The next goal of this series of posts (started here) is to analyze the oriented cobordism spectrum at the prime 2; the main result is that there is a splitting of into a direct sum of copies of (the torsion-free part) and (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 , once we’ve figured out what 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 is called connected if is one-dimensional and for . The next result provides a sufficient criterion for a module over a graded, connected Hopf algebra to be free.
Theorem 5 (Milnor-Moore) Let be a connected, graded Hopf algebra over a field , and let be a graded, connected -module which is simultaneously a coalgebra (in such a way that is an -homomorphism). Let be a generator, and suppose the map of -modules
is a monomorphism. Then is a free graded -module.
This is a pretty surprising result, as a relatively minor hypothesis (coalgebra, and the action on is free) leads to freeness of the whole thing. The idea of the proof is going to be to produce generators of by lifting a vector space basis of . 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 as before, and choose a vector space basis of ; we can assume that the are homogeneous and that their degrees are nondecreasing, for . If we lift the to homogeneous elements , then we have a map
hitting the elements . This map is an isomorphism when tensored with (that’s the definition of the ), so the graded version of Nakayama’s lemma implies that is surjective. If we can show that is injective as well, then it will follow that is an isomorphism and is free.
Suppose to the contrary that we had a nontrivial relation
For convenience, we assume that the are indexed on . We can apply the comultiplication map to this and find that
for the comultiplication. We will now derive a contradiction by reducing modulo a certain subspace: we take the image of this identity in . In other words, we quotient out the second factor of the tensor product by the augmentation ideal as well as everything generated by .
What does the above identity (4) become upon this reduction? We note that . All but the second term have second factors in , which is generated by the and these terms thus go to zero in . Also, the second term itself vanishes in unless . So the identity (4) becomes
But here where the degrees of the are positive; without loss of generality we can assume . When this is multiplied by and one quotients by the augmentation ideal, the only thing that can contribute is the . So we find
But this is impossible, as is nonzero in by assumption and does not reduce to zero in . This contradiction completes the proof.
For the application to , though, we’ll need a generalization of the previous Milnor-Moore result. The observation is that the cohomology of has a unit element (the Thom class) which is not acted upon freely by the Steenrod algebra: it is annihilated by the Bockstein (this is orientability). So we need the following result, which I learned from Peterson’s lectures on cobordism:
Theorem 6 Let be a graded, connected -module and coalgebra. Let be a generator and suppose that the morphism of -modules has kernel . Then is a direct sum of shifts of and .
The proof will be a variant of the previous one, more complicated this time. Observe that is not a two-sided ideal; instead . This is stated specifically for the Steenrod algebra, because the proof will use specific properties of it.
Proof: A useful observation is that acts on , or in fact any -module, as a differential, because it squares to zero. Consequently, we can define the homology and this will be a graded -vector space. Note that
generated by the image of .
Our goal will now be to define a map from a sum of shifts of to so as to induce an isomorphism on homology ( homology, that is). We will then define a map from a sum of shifts of to hit the remaining indecomposable elements; the goal is to show that this together defines a decomposition of as in the statement of the theorem.
Namely, choose classes whose images are a basis for , so the ‘s are annihilated by , and define a map
hitting the ‘s. We can define such a map, precisely because . Consider the image of in , and choose elements which project to a basis of the complement. This gives a map . Together, we have a map
which is a surjection as it is a surjection on indecomposables.
Our goal is now to show that is an isomorphism. By construction and the calculations (5), (6), it is an isomorphism on homology, which is a good start. Let be the module on the right side of (7), so that . We will show that is an isomorphism by an inductive process.
To do so, we will use an inductive argument. We know that is a direct sum of cyclic modules over . Arrange the generators in increasing order, and label them so that for . By our choice, each of the ‘s is either freely acted upon by or generates a submodule isomorphic to (a shift of) .
Let be submodule of generated by the , and let be defined similarly (in terms of the images of the ). We have
so it suffices to show that each is an isomorphism by induction on . (It is clearly surjective.) Note that when , the map is an induction because of our assumption that the kernel of multiplication by is . So the induction can start.
Suppose we have shown inductively that is injective (hence an isomorphism). If we show that is injective, then the five-lemma will show that is an isomorphism, and that will be the inductive step. Note that the map (which we want to prove injective) induces an isomorphism on -homology, because this is true for and it is true for .
Now, is generated by one element , which is either freely acted upon by or acted upon with kernel . We need to show that the image is acted upon similarly. There are two cases:
- Suppose . Then any element of is of the form , . Note that the image of is nonzero in : it is nonzero after reducing mod the augmentation ideal, after which it generates .If , then we find that and taking the coproduct, we find in ,
This is because the terms with primes on them are intermediate terms: for instance, the have smaller degree and are in the submodule generated by . (We note, in fact, that is a coalgebra for this reason.)Now, when we take the image of this in , we find that most of the terms disappear and that
must go to zero in . This is a contradiction, since is not zero (as ), and does not go to zero in . This was the easier case, and it is completely analogous to the Milnor-Moore result above.
- Suppose , so that acts freely on . Then any element of is of the form for .The first claim is that are linearly independent (over ) in ; since they live in different degrees, this equates to saying that . The first is evident because is nonzero in . For the second, if went to zero in , then would be a cycle in lowest degree in which is not hit by a cycle from . This contradicts the fact that is an isomorphism on -homology. The linear independence assertion follows.
Consider an element in , where are not both zero. If such an element goes to zero in , i.e.
then we take coproducts again to get that the element
belongs to .Now we let be the quotient of by elements of degree two greater than , so this is a vector space free on .
When we take the image of the above huge element in , though, all we are left with is
We have seen that this cannot be zero: neither nor is zero in (and they are linearly independent, being in different degrees). Also, at least one of , so one of is nonzero.
With each of these cases established, it follows that is an isomorphism; this completes the inductive step.