We finally have all the computational tools in place to understand Milnor’s computation of , and the goal of this post is to complete it. Let’s recall what we have done so far.
- We described the Adams spectral sequence, which ran
where the hat denotes -adic completion.
- We worked out the homology of . , as a comodule over , is where is a suitable subHopf-algebra of .When , we had that . When is odd, .
- We worked out a general “change-of-rings isomorphism” for groups.
Now it’s time to put these all together. The page of the Adams spectral sequence for is, as a bigraded algebra,
The polynomial ring can just be pulled out, since it’s not relevant to the comodule structure. Consequently, the are in bidegree : the first bidegree comes from the , and the second is because we are in a graded category. The only contribute in the second way.
Now observe that is an exterior algebra on the generators for odd, and on the for . Using the formulas for the codiagonal in in this post, we find that the generators for are primitive.
1. The cohomology of an exterior algebra
So what we need now is compute : that is, compute the cohomology of an exterior algebra. That turns out to be fairly straightforward. We will work it out in the case of an exterior algebra on one generator.
Proposition 1 Let be an exterior algebra over on one primitive generator in an odd degree . Then
is a polynomial algebra on a generator in bidegree .
In fact, this result suffices to prove an analogous result for a tensor product of exterior algebras on primitive generators (by the Künneth theorem).
To prove it, we can write down an explicit resolution of by -free comodules. It looks like
Here denotes shifted by , and the maps are the obvious ones. Taking maps of into each piece, we get the additive structure of the groups: in fact, if is the complex , then is the complex
We can also put a multiplicative structure on this resolution to get the multiplicative structure on . In fact, we can think of the resolution as the algebra
We make this algebra into a differential graded algebra via . Then this is a dga resolution of , and applying and taking cohomology gives , as desired.
2. The page, completely
We saw that the page of the Adams spectral sequence was, by (1),
where is in bidegree . As we’ve seen, the quotient Hopf algebra
is an exterior algebra on primitive generators of degrees:
- When , degrees for .
- When is odd, degrees for .
3. Determination of the structure at a prime
We are now ready to use the spectral sequence to compute the -adic completion . In fact, first let’s note that there is an element given by the element in (2). This element corresponds to an element of Adams filtration one in the associated graded of : in particular, it must represent or a unit times . This we know just from .
With this in mind, choose elements of representing each of the generators and ; let’s call them and . Then must live in degree (where ) and lives in degree .
Proposition 2 is a polynomial ring over on the and .
Proof: In fact, fix , and choose an element . Then the image of in the associated graded of that comes from the Adams filtration is a polynomial in the (where the , admissible are controlled by ). So there exists a polynomial such that
is in one step lower in the Adams filtration. We can keep doing this (subtract from ) to approximate arbitrarily closely by a polynomial, and eventually make equal to a polynomial with -coefficients in the .
It follows that the map
is a surjection, and it is an injection because, again, we can use the associated graded to see algebraic independence of the .
It follows now that for each prime , we have a sequence such that after -adic completion, the become polynomial generators for .
Corollary 3 For any prime , has no -torsion.
In particular, if we consider the Hurewicz map
it must be injective: in fact, it is an isomorphism after tensoring with , and we have just seen that the map is an injection.
4. Determining the global structure
We’re still not completely there. We know that right now is very close to looking like a polynomial ring on even-degree generators. We also have seen that is injective.
Let be the augmentation ideal in , and consider the indecomposable quotient . The degree part maps to the corresponding part of the indecomposable quotient of .
After -adic completion, becomes a free module of rank one (we’ve seen that is a polynomial ring), and as it is finitely generated we conclude that is itself isomorphic to . Moreover, the map
is an injection, since it is an isomorphism mod torsion.
For each , choose an element which generates the indecomposable quotient . Then, we have:
Theorem 4 (Milnor) is a polynomial ring on the , that is
Proof: This is now going to be straightforward from the analysis already done. We have a map
which is a surjection since all the indecomposables are hit. Moreover, it is an injection because it is an injection after tensoring with . After tensoring with , it is a surjection of graded algebras of the same dimension in each degree.