This is the third in the series of posts intended to work through the proof of Lazard’s theorem, that the Lazard ring classifying the universal formal group law is actually a polynomial ring on a countable set of generators. In the first post, we reduced the result to an elementary but tricky “symmetric 2-cocycle lemma.” In the previous post, we proved most of the symmetric 2-cocycle lemma, except in characteristic zero. The case of characteristic zero is not harder than the cases we handled (it’s easier), but in this post we’ll complete the proof of that case by exhibiting a very direct construction of logarithms in characteristic zero. Next, I’ll describe an application in Lazard’s original paper, on “approximate” formal group laws.
After this, I’m going to try to move back to topology, and describe the proof of Quillen’s theorem on the formal group law of complex cobordism. The purely algebraic calculations of the past couple of posts will be necessary, though.
1. Formal group laws in characteristic zero
The last step missing in the proof of Lazard’s theorem was the claim that the map
classifying the formal group law obtained from the additive one by “change-of-coordinates” by the exponential series is an isomorphism mod torsion. In other words, we have an isomorphism
In fact, we didn’t really need this: we could have proved the homological 2-cocycle lemma in all cases, instead of just the finite field case, and it would have been easier. But I’d like to emphasize that the result is really something elementary here. In fact, what it is saying is that to give a formal group law over a -algebra is equivalent to giving a choice of series .
Definition 1 An exponential for a formal group law is a power series such that
The inverse power series is called the logarithm.
That is, a logarithm is an isomorphism of with the additive formal group law.
So another way of phrasing this result is that:
Proposition 2 A formal group law over a -algebra has a unique logarithm (i.e., is uniquely isomorphic to the additive one).
Note that this is very false in characteristic . Formal group laws can be wildly non-isomorphic (we’ll see that the height is an invariant in characteristic which can be used to distinguish FGL’s) and, moreover, the additive group law itself has nontrivial automorphisms.
Proof: In fact, given a formal group law , we want to find a formal power series such that
The exponential series will be the inverse to this. The strategy is to “do calculus” with this equation, just as one defines the usual logarithm as an integral.
In fact, if we differentiate with respect to , we find the differential equation:
(Here the subscript denotes partial differentiation with respect to the first variable.)
Setting , we get
which becomes (since ),
So we should define
where the integral is done formally. Note that in order to define the integral, we need to introduce denominators and thus the fact that we are over a -algebra.
If we do this, we still have to check that the basic equation is actually satisfied. In fact, since both sides have no constant term and are symmetric in both variables, we can check this by differentiation with respect to : we find the derivatives are
So to check our basic equation, we have to prove
This, however, is an infinitesimal version of associativity: differentiate with respect to at .
The last thing we have to see is that the logarithm is unique. But if the logarithm weren’t unique, then we’d have a nontrivial automorphism of the formal additive group in characteristic : that is, a power series satisfying
In particular, this gives , which forces to be the trivial automorphism.
We have, in particular, completed the proof of Lazard’s theorem.
2. Some other corollaries
There are a few other isolated results which we can pick up by virtue of the analysis of Lazard’s theorem. One of them concerns “partial” formal group laws, which Lazard chooses to call “buds.”
Definition 3 Let . A -bud over a ring is a polynomial of degree which satisfies the equations of a formal group law modulo .
Just as with the Lazard ring, there is a “universal” -bud for each . We thus get a series of rings for each , together with maps
which are obtained by restricting terms.
Theorem 4 (Restricted Lazard theorem) Each is a polynomial ring on variables and the maps are the inclusions
In particular, every -bud can be extended to a -bud.
We can prove this in the same way as Lazard’s theorem. Namely, we find that, as before, is a graded ring, and there is an augmentation ideal as before. To understand in relation to , we may as well understand the map on indecomposables
for each , as well as what these indecomposables are. To do so, we might as well understand for each ,
which is identified with the set of -buds in of appropriate grading, so of the form
But when , this is just the set of all symmetric 2-cocycles on . When , the group is zero. In particular, we find (by the symmetric 2-cocycle lemma) that graded pieces of the indecomposables are for and elsewhere, and the map is either an isomorphism or zero.
From this, we want to claim that is actually a polynomial ring, and the map is as claimed. But we can prove this by choosing elements which generate the indecomposables in the appropriate degrees of . Then we get a map
which is an isomorphism on indecomposables. There is a map
and, by the above analysis in terms of cocycles, we find that induces either an isomorphism or zero on indecomposables in the relevant degrees. It follows that must be an isomorphism (i.e., first that is surjective, and second that the are algebraically independent in ). This proves that is a polynomial ring, and even that the map is as claimed.