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_{<n}}, 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

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