Finally, it’s time to try to understand the computation of the cobordism ring . This will be the first step in understanding Quillen’s theorem, that the formal group law associated to is the universal one. We will compute using the Adams spectral sequence.
In this post, I’ll set up what we need for the Adams spectral sequence, which is a little bit of algebraic computation. In the next post, I’ll describe the actual calculation of the spectral sequence, which will complete the description of .
1. The homology of
The starting point for all this is, of course, the homology , which is a ring since is a ring spectrum. (In the past, I had written reduced homology for spectra, but I will omit it now; recall that for a space , we have .)
Anyway, let’s actually do something more general: let be a complex-oriented spectrum (which gives rise to a homology theory). We will compute .
Proposition 1 where each has degree .
The proof of this will be analogous to the computation of . In fact, the idea is essentially that, by the Thom isomorphism theorem,
where the last equality is because is complex-oriented, and consequently the -homology of looks like the ordinary homology of it.
Let’s just review this again, anyway. We recall that is the desuspension of the Thom space of the line bundle (the Thom space is homotopy equivalent to again, though). Then we have
where each is in degree . In general, we have that
because of the Thom isomorphism and the fact that . The multiplicative structure comes along in the same way as the multiplicative structure for , because the Thom isomorphism is multiplicative (this is part of the condition on ). In the limit, we have
where goes to . Observe that becomes the unit element. Anyway, this was more or less the same as before. Observe that this isomorphism is natural in the complex-oriented theory (i.e., for morphisms preserving the complex orientation).
In particular, we have
Since rationally, (stable) homotopy and homology are the same, we find that is a polynomial ring on a countable set of variables in even degrees. We will see that this is in fact true for itself, although the Hurewicz homomorphism is not an isomorphism.
2. The action of the Steenrod algebra
For the purposes of the ASS, we’re now going to specialize to for a prime . We will need not only the structure of as a ring, but also as a comodule over the dual mod -Steenrod algebra .
The structure of the algebra is due to Milnor, and was discussed on this blog in the case :
- When , where
- When , where denotes an exterior algebra. Here and .
The coalgebra structure is somewhat complicated, corresponding to the fact that the Steenrod algebra is messy. One has
- For ,
- For odd,
(I just copied these formulas from Ravenel’s “green book.”) There is a “high-concept” way of looking at at least when ; is precisely the automorphism group scheme of the formal additive group. I don’t know what the analog is for odd: presumably it relies on “super” algebraic geometry.
The algebra is a comodule over . In fact, it is a comodule over something a fair bit smaller. If we dualize to cohomology, we find that the Bockstein has to act trivially, since the cohomology is entirely in even dimensions: in other words, is a module over a quotient of . Dually, is a comoduleover a subcoalgebra (cosubalgebra?) of .
In fact, define the subcoalgebra for as that generated by the , and in case , by the (so ignoring the exterior part of it). In either case, it’s immediate from the formulas given that is an honest subcoalgebra. We are going to see that is a coinduced comodule over .
First, we need:
Claim: is actually a comodule over . Proof: The idea is to use the fact that is entirely in even degrees. In fact, the comodule map
is a graded map, so any term that occurs must have even degree. Choose a homogeneous basis of and, for any homogeneous write
(where the are uniquely determined). The claim is that the live in , though what is only clear right now is that the are in even degrees. But now apply the comodule structure map again, and use coassociativity. One the one hand, we get
and on the other, we get
Using this, we find that the have to be of the form where the live in even degrees. But this means that the have to live in , as one checks from the formulas.
3. The main result
Using this, we will obtain the desired structure of as an -comodule.
Proposition 2 As an -comodule algebra, one has
This is the analog of the computation of as a comodule over .
Maybe it’s worth trying to motivate this result. As with , the point is that the coaction of is essentially determined by the coaction on , which is easy to work out. There are general Hopf-algebra structure theorems that I don’t really understand which can be used to show that the homology is free in this sense. As more motivation, note that if we had an isomorphism as claimed, then because of the structure of as a polynomial algebra, we would get precisely the right number of variables for .
Anyway, note that is not a coinduced comodule anymore, though it’s somewhat close. This is going to make the Adams spectral sequence nontrivial, but it is at least going to degenerate at .
Proof: We deal with the case of odd.
As before, we need to work out the coaction on . This space has a basis for homology where . The coaction is given by
where the indicate terms involving . This reflects the fact that if is the dual to , then one can get from to using an appropriate product of Steenrod powers in precisely one way—and that way is dual to .
Consequently, if as before, we get
This follows from using the formulas in and shifting.
OK, now let’s define the map
The map goes , where the last map sends and annihilates some of the . The first map is the coaction of on .
Now we observe that this map , which is both a map of comodules and of algebras, behaves the right way on indecomposables. We have that
Here the indicate decomposables. Consequently, if we look at indecomposables, we find that we hit all the polynomial generators in , both the ones coming from and the . It follows that the map is an isomorphism, by dimension-counting.