}$ by elements of degree two greater than ${\deg m_k}$, so this is a vector space free on ${m_k, \mathrm{Sq}^1 m_k}$.

When we take the image of the above huge element in ${M \otimes \overline{\overline{M^{(k-1)}}}}$, though, all we are left with is

$\displaystyle a_1.1 \otimes m_k + a_2 .1 \otimes \mathrm{Sq}^1 m_k.$

We have seen that this cannot be zero: neither ${\mathrm{Sq}^1 m_k}$ nor ${m_k}$ is zero in ${\overline{\overline{M/M^{(k-1)}}}}$ (and they are linearly independent, being in different degrees). Also, at least one of ${a_1, a_2 \notin J}$, so one of ${a_1.1, a_2.1}$ is nonzero.

With each of these cases established, it follows that ${N^{(k) } \rightarrow M^{(k)}}$ is an isomorphism; this completes the inductive step. $\Box$